Motion in a Plane

Scalar quantities : volume, mass, speed, density, number of moles and angular frequency.

Vector quantities : acceleration, velocity, displacement, angular velocity and angular acceleration.

Reason — A scalar quantity is described solely by its magnitude and has no associated direction, while a vector quantity has both magnitude and direction and obeys the laws of vector addition. Volume, mass, speed, density, number of moles and angular frequency are completely specified by a number and a unit alone, so they are scalars. Acceleration, velocity, displacement, angular velocity and angular acceleration cannot be specified without stating a direction, so they are vectors.

Work and current are the two scalar quantities.

Reason — Work is the scalar product of force and displacement, $\text W = \vec{\text F} \cdot \vec{\text s}$, and the scalar product of two vectors is always a scalar. Electric current has both magnitude and direction, but it does not obey the laws of vector addition (currents at a junction add algebraically), so it is a scalar. All the remaining quantities in the list have both magnitude and direction and obey the laws of vector addition, so they are vectors.

Impulse is the only vector quantity.

Reason — Impulse is the product of force and time interval, $\vec{\text J} = \vec{\text F}\Delta \text t$. Since force is a vector and time is a scalar, the product is a vector having the same direction as the force. All the other quantities listed are completely specified by their magnitude alone, so they are scalars.

(a) Not always meaningful. Two scalars can be added only if they represent the same physical quantity, that is, if they have the same dimensions. Mass and time cannot be added, but two masses can.

(b) Not meaningful. A scalar has only magnitude while a vector has magnitude as well as direction. Even if the dimensions are the same, the two cannot be added.

(c) Meaningful. On multiplying a vector $\vec{\text A} \space$by a scalar k, a vector $\text k\vec{\text A} \space$is obtained whose magnitude is k times that of $\vec{\text A}$. If k is a pure number the new vector represents the same physical quantity, and if k has a unit the product represents a new physical quantity, as in $\vec{\text p} = \text m\vec{\text v}$.

(d) Meaningful. The product of two scalars is always a scalar, and it represents a new physical quantity, for example power multiplied by time gives work.

(e) Not always meaningful. Two vectors can be added only if they represent the same physical quantity, that is, if they have the same dimensions. Force cannot be added to velocity.

(f) Meaningful. A component of a vector is itself a vector of the same physical quantity, so it can be added to the vector by the laws of vector addition.

(a) True. The magnitude of a vector gives only its size and carries no direction, so it is a scalar.

(b) False. A vector is resolved into components which are themselves vectors along the chosen directions, so each component is a vector and not a scalar.

(c) False. The total path length is the actual length of the path travelled, whereas the magnitude of the displacement is the straight line distance between the initial and the final position. The two are equal only when the particle moves along a straight line without reversing its direction; otherwise the path length is greater.

(d) True. The average speed is the total path length divided by the time interval and the magnitude of the average velocity is the magnitude of the displacement divided by the same time interval. Since the path length is always greater than or equal to the magnitude of the displacement, the average speed is greater than or equal to the magnitude of the average velocity.

(e) True. For three vectors to give a null vector they must form a closed triangle, and a triangle always lies in one plane. Hence three vectors which do not lie in a plane can never add up to give a null vector.

Let two vectors $\vec{\text A} \space$and $\vec{\text B} \space$be inclined to each other at an angle θ. By the parallelogram law of vector addition,

$|\vec{\text A} + \vec{\text B}| = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

and replacing θ by (180° − θ) for the difference,

$|\vec{\text A} - \vec{\text B}| = \sqrt{\text A^2 + \text B^2 - 2\text{AB}\cos \theta}$

(a) Since cos θ can at most be 1,

$|\vec{\text A} + \vec{\text B}| \le \sqrt{\text A^2 + \text B^2 + 2\text{AB}} = \sqrt{(\text A + \text B)^2} = \text A + \text B$

Hence $|\vec{\text A} + \vec{\text B}| \le |\vec{\text A}| + |\vec{\text B}|$. The equality holds when cos θ = 1, that is, when θ = 0° and the two vectors are in the same direction.

(b) Since cos θ is at least −1,

$|\vec{\text A} + \vec{\text B}| \ge \sqrt{\text A^2 + \text B^2 - 2\text{AB}} = \sqrt{(\text A - \text B)^2} = |\text A - \text B|$

Hence $|\vec{\text A} + \vec{\text B}| \ge \left| |\vec{\text A}| - |\vec{\text B}| \right|$. The equality holds when cos θ = −1, that is, when θ = 180° and the two vectors are in opposite directions.

(c) For the difference, the largest value occurs when cos θ = −1,

$|\vec{\text A} - \vec{\text B}| \le \sqrt{\text A^2 + \text B^2 + 2\text{AB}} = \text A + \text B$

Hence $|\vec{\text A} - \vec{\text B}| \le |\vec{\text A}| + |\vec{\text B}|$. The equality holds when θ = 180°, that is, when the two vectors are in opposite directions.

(d) The smallest value of the difference occurs when cos θ = 1,

$|\vec{\text A} - \vec{\text B}| \ge \sqrt{\text A^2 + \text B^2 - 2\text{AB}} = |\text A - \text B|$

Hence $|\vec{\text A} - \vec{\text B}| \ge \left| |\vec{\text A}| - |\vec{\text B}| \right|$. The equality holds when θ = 0°, that is, when the two vectors are in the same direction.

Statements (b), (c) and (d) are correct and statement (a) is incorrect.

Reason —

(a) The four vectors add up to a null vector because they complete a closed polygon when drawn in succession. This does not require any of them to be a null vector, so the statement is incorrect.

(b) Since $\vec{\text A} + \vec{\text B} + \vec{\text C} + \vec{\text D} = 0$,

$(\vec{\text A} + \vec{\text C}) = -(\vec{\text B} + \vec{\text D})$

Two vectors which are the negative of each other have equal magnitudes, so the statement is correct.

(c) From the given relation, $\vec{\text A} = -(\vec{\text B} + \vec{\text C} + \vec{\text D})$. Since the magnitude of the sum of vectors can never exceed the sum of their magnitudes, the magnitude of $\vec{\text A} \space$can never be greater than the sum of the magnitudes of $\vec{\text B}$, $\vec{\text C} \space$and $\vec{\text D}$. The statement is correct.

(d) From the given relation, $(\vec{\text B} + \vec{\text C}) = -(\vec{\text A} + \vec{\text D})$. The sum $(\vec{\text A} + \vec{\text D}) \space$lies in the plane containing $\vec{\text A} \space$and $\vec{\text D}$, and if $\vec{\text A} \space$and $\vec{\text D} \space$are collinear their sum lies along the same line. Hence $(\vec{\text B} + \vec{\text C}) \space$must lie in that plane or along that line, and the statement is correct.

Given,

• Radius of the circular ice ground, r = 200 m
• P and Q are diametrically opposite points

The displacement vector of each girl is the vector drawn from the initial position P to the final position Q. Since all three girls start from P and reach the same point Q, the displacement vector is the same for each of them, and its magnitude is the diameter PQ,

$$\begin{aligned}\text{PQ} = 2\text r = 2 \times 200 \\[1em] = 400\ \text m\end{aligned}$$

Hence, the magnitude of the displacement vector is 400 m for each girl.

The magnitude of the displacement is equal to the actual length of the path only when the path is a straight line. Girl B skates along the straight line PQ, so for girl B the length of the path skated is also 400 m, equal to the magnitude of her displacement.

Given,

• Radius of the circular park, r = 1 km
• Time taken for the round trip, t = 10 min = $\frac{10}{60}$ h = $\frac{1}{6}$ h

(a) The cyclist starts from the centre O and finally returns to the same point O. Since the initial and the final positions are the same,

$\text{Net displacement} = 0$

(b) The average velocity is the net displacement divided by the total time taken,

$$\begin{aligned}\vec{\text v}_{av} = \frac{\text{Net displacement}}{\text{Total time}} = \frac{0}{1/6} \\[1em] = 0\end{aligned}$$

(c) The total path length covered is

$\text{OP} + \text{arc PQ} + \text{QO}$

Here OP = QO = r = 1 km, and PQ is a quarter of the circumference,

$$\begin{aligned}\text{arc PQ} = \frac{1}{4}(2\pi \text r) = \frac{1}{4} \times 2 \times 3.14 \times 1 \\[1em] = 1.57\ \text{km}\end{aligned}$$

Therefore,

$\text{Total path length} = 1 + 1.57 + 1 = 3.57\ \text{km}$

The average speed is

$$\begin{aligned}\text{Average speed} = \frac{\text{Total path length}}{\text{Total time}} = \frac{3.57}{1/6} \\[1em] = 3.57 \times 6 \\[1em] = 21.4\ \text{km h}^{-1}\end{aligned}$$

Hence, the net displacement and the average velocity are both zero, and the average speed is 21.4 km h^-1.

Given,

• Length of each straight stretch = 500 m
• Turn through 60° to the left after every stretch

Since the motorist turns through 60° after every 500 m, the exterior angle of the path is 60° and the track is a regular hexagon of side 500 m.

At the third turn : The motorist has covered three sides of the hexagon and reaches the vertex diametrically opposite to the starting point. Hence the magnitude of the displacement is

$2 \times 500 = 1000\ \text m$ ∵ Diagonal of hexagon = 2 × side of hexagon

The total path length is

$3 \times 500 = 1500\ \text m$

At the sixth turn : The motorist completes the hexagon and returns to the starting point. Hence

$\text{Displacement} = 0, \quad \text{Path length} = 6 \times 500 = 3000\ \text m$

At the eighth turn : After six turns the motorist is back at the starting point, so the position after the eighth turn is the same as that after two turns. The displacement is the diagonal joining the ends of two adjacent sides,

$$\begin{aligned}\text{Displacement} = 2 \times 500 \times \cos 30^\circ \\[1em] = 2 \times 500 \times \frac{\sqrt{3}}{2} \\[1em] = 500\sqrt{3} = 866\ \text m\end{aligned}$$

and it makes an angle of 30° with the initial direction of motion. The total path length is

$8 \times 500 = 4000\ \text m$

Comparison : At the third turn the magnitude of the displacement (1000 m) is less than the path length (1500 m); at the sixth turn the displacement is zero while the path length is 3000 m; and at the eighth turn the displacement (866 m) is much less than the path length (4000 m).

Given,

• Total path length covered = 23 km
• Magnitude of displacement = 10 km
• Time taken, t = 28 min = $\frac{28}{60}$ h

(a) The average speed of the taxi is

$$\begin{aligned}\text{Average speed} = \frac{\text{Total path length}}{\text{Total time}} = \frac{23}{28/60} \\[1em] = \frac{23 \times 60}{28} \\[1em] = 49.3\ \text{km h}^{-1}\end{aligned}$$

(b) The magnitude of the average velocity is

$$\begin{aligned}|\vec{\text v}_{av}| = \frac{\text{Magnitude of displacement}}{\text{Total time}} = \frac{10}{28/60} \\[1em] = \frac{10 \times 60}{28} \\[1em] = 21.4\ \text{km h}^{-1}\end{aligned}$$

No, the two are not equal. The average speed is greater than the magnitude of the average velocity because the path followed by the taxi is not a straight line, so the path length (23 km) is greater than the magnitude of the displacement (10 km).

Given,

• Speed of projection, u = 40 m s^-1
• Maximum height available, h = 25 m
• Acceleration due to gravity, g = 9.8 m s^-2

For the ball to just graze the ceiling, the maximum height of the projectile must be 25 m. The maximum height attained is

$\text h = \frac{\text u^2 \sin^2 \theta}{2\text g}$

Substituting the values,

$$\begin{aligned}25 = \frac{(40)^2 \sin^2 \theta}{2 \times 9.8} \\[1em] \sin^2 \theta = \frac{25 \times 2 \times 9.8}{1600} \\[1em] = 0.30625\end{aligned}$$

Therefore,

$\sin \theta = 0.5534 \quad \text{and} \quad \cos \theta = \sqrt{1 - 0.30625} = 0.8329$

The horizontal range is

$\text R = \frac{\text u^2 \sin 2\theta}{\text g} = \frac{\text u^2 \times 2 \sin \theta \cos \theta}{\text g}$

Substituting the values,

$$\begin{aligned}\text R = \frac{1600 \times 2 \times 0.5534 \times 0.8329}{9.8} \\[1em] = \frac{1474.9}{9.8} \\[1em] = 150.5\ \text m\end{aligned}$$

Hence, the maximum horizontal distance covered without hitting the ceiling is 150.5 m.

Given,

• Maximum horizontal range, R_max = 100 m

The horizontal range is maximum when the angle of projection is 45°, and then

$\text R_{max} = \frac{\text u^2}{\text g}$

Therefore,

$\frac{\text u^2}{\text g} = 100 \quad \Rightarrow \quad \text u^2 = 100\ \text g$

To throw the ball to the greatest height, the cricketer must throw it vertically upwards, that is, at θ = 90°. The maximum height then attained is

$\text h = \frac{\text u^2 \sin^2 90^\circ}{2\text g} = \frac{\text u^2}{2\text g}$

Substituting u² = 100 g,

$$\begin{aligned}\text h = \frac{100\ \text g}{2\text g} \\[1em] = 50\ \text m\end{aligned}$$

Hence, the cricketer can throw the ball to a maximum height of 50 m above the ground.

Given,

• Radius of the circular path, r = 80 cm = 0.80 m
• Number of revolutions = 14 in 25 s

The angular frequency is

$\text n = \frac{14}{25}\ \text{rev s}^{-1}$

The angular velocity is

$$\begin{aligned}\omega = 2\pi \text n = 2 \times 3.14 \times \frac{14}{25} \\[1em] = 3.52\ \text{rad s}^{-1}\end{aligned}$$

The centripetal acceleration is

$\text a = \omega^2 \text r$

Substituting the values,

$$\begin{aligned}\text a = (3.52)^2 \times 0.80 \\[1em] = 12.39 \times 0.80 \\[1em] = 9.9\ \text{m s}^{-2}\end{aligned}$$

Hence, the acceleration of the stone is 9.9 m s^-2, directed along the radius towards the centre of the circular path.

Given,

• Radius of the horizontal loop, r = 1.00 km = 1000 m
• Speed of the aircraft, v = 900 km h^-1

Converting the speed into m s^-1,

$\text v = 900 \times \frac{5}{18} = 250\ \text{m s}^{-1}$

The centripetal acceleration is

$$\begin{aligned}\text a = \frac{\text v^2}{\text r} = \frac{(250)^2}{1000} \\[1em] = \frac{62500}{1000} \\[1em] = 62.5\ \text{m s}^{-2}\end{aligned}$$

Comparing it with g = 9.8 m s^-2,

$\frac{\text a}{\text g} = \frac{62.5}{9.8} = 6.38$

Hence, the centripetal acceleration of the aircraft is 6.38 times the acceleration due to gravity.

(a) False. The net acceleration is directed along the radius towards the centre only in uniform circular motion. If the speed of the particle also changes, a tangential component of acceleration is present, and the net acceleration is then inclined to the radius.

(b) True. The direction of motion of a particle at any point of its path is along the tangent at that point, so the velocity vector is always tangential to the path.

(c) True. In uniform circular motion the centripetal acceleration is constant in magnitude but its direction changes continuously and points towards the centre. Over one complete cycle every direction is paired with an equal and opposite one, so the vector average of the acceleration over one cycle is a null vector.

Given,

• $\vec{\text r} = 3.0\text t\hat{i} - 2.0\text t^2\hat{j} + 4.0\hat{k}$

(a) The velocity vector is obtained by differentiating the position vector with respect to time,

$$\begin{aligned}\vec{\text v} = \frac{\text d\vec{\text r}}{\text{dt}} = \frac{\text d}{\text{dt}}\left(3.0\text t\hat{i} - 2.0\text t^2\hat{j} + 4.0\hat{k}\right) \\[1em] = 3.0\hat{i} - 4.0\text t\hat{j}\ \text{m s}^{-1}\end{aligned}$$

The acceleration vector is obtained by differentiating the velocity vector with respect to time,

$$\begin{aligned}\vec{\text a} = \frac{\text d\vec{\text v}}{\text{dt}} = \frac{\text d}{\text{dt}}\left(3.0\hat{i} - 4.0\text t\hat{j}\right) \\[1em] = -4.0\hat{j}\ \text{m s}^{-2}\end{aligned}$$

(b) At t = 2.0 s,

$$\begin{aligned}\vec{\text v} = 3.0\hat{i} - 4.0 \times 2.0\hat{j} \\[1em] = 3.0\hat{i} - 8.0\hat{j}\ \text{m s}^{-1}\end{aligned}$$

The magnitude of the velocity is

$$\begin{aligned}|\vec{\text v}| = \sqrt{(3.0)^2 + (-8.0)^2} \\[1em] = \sqrt{9 + 64} = \sqrt{73} \\[1em] = 8.54\ \text{m s}^{-1}\end{aligned}$$

If θ is the angle made by the velocity with the X-axis,

$$\begin{aligned}\tan \theta = \frac{\text v_y}{\text v_x} = \frac{-8.0}{3.0} = -2.667 \\[1em] \theta = \tan^{-1}(-2.667) = -69.5^\circ\end{aligned}$$

Hence, the velocity at t = 2.0 s is 8.54 m s^-1, directed at 69.5° below the X-axis (that is, clockwise from the positive X-axis).

Given,

• Initial velocity, $\vec{\text u} = 10.0\hat{j}$ m s^-1, so u_x = 0 and u_y = 10.0 m s^-1
• Acceleration, $\vec{\text a} = 8.0\hat{i} + 2.0\hat{j}$ m s^-2, so a_x = 8.0 m s^-2 and a_y = 2.0 m s^-2

(a) Using the equation of motion along the X-direction,

$$\begin{aligned}\text x = \text u_x \text t + \frac{1}{2}\text a_x \text t^2 \\[1em] = 0 + \frac{1}{2} \times 8.0 \times \text t^2 = 4\text t^2\end{aligned}$$

Putting x = 16 m,

$16 = 4\text t^2 \quad \Rightarrow \quad \text t = 2\ \text s$

The y-coordinate at this instant is

$$\begin{aligned}\text y = \text u_y \text t + \frac{1}{2}\text a_y \text t^2 \\[1em] = 10.0 \times 2 + \frac{1}{2} \times 2.0 \times (2)^2 \\[1em] = 20 + 4 = 24\ \text m\end{aligned}$$

(b) The velocity at t = 2 s is

$$\begin{aligned}\vec{\text v} = \vec{\text u} + \vec{\text a}\text t \\[1em] = 10.0\hat{j} + (8.0\hat{i} + 2.0\hat{j}) \times 2 \\[1em] = 16.0\hat{i} + 14.0\hat{j}\ \text{m s}^{-1}\end{aligned}$$

The speed is

$$\begin{aligned}|\vec{\text v}| = \sqrt{(16.0)^2 + (14.0)^2} \\[1em] = \sqrt{256 + 196} = \sqrt{452} \\[1em] = 21.3\ \text{m s}^{-1}\end{aligned}$$

Hence, at t = 2 s the y-coordinate is 24 m and the speed of the particle is 21.3 m s^-1.

The magnitude of $(\hat{i} + \hat{j})$ is

$|\hat{i} + \hat{j}| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$

and the angle it makes with the X-axis is

$\theta = \tan^{-1}\left(\frac{1}{1}\right) = 45^\circ$

The magnitude of $(\hat{i} - \hat{j})$ is

$|\hat{i} - \hat{j}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{2}$

and the angle it makes with the X-axis is

$\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -45^\circ$

The component of a vector $\vec{\text A} \space$along a direction is the scalar product of $\vec{\text A} \space$with the unit vector along that direction.

The unit vector along $(\hat{i} + \hat{j})$ is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$, so the component of $\vec{\text A} = 2\hat{i} + 3\hat{j}$ along it is

$$\begin{aligned}\vec{\text A} \cdot \frac{(\hat{i} + \hat{j})}{\sqrt{2}} = \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} \\[1em] = \frac{2 + 3}{\sqrt{2}} = \frac{5}{\sqrt{2}} \\[1em] = 3.54\end{aligned}$$

The unit vector along $(\hat{i} - \hat{j})$ is $\frac{\hat{i} - \hat{j}}{\sqrt{2}}$, so the component of $\vec{\text A} \space$along it is

$$\begin{aligned}\vec{\text A} \cdot \frac{(\hat{i} - \hat{j})}{\sqrt{2}} = \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} - \hat{j})}{\sqrt{2}} \\[1em] = \frac{2 - 3}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \\[1em] = -0.71\end{aligned}$$

Hence, both $(\hat{i} + \hat{j})$ and $(\hat{i} - \hat{j})$ have magnitude $\sqrt{2}$, directed at 45° and −45° with the X-axis, and the components of $\vec{\text A} \space$along them are 3.54 and −0.71 respectively.

Relations (b) and (e) are true; relations (a), (c) and (d) are false.

Reason —

(a) False. This relation holds only when the acceleration is uniform. For arbitrary motion the velocity may change in any manner, so the average velocity is not the arithmetic mean of the initial and the final velocities.

(b) True. The average velocity is by definition the displacement divided by the corresponding time interval, and this definition holds for every kind of motion.

(c) False. This is an equation of motion derived for uniform acceleration, so it cannot be applied to arbitrary motion in which the acceleration itself changes.

(d) False. This equation is also derived for uniform acceleration and therefore does not hold for arbitrary motion.

(e) True. The average acceleration is by definition the change in velocity divided by the corresponding time interval, and this definition holds for every kind of motion.

(a) False. A scalar need not be conserved. For example, the kinetic energy of a body is a scalar, but it is not conserved in an inelastic collision.

(b) False. A scalar may be negative. For example, temperature on the Celsius scale and gravitational potential energy can both take negative values.

(c) False. A scalar may have dimensions. For example, mass, work and density are scalars having definite dimensions, while only pure numbers such as the refractive index are dimensionless.

(d) False. A scalar may vary from point to point in space. For example, the temperature or the pressure of the air in a room is different at different points.

(e) True. A scalar has only magnitude and no direction, so its value does not depend on the orientation of the coordinate axes chosen by the observer. Hence all observers, however their axes are oriented, obtain the same value of a scalar.

Given,

• Height of the aircraft above the ground, h = 3400 m
• Angle subtended at the observation point, 2θ = 30°, so θ = 15°
• Time interval, t = 10 s

Let the aircraft move from position A to position B in 10 s, and let O be the observation point on the ground. The perpendicular from O meets AB at its mid-point C, so that

$\angle \text{AOC} = \angle \text{BOC} = 15^\circ$

In the right-angled triangle AOC,

$\tan 15^\circ = \frac{\text{AC}}{\text{OC}} \quad \Rightarrow \quad \text{AC} = \text{OC} \times \tan 15^\circ$

Substituting OC = 3400 m and tan 15° = 0.2679,

$\text{AC} = 3400 \times 0.2679 = 910.9\ \text m$

The total distance travelled by the aircraft is

$\text{AB} = 2 \times \text{AC} = 2 \times 910.9 = 1821.8\ \text m$

Therefore, the speed of the aircraft is

$$\begin{aligned}\text v = \frac{\text{AB}}{\text t} = \frac{1821.8}{10} \\[1em] = 182.2\ \text{m s}^{-1}\end{aligned}$$

Hence, the speed of the aircraft is 182.2 m s^-1.

Given,

• Speed of the rain (vertically downward), v_r = 30 m s^-1
• Speed of the woman (north to south), v_w = 10 m s^-1

To protect herself from the rain, the woman must hold her umbrella along the direction of the relative velocity of the rain with respect to herself,

$\vec{\text v}_{rw} = \vec{\text v}_r - \vec{\text v}_w = \vec{\text v}_r + (-\vec{\text v}_w)$

Since $\vec{\text v}_r$ is vertically downward and $(-\vec{\text v}_w)$ is horizontal (from south to north), the two are mutually perpendicular. If θ is the angle made by $\vec{\text v}_{rw}$ with the vertical,

$$\begin{aligned}\tan \theta = \frac{\text v_w}{\text v_r} = \frac{10}{30} = \frac{1}{3} \\[1em] \theta = \tan^{-1}\left(\frac{1}{3}\right) = 18.4^\circ\end{aligned}$$

The magnitude of the relative velocity is

$$\begin{aligned}\text v_{rw} = \sqrt{\text v_r^2 + \text v_w^2} = \sqrt{(30)^2 + (10)^2} \\[1em] = \sqrt{900 + 100} = \sqrt{1000} \\[1em] = 31.6\ \text{m s}^{-1}\end{aligned}$$

Hence, the woman should hold her umbrella at about 18.4° with the vertical, tilted towards the south, that is, towards her direction of motion, and the rain strikes the umbrella with a speed of 31.6 m s^-1.

Given,

• Speed of the man in still water, v_m = 4 km h^-1
• Speed of the river current, v_r = 3 km h^-1
• Width of the river, d = 1 km

Since the man takes his strokes normal to the current, his full swimming speed is used in crossing the river. The horizontal motion (along the current) and the motion across the river are independent of each other. Hence the time taken to cross the river is

$$\begin{aligned}\text t = \frac{\text{Width of the river}}{\text{Speed of the man}} = \frac{1}{4}\ \text h \\[1em] = \frac{1}{4} \times 60 = 15\ \text{min}\end{aligned}$$

During this time the current carries him downstream through a distance

$$\begin{aligned}\text x = \text v_r \times \text t = 3 \times \frac{1}{4} \\[1em] = 0.75\ \text{km} = 750\ \text m\end{aligned}$$

Hence, the man takes 15 min to cross the river and is carried 750 m down the river.

Given,

• Speed of the plane, v = 720 km h^-1 = $720 \times \frac{5}{18}$ = 200 m s^-1
• Muzzle speed of the shell, u = 600 m s^-1
• Acceleration due to gravity, g = 10 m s^-2

Let the gun be fired at an angle θ from the vertical. For the shell to hit the plane, the horizontal component of the velocity of the shell must be equal to the speed of the plane, so that both cover the same horizontal distance in the same time. Therefore,

$$\begin{aligned}\text u \sin \theta = \text v \\[1em] 600 \sin \theta = 200 \\[1em] \sin \theta = \frac{1}{3} = 0.3333\end{aligned}$$

Hence,

$\theta = \sin^{-1}(0.3333) = 19.5^\circ$

The vertical component of the muzzle velocity is

$$\begin{aligned}\text u \cos \theta = 600 \times \sqrt{1 - \frac{1}{9}} = 600 \times \frac{2\sqrt{2}}{3} \\[1em] = 400\sqrt{2}\ \text{m s}^{-1}\end{aligned}$$

The maximum height reached by the shell is

$$\begin{aligned}\text H = \frac{(\text u \cos \theta)^2}{2\text g} = \frac{(400\sqrt{2})^2}{2 \times 10} \\[1em] = \frac{320000}{20} \\[1em] = 16000\ \text m = 16\ \text{km}\end{aligned}$$

Hence, the gun should be fired at 19.5° from the vertical, and the pilot must fly the plane at a minimum altitude of 16 km to avoid being hit.

Given,

• Speed of the wind, v_w = 72 km h^-1 along the north-east direction
• Speed of the boat, v_B = 51 km h^-1 towards the north

When the boat is anchored, the flag flutters along the direction of the wind, that is, along the north-east direction. Resolving the velocity of the wind along the east and the north directions,

$$\begin{aligned}\text v_{w(E)} = 72 \cos 45^\circ = \frac{72}{\sqrt{2}} = 50.9\ \text{km h}^{-1} \\[1em] \text v_{w(N)} = 72 \sin 45^\circ = \frac{72}{\sqrt{2}} = 50.9\ \text{km h}^{-1}\end{aligned}$$

When the boat moves, the flag flutters along the direction of the relative velocity of the wind with respect to the boat,

$\vec{\text v}_{wB} = \vec{\text v}_w - \vec{\text v}_B$

Its components are

$$\begin{aligned}\text{East component} = 50.9 - 0 = 50.9\ \text{km h}^{-1} \\[1em] \text{North component} = 50.9 - 51 = -0.1\ \text{km h}^{-1}\end{aligned}$$

The north component is almost zero, while the east component is 50.9 km h^-1. Hence the relative velocity of the wind with respect to the boat is directed very nearly along the east.

Hence, the flag on the mast of the moving boat flutters almost exactly along the east direction.

Given,

• Speed of the cyclist, v = 27 km h^-1 = $27 \times \frac{5}{18}$ = 7.5 m s^-1
• Radius of the circular turn, r = 80 m
• Tangential retardation, a_T = 0.5 m s^-2

The centripetal acceleration, directed along the radius towards the centre, is

$$\begin{aligned}\text a_c = \frac{\text v^2}{\text r} = \frac{(7.5)^2}{80} \\[1em] = \frac{56.25}{80} \\[1em] = 0.70\ \text{m s}^{-2}\end{aligned}$$

The tangential acceleration, directed along the tangent opposite to the motion, is a_T = 0.5 m s^-2. Since the two are mutually perpendicular, the magnitude of the net acceleration is

$$\begin{aligned}\text a = \sqrt{\text a_c^2 + \text a_T^2} = \sqrt{(0.70)^2 + (0.5)^2} \\[1em] = \sqrt{0.49 + 0.25} = \sqrt{0.74} \\[1em] = 0.86\ \text{m s}^{-2}\end{aligned}$$

If β is the angle made by the net acceleration with the direction of the tangential acceleration,

$$\begin{aligned}\tan \beta = \frac{\text a_c}{\text a_T} = \frac{0.70}{0.5} = 1.4 \\[1em] \beta = \tan^{-1}(1.4) = 54.5^\circ\end{aligned}$$

Hence, the net acceleration of the cyclist is 0.86 m s^-2, making an angle of 54.5° with the direction of the tangential acceleration.

a quantity with both magnitude and direction, following vector addition rules

Reason — The quantities which have both magnitude and direction and obey the laws of vector addition are called vector quantities. Merely having a magnitude and a direction is not enough, and this is why option 4 is incomplete. For example, electric current has both magnitude and direction but it is a scalar quantity, since it does not obey the laws of vector addition (the triangle or parallelogram law).

another vector

Reason — The subtraction of a vector $\vec{\text B} \space$from a vector $\vec{\text A} \space$means the addition of $(-\vec{\text B}) \space$to $\vec{\text A}$, that is, $\vec{\text A} - \vec{\text B} = \vec{\text A} + (-\vec{\text B})$. Since the sum of two vectors is again a vector, the difference of two vectors is also a vector. It becomes a zero vector only in the special case when the two vectors are equal.

one

Reason — A unit vector is a vector that has a magnitude of exactly one unit. It is obtained by dividing a vector by its own magnitude, $\hat{\text A} = \frac{\vec{\text A}}{|\vec{\text A}|}$, and it retains the direction of the original vector while scaling its magnitude to one. It is used to indicate direction without regard to magnitude.

Sum of two equal vectors

Reason — The position vector of the origin itself is a zero vector, since its initial and terminal points coincide. The acceleration vector of a body moving with constant velocity is a zero vector, since the velocity does not change. The difference of two equal vectors is a zero vector, because $\vec{\text A} - \vec{\text B} = \vec{0} \space$if $\vec{\text A} = \vec{\text B}$. But the sum of two equal vectors is $\vec{\text A} + \vec{\text A} = 2\vec{\text A}$, which has twice the magnitude of $\vec{\text A} \space$and is therefore not a zero vector.

anywhere

Reason — A free vector is a vector that can be moved anywhere in space without changing its properties, that is, its magnitude and direction remain unchanged. It is not tied to a specific point. For example, a wind blowing with a velocity of 10 m s^-1 to the north is the same free vector at every location in a field.

Force

Reason — Bound vectors are associated with a specific point in space, that is, they are functions of the cartesian coordinates of the point of application. A force acts at a definite point of a body and its effect changes if it is shifted to another point, so it is a bound vector. Displacement and velocity are not tied to a particular point and can be drawn anywhere without changing their meaning.

a vector

Reason — On multiplying a vector $\vec{\text A} \space$by a scalar k, a vector $\vec{\text R} = \text k\vec{\text A}$is obtained whose magnitude is k times the magnitude of $\vec{\text A} \space$and whose direction is the same as that of $\vec{\text A}$(opposite if k is negative). A vector scaled by a scalar therefore remains a vector. For example, multiplying velocity $\vec{\text v} \space$by mass m gives the vector quantity momentum $\vec{\text p} = \text m\vec{\text v}$.

a zero vector

Reason — One of the properties of the zero vector is that the multiplication of a finite vector $\vec{\text A} \space$by zero is equal to the zero vector, that is, $0\vec{\text A} = \vec{0}$. The resulting vector has zero magnitude and no specific direction, and its initial and terminal points are coincident.

parabolic

Reason — For a body projected with velocity u at an angle θ₀ with the horizontal, the equation of the trajectory is

$\text y = \text x \tan \theta_0 - \frac{\text{gx}^2}{2\text u^2 \cos^2 \theta_0}$

This equation is quadratic in x and linear in y, so it represents a parabola. Hence the path of a projectile, called its trajectory, is parabolic.

remains constant

Reason — The horizontal and the vertical motions of a projectile take place independent of each other. No force acts on the projectile along the horizontal direction (air resistance being neglected), and the force of gravity is vertical, so it changes only the vertical component of velocity. Hence the horizontal component u_x = u cos θ remains unchanged throughout the flight, including at the highest point.

45°

Reason — The horizontal range of a projectile is

$\text R = \frac{\text u^2 \sin 2\theta_0}{\text g}$

For a given speed of projection u, the range is maximum when sin 2θ₀ is maximum, that is, when sin 2θ₀ = 1 or 2θ₀ = 90°, giving θ₀ = 45°. This is why a long-jumper takes his jump at an angle of about 45°.

both initial velocity and angle of projection

Reason — The horizontal range is given by $\text R = \frac{\text u^2 \sin 2\theta_0}{\text g}$. A higher initial velocity allows the projectile to cover a greater distance, and the angle of projection decides the trajectory and hence how long the projectile stays in the air and how far it travels horizontally. The range also depends on g, but not on g alone.

square of the initial velocity

Reason — The maximum height attained by a projectile is

$\text h = \frac{\text u^2 \sin^2 \theta_0}{2\text g}$

For a fixed angle of projection, h ∝ u². The greatest height for a given speed is obtained by putting θ₀ = 90°, which gives $\text h_{max} = \frac{\text u^2}{2\text g}$, again directly proportional to the square of the velocity of projection.

parabolic path

Reason — When a projectile is launched from a height, its initial velocity is resolved into a horizontal component and a vertical component. The horizontal motion is uniform because no horizontal force acts on it, while the vertical motion is uniformly accelerated due to gravity. The combination of a uniform horizontal motion and a uniformly accelerated vertical motion produces a parabolic trajectory, whatever the launch height may be.

zero

Reason — The scalar product of two vectors is $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$. If the two vectors are perpendicular, θ = 90° and cos 90° = 0, so

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos 90^\circ = 0$

Thus the scalar product of two mutually perpendicular vectors is zero even though neither of them is a null vector. For example, the work done is zero when the force is perpendicular to the displacement.

$6\hat{i} - 6\hat{j}$

Reason — The relative velocity of a body B with respect to a body A is

$\vec{\text v}_{BA} = \vec{\text v}_B - \vec{\text v}_A$

Substituting the values,

$$\begin{aligned}\vec{\text v}_{BA} = (11\hat{i} + 5\hat{j}) - (5\hat{i} + 11\hat{j}) \\[1em] = (11 - 5)\hat{i} + (5 - 11)\hat{j} \\[1em] = 6\hat{i} - 6\hat{j}\end{aligned}$$

Note: The textbook’s printed answer key gives option 2, which represents only the velocity of car A. However, book's own solution hint correctly calculates the relative velocity as $6\hat{i} - 6\hat{j}$. Therefore, the solution given above is correct.

206 m/s

Reason — The velocity of the plane relative to the ground is the vector sum of the velocity of the plane and the velocity of the wind. Since the two are mutually perpendicular,

$$\begin{aligned}\text v = \sqrt{(200)^2 + (50)^2} \\[1em] = \sqrt{40000 + 2500} = \sqrt{42500} \\[1em] = 206\ \text{m s}^{-1}\end{aligned}$$

0.8

Reason — On a flat curved road the necessary centripetal force is provided entirely by the force of friction between the tyres and the road. Hence

$\mu \text{mg} = \frac{\text{mv}^2}{\text r} \quad \Rightarrow \quad \mu = \frac{\text v^2}{\text{rg}}$

Substituting the values,

$$\begin{aligned}\mu = \frac{(20)^2}{50 \times 9.8} \\[1em] = \frac{400}{490} \\[1em] = 0.82 \approx 0.8\end{aligned}$$

7.5 m/s²

Reason — The centripetal acceleration of a body moving on a circular path of radius r with speed v is

$\text a = \frac{\text v^2}{\text r}$

Substituting the values,

$$\begin{aligned}\text a = \frac{(15)^2}{30} = \frac{225}{30} \\[1em] = 7.5\ \text{m s}^{-2}\end{aligned}$$

Note: The textbook’s printed answer key gives option 2, which is $2.5\ \text{m s}^{-2}$. However, the book's own solution hint correctly calculates the acceleration as $7.5\ \text{m s}^{-2}$. Therefore, the solution given above is correct.

13 units

Reason — By the parallelogram law of vector addition, the magnitude of the resultant is

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

Substituting the values,

$$\begin{aligned}\text R = \sqrt{(5)^2 + (10)^2 + 2 \times 5 \times 10 \times \cos 60^\circ} \\[1em] = \sqrt{25 + 100 + 100 \times \frac{1}{2}} \\[1em] = \sqrt{175} = 13.2 \approx 13\ \text{units}\end{aligned}$$

5

Reason — If a vector is written in terms of its rectangular components as $\vec{\text A} = \text A_x\hat{i} + \text A_y\hat{j}$, then its magnitude is

$\text A = \sqrt{\text A_x^2 + \text A_y^2}$

Substituting the values,

$$\begin{aligned}\text A = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} \\[1em] = \sqrt{25} = 5\end{aligned}$$

5

Reason — The scalar product of two vectors is equal to the sum of the products of their corresponding x-, y- and z-components,

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

Substituting the values,

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (2)(4) + (3)(-1) \\[1em] = 8 - 3 = 5\end{aligned}$$

only vertical velocity

Reason — The time taken by a projectile to reach the highest point is obtained from v = u − gt by putting v = 0 and u = u sin θ₀, which gives $\text t = \frac{\text u \sin \theta_0}{\text g}$. Hence the time of flight is

$\text T = 2\text t = \frac{2\text u \sin \theta_0}{\text g}$

Only the vertical component u sin θ₀ appears in this expression, so the time of flight depends on the vertical component of velocity alone.

$(0, v_0 \cos θ)$

Reason — At the highest point of the trajectory the projectile stops rising, so the vertical component of its velocity becomes zero. The horizontal component is unaffected by gravity and remains unaltered at v₀ cos θ throughout the flight. Hence at the peak the vertical and horizontal components are 0 and v₀ cos θ respectively.

To balance the centrifugal force acting outward.

Reason — When a cyclist takes a turn, an outward centrifugal force acts on him due to his inertia. By leaning inward he counteracts this force, so that the combined effect of the gravitational force and the normal reaction points towards the centre of the curve and supplies the necessary centripetal force. This keeps him in balance while negotiating the turn.

To provide centripetal force for circular motion.

Reason — A body moving on a circular path needs a net inward force directed towards the centre. On a flat curved road this centripetal force is provided entirely by the friction between the tyres and the road. In the absence of this force the car would continue to move along a straight line because of its inertia, and would not be able to follow the curve.

Because of the object's inertia resisting the centripetal force.

Reason — In a non-inertial frame, such as the rotating frame of an observer moving with the object, an outward force appears to act on the object. This centrifugal force is not a real force; it arises from the inertia of the object, which resists the centripetal acceleration directed towards the centre. It is therefore called a pseudo force and has no existence in an inertial frame.

All of these.

Reason — The centripetal force required to keep a body of mass m moving with speed v on a circular path of radius r is

$\text F = \frac{\text{mv}^2}{\text r}$

The expression contains the mass, the velocity as well as the radius, so the magnitude of the centripetal force depends on all three of them.

provide additional centripetal force

Reason — When a road is banked, the outer edge is raised above the inner edge so that the normal reaction becomes inclined to the vertical. Its horizontal component then acts towards the centre of the circular path and supplies an additional centripetal force. The vehicle can therefore take the turn safely without depending on friction alone.

only friction to maintain circular motion

Reason — A flat road has no banking, so the normal reaction is vertical and has no component towards the centre of the circular path. The whole of the centripetal force needed for the circular motion must therefore be supplied by the friction between the tyres and the road.

to balance the forces completely

Reason — At the designed (optimum) speed the horizontal component of the normal reaction by itself provides exactly the centripetal force required for the circular motion. Friction is then not needed for turning, and it only serves to balance the remaining forces, coming into play to correct any excess whenever the speed departs from the designed value.

slide outward

Reason — If the speed exceeds the designed speed for the given banking angle, the centripetal force required becomes greater than that supplied by the horizontal component of the normal reaction. The friction available is then not enough to counteract the outward centrifugal effect, and the vehicle slides outward, away from the centre of the curve.

increase by four times

Reason — On a flat road the friction supplies the whole centripetal force,

$\text F = \frac{\text{mv}^2}{\text r}$

so F ∝ v² for a given mass and radius. If the speed is doubled, the required force becomes

$\text F' = \frac{\text m(2\text v)^2}{\text r} = 4 \times \frac{\text{mv}^2}{\text r} = 4\text F$

that is, four times the original frictional force.

act up the slope

Reason — At a speed lower than the designed speed, the horizontal component of the normal reaction is greater than the centripetal force actually required, so the vehicle tends to slide down the banked surface towards the inner edge. Friction opposes this tendency and therefore acts up the slope, keeping the vehicle moving on its circular path.

The vehicle must have a net inward force to maintain circular motion.

Reason — For circular motion a net inward force, that is, a centripetal force directed towards the centre of the curve, is always required. It can be provided by friction, by the banking of the road, or by both together. Without such an inward force the vehicle would move along a straight line because of its inertia.

zero

Reason — The magnitude of the vector product of two vectors is $|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin \theta$. When the two vectors are parallel the angle between them is 0° (or 180°), and sin 0° = 0, so

$|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin 0^\circ = 0$

Thus the vector product of two parallel vectors is a null vector.

complementary angles

Reason — Substituting (90° − θ₀) in place of θ₀ in the expression for the range,

$$\begin{aligned}\text R = \frac{\text u^2 \sin 2(90^\circ - \theta_0)}{\text g} = \frac{\text u^2 \sin (180^\circ - 2\theta_0)}{\text g} \\[1em] = \frac{\text u^2 \sin 2\theta_0}{\text g}\end{aligned}$$

which is the same as the range for θ₀. Hence the horizontal range is the same for two complementary angles of projection, and a football kicked at 30° or at 60° strikes the ground at the same place.

Z-component

Reason — A vector lying in the XY-plane can be written in terms of its rectangular components as $\vec{\text A} = \text A_x\hat{i} + \text A_y\hat{j}$. Since the vector has no extension along the Z-direction, its component along the Z-axis is necessarily zero, that is, A_z = 0. The X- and Y-components may have any values.

4

Reason — For a projectile,

$\text R = \frac{\text u^2 \sin 2\theta}{\text g} \quad \text{and} \quad \text H = \frac{\text u^2 \sin^2 \theta}{2\text g}$

Therefore,

$\frac{\text R}{\text H} = \frac{\frac{\text u^2 \sin 2\theta}{\text g}}{\frac{\text u^2 \sin^2 \theta}{2\text g}} = \frac{2 \sin 2\theta}{\sin^2 \theta}$

Putting θ = 45°,

$$\begin{aligned}\frac{\text R}{\text H} = \frac{2 \times \sin 90^\circ}{(\sin 45^\circ)^2} = \frac{2}{1/2} \\[1em] = 4\end{aligned}$$

90°

Reason — If α, β and γ are the angles made by a vector with the X-, Y- and Z-axes, the direction cosines satisfy

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

Putting α = β = 45°,

$\frac{1}{2} + \frac{1}{2} + \cos^2 \gamma = 1 \quad \Rightarrow \quad \cos \gamma = 0 \quad \Rightarrow \quad \gamma = 90^\circ$

Thus the vector lies in the XY-plane and is perpendicular to the Z-axis.

$\hat{k}$

Reason — The vector product $\vec{\text A} \times \vec{\text B} \space$gives a vector perpendicular to the plane containing $\vec{\text A} \space$and $\vec{\text B}$. Here both vectors lie in the XY-plane, so their cross product must be along the Z-axis,

$$\vec{\text A} \times \vec{\text B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 3 & 1 & 0 \end{vmatrix} \\[1em] = \hat{k}(1 \times 1 - 2 \times 3) = -5\hat{k}$$

The unit vector perpendicular to both is therefore $\hat{k}$ (directed along the Z-axis).

76°

Reason — Equating the horizontal range and the maximum height,

$\frac{\text u^2 \sin 2\theta}{\text g} = \frac{\text u^2 \sin^2 \theta}{2\text g}$

so that

$$\begin{aligned}2 \sin 2\theta = \sin^2 \theta \\[1em] 2 \times 2 \sin \theta \cos \theta = \sin^2 \theta \\[1em] \tan \theta = 4\end{aligned}$$

Since tan 76° = 4, the angle of projection is θ ≈ 76°.

At an angle of 60° upstream

Reason — For the boat to move directly across the river, the component of its velocity along the flow must exactly cancel the velocity of the river. If θ is the angle made by the heading of the boat with the line joining the two banks (that is, with the direction straight across),

$$\begin{aligned}\text v_{river} = \text v_{boat} \sin \theta \\[1em] 5 = 10 \sin \theta \quad \Rightarrow \quad \sin \theta = 0.5 \quad \Rightarrow \quad \theta = 30^\circ\end{aligned}$$

So the boat must head 30° upstream of the line straight across, which is the same as making an angle of 60° with the upstream bank. Measured from the bank, the required heading is therefore 60° upstream.

20 m/s

Reason — The horizontal component of the velocity of a projectile remains constant, so the horizontal motion is uniform. Hence

$\text x = \text u_x \text t \quad \Rightarrow \quad \text u_x = \frac{\text x}{\text t}$

Substituting the values,

$\text u_x = \frac{40}{2} = 20\ \text{m s}^{-1}$

Due North

Reason — The time taken to cross the river depends only on the component of the swimmer's velocity perpendicular to the banks. This component is greatest when the man swims directly due north, since then his full speed of 10 m min^-1 is used in crossing. The current will carry him some distance to the west, but the drift does not affect the time of crossing, so the time taken is the shortest.

Note: The textbook’s printed answer key gives option 2. However, the book's own solution hint correctly states that the man should swim directly towards the north. Therefore, the solution given above is correct.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a zero resultant it must be possible to represent the given vectors by the sides of a closed polygon taken in the same order. Since two vectors of different magnitudes cannot be added to get a zero resultant, and the minimum number of sides of a closed polygon is three, at least three unequal vectors are needed in a plane to give a zero resultant.

Reason (R) is also correct: If $\vec{\text A} + \vec{\text B} + \vec{\text C} = 0$, the three vectors form a closed triangle, and a triangle always lies in one plane. Hence the three vectors must be coplanar.

However, the Reason only states that three such vectors are coplanar. It does not explain why the minimum number required is three.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The vector product $\vec{\text A} \times \vec{\text {B }}$is perpendicular to both the sum and the difference of $\vec{\text {A }}$and $\vec{\text B}$.

Reason (R) is also correct: The vector product of two vectors has the direction perpendicular to the plane containing the two vectors, its sense being given by the right-hand screw rule. Both $\vec{\text A} + \vec{\text {B }}$and $\vec{\text A} - \vec{\text {B }}$are obtained by the laws of vector addition from $\vec{\text {A }}$and $\vec{\text B}$, so they lie in the same plane as $\vec{\text {A }}$and $\vec{\text B}$. A vector perpendicular to this plane is therefore perpendicular to every vector lying in it.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Taking the scalar product,

$\cos \theta = \frac{(\hat{i} + \hat{j}) \cdot \hat{i}}{|\hat{i} + \hat{j}||\hat{i}|} = \frac{1}{\sqrt{2} \times 1} = \frac{1}{\sqrt{2}}$

so that θ = 45°.

Reason (R) is also correct: Since $\hat{i}$ and $\hat{j}$ are unit vectors of equal magnitude, their sum lies along the diagonal of the square formed by them and is therefore equally inclined to both. As the angle between $\hat{i}$ and $\hat{j}$ is 90°, this diagonal bisects it, giving an angle of $\frac{90^\circ}{2} = 45^\circ$ with each. This is exactly why the angle in the Assertion is 45°.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A vector which is perpendicular to both $\vec{\text {A }}$and $\vec{\text {B }}$must lie along the normal to the plane containing them, and so must be parallel (or antiparallel) to $\vec{\text A} \times \vec{\text B}$.

Reason (R) is also correct: By definition, the vector product of two vectors has the direction perpendicular to the plane containing the two vectors, its sense being decided by the right-hand screw rule. Since there is only one direction normal to a given plane, any vector perpendicular to both $\vec{\text {A }}$and $\vec{\text {B }}$must be along that normal, which is the direction of $\vec{\text A} \times \vec{\text B}$.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: In javelin throw and discus throw the projectile is released from a place above the surface of the earth, and for such a projection the maximum range is obtained for an angle of projection slightly less than 45°, not more than 45°. Hence the player throws the projectile at an angle slightly less than 45° to the horizontal.

Reason (R) is also false: The horizontal range $\text R = \frac{\text u^2 \sin 2\theta_0}{\text g}$ clearly depends on the angle of projection, and it becomes maximum only for a particular angle of projection (45° for projection from the ground). Hence the maximum range does depend upon the angle of projection.

Therefore, both assertion and reason are false.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: At the highest point of the trajectory it is the vertical component of velocity that becomes zero, while the horizontal component u cos θ₀ remains unaltered. Hence the projectile has only the horizontal component of velocity there, not the vertical component.

Reason (R) is correct: At the highest point only one component of velocity, namely the horizontal component, is present, since the vertical component has been reduced to zero by gravity.

Therefore, assertion is false but reason is true.

(i) Scalars : mass, pressure, work. Vectors : displacement, acceleration.

(ii) Scalars : energy, density, temperature, speed. Vectors : momentum, impulse.

(iii) Scalars : surface tension, temperature. Vectors : moment of a force, momentum.

(iv) Pressure is a scalar quantity, since it is completely specified by its magnitude alone and does not obey the laws of vector addition.

If the direction of a vector is reversed, the sign of the vector is reversed, and on multiplying a vector by a scalar k the magnitude becomes k times while the direction remains the same (opposite if k is negative).

(a) $-\vec{\text A}$: magnitude 2.5 m, directed towards south.

(b) $\frac{\vec{\text A}}{2.0}$ : magnitude 1.25 m, directed towards north.

(c) $-2.5\vec{\text A}$: magnitude 6.25 m, directed towards south.

(d) $4.0\vec{\text A}$: magnitude 10 m, directed towards north.

No, a scalar quantity cannot be added to a vector quantity, because a scalar has only magnitude whereas a vector has magnitude as well as direction. Even if the two have the same dimensions, they cannot be added.

Yes, their product is possible. On multiplying a vector $\vec{\text {A }}$by a scalar k, a vector $\text k\vec{\text {A }}$is obtained whose magnitude is k times that of $\vec{\text A}$. For example, multiplying the velocity $\vec{\text {v }}$by the mass m gives the momentum $\vec{\text p} = \text m\vec{\text v}$.

Yes. The addition of vectors is commutative, that is,

$\vec{\text A} + \vec{\text B} = \vec{\text B} + \vec{\text A}$

The two sums are represented by vectors which are parallel and of equal length, so their magnitudes as well as their directions are the same.

Yes, the magnitudes are the same, but the directions are opposite. Since

$\vec{\text A} - \vec{\text B} = -(\vec{\text B} - \vec{\text A})$

the two difference vectors are parallel and equal in length, but they point in opposite directions.

The magnitude of the sum of two vectors is

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

The sum is maximum when the two vectors are in the same direction, that is, θ = 0°, and then R = A + B.

The sum is minimum when the two vectors are in opposite directions, that is, θ = 180°, and then R = A − B (or B − A, whichever is positive).

Yes, this is possible when the angle between the two given vectors is more than 90°. For example, if two vectors of magnitudes 3 units and 4 units are oppositely directed (θ = 180°), the magnitude of the resultant is

$\text R = 4 - 3 = 1\ \text{unit}$

which is less than the magnitude of either of the given vectors.

The resultant will be a vector $\text m\vec{\text A} \space$whose magnitude is m times the magnitude of $\vec{\text A}$. Its direction is the same as that of $\vec{\text A} \space$if m is positive, and opposite to that of $\vec{\text A} \space$if m is negative. If m is a pure number, the resultant represents the same physical quantity as $\vec{\text A}$; if m carries a unit, the resultant represents a new physical quantity.

Work is a scalar quantity. It is the scalar (dot) product of the two vectors $\vec{\text F} \space$and $\vec{\text s}$, and the scalar product of two vectors is always a scalar,

$\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$

Yes. The scalar product is $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$. When the angle between the two vectors is obtuse, that is, when 90° < θ ≤ 180°, cos θ is negative and hence the scalar product is negative. For example, the work done by the force of friction is negative because the friction acts opposite to the displacement.

The scalar product $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$ is maximum when cos θ is maximum, that is, when θ = 0° and the two vectors are parallel (in the same direction). Its maximum value is then

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos 0^\circ = \text{AB}$

The scalar product is numerically minimum, that is, zero, when the two vectors are mutually perpendicular (θ = 90°), because

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos 90^\circ = 0$

Thus the scalar product vanishes even though neither of the two vectors is a null vector.

90°. At the highest point of the trajectory the vertical component of the velocity is zero and only the horizontal component u cos θ₀ remains, so the velocity is directed horizontally. The acceleration of the projectile is the acceleration due to gravity, which acts vertically downwards. Hence the angle between the velocity and the acceleration there is 90°.

The speed is minimum at the highest point of the path, since there the vertical component of velocity is zero and only the horizontal component u cos θ₀ is left.

The speed is maximum at the point of projection (and again at the point where the projectile strikes the ground, since it returns to the ground with the same speed with which it was projected).

Yes, it is important. The horizontal range of the jumper is

$\text R = \frac{\text u^2 \sin 2\theta_0}{\text g}$

which is maximum for θ₀ = 45°. To be projected at 45° the jumper must gain a certain height, since the height attained is $\text h = \frac{\text u^2 \sin^2 \theta_0}{2\text g}$. If he takes too little height the angle of projection is too small, and if he takes too much height the angle is too large; in both cases the range becomes less than the maximum.

Both bombs will reach the ground at the same time. The time of flight is

$\text T = \frac{2\text u \sin \theta_0}{\text g}$

which does not contain the mass of the projectile. Since both bombs are thrown with the same velocity in the same direction, their time of flight is the same, whatever their masses.

If the bombs are thrown in the same direction but with different velocities, their vertical components of velocity are different. The bomb having the greater vertical component u sin θ₀ will have the longer time of flight, so the bomb thrown with the smaller velocity will reach the ground first.

The magnitude of the resultant of two vectors is $\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$, with A = 3 m and B = 4 m.

(a) For R = 7 m, the resultant is maximum, so the two vectors must be added in the same direction, that is, θ = 0°.

(b) For R = 1 m, the resultant is minimum, so the two vectors must be added in opposite directions, that is, θ = 180°.

(c) For R = 5 m,

$$\begin{aligned}5 = \sqrt{(3)^2 + (4)^2 + 2 \times 3 \times 4 \times \cos \theta} \\[1em] 25 = 25 + 24 \cos \theta \quad \Rightarrow \quad \cos \theta = 0 \quad \Rightarrow \quad \theta = 90^\circ\end{aligned}$$

So the two vectors must be added at right angles to each other.

No. The minimum magnitude of the sum of two vectors is (A − B), which is obtained when they are oppositely directed. This is zero only if A = B. Hence two vectors of different magnitudes cannot be added to get a zero resultant.

Yes. If three vectors can be represented in magnitude and direction by the three sides of a triangle taken in the same (cyclic) order, they themselves complete a closed figure and their sum-vector cannot be drawn. This means that the sum of these three vectors is zero.

Both the magnitude and the direction of the resultant will change. Since

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta} \quad \text{and} \quad \tan \alpha = \frac{\text B \sin \theta}{\text A + \text B \cos \theta}$

the magnitude of the resultant decreases from its maximum value (A + B) at θ = 0° to its minimum value (A − B) at θ = 180°, and for other angles it lies between these two values. The angle α made by the resultant with $\vec{\text A} \space$also changes.

Statement (a) is true. The scalar product of two vectors is commutative, since

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta = \text{BA} \cos \theta = \vec{\text B} \cdot \vec{\text A}$

(It is the vector product that is not commutative, for which $\vec{\text A} \times \vec{\text B} = -\vec{\text B} \times \vec{\text A}$.)

(a) Since $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta = 0$ and neither vector is a null vector, cos θ = 0, so θ = 90°. Hence the two vectors are mutually perpendicular.

(b) Since $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta = \text{AB}$, we get cos θ = 1, so θ = 0°. Hence the two vectors are parallel, that is, they are in the same direction.

No. From the given relation,

$\vec{\text A} \cdot \vec{\text C} - \vec{\text B} \cdot \vec{\text C} = 0 \quad \Rightarrow \quad (\vec{\text A} - \vec{\text B}) \cdot \vec{\text C} = 0$

This is satisfied either when $\vec{\text A} = \vec{\text B}$, or when $(\vec{\text A} - \vec{\text B}) \space$is perpendicular to $\vec{\text C}$. Hence $\vec{\text A} \space$and $\vec{\text B} \space$need not always be equal.

Equating the magnitudes,

$\sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta} = \sqrt{\text A^2 + \text B^2 - 2\text{AB}\cos \theta}$

which gives 4AB cos θ = 0, so cos θ = 0 and θ = 90°.

Hence the sum and the difference of two vectors are equal in magnitude when the two vectors are mutually perpendicular.

The sum $\vec{\text A} + \vec{\text B} \space$and the difference $\vec{\text A} - \vec{\text B} \space$can point in the same direction only when $\vec{\text B} \space$lies along the same line as $\vec{\text A}$, that is, when the two vectors are collinear (θ = 0° or θ = 180°).

If θ = 0°, the sum has magnitude A + B and the difference has magnitude A − B, both directed along $\vec{\text A} \space$provided A > B.

If θ = 180°, the sum has magnitude A − B and the difference has magnitude A + B, and again both are directed along $\vec{\text A} \space$provided A > B.

In either case, if A < B the difference reverses relative to the sum.

Hence the directions of the sum and the difference are the same when the two vectors are collinear, that is θ = 0° or θ = 180°, and the magnitude of the vector being subtracted is the smaller of the two, that is A > B.

(i) Work is the scalar product of force and displacement, $\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$.

(ii) Power is the scalar product of force and velocity, $\text P = \vec{\text F} \cdot \vec{\text v} = \text{Fv} \cos \theta$.

(i) Torque is the vector product of the position vector and the force, $\vec{\tau} = \vec{\text r} \times \vec{\text F}$, and its magnitude is τ = rF sin θ.

(ii) Angular momentum is the vector product of the position vector and the linear momentum, $\vec{\text L} = \vec{\text r} \times \vec{\text p}$, and its magnitude is L = rp sin θ.

The work done by the force is the scalar product of the force and the displacement,

$\text W = \vec{\text F} \cdot \vec{\text d} = \text{Fd} \cos \theta$

where θ is the angle between the directions of $\vec{\text F} \space$and $\vec{\text d}$.

Given,

• Horizontal range = maximum height, that is, R = h

Equating the two,

$\frac{\text u^2 \sin 2\theta_0}{\text g} = \frac{\text u^2 \sin^2 \theta_0}{2\text g}$

so that

$$\begin{aligned}2 \sin 2\theta_0 = \sin^2 \theta_0 \\[1em] 2 \times 2 \sin \theta_0 \cos \theta_0 = \sin^2 \theta_0 \\[1em] \tan \theta_0 = 4\end{aligned}$$

Since tan 76° = 4,

$\theta_0 = 76^\circ$

Hence, the angle of projection is 76°.

Work is the scalar product of force and displacement,

$\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$

The scalar product of two vectors is always a scalar, because the product Fs cos θ has magnitude only and no direction is associated with it. This is why work is a scalar even though both force and displacement are vectors.

If $\vec{\text F} \space$and $\vec{\text s} \space$are both non-zero, the work can still be zero when θ = 90°, since cos 90° = 0. Thus when the force acting on a body is perpendicular to its displacement, the work done by that force is zero. For example, the centripetal force does no work on a body in circular motion, since it is always perpendicular to the displacement.

No. The vector product of two vectors is itself a vector, and a vector cannot be called positive or negative. Its magnitude AB sin θ is non-negative, since sin θ is never negative for any angle from 0° to 180°. Only the direction of the vector product is reversed when the order of the vectors is reversed, because

$\vec{\text A} \times \vec{\text B} = -\vec{\text B} \times \vec{\text A}$

For a given velocity of projection u, the range

$\text R = \left(\frac{\text u^2}{\text g}\right) \sin 2\theta$

is maximum when sin 2θ has its greatest value, that is, when

$\sin 2\theta = 1 \quad \Rightarrow \quad 2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ$

Hence, the range is maximum for an angle of projection of 45°, and then $\text R_{max} = \frac{\text u^2}{\text g}$.

The player should throw the ball at an angle of 45° with the horizontal. For a given speed of projection the horizontal range $\text R = \frac{\text u^2 \sin 2\theta_0}{\text g}$ is maximum when sin 2θ₀ = 1, that is, when θ₀ = 45°. This is also the reason why a long-jumper takes his jump at an angle of about 45°.

(i) Here the vector (6 km/h, west) is multiplied by the pure number 5. Since a pure number has no unit, the product represents the same physical quantity, that is, a velocity,

$5 \times (6\ \text{km h}^{-1},\ \text{west}) = 30\ \text{km h}^{-1},\ \text{west}$

(ii) Here the vector is multiplied by the scalar quantity 5 h, which carries a unit. The product therefore represents a new physical quantity, that is, a displacement,

$(5\ \text h) \times (6\ \text{km h}^{-1},\ \text{west}) = 30\ \text{km},\ \text{west}$

Hence, the first is a velocity of 30 km h^-1 towards the west, while the second is a displacement of 30 km towards the west.

Yes, this happens when the angle between the two given vectors is more than 90°, because then the term 2AB cos θ in

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

is negative and it reduces the magnitude of the resultant.

For example, let A = 3 units and B = 4 units.

If θ = 180°, the magnitude of the resultant is

$\text R = \text B - \text A = 4 - 3 = 1\ \text{unit}$

which is less than the magnitude of either of the given vectors.

Again, if θ = 120°,

$$\begin{aligned}\text R = \sqrt{(3)^2 + (4)^2 + 2 \times 3 \times 4 \times \left(-\frac{1}{2}\right)} \\[1em] = \sqrt{9 + 16 - 12} = \sqrt{13} \\[1em] = 3.6\ \text{units}\end{aligned}$$

which is less than 4 units, the magnitude of the larger vector.

Given,

• Magnitude of each force, A = B = 3 N
• Magnitude of the resultant, R = 3 N

By the parallelogram law of vector addition,

$\text R^2 = \text A^2 + \text B^2 + 2\text{AB}\cos \theta$

Substituting the values,

$$\begin{aligned}(3)^2 = (3)^2 + (3)^2 + 2 \times 3 \times 3 \times \cos \theta \\[1em] 9 = 9 + 9 + 18 \cos \theta \\[1em] 18 \cos \theta = -9 \\[1em] \cos \theta = -\frac{1}{2}\end{aligned}$$

Therefore,

$\theta = 120^\circ$

Hence, the angle between the two forces is 120°.

(a) Yes. The horizontal and vertical motions of a projectile are independent of each other. The initial vertical velocity of both the balls is zero, and both fall through the same height under the same vertical acceleration g. Since the time of fall is $\text T = \sqrt{\frac{2\text h}{\text g}}$, which does not depend on the horizontal velocity, both balls reach the ground at the same time.

(b) No. On striking the ground the vertical components of the velocities of the two balls are equal, but the ball thrown horizontally also has a horizontal component of velocity which remains unchanged throughout. Hence its resultant velocity $\text v = \sqrt{\text u^2 + \text g^2\text T^2}$ is greater in magnitude and is inclined to the vertical, whereas the dropped ball strikes the ground vertically.

The statement is true.

The equation of the trajectory of a projectile is

$\text y = \text x \tan \theta_0 - \frac{\text{gx}^2}{2\text u^2 \cos^2 \theta_0}$

which is quadratic in x and linear in y, and therefore represents a parabola. Hence the path of a projectile is parabolic.

The horizontal component of the velocity u cos θ₀ remains constant throughout the flight, while the vertical component goes on decreasing as the projectile rises and becomes zero at the highest point. Hence at the highest point only the horizontal component is left, and the speed there,

$\text v = \text u \cos \theta_0$

is the least value of the speed anywhere on the path.

(i) Yes, the speed can remain constant while the velocity changes, since the velocity may change on account of a change of direction alone, as in uniform circular motion. No, if the speed is changing the velocity cannot remain constant, because the speed is the magnitude of the velocity.

(ii) Yes. It is possible if the body is moving on a curved path. Then the direction of the velocity is changing and the motion is accelerated, although the magnitude of the velocity remains constant, as in uniform circular motion.

(iii) Yes, it has an acceleration. Although its speed is constant, the direction of its velocity changes continuously, so its velocity is changing. This gives rise to the centripetal acceleration $\text a = \frac{\text v^2}{\text r}$, directed along the radius towards the centre.

(iv) No. A curved path means that the direction of the velocity is continuously changing, and a change of velocity necessarily means acceleration.

(v) The acceleration is either zero, when the body moves with a uniform velocity, or its direction is always along the line of motion, when the speed is changing. There can be no component of acceleration perpendicular to the path.

(vi) The acceleration is perpendicular to the direction of motion, that is, it is directed along the radius towards the centre of curvature. It changes only the direction of the velocity and not its magnitude, and is therefore called centripetal acceleration.

(vii) Yes, all curved paths are possible. If the acceleration is zero the body moves with constant speed along a straight line; if the magnitude of the acceleration is constant and its direction remains perpendicular to the motion, the body moves with constant speed on a circle; and if the magnitude of the perpendicular acceleration changes, the body moves with constant speed along some other curved path.

(viii) The centripetal force is $\text F = \frac{\text{mv}^2}{\text r}$. On doubling both v and r,

$\text F' = \frac{\text m(2\text v)^2}{2\text r} = \frac{4\text{mv}^2}{2\text r} = 2\text F$

so the centripetal force becomes twice its original value.

(ix) The kinetic energy remains unchanged, since the speed is constant. The velocity and the acceleration both change continuously, as their directions keep changing.

The scalar product (or dot product) of two vectors $\vec{\text A} \space$and $\vec{\text B} \space$is

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$

where,

• A is the magnitude of the vector $\vec{\text A}$
• B is the magnitude of the vector $\vec{\text B}$
• θ is the angle between the directions of $\vec{\text A} \space$and $\vec{\text B}$

The product AB cos θ is a scalar, so the scalar product of two vectors is always a scalar quantity.

In terms of the rectangular components of the two vectors, the same product is written as

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

The vector product (or cross product) of two vectors $\vec{\text A} \space$and $\vec{\text B} \space$is

$\vec{\text A} \times \vec{\text B} = \text{AB} \sin \theta\ \hat{n}$

where,

• A and B are the magnitudes of $\vec{\text A} \space$and $\vec{\text B}$
• θ is the angle between the two vectors
• $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{\text A} \space$and $\vec{\text B}$, its sense being decided by the right-hand screw rule

Example : The torque acting on a particle about the origin O is the vector product of the position vector $\vec{\text r} \space$and the force $\vec{\text F}$,

$\vec{\tau} = \vec{\text r} \times \vec{\text F}$

whose magnitude is τ = rF sin θ and whose direction is perpendicular to the plane containing $\vec{\text r} \space$and $\vec{\text F}$.

The angular velocity is the time rate of change of angular displacement,

$\omega = \frac{\text d\theta}{\text{dt}}$

S.I. unit : radian per second (rad s^-1)

Dimensional formula : Since the angular displacement is a dimensionless quantity,

$[\omega] = \frac{[\text M^0\text L^0\text T^0]}{[\text T]} = [\text M^0\text L^0\text T^{-1}]$

If a particle moves on a circular path of radius r with angular velocity ω, its linear velocity is

$\text v = \text r\omega$

In vector form the relation is written as

$\vec{\text v} = \vec{\omega} \times \vec{\text r}$

Thus, for a given angular velocity the linear velocity of the particle is directly proportional to its distance from the centre; the greater the distance of the particle from the centre, the greater is its linear velocity.

The centripetal acceleration of a particle moving on a circular path of radius r is

$\text a = \frac{\text v^2}{\text r}$

Substituting v = rω,

$$\begin{aligned}\text a = \frac{(\text r\omega)^2}{\text r} \\[1em] = \omega^2 \text r\end{aligned}$$

Hence, in terms of the angular velocity the centripetal acceleration is a = ω²r, directed along the radius towards the centre of the circular path.

The angular velocity is related to the frequency n by

$\omega = 2\pi \text n$

The centripetal acceleration is

$\text a = \omega^2 \text r$

Substituting the value of ω,

$$\begin{aligned}\text a = (2\pi \text n)^2 \text r \\[1em] = 4\pi^2 \text n^2 \text r\end{aligned}$$

Hence, the centripetal acceleration in terms of n and r is a = 4π²n²r.

Given,

• Angle of the force with the horizontal, θ = 30°
• Vertical component of the force, F_y = 200 N

If a force F makes an angle θ with the horizontal, its vertical component is

$\text F_y = \text F \sin \theta$

Substituting the values,

$$\begin{aligned}200 = \text F \sin 30^\circ \\[1em] 200 = \text F \times \frac{1}{2} \\[1em] \text F = 400\ \text N\end{aligned}$$

Hence, the magnitude of the force is 400 N.

Expanding the determinant along the first row,

$\vec{\text A} \times \vec{\text B} = \hat{i}(\text A_y\text B_z - \text A_z\text B_y) - \hat{j}(\text A_x\text B_z - \text A_z\text B_x) + \hat{k}(\text A_x\text B_y - \text A_y\text B_x)$

which may also be written as

$\vec{\text A} \times \vec{\text B} = (\text A_y\text B_z - \text A_z\text B_y)\hat{i} + (\text A_z\text B_x - \text A_x\text B_z)\hat{j} + (\text A_x\text B_y - \text A_y\text B_x)\hat{k}$

The magnitude of the resultant of two vectors $\vec{\text A} \space$and $\vec{\text B} \space$is

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

Here the two vectors are equal in magnitude, so B = A and

$\text R = \sqrt{2\text A^2(1 + \cos \theta)}$

(i) For R = 0 :

$$\begin{aligned}2\text A^2(1 + \cos \theta) = 0 \\[1em] 1 + \cos \theta = 0 \quad \Rightarrow \quad \cos \theta = -1 \quad \Rightarrow \quad \theta = 180^\circ\end{aligned}$$

Hence the resultant is zero when the two vectors are oppositely directed.

(ii) For R = A :

$$\begin{aligned}\text A^2 = 2\text A^2(1 + \cos \theta) \\[1em] \frac{1}{2} = 1 + \cos \theta \quad \Rightarrow \quad \cos \theta = -\frac{1}{2} \quad \Rightarrow \quad \theta = 120^\circ\end{aligned}$$

Hence the resultant is equal in magnitude to each of them when the angle between the two vectors is 120°.

The scalar product of the two vectors is $\vec{\text a} \cdot \vec{\text b} = \text{ab} \cos \theta$. Therefore,

$(\vec{\text a} \cdot \vec{\text b})^2 = \text a^2\text b^2 \cos^2 \theta$

Putting this equal to a²b²,

$$\begin{aligned}\text a^2\text b^2 \cos^2 \theta = \text a^2\text b^2 \\[1em] \cos^2 \theta = 1 \quad \Rightarrow \quad \cos \theta = \pm 1 \quad \Rightarrow \quad \theta = 0^\circ \text{ or } 180^\circ\end{aligned}$$

Hence, the given condition holds when the two vectors are parallel or antiparallel, that is, when they are collinear.

The vector product of two parallel vectors is a null vector, since

$\vec{\text A} \times \vec{\text B} = \text{AB} \sin \theta\ \hat{n} = 0 \quad \text{when } \theta = 0^\circ \text{ or } 180^\circ$

Given that $\vec{\text A} \times \vec{\text B} = 0$ and neither is a null vector, $\vec{\text A} \space$is parallel to $\vec{\text B}$.

Given that $\vec{\text B} \times \vec{\text C} = 0$, similarly $\vec{\text B} \space$is parallel to $\vec{\text C}$.

Since $\vec{\text A} \space$is parallel to $\vec{\text B} \space$and $\vec{\text B} \space$is parallel to $\vec{\text C}$, the vectors $\vec{\text A} \space$and $\vec{\text C} \space$must also be parallel to each other. Hence the angle between them is 0° (or 180°), and

$\vec{\text A} \times \vec{\text C} = \text{AC} \sin 0^\circ\ \hat{n} = 0$

Hence proved.

Let θ be the angle between the two vectors. Then

$|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin \theta \quad \text{and} \quad |\vec{\text A} \cdot \vec{\text B}| = \text{AB} \cos \theta$

Squaring and adding,

$$\begin{aligned}|\vec{\text A} \times \vec{\text B}|^2 + |\vec{\text A} \cdot \vec{\text B}|^2 = \text A^2\text B^2 \sin^2 \theta + \text A^2\text B^2 \cos^2 \theta \\[1em] = \text A^2\text B^2(\sin^2 \theta + \cos^2 \theta) \\[1em] = \text A^2\text B^2 = (\text{AB})^2\end{aligned}$$

Hence proved.

Given,

• Maximum height = horizontal range, that is, h = R

Equating the two,

$\frac{\text u^2 \sin^2 \theta_0}{2\text g} = \frac{\text u^2 \sin 2\theta_0}{\text g}$

so that

$$\begin{aligned}\sin^2 \theta_0 = 2 \sin 2\theta_0 \\[1em] \sin^2 \theta_0 = 2 \times 2 \sin \theta_0 \cos \theta_0 \\[1em] \frac{\sin \theta_0}{\cos \theta_0} = 4 \quad \Rightarrow \quad \tan \theta_0 = 4\end{aligned}$$

Since tan 76° = 4,

$\theta_0 = 76^\circ$

Hence, the angle of projection with the horizontal is 76°.

A geostationary satellite always remains above the same point of the earth's surface, so its period of revolution around the earth is equal to the period of the axial rotation of the earth.

Given,

• Time period, T = 24 h = 24 × 3600 = 86400 s

The angular velocity is

$\omega = \frac{2\pi}{\text T}$

Substituting the value,

$$\begin{aligned}\omega = \frac{2 \times 3.14}{86400} \\[1em] = 7.27 \times 10^{-5}\ \text{rad s}^{-1}\end{aligned}$$

Hence, the angular velocity of a geostationary satellite is 7.27 × 10^-5 rad s^-1, which is the same as $\frac{\pi}{12}$ rad h^-1.

(i) Correct. If three vectors are represented in magnitude and direction by the three sides of a triangle taken in the same (cyclic) order, they complete a closed figure. Their sum-vector cannot then be drawn, which means that their resultant is zero.

(ii) Correct. The volume of a sphere is a scalar quantity and no unique direction can be associated with it, because the outward normal to the surface of the sphere points in a different direction at every point of the surface.

(iii) Incorrect. The scalar product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them, that is,

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$

The sine of the angle occurs in the magnitude of the vector product, $|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin \theta$.

(i) The magnitude of the resultant of two vectors is

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

Here R = A = B, so

$$\begin{aligned}\text A^2 = \text A^2 + \text A^2 + 2\text A^2 \cos \theta \\[1em] \text A^2 = 2\text A^2(1 + \cos \theta) \\[1em] \frac{1}{2} = 1 + \cos \theta \quad \Rightarrow \quad \cos \theta = -\frac{1}{2}\end{aligned}$$

Therefore,

$\theta = 120^\circ$

Hence, the angle between the two vectors is 120°.

(ii) Given,

• Forces A = 6 N and B = 8 N, with θ = 90°

The magnitude of the resultant force is

$$\begin{aligned}\text F = \sqrt{(6)^2 + (8)^2 + 2 \times 6 \times 8 \times \cos 90^\circ} \\[1em] = \sqrt{36 + 64 + 0} = \sqrt{100} \\[1em] = 10\ \text N\end{aligned}$$

Hence, the magnitude of the resultant force is 10 N.

(iii) A few examples of vector quantities are displacement, velocity, acceleration, force, weight, momentum, impulse, electric field and magnetic field.

(i) The scalar product of the two vectors is

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (1)(2) + (2)(-1) + (3)(0) \\[1em] = 2 - 2 + 0 = 0\end{aligned}$$

Since neither vector is a null vector and their scalar product is zero, cos θ = 0 and θ = 90°.

Hence, the two vectors are perpendicular to each other.

(ii) Given,

• A = 6, B = 8 and $\vec{\text A} \cdot \vec{\text B}$= 24

Using $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$,

$$\begin{aligned}24 = 6 \times 8 \times \cos \theta \\[1em] \cos \theta = \frac{24}{48} = \frac{1}{2}\end{aligned}$$

Therefore,

$\theta = 60^\circ$

Hence, the angle between the two vectors is 60°.

(iii) The scalar product is the sum of the products of the corresponding components,

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (3)(2) + (4)(-5) + (6)(0) \\[1em] = 6 - 20 + 0 \\[1em] = -14\end{aligned}$$

Hence, $\vec{\text A} \cdot \vec{\text B} = -14$.

(i) $R = \sqrt{A^2 + B^2 + 2AB \cos θ}$

Reason — By the triangle (or parallelogram) law of vector addition, the magnitude of the resultant of two vectors inclined at an angle θ is $\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$, and the direction of the resultant is given by $\tan \alpha = \frac{\text B \sin \theta}{\text A + \text B \cos \theta}$.

(ii) When the two vectors are in the same direction.

Reason — Putting θ = 0° in the expression for the resultant,

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}} = \sqrt{(\text A + \text B)^2} = \text A + \text B$

which is the sum of the magnitudes. This is also the maximum possible value of the resultant.

(iii) θ = 90°

Reason — Equating the two magnitudes,

$$\begin{aligned}\sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta} = \sqrt{\text A^2 + \text B^2 - 2\text{AB}\cos \theta} \\[1em] 4\text{AB} \cos \theta = 0 \quad \Rightarrow \quad \cos \theta = 0 \quad \Rightarrow \quad \theta = 90^\circ\end{aligned}$$

(iv) $\vec{A} + \vec{B} + \vec{C} = 0$

Reason — If three vectors are represented by the three sides of a triangle taken in the same order, they complete a closed figure. Their sum-vector cannot be drawn, which means that their resultant is a null vector.

(v) 120°

Reason — With A = B and R = A,

$\text A^2 = 2\text A^2(1 + \cos \theta) \quad \Rightarrow \quad \cos \theta = -\frac{1}{2} \quad \Rightarrow \quad \theta = 120^\circ$

(i) two dimensional motion

Reason — The motion of a projectile takes place in a plane, since the projectile moves simultaneously along the horizontal and the vertical directions. It is therefore a two-dimensional motion, and by the principle of physical independence of motions the two component motions can be treated separately.

(ii) 8s

Reason — The two walls are of the same height and are situated symmetrically about the highest point of the trajectory. Hence the projectile is above the level of the walls between t₁ and t₂, and the total time of flight is

$\text T = \text t_1 + \text t_2 = 2 + 6 = 8\ \text s$

(iii) $\tan^{-1} (4/3)$

Reason — From the time of flight,

$\text T = \frac{2\text u \sin \theta}{\text g} \quad \Rightarrow \quad 8 = \frac{2\text u \sin \theta}{10} \quad \Rightarrow \quad \text u \sin \theta = 40\ \text{m s}^{-1}$

The horizontal distance between the walls is covered in (t₂ − t₁) = 4 s with the constant horizontal velocity,

$\text d = \text u \cos \theta \times (\text t_2 - \text t_1) \quad \Rightarrow \quad 120 = \text u \cos \theta \times 4 \quad \Rightarrow \quad \text u \cos \theta = 30\ \text{m s}^{-1}$

Therefore,

$\tan \theta = \frac{\text u \sin \theta}{\text u \cos \theta} = \frac{40}{30} = \frac{4}{3} \quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{4}{3}\right)$

(iv) 50 ms^-1

Reason — Squaring and adding the two components obtained above,

$$\begin{aligned}\text u = \sqrt{(\text u \sin \theta)^2 + (\text u \cos \theta)^2} = \sqrt{(40)^2 + (30)^2} \\[1em] = \sqrt{1600 + 900} = \sqrt{2500} \\[1em] = 50\ \text{m s}^{-1}\end{aligned}$$

(v) 60 m

Reason — Considering the vertical motion up to t₁ = 2 s,

$$\begin{aligned}\text h = (\text u \sin \theta)\text t_1 - \frac{1}{2}\text g\text t_1^2 \\[1em] = 40 \times 2 - \frac{1}{2} \times 10 \times (2)^2 \\[1em] = 80 - 20 \\[1em] = 60\ \text m\end{aligned}$$

(i) along the tangent to a point on circular path.

Reason — In uniform circular motion the velocity vector acts along the tangent to the circular path at any point. Its magnitude remains constant but its direction changes continuously as the particle moves round the circle.

(ii) along the radius towards the centre.

Reason — The acceleration in uniform circular motion is always directed along the radius towards the centre of the circular path, and is therefore called centripetal acceleration. It is responsible only for the change in the direction of the velocity and not for a change in its magnitude.

(iii) 41.67 ms^-2

Reason — Given r = 1.5 km = 1500 m and v = 900 km h^-1 = $900 \times \frac{5}{18}$ = 250 m s^-1. Therefore,

$$\begin{aligned}\text a = \frac{\text v^2}{\text r} = \frac{(250)^2}{1500} \\[1em] = \frac{62500}{1500} \\[1em] = 41.67\ \text{m s}^{-2}\end{aligned}$$

(iv) 160 π² cm s^-2

Reason — Given r = 10 cm and n = 120 rev per minute = 2 rev s^-1. The angular velocity is

$\omega = 2\pi \text n = 2\pi \times 2 = 4\pi\ \text{rad s}^{-1}$

Therefore,

$$\begin{aligned}\text a = \omega^2 \text r = (4\pi)^2 \times 10 \\[1em] = 16\pi^2 \times 10 \\[1em] = 160\pi^2\ \text{cm s}^{-2}\end{aligned}$$

(v) 90°

Reason — In uniform circular motion the velocity vector is always tangential to the circular path while the centripetal acceleration is always directed along the radius towards the centre. Since the radius is perpendicular to the tangent at every point, the angle between the two vectors is always 90°.

The physical quantities are of two types : scalar and vector.

Scalar quantity : A scalar quantity is one which is described solely by its magnitude (size) and does not have any associated direction. A scalar quantity can be completely defined by a number together with a unit. For example, when we say that the mass of a truck is 200 kg, or that my college is at a distance of 2 km, the statement gives complete information about the quantity. The summation, subtraction, multiplication and division of scalar quantities can be done by ordinary algebra.

Examples of scalars : mass, distance, time, speed, volume, density, pressure, work, energy, power, charge, electric current, temperature, potential, specific heat and frequency.

Vector quantity : The quantities which have both magnitude and direction and obey the laws of vector addition are called vector quantities. To express a vector quantity completely, the direction must also be specified along with the magnitude. If we say that a college is 2 km away from the railway station, the statement is incomplete, since the college cannot be located until we also say that it is 2 km towards the north of the station. Thus position is a vector quantity.

Examples of vectors : position, displacement, velocity, acceleration, force, weight, momentum, impulse, electric field, magnetic field and current density.

It should be noted that a quantity having magnitude and direction is not necessarily a vector. For example, electric current has both magnitude and direction, but it is a scalar because it does not obey the laws of vector addition.

Scalar product : The scalar product (or dot product) of two vectors is a scalar quantity equal to the product of the magnitudes of the two vectors and the cosine of the angle between them. It is written by placing a dot between the two vectors, and hence the name dot product.

The formula for the scalar product of two vectors $\vec{\text A} \space$and $\vec{\text B} \space$is

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$

where,

• A is the magnitude of the vector $\vec{\text A}$
• B is the magnitude of the vector $\vec{\text B}$
• θ is the angle between the directions of $\vec{\text A} \space$and $\vec{\text B}$

Properties of the scalar product :

(i) The scalar product is commutative, that is, $\vec{\text A} \cdot \vec{\text B} = \vec{\text B} \cdot \vec{\text A}$.

(ii) The scalar product is distributive, that is, $\vec{\text A} \cdot (\vec{\text B} + \vec{\text C}) = \vec{\text A} \cdot \vec{\text B} + \vec{\text A} \cdot \vec{\text C}$.

(iii) The scalar product of two mutually perpendicular vectors is zero, since cos 90° = 0.

(iv) The scalar product of two parallel vectors is equal to the product of their magnitudes, since cos 0° = 1.

(v) The self-product of a vector is the square of its magnitude, $\vec{\text A} \cdot \vec{\text A} = \text A^2$.

In terms of the rectangular components of the two vectors,

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

Example : The work done by a force $\vec{\text F} \space$producing a displacement $\vec{\text s} \space$is the scalar product $\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$.

By definition, the scalar product of the two vectors is

$\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta \quad \text{...............(1)}$

Writing the two vectors in terms of their rectangular components and taking their scalar product,

$\vec{\text A} \cdot \vec{\text B} = (\text A_x\hat{i} + \text A_y\hat{j} + \text A_z\hat{k}) \cdot (\text B_x\hat{i} + \text B_y\hat{j} + \text B_z\hat{k})$

Using the relations for the unit orthogonal vectors,

$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \quad \text{and} \quad \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

we get

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z \quad \text{...............(2)}$

The magnitudes of the two vectors are

$\text A = \sqrt{\text A_x^2 + \text A_y^2 + \text A_z^2} \quad \text{and} \quad \text B = \sqrt{\text B_x^2 + \text B_y^2 + \text B_z^2} \quad \text{...............(3)}$

Comparing equations (1) and (2),

$\text{AB} \cos \theta = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

Substituting the magnitudes from equation (3),

$\cos \theta = \frac{\text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z}{\sqrt{\text A_x^2 + \text A_y^2 + \text A_z^2}\sqrt{\text B_x^2 + \text B_y^2 + \text B_z^2}}$

Hence proved.

Let a body be projected from the point O with an initial velocity u in a direction making an angle θ with the horizontal. Let OX and OY be the horizontal and the vertical axes. Resolving u into its rectangular components,

$\text u_x = \text u \cos \theta \quad \text{and} \quad \text u_y = \text u \sin \theta$

The body moves under a constant gravitational acceleration g acting vertically downwards. By the principle of physical independence of motions, the horizontal velocity u_x remains unchanged throughout the motion (air resistance being negligible), while the vertical velocity u_y changes continuously.

Horizontal motion : The horizontal displacement after time t is

$$\begin{aligned}\text x = \text u_x \times \text t \\[1em] \text x = (\text u \cos \theta)\text t \\[1em] \Rightarrow \quad \text t = \frac{\text x}{\text u \cos \theta} \quad \text{...............(1)}\end{aligned}$$

Vertical motion : Using $\text s_y = \text{u}_y{\text t} - \frac{1}{2}\text{gt}^2$, the vertical displacement after time t is

$\text y = (\text u \sin \theta)\text t - \frac{1}{2}\text{gt}^2 \quad \text{...............(2)}$

Substituting the value of t from equation (1) in equation (2),

$$\begin{aligned}\text y = (\text u \sin \theta)\left(\frac{\text x}{\text u \cos \theta}\right) - \frac{1}{2}\text g\left(\frac{\text x}{\text u \cos \theta}\right)^2 \\[1em] = \text x \tan \theta - \frac{\text{gx}^2}{2\text u^2 \cos^2 \theta}\end{aligned}$$

For a given u and θ, the quantities tan θ and $\frac{\text g}{2\text u^2 \cos^2 \theta}$ are constants. Hence the equation is of the form

$\text y = \text{ax} - \text{bx}^2$

which is quadratic in x and linear in y, and therefore represents a parabola.

Hence, the path of the projectile is parabolic.

Can the path be rectilinear? Yes. The path of a projectile is parabolic only when the direction of the acceleration is different from the direction of the velocity of the projectile. If the body is projected vertically upward (θ = 90°) or vertically downward, the velocity and the acceleration are along the same line, so there is no horizontal motion and the path becomes a straight line.

Let a body be projected with velocity u at an angle θ with the horizontal. Resolving the velocity of projection,

$\text u_x = \text u \cos \theta \quad \text{and} \quad \text u_y = \text u \sin \theta$

At the highest point of the trajectory the vertical component of the velocity becomes zero. Applying the equation $\text v^2 = \text u^2 - 2\text{gh}$ to the vertical motion, with v = 0 and u = u sin θ,

$0 = (\text u \sin \theta)^2 - 2\text{gh}$

so that

$$\begin{aligned}\text u^2 \sin^2 \theta = 2\text{gh} \\[1em] \text u^2 = \frac{2\text{gh}}{\sin^2 \theta}\end{aligned}$$

Taking the square root,

$\text u = \frac{\sqrt{2\text{gh}}}{\sin \theta}$

Hence proved.

Let the body be projected from O with velocity u at an angle θ with the horizontal. Resolving the velocity of projection,

$\text u_x = \text u \cos \theta \quad \text{and} \quad \text u_y = \text u \sin \theta$

The horizontal component remains unchanged, while the vertical component changes under the constant acceleration g acting vertically downwards.

(a) Time of flight : Let the body take time t to reach the highest point P, where the vertical velocity is zero. Using v = u − gt with v = 0 and u = u sin θ,

$0 = \text u \sin \theta - \text{gt} \quad \Rightarrow \quad \text t = \frac{\text u \sin \theta}{\text g}$

The body takes the same time in coming down from P to the ground. Hence the time of flight is

$\text T = 2\text t = \frac{2\text u \sin \theta}{\text g}$

(b) Maximum height : Let h be the vertical height of the highest point P, where the vertical velocity is zero. Using v² = u² − 2gh with v = 0 and u = u sin θ,

$0 = (\text u \sin \theta)^2 - 2\text{gh} \quad \Rightarrow \quad \text h = \frac{\text u^2 \sin^2 \theta}{2\text g}$

(c) Horizontal range : The horizontal distance covered during the flight is the product of the constant horizontal velocity and the time of flight,

$$\begin{aligned}\text R = \text u_x \times \text T = (\text u \cos \theta) \times \frac{2\text u \sin \theta}{\text g} \\[1em] = \frac{\text u^2(2 \sin \theta \cos \theta)}{\text g} \\[1em] = \frac{\text u^2 \sin 2\theta}{\text g}\end{aligned}$$

Hence, the time of flight is $\frac{2\text u \sin \theta}{\text g}$, the maximum height attained is $\frac{\text u^2 \sin^2 \theta}{2\text g}$ and the horizontal range is $\frac{\text u^2 \sin 2\theta}{\text g}$.

Let a body be projected from O with speed u at an angle θ with the horizontal. Resolving the speed of projection,

$\text u_x = \text u \cos \theta \quad \text{and} \quad \text u_y = \text u \sin \theta$

The vertical velocity becomes zero at the highest point, so the time taken to reach it is obtained from v = u − gt as $\text t = \frac{\text u \sin \theta}{\text g}$, and the time of flight is

$\text T = \frac{2\text u \sin \theta}{\text g}$

Since the horizontal component of the velocity remains constant, the horizontal range is

$$\begin{aligned}\text R = \text u_x \times \text T = (\text u \cos \theta) \times \frac{2\text u \sin \theta}{\text g} \\[1em] = \frac{\text u^2(2 \sin \theta \cos \theta)}{\text g} \\[1em] = \frac{\text u^2 \sin 2\theta}{\text g}\end{aligned}$$

Hence proved.

Angle for maximum range : For a given initial speed u, the range is maximum when sin 2θ has its greatest value, that is,

$\sin 2\theta = 1 \quad \Rightarrow \quad 2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ$

Maximum range : Substituting θ = 45° in the expression for the range,

$\text R_{max} = \frac{\text u^2 \sin 90^\circ}{\text g} = \frac{\text u^2}{\text g}$

Hence, the range is maximum for an angle of projection of 45°, and its maximum value is $\frac{\text u^2}{\text g}$. This is why a long-jumper takes his jump at an angle of about 45°.

Given,

• P = 5, Q = 7 and the angle between them, θ = 90°

For $\vec{\text P} + \vec{\text Q}$ : Draw $\vec{\text P} \space$and then, starting from its arrow-head, draw $\vec{\text Q} \space$perpendicular to it. The vector joining the initial point of $\vec{\text P} \space$to the arrow-head of $\vec{\text Q} \space$is the sum. Its magnitude is

$$\begin{aligned}|\vec{\text P} + \vec{\text Q}| = \sqrt{\text P^2 + \text Q^2 + 2\text{PQ}\cos 90^\circ} \\[1em] = \sqrt{(5)^2 + (7)^2} = \sqrt{74} \\[1em] = 8.6\end{aligned}$$

and it makes an angle α with $\vec{\text P} \space$given by

$\tan \alpha = \frac{\text Q}{\text P} = \frac{7}{5} \quad \Rightarrow \quad \alpha = 54.5^\circ$

For $\vec{\text P} - \vec{\text Q}$ : Reverse $\vec{\text Q} \space$to get $(-\vec{\text Q}) \space$and add it to $\vec{\text P}$. Its magnitude is

$$\begin{aligned}|\vec{\text P} - \vec{\text Q}| = \sqrt{\text P^2 + \text Q^2 - 2\text{PQ}\cos 90^\circ} \\[1em] = \sqrt{(5)^2 + (7)^2} = \sqrt{74} \\[1em] = 8.6\end{aligned}$$

and it makes an angle of 54.5° with $\vec{\text P} \space$on the other side.

Hence, the magnitude of each is 8.6, but their directions are different.

Given,

• Initial velocity, v₁ = 20 m s^-1 towards north
• Final velocity, v₂ = 20 m s^-1 towards south

The change in velocity is

$\Delta \vec{\text v} = \vec{\text v}_2 - \vec{\text v}_1 = \vec{\text v}_2 + (-\vec{\text v}_1)$

Taking the northward direction as positive,

$$\begin{aligned}\Delta \text v = (-20) - (+20) \\[1em] = -40\ \text{m s}^{-1}\end{aligned}$$

Hence, the change in the velocity of the car is 40 m s^-1 directed towards the south.

Given,

• A = 6 N, B = 8 N and θ = 90°

By the parallelogram law of vector addition, the magnitude of the resultant is

$$\begin{aligned}\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta} \\[1em] = \sqrt{(6)^2 + (8)^2 + 2 \times 6 \times 8 \times \cos 90^\circ} \\[1em] = \sqrt{36 + 64 + 0} = \sqrt{100} \\[1em] = 10\ \text N\end{aligned}$$

If α is the angle made by the resultant with the 6 N force,

$$\begin{aligned}\tan \alpha = \frac{\text B \sin \theta}{\text A + \text B \cos \theta} = \frac{8 \sin 90^\circ}{6 + 8 \cos 90^\circ} \\[1em] = \frac{8}{6} = 1.333\end{aligned}$$

so that

$\alpha = \tan^{-1}(1.333) = 53.1^\circ$

Hence, the resultant force is 10 N, making an angle of 53.1° with the 6 N force.

Given,

• $\vec{\text F}_1$ = 500 N towards east
• $\vec{\text F}_2$ = 600 N towards north

To subtract $\vec{\text F}_1$ from $\vec{\text F}_2$, we reverse $\vec{\text F}_1$ to get 500 N towards the west and add it to $\vec{\text F}_2$. Since the two are mutually perpendicular, the magnitude of the difference is

$$\begin{aligned}|\vec{\text F}_2 - \vec{\text F}_1| = \sqrt{(600)^2 + (500)^2} \\[1em] = \sqrt{360000 + 250000} = \sqrt{610000} \\[1em] = 781\ \text N\end{aligned}$$

If α is the angle made by this resultant with the north direction,

$\tan \alpha = \frac{500}{600} = 0.833 \quad \Rightarrow \quad \alpha = 39.8^\circ$

Hence, the difference is a force of 781 N directed at 39.8° west of north.

Given,

• Force applied, F = 50 N
• Angle with the horizontal, θ = 30°

Resolving the force into its rectangular components,

Horizontal component :

$$\begin{aligned}\text F_x = \text F \cos \theta = 50 \times \cos 30^\circ \\[1em] = 50 \times \frac{\sqrt{3}}{2} \\[1em] = 43.3\ \text N\end{aligned}$$

Vertical component :

$$\begin{aligned}\text F_y = \text F \sin \theta = 50 \times \sin 30^\circ \\[1em] = 50 \times \frac{1}{2} \\[1em] = 25\ \text N\end{aligned}$$

Hence, the horizontal component is 43.3 N and the vertical component is 25 N.

Given,

• Force, F = 150 N
• Displacement, s = 15 m
• Angle between the force and the displacement, θ = 60°

The work done is the scalar product of the force and the displacement,

$\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$

Substituting the values,

$$\begin{aligned}\text W = 150 \times 15 \times \cos 60^\circ \\[1em] = 150 \times 15 \times \frac{1}{2} \\[1em] = 1125\ \text J\end{aligned}$$

Hence, the work done by the force is 1125 J.

Given,

• Force, F = 50 N
• Displacement, s = 5 m
• Work done, W = 125 J

Using W = Fs cos θ,

$$\begin{aligned}125 = 50 \times 5 \times \cos \theta \\[1em] \cos \theta = \frac{125}{250} = \frac{1}{2}\end{aligned}$$

Therefore,

$\theta = 60^\circ$

Hence, the angle between the directions of the force and the displacement is 60°.

The magnitude of a vector in terms of its rectangular components is

$\text A = \sqrt{\text A_x^2 + \text A_y^2 + \text A_z^2}$

Substituting the values,

$$\begin{aligned}\text A = \sqrt{(2)^2 + (3)^2 + (-5)^2} \\[1em] = \sqrt{4 + 9 + 25} \\[1em] = \sqrt{38} = 6.16\end{aligned}$$

Hence, the magnitude of $\vec{\text A} \space$is $\sqrt{38}$ = 6.16.

The unit vector in the direction of a vector $\vec{\text A} \space$is

$\hat{\text A} = \frac{\vec{\text A}}{|\vec{\text A}|}$

The magnitude of $\vec{\text A} \space$is

$$\begin{aligned}|\vec{\text A}| = \sqrt{(3)^2 + (4)^2 + (-5)^2} \\[1em] = \sqrt{9 + 16 + 25} = \sqrt{50} \\[1em] = 5\sqrt{2}\end{aligned}$$

Therefore,

$\hat{\text A} = \frac{3\hat{i} + 4\hat{j} - 5\hat{k}}{\sqrt{50}}$

Hence, the unit vector of $\vec{\text A} \space$is $\frac{3\hat{i} + 4\hat{j} - 5\hat{k}}{\sqrt{50}}$.

Adding the corresponding components,

$$\begin{aligned}\vec{\text A} + \vec{\text B} = (3 + 1)\hat{i} + (2 - 2)\hat{j} + (0 + 3)\hat{k} \\[1em] = 4\hat{i} + 0\hat{j} + 3\hat{k}\end{aligned}$$

Its magnitude is

$$\begin{aligned}|\vec{\text A} + \vec{\text B}| = \sqrt{(4)^2 + (0)^2 + (3)^2} \\[1em] = \sqrt{16 + 9} = \sqrt{25} = 5\end{aligned}$$

Subtracting the corresponding components,

$$\begin{aligned}\vec{\text A} - \vec{\text B} = (3 - 1)\hat{i} + (2 + 2)\hat{j} + (0 - 3)\hat{k} \\[1em] = 2\hat{i} + 4\hat{j} - 3\hat{k}\end{aligned}$$

Its magnitude is

$$\begin{aligned}|\vec{\text A} - \vec{\text B}| = \sqrt{(2)^2 + (4)^2 + (-3)^2} \\[1em] = \sqrt{4 + 16 + 9} = \sqrt{29} = 5.39\end{aligned}$$

Hence, the magnitudes are 5 and $\sqrt{29}$ = 5.39 respectively.

Given,

$\vec{A} + \vec{B} = 2\hat{i} + 6\hat{j} + \hat{k}$

$\vec{A} - \vec{B} = 4\hat{i} + 2\hat{j} - 11\hat{k}$

Adding the two given relations,

$$\begin{aligned}2\vec{\text A} = (2\hat{i} + 6\hat{j} + \hat{k}) + (4\hat{i} + 2\hat{j} - 11\hat{k}) \\[1em] = 6\hat{i} + 8\hat{j} - 10\hat{k}\end{aligned}$$

so that

$\vec{\text A} = 3\hat{i} + 4\hat{j} - 5\hat{k}$

Subtracting the two given relations,

$$\begin{aligned}2\vec{\text B} = (2\hat{i} + 6\hat{j} + \hat{k}) - (4\hat{i} + 2\hat{j} - 11\hat{k}) \\[1em] 2\vec{\text B}= -2\hat{i} + 4\hat{j} + 12\hat{k}\end{aligned}$$

so that

$\vec{\text B} = -\hat{i} + 2\hat{j} + 6\hat{k}$

The magnitudes are

$$\begin{aligned}|\vec{\text A}| = \sqrt{(3)^2 + (4)^2 + (-5)^2} = \sqrt{50} = 7.07 \\[1em] |\vec{\text B}| = \sqrt{(-1)^2 + (2)^2 + (6)^2} = \sqrt{41} = 6.40\end{aligned}$$

The scalar product is

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (3)(-1) + (4)(2) + (-5)(6) \\[1em] = -3 + 8 - 30 \\[1em] = -25\end{aligned}$$

Hence, $|\vec{\text A}| = \sqrt{50}$, $|\vec{\text B}| = \sqrt{41}$ and $\vec{\text A} \cdot \vec{\text B} = -25$.

The scalar product is equal to the sum of the products of the corresponding components,

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y$

Substituting the values,

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (6)(3) + (-8)(0) \\[1em] = 18 + 0 \\[1em] = 18\end{aligned}$$

Hence, the scalar product of $\vec{\text A} \space$and $\vec{\text B} \space$is 18.

(i) The magnitude of $\vec{\text A} \space$is

$$\begin{aligned}|\vec{\text A}| = \sqrt{(3)^2 + (-6)^2 + (-2)^2} \\[1em] = \sqrt{9 + 36 + 4} = \sqrt{49} \\[1em] = 7\end{aligned}$$

(ii) The scalar product is

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (3)(1) + (-6)(2) + (-2)(-2) \\[1em] = 3 - 12 + 4 \\[1em] = -5\end{aligned}$$

Hence, the magnitude of $\vec{\text A} \space$is 7 and $\vec{\text A} \cdot \vec{\text B} = -5$.

Using the relations for the unit orthogonal vectors,

$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \quad \text{and} \quad \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

(i)

$$\begin{aligned}\hat{j} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = 2(\hat{j} \cdot \hat{i}) - 3(\hat{j} \cdot \hat{j}) + (\hat{j} \cdot \hat{k}) \\[1em] = 0 - 3 + 0 \\[1em] = -3\end{aligned}$$

(ii)

$$\begin{aligned}(2\hat{i} - \hat{j}) \cdot (3\hat{i} + \hat{k}) = (2)(3) + (-1)(0) + (0)(1) \\[1em] = 6\end{aligned}$$

Hence, the values are −3 and 6 respectively.

The scalar product of the two vectors is

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (2)(6) + (2)(-3) + (-1)(2) \\[1em] = 12 - 6 - 2 = 4\end{aligned}$$

The magnitudes are

$$\begin{aligned}|\vec{\text A}| = \sqrt{(2)^2 + (2)^2 + (-1)^2} = \sqrt{9} = 3 \\[1em] |\vec{\text B}| = \sqrt{(6)^2 + (-3)^2 + (2)^2} = \sqrt{49} = 7\end{aligned}$$

Using $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$,

$$\begin{aligned}\cos \theta = \frac{\vec{\text A} \cdot \vec{\text B}}{\text{AB}} = \frac{4}{3 \times 7} \\[1em] = \frac{4}{21} = 0.1905\end{aligned}$$

Therefore,

$\theta = \cos^{-1}(0.1905) = 79^\circ$

Hence, the angle between the two vectors is 79°.

The work done is the scalar product of the force and the displacement,

$\text W = \vec{\text F} \cdot \vec{\text s}$

Substituting the values,

$$\begin{aligned}\text W = (5)(10) + (2)(12) + (3)(13) \\[1em] = 50 + 24 + 39 \\[1em] = 113\ \text J\end{aligned}$$

Hence, the work done by the force on the body is 113 J.

Writing the vector product as a determinant,

$$\vec{\text A} \times \vec{\text B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 2 \\ 3 & 1 & 2 \end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned}\vec{\text A} \times \vec{\text B} = \hat{i}(3 \times 2 - 2 \times 1) - \hat{j}(1 \times 2 - 2 \times 3) + \hat{k}(1 \times 1 - 3 \times 3) \\[1em] = \hat{i}(6 - 2) - \hat{j}(2 - 6) + \hat{k}(1 - 9) \\[1em] = 4\hat{i} + 4\hat{j} - 8\hat{k}\end{aligned}$$

Hence, $\vec{\text A} \times \vec{\text B} = 4\hat{i} + 4\hat{j} - 8\hat{k}$.

(i) Writing the vector product as a determinant,

$$\vec{\text A} \times \vec{\text B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -2 \\ 1 & 2 & -2 \end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned}\vec{\text A} \times \vec{\text B} = \hat{i}[(-6)(-2) - (-2)(2)] - \hat{j}[(3)(-2) - (-2)(1)] + \hat{k}[(3)(2) - (-6)(1)] \\[1em] = \hat{i}(12 + 4) - \hat{j}(-6 + 2) + \hat{k}(6 + 6) \\[1em] = 16\hat{i} + 4\hat{j} + 12\hat{k}\end{aligned}$$

(ii) The magnitude is

$$\begin{aligned}|\vec{\text A} \times \vec{\text B}| = \sqrt{(16)^2 + (4)^2 + (12)^2} \\[1em] = \sqrt{256 + 16 + 144} = \sqrt{416} \\[1em] = 20.4\end{aligned}$$

Hence, $\vec{\text A} \times \vec{\text B} = 16\hat{i} + 4\hat{j} + 12\hat{k}$ and its magnitude is $\sqrt{416}$ = 20.4.

Given,

• Displacement (position) vector of the point, $\vec{\text r} = (3\hat{i} + 2\hat{j} + 3\hat{k})$ m
• Force acting at this point, $\vec{\text F} = (2\hat{i} - 3\hat{j} + 4\hat{k})$ N

The vector moment $\vec{\tau} \space$of the force about the origin and the moment of force $|\vec{\tau}|$ have to be calculated.

The moment of a force about a point, also called the torque, is the vector product of the position vector of the point of application and the force,

$\vec{\tau} = \vec{\text r} \times \vec{\text F}$

Its magnitude is τ = rF sin θ and its direction is perpendicular to the plane containing $\vec{\text r} \space$and $\vec{\text F}$, the sense being given by the right-hand screw rule.

Writing the vector product as a determinant,

$$\vec{\tau} = \vec{\text r} \times \vec{\text F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned}\vec{\tau} = \hat{i}[(2)(4) - (3)(-3)] - \hat{j}[(3)(4) - (3)(2)] + \hat{k}[(3)(-3) - (2)(2)] \\[1em] = \hat{i}(8 + 9) - \hat{j}(12 - 6) + \hat{k}(-9 - 4) \\[1em] = 17\hat{i} - 6\hat{j} - 13\hat{k}\ \text{N m}\end{aligned}$$

The moment of the force is the magnitude of this vector,

$$\begin{aligned}|\vec{\tau}| = \sqrt{\tau_x^2 + \tau_y^2 + \tau_z^2} \\[1em] = \sqrt{(17)^2 + (-6)^2 + (-13)^2} \\[1em] = \sqrt{289 + 36 + 169} \\[1em] = \sqrt{494} \\[1em] = 22.2\ \text{N m}\end{aligned}$$

Hence, the vector moment is $17\hat{i} - 6\hat{j} - 13\hat{k}$ N m and the moment of the force is $\sqrt{494}$ = 22.2 N m.

Given,

• $\vec{\text A} = 2\hat{i} + 2\hat{j} - 2\hat{k}$, so A_x = 2, A_y = 2 and A_z = −2
• $\vec{\text B} = 7\hat{i} - 5\hat{j} + 2\hat{k}$, so B_x = 7, B_y = −5 and B_z = 2

The scalar product $\vec{\text A} \cdot \vec{\text B} \space$and the vector product $\vec{\text A} \times \vec{\text B} \space$have to be calculated.

Scalar product : In terms of the rectangular components of the two vectors, the scalar product is equal to the sum of the products of their corresponding components,

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

Substituting the values,

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (2)(7) + (2)(-5) + (-2)(2) \\[1em] = 14 - 10 - 4 \\[1em] = 0\end{aligned}$$

Since $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$ and neither vector is a null vector, cos θ = 0, so θ = 90°. Thus the two vectors are mutually perpendicular.

Vector product : Writing the vector product as a determinant,

$$\vec{\text A} \times \vec{\text B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \text A_x & \text A_y & \text A_z \\ \text B_x & \text B_y & \text B_z \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -2 \\ 7 & -5 & 2 \end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned}\vec{\text A} \times \vec{\text B} = \hat{i}[(2)(2) - (-2)(-5)] - \hat{j}[(2)(2) - (-2)(7)] + \hat{k}[(2)(-5) - (2)(7)] \\[1em] = \hat{i}(4 - 10) - \hat{j}(4 + 14) + \hat{k}(-10 - 14) \\[1em] = -6\hat{i} - 18\hat{j} - 24\hat{k}\end{aligned}$$

Hence, $\vec{\text A} \cdot \vec{\text B} = 0$ and $\vec{\text A} \times \vec{\text B} = -6\hat{i} - 18\hat{j} - 24\hat{k}$.

Given,

• Velocity of projection, u = 30 m s^-1
• Acceleration due to gravity, g = 10 m s^-2

The maximum horizontal range is obtained for θ = 45°,

$$\begin{aligned}\text R_{max} = \frac{\text u^2}{\text g} = \frac{(30)^2}{10} \\[1em] = \frac{900}{10} = 90\ \text m\end{aligned}$$

For a range of 45 m,

$$\begin{aligned}\text R = \frac{\text u^2 \sin 2\theta}{\text g} \\[1em] 45 = \frac{900 \times \sin 2\theta}{10} \\[1em] \sin 2\theta = \frac{45 \times 10}{900} = \frac{1}{2}\end{aligned}$$

Therefore,

$$\begin{aligned}2\theta = 30^\circ \quad \text{or} \quad 2\theta = 150^\circ \\[1em] \theta = 15^\circ \quad \text{or} \quad \theta = 75^\circ\end{aligned}$$

Hence, the maximum horizontal range is 90 m, and a range of 45 m is obtained for angles of projection of 15° and 75°, which are complementary angles.

Given,

• Velocity of projection, u = 15 m s^-1
• Angle of projection, θ = 30°
• Acceleration due to gravity, g = 10 m s^-2

(i) The time of flight is

$$\begin{aligned}\text T = \frac{2\text u \sin \theta}{\text g} = \frac{2 \times 15 \times \sin 30^\circ}{10} \\[1em] = \frac{2 \times 15 \times 0.5}{10} \\[1em] = 1.5\ \text s\end{aligned}$$

(ii) The maximum height attained is

$$\begin{aligned}\text h = \frac{\text u^2 \sin^2 \theta}{2\text g} = \frac{(15)^2 \times (0.5)^2}{2 \times 10} \\[1em] = \frac{225 \times 0.25}{20} \\[1em] = 2.8\ \text m\end{aligned}$$

Hence, the time of flight is 1.5 s and the maximum height attained is 2.8 m.

Given,

• Horizontal range, R = 500 m
• Velocity of projection, u = 100 m s^-1
• Acceleration due to gravity, g = 10 m s^-2

Using the expression for the horizontal range,

$\text R = \frac{\text u^2 \sin 2\theta}{\text g}$

Substituting the values,

$$\begin{aligned}500 = \frac{(100)^2 \times \sin 2\theta}{10} \\[1em] \sin 2\theta = \frac{500 \times 10}{10000} = \frac{1}{2}\end{aligned}$$

Therefore,

$$\begin{aligned}2\theta = 30^\circ \quad \text{or} \quad 2\theta = 150^\circ \\[1em] \theta = 15^\circ \quad \text{or} \quad \theta = 75^\circ\end{aligned}$$

Hence, the angle of projection should be 15° or 75°.

Given,

• Initial velocity, v₁ = 5 cm s^-1 towards north
• Final velocity, v₂ = 5 cm s^-1 towards east

The change in velocity is

$\Delta \vec{\text v} = \vec{\text v}_2 - \vec{\text v}_1 = \vec{\text v}_2 + (-\vec{\text v}_1)$

Here $\vec{\text v}_2$ is towards the east and $(-\vec{\text v}_1)$ is towards the south, and the two are mutually perpendicular. Hence

$$\begin{aligned}|\Delta \vec{\text v}| = \sqrt{\text v_2^2 + \text v_1^2} = \sqrt{(5)^2 + (5)^2} \\[1em] = 5\sqrt{2} \\[1em] = 7.1\ \text{cm s}^{-1}\end{aligned}$$

Since the two components are equal, the change in velocity makes an angle of 45° with each of them.

Hence, the change in velocity is 7.1 cm s^-1, directed 45° south of east.

Given,

• Velocity of the body, v = 100 km h^-1, directed 30° west of south

Resolving the velocity along the north-south and east-west directions,

Component along the south direction :

$\text v \cos 30^\circ = 100 \times \frac{\sqrt{3}}{2} = 86.6\ \text{km h}^{-1}$

Component along the west direction :

$\text v \sin 30^\circ = 100 \times \frac{1}{2} = 50\ \text{km h}^{-1}$

Since the north direction is opposite to the south and the east is opposite to the west,

$$\begin{aligned}\text{North component} = -86.6\ \text{km h}^{-1} \\[1em] \text{East component} = -50\ \text{km h}^{-1}\end{aligned}$$

Hence, the north component is −86.6 km h^-1 and the east component is −50 km h^-1, the negative signs showing that the components are actually directed towards the south and the west respectively.

The given relation can be rewritten as

$\vec{\text C} = \vec{\text B} - \vec{\text A}$

Taking the magnitude of the difference of two vectors inclined at an angle θ,

$\text C^2 = \text A^2 + \text B^2 - 2\text{AB}\cos \theta$

where θ is the angle between $\vec{\text A} \space$and $\vec{\text B}$. Solving for cos θ,

$$\begin{aligned}2\text{AB}\cos \theta = \text A^2 + \text B^2 - \text C^2 \\[1em] \cos \theta = \frac{\text A^2 + \text B^2 - \text C^2}{2\text{AB}}\end{aligned}$$

Therefore,

$\theta = \cos^{-1}\left(\frac{\text A^2 + \text B^2 - \text C^2}{2\text{AB}}\right)$

Hence, the angle between $\vec{\text A} \space$and $\vec{\text B} \space$is $\cos^{-1}\left(\frac{\text A^2 + \text B^2 - \text C^2}{2\text{AB}}\right)$.

Given,

• A = 3, B = 4 and $\vec{\text A} \cdot \vec{\text B}$= 6

Using $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$,

$$\begin{aligned}6 = 3 \times 4 \times \cos \theta \\[1em] \cos \theta = \frac{6}{12} = \frac{1}{2}\end{aligned}$$

Therefore,

$\theta = 60^\circ$

The scalar product will be zero when cos θ = 0, that is, when θ = 90°.

Hence, the angle between the two vectors is 60°, and the scalar product becomes zero when the two vectors are mutually perpendicular (θ = 90°).

Given,

• Force, $\vec{\text F} = 10\hat{i} - 3\hat{j} + 6\hat{k}$ N
• Initial position, $\vec{\text r}_1 = 6\hat{i} + 5\hat{j} - 3\hat{k}$ m
• Final position, $\vec{\text r}_2 = 10\hat{i} - 2\hat{j} + 7\hat{k}$ m

The displacement is the difference between the final and the initial position vectors,

$$\begin{aligned}\vec{\text s} = \vec{\text r}_2 - \vec{\text r}_1 = (10 - 6)\hat{i} + (-2 - 5)\hat{j} + (7 + 3)\hat{k} \\[1em] = 4\hat{i} - 7\hat{j} + 10\hat{k}\ \text m\end{aligned}$$

The work done is

$$\begin{aligned}\text W = \vec{\text F} \cdot \vec{\text s} = (10)(4) + (-3)(-7) + (6)(10) \\[1em] = 40 + 21 + 60 \\[1em] = 121\ \text J\end{aligned}$$

Hence, the work done is 121 J.

(i) Writing the vector product as a determinant,

$$\vec{\text A} \times \vec{\text B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 2 & 4 \end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned}\vec{\text A} \times \vec{\text B} = \hat{i}(3 \times 4 - 1 \times 2) - \hat{j}(2 \times 4 - 1 \times 3) + \hat{k}(2 \times 2 - 3 \times 3) \\[1em] = \hat{i}(12 - 2) - \hat{j}(8 - 3) + \hat{k}(4 - 9) \\[1em] = 10\hat{i} - 5\hat{j} - 5\hat{k}\end{aligned}$$

(ii) Since the vector product is not commutative,

$$\begin{aligned}\vec{\text B} \times \vec{\text A} = -(\vec{\text A} \times \vec{\text B}) \\[1em] = -10\hat{i} + 5\hat{j} + 5\hat{k}\end{aligned}$$

(iii) Expanding the product using the distributive law,

$(\vec{\text A} + \vec{\text B}) \times (\vec{\text A} - \vec{\text B}) = \vec{\text A} \times \vec{\text A} - \vec{\text A} \times \vec{\text B} + \vec{\text B} \times \vec{\text A} - \vec{\text B} \times \vec{\text B}$

Since $\vec{\text A} \times \vec{\text A} = 0$, $\vec{\text B} \times \vec{\text B} = 0$ and $\vec{\text B} \times \vec{\text A} = -\vec{\text A} \times \vec{\text B}$,

$$\begin{aligned}(\vec{\text A} + \vec{\text B}) \times (\vec{\text A} - \vec{\text B}) = -2(\vec{\text A} \times \vec{\text B}) \\[1em] = -2(10\hat{i} - 5\hat{j} - 5\hat{k}) \\[1em] = -20\hat{i} + 10\hat{j} + 10\hat{k}\end{aligned}$$

Hence, the three results are $10\hat{i} - 5\hat{j} - 5\hat{k}$, $-10\hat{i} + 5\hat{j} + 5\hat{k}$ and $-20\hat{i} + 10\hat{j} + 10\hat{k}$ respectively.

Given,

• Height of projection, h = 19.6 m
• Horizontal velocities, u₁ = 50 m s^-1 and u₂ = 100 m s^-1
• Acceleration due to gravity, g = 9.8 m s^-2

(i) Yes, both the bullets will strike the ground, since both are projected horizontally from a height and gravity acts on both of them.

(ii) For a body projected horizontally, the initial vertical velocity is zero, so the time of flight is

$\text T = \sqrt{\frac{2\text h}{\text g}}$

Substituting the values,

$$\begin{aligned}\text T = \sqrt{\frac{2 \times 19.6}{9.8}} = \sqrt{4} \\[1em] = 2.0\ \text s\end{aligned}$$

The time of flight is independent of the velocity of projection and depends only on the height. Hence both the bullets strike the ground together, after 2.0 s.

(iii) The horizontal motion is uniform and the vertical motion is uniformly accelerated. Their combined effect gives an equation of the form y = kx², so the path of each bullet is a parabola. The bullet fired with the greater velocity travels a longer horizontal range and therefore describes a wider parabola.

Given,

• Time of flight, T = 5 s
• Horizontal range, R = 200 m
• Acceleration due to gravity, g = 9.8 m s^-2

(i) The time of flight is $\text T = \frac{2\text u \sin \theta}{\text g}$, so the vertical component is

$$\begin{aligned}\text u \sin \theta = \frac{\text{gT}}{2} = \frac{9.8 \times 5}{2} \\[1em] = 24.5\ \text{m s}^{-1}\end{aligned}$$

(ii) Since the horizontal velocity remains constant, R = (u cos θ)T, so the horizontal component is

$$\begin{aligned}\text u \cos \theta = \frac{\text R}{\text T} = \frac{200}{5} \\[1em] = 40\ \text{m s}^{-1}\end{aligned}$$

(iii) The maximum height attained is

$$\begin{aligned}\text h = \frac{(\text u \sin \theta)^2}{2\text g} = \frac{(24.5)^2}{2 \times 9.8} \\[1em] = \frac{600.25}{19.6} \\[1em] = 30.6\ \text m\end{aligned}$$

(iv) The angle of projection is given by

$$\begin{aligned}\tan \theta = \frac{\text u \sin \theta}{\text u \cos \theta} = \frac{24.5}{40} = 0.6125 \\[1em] \theta = \tan^{-1}(0.6125) = 31.5^\circ\end{aligned}$$

Hence, the vertical component is 24.5 m s^-1, the horizontal component is 40 m s^-1, the maximum height is 30.6 m and the angle of projection is 31.5° with the horizontal.

Given,

• Velocity of projection, u = 25 m s^-1, directed 30° below the horizontal
• Time taken, t = 2.5 s
• Acceleration due to gravity, g = 9.8 m s^-2

The vertical component of the velocity of projection, directed downwards, is

$\text u_y = \text u \sin 30^\circ = 25 \times \frac{1}{2} = 12.5\ \text{m s}^{-1}$

Considering the vertical motion downwards and using $\text h = \text u_y\text t + \frac{1}{2}\text{gt}^2$,

$$\begin{aligned}\text h = 12.5 \times 2.5 + \frac{1}{2} \times 9.8 \times (2.5)^2 \\[1em] = 31.25 + 4.9 \times 6.25 \\[1em] = 31.25 + 30.625 \\[1em] = 61.9\ \text m\end{aligned}$$

Hence, the height of the bridge from the water surface is 61.9 m.

Given,

• $\vec{\text A} = \hat{i} + \text c\hat{j} + 5\hat{k}$, so A_x = 1, A_y = c and A_z = 5
• $\vec{\text B} = 2\hat{i} + \hat{j} - \hat{k}$, so B_x = 2, B_y = 1 and B_z = −1
• The two vectors are perpendicular to each other, that is, θ = 90°

The value of c has to be calculated.

The scalar product of two vectors is $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$. For perpendicular vectors θ = 90° and cos 90° = 0, so their scalar product is zero,

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z = 0$

Substituting the values,

$$\begin{aligned}(1)(2) + (\text c)(1) + (5)(-1) = 0 \\[1em] 2 + \text c - 5 = 0 \\[1em] \text c - 3 = 0 \\[1em] \text c = 3\end{aligned}$$

Hence, the value of c is 3.

Given,

• $\vec{\text A} = 2\hat{i} + 2\hat{j} - 2\hat{k}$, so A_x = 2, A_y = 2 and A_z = −2
• $\vec{\text B} = 7\hat{i} - 5\hat{j} + 2\hat{k}$, so B_x = 7, B_y = −5 and B_z = 2

It has to be shown that the two vectors are mutually perpendicular.

Two non-zero vectors are mutually perpendicular if the angle between them is 90°. Since the scalar product is $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$ and cos 90° = 0, the condition for perpendicularity is that the scalar product of the two vectors must vanish.

Taking the scalar product in terms of the rectangular components,

$\vec{\text A} \cdot \vec{\text B} = \text A_x\text B_x + \text A_y\text B_y + \text A_z\text B_z$

Substituting the values,

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (2)(7) + (2)(-5) + (-2)(2) \\[1em] = 14 - 10 - 4 \\[1em] = 0\end{aligned}$$

Since $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$ and neither of the vectors is a null vector, that is, A ≠ 0 and B ≠ 0,

$\text{AB} \cos \theta = 0 \quad \Rightarrow \quad \cos \theta = 0 \quad \Rightarrow \quad \theta = 90^\circ$

Hence, the two vectors are mutually perpendicular.

Example of a scalar product : The work done by a force $\vec{\text F} \space$producing a displacement $\vec{\text s} \space$is the scalar product of the two vectors,

$\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs} \cos \theta$

where θ is the angle between the directions of the force and the displacement. Although both $\vec{\text F} \space$and $\vec{\text s} \space$are vectors, the work done is a scalar quantity.

Given,

• $\vec{\text A} = 4\hat{i} + 3\hat{j} + \hat{k}$, so A_x = 4, A_y = 3 and A_z = 1
• $\vec{\text B} = 12\hat{i} + 9\hat{j} + 3\hat{k}$, so B_x = 12, B_y = 9 and B_z = 3

It has to be proved that the two vectors are parallel to each other.

Two vectors are parallel if one of them can be obtained by multiplying the other by a positive scalar, since on multiplying a vector by a positive scalar only the magnitude changes while the direction remains unaltered.

Comparing the corresponding components of the two vectors,

$\frac{\text B_x}{\text A_x} = \frac{12}{4} = 3, \quad \frac{\text B_y}{\text A_y} = \frac{9}{3} = 3, \quad \frac{\text B_z}{\text A_z} = \frac{3}{1} = 3$

Since all the three ratios are equal,

$\vec{\text B} = 3\vec{\text A}$

Here the scalar 3 is positive, so $\vec{\text B} \space$has the same direction as $\vec{\text A} \space$and its magnitude is 3 times that of $\vec{\text A}$.

This can also be verified by taking the vector product, since the vector product of two parallel vectors is a null vector,

$$\vec{\text A} \times \vec{\text B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ 12 & 9 & 3 \end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned}\vec{\text A} \times \vec{\text B} = \hat{i}[(3)(3) - (1)(9)] - \hat{j}[(4)(3) - (1)(12)] + \hat{k}[(4)(9) - (3)(12)] \\[1em] = \hat{i}(9 - 9) - \hat{j}(12 - 12) + \hat{k}(36 - 36) \\[1em] = 0\end{aligned}$$

Since $|\vec{\text A} \times \vec{\text B}| = \text{AB} \sin \theta = 0$ and neither vector is a null vector, sin θ = 0, so θ = 0°. The result confirms that the two vectors are parallel.

Hence proved.

Given,

• Horizontal velocity of projection, u = 20 m s^-1
• Height of the tower, h = 40 m
• Acceleration due to gravity, g = 9.8 m s^-2

For a body projected horizontally, the initial vertical velocity is zero, so the time of flight is

$\text T = \sqrt{\frac{2\text h}{\text g}}$

Substituting the values,

$$\begin{aligned}\text T = \sqrt{\frac{2 \times 40}{9.8}} = \sqrt{8.163} \\[1em] = 2.86\ \text s\end{aligned}$$

Since the horizontal velocity remains constant, the horizontal range is

$$\begin{aligned}\text R = \text u_x \times \text T = 20 \times 2.86 \\[1em] = 57.2\ \text m\end{aligned}$$

Hence, the bomb strikes the ground after 2.86 s at a horizontal distance of 57.2 m from the tower.

Given,

• Velocity of projection, u = 16 m s^-1, at 30° above the horizontal
• Time of flight, t = 4 s
• Acceleration due to gravity, g = 9.8 m s^-2

Resolving the velocity of projection,

$$\begin{aligned}\text u_x = \text u \cos 30^\circ = 16 \times \frac{\sqrt{3}}{2} = 13.86\ \text{m s}^{-1} \\[1em] \text u_y = \text u \sin 30^\circ = 16 \times \frac{1}{2} = 8\ \text{m s}^{-1}\ \text{(upward)}\end{aligned}$$

Taking the downward direction as positive for the vertical motion, the vertical component of the velocity of projection is u_y = −8 m s^-1. Using $\text h = \text u_y\text t + \frac{1}{2}\text{gt}^2$,

$$\begin{aligned}\text h = (-8) \times 4 + \frac{1}{2} \times 9.8 \times (4)^2 \\[1em] = -32 + 78.4 \\[1em] = 46.4\ \text m\end{aligned}$$

The horizontal range is

$$\begin{aligned}\text R = \text u_x \times \text t = 13.86 \times 4 \\[1em] = 55.4\ \text m\end{aligned}$$

Hence, the height of the tower is 46.4 m and the horizontal range of the stone is 55.4 m.

Given,

• Velocity of projection, u = 14 m s^-1, at 30° above the horizontal
• Height of the point of projection, h = 12 m
• Acceleration due to gravity, g = 10 m s^-2

Resolving the velocity of projection,

$$\begin{aligned}\text u_x = 14 \cos 30^\circ = 14 \times \frac{\sqrt{3}}{2} = 12.12\ \text{m s}^{-1} \\[1em] \text u_y = 14 \sin 30^\circ = 14 \times \frac{1}{2} = 7\ \text{m s}^{-1}\ \text{(upward)}\end{aligned}$$

Taking the downward direction as positive for the vertical motion and using $\text h = -\text u_y\text t + \frac{1}{2}\text{gt}^2$,

$$\begin{aligned}12 = -7\text t + \frac{1}{2} \times 10 \times \text t^2 \\[1em] 5\text t^2 - 7\text t - 12 = 0\end{aligned}$$

Solving this quadratic equation in t,

$$\begin{aligned}\text t = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 5 \times (-12)}}{2 \times 5} \\[1em] = \frac{7 \pm \sqrt{49 + 240}}{10} \\[1em] = \frac{7 \pm 17}{10}\end{aligned}$$

Taking the positive value, t = 2.4 s.

The horizontal distance covered is

$$\begin{aligned}\text R = \text u_x \times \text t = 12.12 \times 2.4 \\[1em] = 29.1\ \text m\end{aligned}$$

Hence, the ball is thrown up to a distance of 29.1 m.

$\vec{S} = (1 - b)\vec{P} + b\vec{Q}$

Reason —

From the figure, the vector $\vec{\text{{R }}}$joins the tip of $\vec{\text{{P }}}$to the tip of $\vec{\text Q}$, so by the triangle law of vector addition,

$\vec{\text R} = \vec{\text Q} - \vec{\text P}$

Since S lies on $\vec{\text{R }}$and the displacement from P to S is $\text b\vec{\text {R }}$, the position vector of S is

$$\begin{aligned}\vec{\text S} = \vec{\text P} + \text b\vec{\text R} \\[1em] = \vec{\text P} + \text b(\vec{\text Q} - \vec{\text P}) \\[1em] = (1 - \text b)\vec{\text P} + \text b\vec{\text Q}\end{aligned}$$

$-7\hat{i} - 4\hat{j} - 8\hat{k}$

Reason —

Given,

• Force, $\vec{\text F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$
• Point of application of the force, P(2, 0, −3)
• Point about which the moment is taken, Q(2, −2, −2)

The moment of a force about a point is the vector product of the position vector of the point of application with respect to that point and the force,

$\vec{\tau} = \vec{\text r} \times \vec{\text F}$

The position vector of P with respect to Q is

$$\begin{aligned}\vec{\text r} = (\text x_P - \text x_Q)\hat{i} + (\text y_P - \text y_Q)\hat{j} + (\text z_P - \text z_Q)\hat{k} \\[1em] = (2 - 2)\hat{i} + [0 - (-2)]\hat{j} + [-3 - (-2)]\hat{k} \\[1em] = 0\hat{i} + 2\hat{j} - \hat{k}\end{aligned}$$

Writing the vector product as a determinant,

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned}\vec{\tau} = \hat{i}[(2)(-6) - (-1)(5)] - \hat{j}[(0)(-6) - (-1)(4)] + \hat{k}[(0)(5) - (2)(4)] \\[1em] = \hat{i}(-12 + 5) - \hat{j}(0 + 4) + \hat{k}(0 - 8) \\[1em] = -7\hat{i} - 4\hat{j} - 8\hat{k}\end{aligned}$$

remain constant

Reason —

The three forces are represented in magnitude and direction by the three sides of the triangle PQR taken in the same order. Such vectors complete a closed figure, so their sum-vector cannot be drawn, which means that their resultant is a null vector.

$\vec{\text F}_{net} = \vec{\text{PQ}} + \vec{\text{QR}} + \vec{\text{RP}} = 0$

By Newton's second law, the acceleration produced in the particle is

$\vec{\text a} = \frac{\vec{\text F}_{net}}{\text m} = \frac{0}{\text m} = 0$

Since no net force acts on the particle, its acceleration is zero and hence its velocity remains constant in both magnitude and direction.

−118.5

Reason —

Given,

• $|\vec{\text A}_1|$ = 3, so $\text A_1^2$ = 9
• $|\vec{\text A}_2|$ = 5, so $\text A_2^2$ = 25
• $|\vec{\text A}_1 + \vec{\text A}_2|$ = 5

Taking the self-product of the sum of the two vectors,

$$\begin{aligned}|\vec{\text A}_1 + \vec{\text A}_2|^2 = (\vec{\text A}_1 + \vec{\text A}_2) \cdot (\vec{\text A}_1 + \vec{\text A}_2) \\[1em] = \text A_1^2 + \text A_2^2 + 2\vec{\text A}_1 \cdot \vec{\text A}_2\end{aligned}$$

Substituting the values,

$$\begin{aligned}(5)^2 = (3)^2 + (5)^2 + 2\vec{\text A}_1 \cdot \vec{\text A}_2 \\[1em] 25 = 9 + 25 + 2\vec{\text A}_1 \cdot \vec{\text A}_2 \\[1em] 2\vec{\text A}_1 \cdot \vec{\text A}_2 = 25 - 34 = -9 \\[1em] \vec{\text A}_1 \cdot \vec{\text A}_2 = -\frac{9}{2} = -4.5\end{aligned}$$

Expanding the required product by the distributive law, and using $\vec{\text A}_2 \cdot \vec{\text A}_1 = \vec{\text A}_1 \cdot \vec{\text A}_2$,

$$\begin{aligned}(2\vec{\text A}_1 + 3\vec{\text A}_2) \cdot (3\vec{\text A}_1 - 2\vec{\text A}_2) = 6\vec{\text A}_1 \cdot \vec{\text A}_1 - 4\vec{\text A}_1 \cdot \vec{\text A}_2 + 9\vec{\text A}_2 \cdot \vec{\text A}_1 - 6\vec{\text A}_2 \cdot \vec{\text A}_2 \\[1em] = 6\text A_1^2 + 5(\vec{\text A}_1 \cdot \vec{\text A}_2) - 6\text A_2^2\end{aligned}$$

Substituting the values,

$$\begin{aligned}= 6(9) + 5(-4.5) - 6(25) \\[1em] = 54 - 22.5 - 150 \\[1em] = -118.5\end{aligned}$$

$\cos^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$

Reason —

Given,

• The two vectors have equal magnitudes, that is, A = B
• $|\vec{\text A} + \vec{\text B}| = \text n|\vec{\text A} - \vec{\text B}|$

Let θ be the angle between the two vectors. By the parallelogram law of vector addition,

$|\vec{\text A} + \vec{\text B}|^2 = \text A^2 + \text B^2 + 2\text{AB} \cos \theta$

Putting B = A,

$|\vec{\text A} + \vec{\text B}|^2 = \text A^2 + \text A^2 + 2\text A^2 \cos \theta = 2\text A^2(1 + \cos \theta)$

Similarly, replacing θ by (180° − θ) for the difference,

$|\vec{\text A} - \vec{\text B}|^2 = \text A^2 + \text A^2 - 2\text A^2 \cos \theta = 2\text A^2(1 - \cos \theta)$

Squaring both sides of the given condition,

$|\vec{\text A} + \vec{\text B}|^2 = \text n^2|\vec{\text A} - \vec{\text B}|^2$

Substituting the two expressions obtained above,

$$\begin{aligned}2\text A^2(1 + \cos \theta) = \text n^2 \times 2\text A^2(1 - \cos \theta) \\[1em] 1 + \cos \theta = \text n^2 - \text n^2 \cos \theta \\[1em] \cos \theta + \text n^2 \cos \theta = \text n^2 - 1 \\[1em] \cos \theta(1 + \text n^2) = \text n^2 - 1\end{aligned}$$

Therefore,

$\cos \theta = \frac{\text n^2 - 1}{\text n^2 + 1} \quad \Rightarrow \quad \theta = \cos^{-1}\left(\frac{\text n^2 - 1}{\text n^2 + 1}\right)$

y² = x² + constant

Reason —

Given,

• Velocity of the particle, $\vec{\text v} = \text K(\text y\hat{i} + \text x\hat{j})$, where K is a constant

The velocity of a particle is the time rate of change of its position, so its rectangular components are

$\text v_x = \frac{\text{dx}}{\text{dt}} \quad \text{and} \quad \text v_y = \frac{\text{dy}}{\text{dt}}$

Comparing these with the components of the given velocity,

$\frac{\text{dx}}{\text{dt}} = \text{Ky} \quad \text{and} \quad \frac{\text{dy}}{\text{dt}} = \text{Kx}$

Dividing the second equation by the first, the variable t is eliminated,

$$\begin{aligned}\frac{\text{dy}/\text{dt}}{\text{dx}/\text{dt}} = \frac{\text{Kx}}{\text{Ky}} \\[1em] \frac{\text{dy}}{\text{dx}} = \frac{\text x}{\text y}\end{aligned}$$

Separating the variables,

$\text y\text{dy} = \text x\text{dx}$

Integrating both sides,

$$\begin{aligned}\int \text y\text{dy} = \int \text x\text{dx} \\[1em] \frac{\text y^2}{2} = \frac{\text x^2}{2} + \text C\end{aligned}$$

where C is the constant of integration. Multiplying throughout by 2,

$$\begin{aligned}\text y^2 = \text x^2 + 2\text C \\[1em] \text y^2 = \text x^2 + \text{constant}\end{aligned}$$

30° west

Reason —

Given,

• Speed of the swimmer in still water, v_s = 20 m s^-1
• Speed of the river water, v_r = 10 m s^-1, flowing due east

To cross the river along the shortest path, the swimmer must move straight across, so the component of his velocity (v_s sinθ) along the direction of the flow must exactly cancel the velocity of the river (v_r ). If θ is the angle made by his strokes with the north,

$\text v_{r} = \text v_{s} \sin \theta$

Substituting the values,

$$\begin{aligned}10 = 20 \sin \theta \\[1em] \sin \theta = \frac{10}{20} = \frac{1}{2} \\[1em] \theta = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ\end{aligned}$$

Since the river flows due east, the swimmer must head 30° west of north.

120°

Reason —

Given,

• Speed of the stream of the river, v_r = 2 km h^-1
• Speed of the swimmer in still water, v_s = 4 km h^-1

For the swimmer to cross the river straight, the component of his velocity (v_s sinθ) along the flow must cancel the velocity of the river (v_r). Let θ be the angle made by his heading with the direction straight across the river. Then

$$\begin{aligned}\text v_{r} = \text v_{s} \sin \theta \\[1em] 2 = 4 \sin \theta \\[1em] \sin \theta = \frac{2}{4} = \frac{1}{2} \quad \Rightarrow \quad \theta = 30^\circ\end{aligned}$$

The direction straight across makes an angle of 90° with the flow of the river, and the swimmer must head 30° upstream of it. Hence the angle made with the direction of the flow is

$\alpha = 90^\circ + 30^\circ = 120^\circ$

$θ_0 = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$ and $v_0 = \frac{5}{3}$ m/s

Reason —

Given,

• Equation of the trajectory, y = 2x − 9x²
• Acceleration due to gravity, g = 10 m s^-2

The equation of the trajectory of a projectile launched with speed v₀ at an angle θ₀ with the horizontal is

$\text y = \text x \tan \theta_0 - \frac{\text{gx}^2}{2\text v_0^2 \cos^2 \theta_0}$

Comparing the coefficients of x in this equation with those in y = 2x − 9x²,

$\tan \theta_0 = 2$

Constructing a right-angled triangle with the side opposite to θ₀ equal to 2 and the adjacent side equal to 1, the hypotenuse is $\sqrt{(2)^2 + (1)^2} = \sqrt{5}$. Therefore,

$\sin \theta_0 = \frac{2}{\sqrt{5}} \quad \text{and} \quad \cos \theta_0 = \frac{1}{\sqrt{5}}$

that is, $\theta_0 = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$.

Comparing the coefficients of x²,

$\frac{\text g}{2\text v_0^2 \cos^2 \theta_0} = 9$

Substituting g = 10 m s^-2 and $\cos^2 \theta_0 = \frac{1}{5}$,

$$\begin{aligned}\frac{10}{2\text v_0^2 \times \frac{1}{5}} = 9 \\[1em] \frac{10 \times 5}{2\text v_0^2} = 9 \\[1em] \frac{25}{\text v_0^2} = 9 \\[1em] \text v_0^2 = \frac{25}{9} \quad \Rightarrow \quad \text v_0 = \frac{5}{3}\ \text{m s}^{-1}\end{aligned}$$

2.8 m

Reason —

Given,

• Velocity of projection, u = 10 m s^-1
• Angle of projection, θ = 60° with the horizontal
• Time at which the radius of curvature is required, t = 1 s
• Acceleration due to gravity, g = 10 m s^-2

Resolving the velocity of projection,

$$\begin{aligned}\text u_x = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5\ \text{m s}^{-1} \\[1em] \text u_y = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\ \text{m s}^{-1}\end{aligned}$$

At t = 1 s, the horizontal component is unchanged, that is, v_x = 5 m s^-1, while the vertical component is obtained from :

$$\begin{aligned}\text v_y = \text u_y - \text{gt} \\[1em] = 5\sqrt{3} - 10 \times 1 \\[1em] = 8.66 - 10 \\[1em] = -1.34\ \text{m s}^{-1}\end{aligned}$$

The negative sign shows that the body is moving downwards at this instant. The speed at that instant is

$$\begin{aligned}\text v^2 = \text v_x^2 + \text v_y^2 \\[1em] = (5)^2 + (-1.34)^2 \\[1em] = 25 + 1.8 \\[1em] = 26.8\ \text{m}^2\text{s}^{-2}\end{aligned}$$

If α is the angle made by the velocity with the horizontal,

$$\begin{aligned}\tan \alpha = \frac{|\text v_y|}{|\text v_x|} \\[1em] = \frac{1.34}{5} = 0.268 \quad \Rightarrow \quad \alpha = 15^\circ\end{aligned}$$

The component of the weight mg perpendicular to the velocity supplies the centripetal force needed to keep the body on its curved path, so

$$\begin{aligned}{\text {mg} \cos \alpha}= \frac{\text {mv}^2}{\text R} \\[1em] \text R = \frac{\text v^2}{\text g \cos \alpha}\\[1em] = \frac{26.8}{10 \times \cos 15^\circ} \\[1em] = \frac{26.8}{10 \times 0.97} \\[1em] = 2.8\ \text m\end{aligned}$$

$R^2 = 16 h_1 h_2$

Reason —

Given,

• Both particles are projected with the same speed u
• Both have the same horizontal range R
• Their maximum heights are h₁ and h₂

Two projectiles thrown with the same speed have the same range when their angles of projection are complementary, that is, θ and (90° − θ). Using $\text h = \frac{\text u^2 \sin^2 \theta}{2\text g}$ for each of them,

$\text h_1 = \frac{\text u^2 \sin^2 \theta}{2\text g} \quad \text{and} \quad \text h_2 = \frac{\text u^2 \sin^2(90^\circ - \theta)}{2\text g} = \frac{\text u^2 \cos^2 \theta}{2\text g}$

Multiplying the two,

$$\begin{aligned}\text h_1\text h_2 = \frac{\text u^2 \sin^2 \theta}{2\text g} \times \frac{\text u^2 \cos^2 \theta}{2\text g} \\[1em] = \frac{\text u^4 \sin^2 \theta \cos^2 \theta}{4\text g^2}\end{aligned}$$

Using sin 2θ = 2 sin θ cos θ, so that $\sin^2 \theta \cos^2 \theta = \frac{\sin^2 2\theta}{4}$,

$$\begin{aligned}\text h_1\text h_2 = \frac{\text u^4}{4\text g^2} \times \frac{\sin^2 2\theta}{4} \\[1em] \text h_1\text h_2= \frac{\text u^4 \sin^2 2\theta}{16\text g^2}\end{aligned}$$

The common range is $\text R = \frac{\text u^2 \sin 2\theta}{\text g}$, so on squaring,

$$\begin{aligned}\text R^2 = \frac{\text u^4 \sin^2 2\theta}{\text g^2} \\[1em] = \frac{16}{16} \times \frac{\text u^4 \sin^2 2\theta}{\text g^2} \\[1em] = 16\text h_1\text h_2\end{aligned}$$

$\sqrt{2}aω$

Reason —

Given,

• x = a cos ωt, y = a sin ωt and z = aωt, where a and ω are constants

The components of the velocity are obtained by differentiating the corresponding coordinates with respect to time.

Differentiating x with respect to t, and using the chain rule with $\frac{\text d(\omega \text t)}{\text{dt}} = \omega$,

$$\begin{aligned}\text v_x = \frac{\text{dx}}{\text{dt}} = \text a\frac{\text d}{\text{dt}}(\cos \omega \text t) \\[1em] = \text a(-\sin \omega \text t) \times \omega \\[1em] = -\text a\omega \sin \omega \text t\end{aligned}$$

Differentiating y with respect to t,

$$\begin{aligned}\text v_y = \frac{\text{dy}}{\text{dt}} = \text a\frac{\text d}{\text{dt}}(\sin \omega \text t) \\[1em] = \text a(\cos \omega \text t) \times \omega \\[1em] = \text a\omega \cos \omega \text t\end{aligned}$$

Differentiating z with respect to t,

$\text v_z = \frac{\text{dz}}{\text{dt}} = \text a\omega\frac{\text{dt}}{\text{dt}} = \text a\omega$

The magnitude of the velocity is

$$\begin{aligned}\text v = \sqrt{\text v_x^2 + \text v_y^2 + \text v_z^2} \\[1em] = \sqrt{\text a^2\omega^2 \sin^2 \omega \text t + \text a^2\omega^2 \cos^2 \omega \text t + \text a^2\omega^2} \\[1em] = \sqrt{\text a^2\omega^2(\sin^2 \omega \text t + \cos^2 \omega \text t) + \text a^2\omega^2}\end{aligned}$$

Using sin² ωt + cos² ωt = 1,

$$\begin{aligned}\text v = \sqrt{\text a^2\omega^2 + \text a^2\omega^2} \\[1em] = \sqrt{2\text a^2\omega^2} \\[1em] = \sqrt{2}\text a\omega\end{aligned}$$

$\frac{1}{2}a(\hat{j} - \hat{i})$

Reason —

Given,

• Side of the cube = a

From the figure, the origin O is at a corner of the cube, with OD along the X-axis, OF along the Y-axis and OB along the Z-axis. Hence the coordinates of the vertices are

$\text O(0, 0, 0), \quad \text D(\text a, 0, 0), \quad \text F(0, \text a, 0), \quad \text B(0, 0, \text a), \quad \text A(\text a, 0, \text a), \quad \text E(0, \text a, \text a)$

The central point of a face is the mid-point of its diagonal, so its coordinates are the averages of the coordinates of the opposite corners of that face.

The face ABOD lies in the plane y = 0, and its diagonal joins A(a, 0, a) and O(0, 0, 0). Hence the coordinates of its central point G are

$\text G\left(\frac{\text a + 0}{2},\ \frac{0 + 0}{2},\ \frac{\text a + 0}{2}\right) = \text G\left(\frac{\text a}{2},\ 0,\ \frac{\text a}{2}\right)$

The face BEFO lies in the plane x = 0, and its diagonal joins E(0, a, a) and O(0, 0, 0). Hence the coordinates of its central point H are

$\text H\left(\frac{0 + 0}{2},\ \frac{\text a + 0}{2},\ \frac{\text a + 0}{2}\right) = \text H\left(0,\ \frac{\text a}{2},\ \frac{\text a}{2}\right)$

The required vector is

$$\begin{aligned}\vec{\text{GH}} = (\text x_2 - \text x_1)\hat{i} + (\text y_2 - \text y_1)\hat{j} + (\text z_2 - \text z_1)\hat{k} \\[1em] = \left(0 - \frac{\text a}{2}\right)\hat{i} + \left(\frac{\text a}{2} - 0\right)\hat{j} + \left(\frac{\text a}{2} - \frac{\text a}{2}\right)\hat{k} \\[1em] = -\frac{\text a}{2}\hat{i} + \frac{\text a}{2}\hat{j} + 0\hat{k} \\[1em] = \frac{1}{2}\text a(\hat{j} - \hat{i})\end{aligned}$$

$\frac{1}{\sqrt{γg}}\tan^{-1}\left(\sqrt{\frac{γ}{g}}v_0\right)$

Reason —

Given,

• Initial velocity of the ball, v₀, directed vertically upwards
• Drag force on the ball, F = m$\gamma \text v^2$

Comparing it with F = ma, the retardation produced by the drag force alone is

$\text a = \frac{\text F}{\text m} = \frac{\text m\gamma \text v^2}{\text m} = \gamma \text v^2$

While the ball rises, the drag force opposes the motion and therefore acts downwards along with the force of gravity. Hence the net retardation is

$\text a_{net} = \frac{\text{dv}}{\text{dt}} = -(\text g + \gamma \text v^2)$

Separating the variables,

$\text{dt} = -\frac{\text{dv}}{\text g + \gamma \text v^2}$

At the zenith the ball comes to rest momentarily. Integrating from v = v₀ at t = 0 to v = 0 at t = T,

$\int_0^{\text T} \text{dt} = -\int_{\text v_0}^{0} \frac{\text{dv}}{\text g + \gamma \text v^2}$

Interchanging the limits of the integral on the right removes the negative sign,

$\text T = \int_0^{\text v_0} \frac{\text{dv}}{\text g + \gamma \text v^2}$

Taking γ common from the denominator,

$\text T = \frac{1}{\gamma}\int_0^{\text v_0} \frac{\text{dv}}{\frac{\text g}{\gamma} + \text v^2}$

Using the standard integral $\displaystyle\int \frac{\text{dv}}{\text b^2 + \text v^2} = \frac{1}{\text b}\tan^{-1}\left(\frac{\text v}{\text b}\right)$, with $\text b = \sqrt{\frac{\text g}{\gamma}}$

$$\begin{aligned}\text T = \frac{1}{\gamma} \times \sqrt{\frac{\gamma}{\text g}}\left[\tan^{-1}\left(\text v\sqrt{\frac{\gamma}{\text g}}\right)\right]_0^{\text v_0} \\[1em] = \frac{1}{\sqrt{\gamma \text g}}\left[\tan^{-1}\left(\text v\sqrt{\frac{\gamma}{\text g}}\right)\right]_0^{\text v_0}\end{aligned}$$

Applying the limits, and using tan^-1(0) = 0,

$$\begin{aligned}\text T = \frac{1}{\sqrt{\gamma \text g}}\left[\tan^{-1}\left(\text v_0\sqrt{\frac{\gamma}{\text g}}\right) - 0\right] \\[1em] = \frac{1}{\sqrt{\gamma \text g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{\text g}}\text v_0\right)\end{aligned}$$

$20\sqrt{2}$ m

Reason —

Given,

• Initial position vector, $\vec{\text r}_0 = (2.0\hat{i} + 4.0\hat{j})$ m at t = 0
• Initial velocity, $\vec{\text u} = (5.0\hat{i} + 4.0\hat{j})$ m s^-1
• Constant acceleration, $\vec{\text a} = (4.0\hat{i} + 4.0\hat{j})$ m s^-2
• Time, t = 2 s

Since the acceleration is constant, the equation of motion for the position vector at time t is

$\vec{\text r} = \vec{\text r}_0 + \vec{\text u}\text t + \frac{1}{2}\vec{\text a}\text t^2$

Substituting the values at t = 2 s,

$$\begin{aligned}\vec{\text r} = (2.0\hat{i} + 4.0\hat{j}) + (5.0\hat{i} + 4.0\hat{j})(2) + \frac{1}{2}(4.0\hat{i} + 4.0\hat{j})(2)^2 \\[1em] = (2\hat{i} + 4\hat{j}) + (10\hat{i} + 8\hat{j}) + \frac{1}{2}(4\hat{i} + 4\hat{j})(4) \\[1em] = (2\hat{i} + 4\hat{j}) + (10\hat{i} + 8\hat{j}) + (8\hat{i} + 8\hat{j})\end{aligned}$$

Adding the corresponding components,

$$\begin{aligned}\vec{\text r} = (2 + 10 + 8)\hat{i} + (4 + 8 + 8)\hat{j} \\[1em] = 20\hat{i} + 20\hat{j}\end{aligned}$$

The distance from the origin is the magnitude of this position vector,

$$\begin{aligned}|\vec{\text r}| = \sqrt{(20)^2 + (20)^2} \\[1em] = \sqrt{400 + 400} = \sqrt{800} \\[1em] = 20\sqrt{2}\ \text m\end{aligned}$$

$2 : \sqrt{3}$

Reason —

Given,

• Both projectiles are thrown with the same initial velocity u
• Angles of projection, θ₁ = 45° and θ₂ = 30°

The horizontal range of a projectile is

$\text R = \frac{\text u^2 \sin 2\theta}{\text g}$

For the same initial velocity u and the same g, R ∝ sin 2θ. Therefore,

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \frac{\sin 2\theta_1}{\sin 2\theta_2} = \frac{\sin(2 \times 45^\circ)}{\sin(2 \times 30^\circ)} \\[1em] = \frac{\sin 90^\circ}{\sin 60^\circ} \\[1em] = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\end{aligned}$$

Hence the ratio of the ranges is $2 : \sqrt{3}$.

$\sqrt{2} : 1$

Reason —

Given,

• Angles of projection, θ₁ = 30° and θ₂ = 45°
• Both projectiles reach the maximum height in the same time, that is, t₁ = t₂

At the maximum height the vertical component of the velocity becomes zero. Using v = u sin θ − gt with v = 0, the time taken to reach the maximum height is

$\text t = \frac{\text u \sin \theta}{\text g}$

Since the two times are equal,

$$\begin{aligned}\frac{\text u_1 \sin 30^\circ}{\text g} = \frac{\text u_2 \sin 45^\circ}{\text g} \\[1em] \text u_1 \sin 30^\circ = \text u_2 \sin 45^\circ\end{aligned}$$

Therefore,

$$\begin{aligned}\frac{\text u_1}{\text u_2} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} \\[1em] = \frac{1}{\sqrt{2}} \times \frac{2}{1} \\[1em] = \frac{2}{\sqrt{2}} = \sqrt{2}\end{aligned}$$

Hence the ratio of the initial velocities is $\sqrt{2} : 1$.

1 : 3 : 5 : 7

Reason —

Given,

• The body falls freely from rest, so its initial velocity u = 0 and its acceleration is g

For a body falling freely from rest (u = 0), the distance covered in the n^th second is

$\text s_n = u+ \frac{1}{2}\text g(2\text n - 1)$

Putting n = 1, 2, 3 and 4, and u = 0

$$\begin{aligned}\text s_1 = \frac{\text g}{2}(2 \times 1 - 1) = \frac{\text g}{2} \\[1em] \text s_2 = \frac{\text g}{2}(2 \times 2 - 1) = \frac{3\text g}{2} \\[1em] \text s_3 = \frac{\text g}{2}(2 \times 3 - 1) = \frac{5\text g}{2} \\[1em] \text s_4 = \frac{\text g}{2}(2 \times 4 - 1) = \frac{7\text g}{2}\end{aligned}$$

Taking the ratio and cancelling the common factor $\frac{\text g}{2}$,

$$\begin{aligned}\text s_1 : \text s_2 : \text s_3 : \text s_4 = \frac{\text g}{2} : \frac{3\text g}{2} : \frac{5\text g}{2} : \frac{7\text g}{2} \\[1em] = 1 : 3 : 5 : 7\end{aligned}$$

that is, the distances covered in successive seconds are in the ratio of the odd numbers.

$5\sqrt{3}$ ms^-1

Reason —

Given,

• Velocity of projection, u = 10 m s^-1
• Angle of projection with the vertical direction = 60°

Since the angle of projection is measured from the vertical, the angle made with the horizontal is

$\theta = 90^\circ - 60^\circ = 30^\circ$

The horizontal and the vertical motions of a projectile are independent of each other. No force acts along the horizontal direction, so the horizontal component of the velocity remains unaltered throughout the flight, while the vertical component becomes zero at the highest point.

Hence at the highest point only the horizontal component remains,

$$\begin{aligned}\text v = \text u \cos 30^\circ \\[1em] = 10 \times \frac{\sqrt{3}}{2} \\[1em] = 5\sqrt{3}\ \text{m s}^{-1}\end{aligned}$$

1 : 1

Reason —

Given,

• Radii of the concentric circular paths, r_A and r_B
• Speeds of the two particles, v_A and v_B
• The time periods of rotation are the same, that is, T_A = T_B

In one complete revolution a particle turns through an angle 2π in time T, so the angular velocity is related to the time period by

$\omega = \frac{2\pi}{\text T}$

Since the time periods of the two particles are equal, T_A = T_B, and therefore

$\frac{\omega_A}{\omega_B} = \frac{2\pi / \text T_A}{2\pi / \text T_B} = \frac{\text T_B}{\text T_A} = 1$

Thus the ratio of their angular speeds is 1 : 1, whatever the radii of their paths may be. This agrees with the fact that the angular speed does not change with radius, although the linear speed does, since v = rω.

$ω(R_2 - R_1)\hat{i}$

Reason —

Given,

• Radii of the two concentric circles, R₁ (inner) and R₂ (outer)
• Both particles have the same angular speed ω
• Time at which the relative velocity is required, $\text t = \frac{\pi}{2\omega}$

From the figure, at t = 0 the particle A is on the inner circle at (R₁, 0) moving along the positive Y-direction, so it moves anticlockwise; and the particle B is on the outer circle at (R₂, 0) moving along the negative Y-direction, so it moves clockwise.

Writing the position vectors at time t, the angle turned through in time t being ωt,

$$\begin{aligned}\vec{\text r}_A = \text R_1 \cos \omega \text t\ \hat{i} + \text R_1 \sin \omega \text t\ \hat{j} \\[1em] \vec{\text r}_B = \text R_2 \cos \omega \text t\ \hat{i} - \text R_2 \sin \omega \text t\ \hat{j}\end{aligned}$$

The negative sign in the second relation appears because B turns clockwise.

Differentiating each position vector with respect to time, and using the chain rule with $\frac{\text d(\omega \text t)}{\text{dt}} = \omega$,

$$\begin{aligned}\vec{\text v}_A = \frac{\text d\vec{\text r}_A}{\text{dt}} = -\text R_1\omega \sin \omega \text t\ \hat{i} + \text R_1\omega \cos \omega \text t\ \hat{j} \\[1em] \vec{\text v}_B = \frac{\text d\vec{\text r}_B}{\text{dt}} = -\text R_2\omega \sin \omega \text t\ \hat{i} - \text R_2\omega \cos \omega \text t\ \hat{j}\end{aligned}$$

At $\text t = \frac{\pi}{2\omega}$ that is ωt = 90°, we have sin ωt = 1 and cos ωt = 0. Therefore,

$\vec{\text v}_A = -\text R_1\omega\ \hat{i} \quad \text{and} \quad \vec{\text v}_B = -\text R_2\omega\ \hat{i}$

Hence the relative velocity is

$$\begin{aligned}\vec{\text v}_A - \vec{\text v}_B = -\text R_1\omega\ \hat{i} - (-\text R_2\omega\ \hat{i}) \\[1em] = -\text R_1\omega\ \hat{i} + \text R_2\omega\ \hat{i} \\[1em] = \omega(\text R_2 - \text R_1)\hat{i}\end{aligned}$$

10 m/s

Reason —

Given,

• Constant speed of the particle, v = 10 m s^-1
• Angle turned through around the centre of the circle = 60°

The velocity of a particle in circular motion is always tangential to the path. When the particle turns through 60° around the centre, its velocity vector also turns through 60°, so the angle between $\vec{\text v}_1$ and $\vec{\text v}_2$ is 60°.

The change in velocity is $\Delta \vec{\text v} = \vec{\text v}_2 - \vec{\text v}_1 = \vec{\text v}_2 + (-\vec{\text v}_1)$. The two velocity vectors are equal in magnitude (v = 10 m s^-1) and the angle between $\vec{\text v}_2$ and $(-\vec{\text v}_1)$ is (180° − 60°) = 120°. Therefore,

$$\begin{aligned}|\Delta \vec{\text v}|^2 = \text v^2 + \text v^2 + 2\text v^2 \cos 120^\circ \\[1em] = 2\text v^2 + 2\text v^2\left(-\frac{1}{2}\right) \\[1em] = 2\text v^2 - \text v^2 \\[1em] = \text v^2\end{aligned}$$

Hence

$|\Delta \vec{\text v}| = \text v = 10\ \text{m s}^{-1}$

varying velocity and varying acceleration.

Reason — In uniform circular motion the speed remains constant, but the direction of the velocity changes continuously as the velocity is always tangential to the path. Hence the velocity is varying. The centripetal acceleration $\text a = \frac{\text v^2}{\text r}$ is constant in magnitude but is always directed along the radius towards the centre, so its direction also changes continuously. Since both are vectors, and a vector changes if either its magnitude or its direction changes, both the velocity and the acceleration are varying.

112.5 m

Reason —

Given,

• Speed of the river, v_r = 9 km h^-1, from west to east
• Maximum speed of the boat in still water, v_b = 27 km h^-1
• Angle of the boat's heading with the direction of the river flow = 150°
• Time taken to cross the river, t = half a minute = 30 s

The boat moves at an angle of 150° to the direction of the river flow, so the component of the boat's velocity perpendicular to the banks is

$$\begin{aligned}\text v_\perp = \text v_b \sin 150^\circ = 27 \times \frac{1}{2} \\[1em] = 13.5\ \text{km h}^{-1}\end{aligned}$$

The velocity of the river is along the banks and therefore does not help in crossing; it only carries the boat downstream. Converting into m s^-1,

$$\begin{aligned}\text v_\perp = 13.5 \times \frac{5}{18} \\[1em] = 3.75\ \text{m s}^{-1}\end{aligned}$$

The time taken to cross is 30 s, so the width of the river is

$$\begin{aligned}\text d = \text v_\perp \times \text t \\[1em] = 3.75 \times 30 \\[1em] = 112.5\ \text m\end{aligned}$$

1. the path of the particle is an ellipse
2. the velocity and acceleration of the particle are normal to each other at t = π / 2p

Reason —

Given,

• x = a cos pt and y = b sin pt, where a, b (< a) and p are positive constants

Option 1 : From the given coordinates,

$\frac{\text x}{\text a} = \cos \text{pt} \quad \text{and} \quad \frac{\text y}{\text b} = \sin \text{pt}$

Squaring and adding, and using sin² pt + cos² pt = 1,

$$\begin{aligned}\frac{\text x^2}{\text a^2} + \frac{\text y^2}{\text b^2} = \cos^2 \text{pt} + \sin^2 \text{pt} \\[1em] \frac{\text x^2}{\text a^2} + \frac{\text y^2}{\text b^2} = 1\end{aligned}$$

which is the equation of an ellipse with semi-axes a and b. Hence option 1 is correct.

Option 2 : Differentiating the coordinates with respect to time, and using the chain rule with $\frac{\text d(\text{pt})}{\text{dt}} = \text p$,

$$\begin{aligned}\text v_x = \frac{\text{dx}}{\text{dt}} = \text a(-\sin \text{pt}) \times \text p = -\text{ap} \sin \text{pt} \\[1em] \text v_y = \frac{\text{dy}}{\text{dt}} = \text b(\cos \text{pt}) \times \text p = \text{bp} \cos \text{pt}\end{aligned}$$

so that

$\vec{\text v} = -\text{ap} \sin \text{pt}\ \hat{i} + \text{bp} \cos \text{pt}\ \hat{j}$

Differentiating the velocity components once more,

$$\begin{aligned}\text a_x = \frac{\text{dv}_x}{\text{dt}} = -\text{ap}(\cos \text{pt}) \times \text p = -\text{ap}^2 \cos \text{pt} \\[1em] \text a_y = \frac{\text{dv}_y}{\text{dt}} = \text{bp}(-\sin \text{pt}) \times \text p = -\text{bp}^2 \sin \text{pt}\end{aligned}$$

so that

$\vec{\text a} = -\text{ap}^2 \cos \text{pt}\ \hat{i} - \text{bp}^2 \sin \text{pt}\ \hat{j}$

At $\text t = \frac{\pi}{2\text p}$, that is pt = 90°, we have sin pt = 1 and cos pt = 0. Therefore,

${\vec{\text v}} = -\text{ap}\ \hat{i} \quad \text{and} \quad \vec{\text a} = -\text{bp}^2\ \hat{j}$

Taking their scalar product,

$\vec{\text v} \cdot \vec{\text a} = (-\text{ap})(0) + (0)(-\text{bp}^2) = 0$

Since the scalar product is zero and neither vector is a null vector, the two are normal to each other. Hence option 2 is correct.

Option 3 : The acceleration can be written as

$$\begin{aligned}\vec{\text a} = -\text p^2(\text a \cos \text{pt}\ \hat{i} + \text b \sin \text{pt}\ \hat{j}) \\[1em] = -\text p^2(\text x\hat{i} + \text y\hat{j}) = -\text p^2\vec{\text r}\end{aligned}$$

The negative sign shows that the acceleration is always directed opposite to the position vector, that is, towards the centre of the ellipse and not towards a focus. Hence option 3 is incorrect.

Option 4 : In the interval t = 0 to $\text t = \frac{\pi}{2\text p}$ the particle moves from the point (a, 0) to the point (0, b) along a quarter of the elliptical path. The distance travelled is the length of this arc, which is greater than a and is not equal to a. Hence option 4 is incorrect.

Given,

• $\vec{\text A} = \text a\hat{i}$ and $\vec{\text B} = \text a(\cos \omega \text t\ \hat{i} + \sin \omega \text t\ \hat{j})$
• $\omega = \frac{\pi}{6}$ rad s^-1

Both vectors have the same magnitude a, and the angle between them is ωt. Therefore,

$$\begin{aligned}|\vec{\text A} + \vec{\text B}|^2 = \text a^2 + \text a^2 + 2\text a^2 \cos \omega \text t = 2\text a^2(1 + \cos \omega \text t) \\[1em] |\vec{\text A} - \vec{\text B}|^2 = \text a^2 + \text a^2 - 2\text a^2 \cos \omega \text t = 2\text a^2(1 - \cos \omega \text t)\end{aligned}$$

Given that $|\vec{\text A} + \vec{\text B}| = \sqrt{3}|\vec{\text A} - \vec{\text B}|$, squaring both sides,

$$\begin{aligned}2\text a^2(1 + \cos \omega \text t) = 3 \times 2\text a^2(1 - \cos \omega \text t) \\[1em] 1 + \cos \omega \text t = 3 - 3\cos \omega \text t \\[1em] 4\cos \omega \text t = 2 \quad \Rightarrow \quad \cos \omega \text t = \frac{1}{2}\end{aligned}$$

For the first time this happens when

$\omega \text t = \frac{\pi}{3}$

Substituting $\omega = \frac{\pi}{6}$,

$\frac{\pi}{6}\tau = \frac{\pi}{3} \quad \Rightarrow \quad \tau = 2\ \text s$

Hence, the value of τ is 2.00 s.

Given,

• Angle of projection, θ₁ = 45°
• Maximum height before the bounce, h₁ = 120 m
• Angle after the bounce, θ₂ = 30°
• Kinetic energy after the bounce = half the kinetic energy before the bounce

Before the bounce, the maximum height is

$\text h_1 = \frac{\text u^2 \sin^2 45^\circ}{2\text g} = \frac{\text u^2}{4\text g} = 120 \quad \Rightarrow \quad \text u^2 = 480\ \text g$

The ball returns to the ground with the same speed u. Since it loses half of its kinetic energy on hitting the ground,

$\frac{1}{2}\text{mv}^2 = \frac{1}{2}\left(\frac{1}{2}\text{mu}^2\right) \quad \Rightarrow \quad \text v^2 = \frac{\text u^2}{2} = 240\ \text g$

The maximum height reached after the bounce is

$$\begin{aligned}\text h_2 = \frac{\text v^2 \sin^2 30^\circ}{2\text g} = \frac{240\text g \times \frac{1}{4}}{2\text g} \\[1em] = \frac{60}{2} \\[1em] = 30\ \text m\end{aligned}$$

Hence, the maximum height reached after the bounce is 30.00 m.

Given,

• $\vec{\text A} = (2\hat{i} + 3\hat{j} - \hat{k})$ m
• $\vec{\text B} = (\hat{i} + 2\hat{j} + 2\hat{k})$ m

The magnitude of the component of $\vec{\text A} \space$along $\vec{\text B} \space$has to be calculated.

If θ is the angle between the two vectors, the component (projection) of $\vec{\text A} \space$along $\vec{\text B} \space$is A cos θ. Since $\vec{\text A} \cdot \vec{\text B} = \text{AB} \cos \theta$,

$\text A \cos \theta = \frac{\vec{\text A} \cdot \vec{\text B}}{\text B} = \frac{\vec{\text A} \cdot \vec{\text B}}{|\vec{\text B}|}$

The scalar product is equal to the sum of the products of the corresponding components,

$$\begin{aligned}\vec{\text A} \cdot \vec{\text B} = (2)(1) + (3)(2) + (-1)(2) \\[1em] = 2 + 6 - 2 \\[1em] = 6\ \text m^2\end{aligned}$$

The magnitude of $\vec{\text B} \space$is

$$\begin{aligned}|\vec{\text B}| = \sqrt{(1)^2 + (2)^2 + (2)^2} \\[1em] = \sqrt{1 + 4 + 4} = \sqrt{9} \\[1em] = 3\ \text m\end{aligned}$$

Substituting these values,

$$\begin{aligned}\text A \cos \theta = \frac{6}{3} \\[1em] = 2\ \text m\end{aligned}$$

Hence, the magnitude of the component of $\vec{\text A} \space$along $\vec{\text B} \space$is 2 m.

Given,

• The ball is thrown from (0, 0) with speed v₀ at an angle θ₀ with the x-direction
• The stone is thrown from (L, 0) with speed v₁ at an angle (180° − θ₁) with the x-direction
• Case (i) : (θ₀, θ₁) = (45°, 45°), time of collision T₁
• Case (ii) : (θ₀, θ₁) = (60°, 30°), time of collision T₂

The value of $\left(\frac{\text T_1}{\text T_2}\right)^2$ has to be calculated.

Both the ball and the stone move under the same acceleration due to gravity, so their relative acceleration is zero and the relative motion between them is uniform along a straight line.

Resolving the velocities of the two bodies, the ball has components (v₀ cos θ₀, v₀ sin θ₀) and the stone, thrown at (180° − θ₁), has components (−v₁ cos θ₁, v₁ sin θ₁).

Since the two must meet, their vertical displacements must be equal at the instant of collision,

$\text v_0 \sin \theta_0\text T - \frac{1}{2}\text{gT}^2 = \text v_1 \sin \theta_1\text T - \frac{1}{2}\text{gT}^2$

The terms containing g cancel, giving

$\text v_0 \sin \theta_0 = \text v_1 \sin \theta_1 \quad \text{...............(1)}$

Along the horizontal direction the two bodies approach each other, so the separation L is closed by the sum of the horizontal components of the two velocities,

$$\begin{aligned}(\text v_0 \cos \theta_0 + \text v_1 \cos \theta_1)\text T = \text L \\[1em] \text T = \frac{\text L}{\text v_0 \cos \theta_0 + \text v_1 \cos \theta_1} \quad \text{...............(2)}\end{aligned}$$

Case (i) : (θ₀, θ₁) = (45°, 45°)

From equation (1),

$\text v_0 \sin 45^\circ = \text v_1 \sin 45^\circ \quad \Rightarrow \quad \text v_1 = \text v_0$

Substituting in equation (2),

$$\begin{aligned}\text T_1 = \frac{\text L}{\text v_0 \cos 45^\circ + \text v_0 \cos 45^\circ} \\[1em] = \frac{\text L}{\text v_0 \times \frac{1}{\sqrt{2}} + \text v_0 \times \frac{1}{\sqrt{2}}} \\[1em] = \frac{\text L}{\frac{2\text v_0}{\sqrt{2}}} = \frac{\text L}{\sqrt{2}\text v_0}\end{aligned}$$

Case (ii) : (θ₀, θ₁) = (60°, 30°)

From equation (1),

$$\begin{aligned}\text v_0 \sin 60^\circ = \text v_1 \sin 30^\circ \\[1em] \text v_0 \times \frac{\sqrt{3}}{2} = \text v_1 \times \frac{1}{2} \\[1em] \text v_1 = \sqrt{3}\text v_0\end{aligned}$$

Substituting in equation (2),

$$\begin{aligned}\text T_2 = \frac{\text L}{\text v_0 \cos 60^\circ + \sqrt{3}\text v_0 \cos 30^\circ} \\[1em] = \frac{\text L}{\text v_0 \times \frac{1}{2} + \sqrt{3}\text v_0 \times \frac{\sqrt{3}}{2}} \\[1em] = \frac{\text L}{\frac{\text v_0}{2} + \frac{3\text v_0}{2}} = \frac{\text L}{2\text v_0}\end{aligned}$$

Taking the ratio of the two times,

$$\begin{aligned}\frac{\text T_1}{\text T_2} = \frac{\text L}{\sqrt{2}\text v_0} \times \frac{2\text v_0}{\text L} \\[1em] = \frac{2}{\sqrt{2}} = \sqrt{2}\end{aligned}$$

Therefore,

$$\begin{aligned}\left(\frac{\text T_1}{\text T_2}\right)^2 = (\sqrt{2})^2 \\[1em] = 2\end{aligned}$$

Hence, the value of $\left(\frac{\text T_1}{\text T_2}\right)^2$ is 2.

Note: In the question, both $t_1$ and $T_1$ are used to represent the same time, while both $\theta$ and $\theta_1$ are used to represent the same angle. To avoid confusion and keep the notation consistent, the solution uses $T_1$ for time and $\theta_1$ for the angle throughout.

Given,

• Mass of the projectile, m = 200 g = 0.2 kg
• Initial velocity, u = 270 m s^-1 at 60° with the horizontal
• Drag coefficient, C = 0.1 kg s^-1
• Time of flight up to the wall, t = 2 s
• e = 2.7

The horizontal distance of the wall from the point of projection has to be calculated.

The horizontal component of the velocity of projection is

$$\begin{aligned}\text u_x = \text u \cos 60^\circ = 270 \times \frac{1}{2} \\[1em] = 135\ \text{m s}^{-1}\end{aligned}$$

Along the horizontal direction the only force acting is the viscous drag, which opposes the motion. Applying Newton's second law,

$\text m\frac{\text{dv}}{\text{dt}} = -\text{Cv}$

Separating the variables,

$\frac{\text{dv}}{\text v} = -\frac{\text C}{\text m}\text{dt}$

Integrating from v = u_x at t = 0 to v = v_x at time t,

$$\begin{aligned}\int_{\text u_x}^{\text v_x} \frac{\text{dv}}{\text v} = -\frac{\text C}{\text m}\int_0^{\text t} \text{dt} \\[1em] \left[\ln \text v\right]_{\text u_x}^{\text v_x} = -\frac{\text C}{\text m}\left[\text t\right]_0^{\text t} \\[1em] \ln\left(\frac{\text v_x}{\text u_x}\right) = -\frac{\text{Ct}}{\text m}\end{aligned}$$

Taking the antilogarithm,

$\text v_x = \text u_x\text e^{-\text{Ct}/\text m}$

Here $\frac{\text C}{\text m} = \frac{0.1}{0.2} = 0.5$ s^-1, so v_x = u_x e^-0.5t. The horizontal distance covered in time t is obtained by integrating the velocity,

$$\begin{aligned}\text x = \int_0^{\text t} \text v_x\text{dt} = \int_0^{\text t} \text u_x\text e^{-0.5\text t}\text{dt} \\[1em] = \text u_x\left[\frac{\text e^{-0.5\text t}}{-0.5}\right]_0^{\text t} \\[1em] = -\frac{\text u_x}{0.5}\left(\text e^{-0.5\text t} - \text e^{0}\right) \\[1em] = \frac{\text u_x}{0.5}\left(1 - \text e^{-0.5\text t}\right)\end{aligned}$$

Substituting u_x = 135 m s^-1 and t = 2 s,

$$\begin{aligned}\text x = \frac{135}{0.5}\left(1 - \text e^{-0.5 \times 2}\right) \\[1em] = 270\left(1 - \text e^{-1}\right) \\[1em] = 270\left(1 - \frac{1}{2.7}\right) \\[1em] = 270 \times \frac{2.7 - 1}{2.7} \\[1em] = 270 \times \frac{1.7}{2.7} \\[1em] = 170\ \text m\end{aligned}$$

Hence, the horizontal distance of the wall from the point of projection is 170 m.

Note: In the question, the drag force is incorrectly printed as $\vec{F} - C\vec{v}$. The correct formula is $\vec{F} = -C\vec{v}$, where the negative sign shows that the drag force acts in the direction opposite to the velocity. Therefore, the correct formula has been used in the solution.