Motion

a body at rest

Reason — A book lying on a table will not change its position if it is not disturbed, so it is said to be at rest.

oscillatory

Reason — The motion of pendulum is to and fro motion which is called oscillatory motion.

rectilinear motion

Reason — As the car is moving on a straight road so it is in rectilinear motion.

linear

Reason — A ball falls down vertically i.e. in a straight line, so its motion is linear.

uniform

Reason — If a moving body travels equal distances in equal intervals of time, then the motion is said to be uniform.

24 km h^-1

Reason — Suppose the distance between school and house is d km.

Case 1 : When the boy goes to the school

• Speed of the bus = 20 km h^-1.

Then

Time taken to reach the school is given by,

$$\begin{aligned}\text {t}_1 = \frac{\text{Distance between the house and the school}}{\text{Speed of the bus}} \\[1em] = \frac{\text d}{\text{20}}\end{aligned}$$

Case 2 : When the boy returns from the school

• Speed of the bus = 30 km h^-1.

Then

Time taken to return to the house is given by,

$$\begin{aligned}\text {t}_2 = \frac{\text{Distance between the house and the school}}{\text{Speed of the bus}} \\[1em] = \frac{\text d}{\text{30}}\end{aligned}$$

Now

• Total distance covered = d + d = 2d
• Total time taken = $\text {t}_1 + \text {t}_2$

And

$$\begin{aligned}\Rightarrow \text {t}_1 + \text {t}_2 = \frac{\text d}{\text{20}} + \frac{\text d}{\text{30}} \\[1em] = \frac{3\text d + 2\text d}{60} \\[1em] = \frac{5\text d}{60} \\[1em] = \frac{\text d}{12}\end{aligned}$$

Also,

$$\begin{aligned}\text {Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}} \\[1em] = \frac{2\text d}{\frac{\text d}{12}} \\[1em] = \frac{2\times 12}{1} \\[1em] = 2\times 12\\[1em] = 24 \text{ km h}^{-1}\end{aligned}$$

Hence, the average speed of the journey is 24 km h^-1.

500 N

Reason — Weight = 50 x 10 = 500 N

curvilinear

Reason — When a cyclist takes a turn, the path followed is curved, so the motion is called curvilinear motion.

• It is not rectilinear motion because rectilinear motion occurs along a straight line, whereas the cyclist is moving along a curved path.
• It is not translatory motion as the best description here, although the cyclist does move from one place to another; this term does not specify the nature of the path (straight or curved), so it is not the most appropriate answer.
• It is not circular motion because circular motion requires movement along a fixed circular path with a constant radius and a definite centre, which is not the case when a cyclist takes a turn.

Hence, the cyclist taking a turn on a road is an example of curvilinear motion.

circular

Reason — In circular motion, a body moves along a circular path such that the distance of any point on the body from a fixed reference point (the centre) remains constant.

The other options are not correct because

• In vibratory motion, the body moves to and fro about a mean position
• In rectilinear motion, the motion is along a straight line
• In curvilinear motion, the path is curved but the distance from a fixed point is not constant.

Hence, in circular motion, the distance of a point on the body remains constant from a fixed reference point.

A is false but R is true

Explanation

Assertion (A) is false because weight is not equal to mass in deep space but the gravitational force is extremely small there, so the weight of an object becomes nearly zero, while its mass remains unchanged.

Reason (R) is true because the weight of a body is the force with which earth attracts the body i.e. the weight of a body is the force of gravity.

Therefore, assertion is false but reason is true.

Both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because when a moving body covers equal distances in equal intervals of time, its motion is defined as uniform motion.

Reason (R) is true because in uniform motion, the speed remains constant, meaning the body covers the same distance in each equal time interval.

The Reason justifies the Assertion because constant speed directly leads to equal distance covered in equal time intervals which is nothing but definition of uniform motion.

Therefore, both A and R are true and R is the correct explanation of A.

(a) Two boys cycling on the road with the same speed are at rest relative to each other.

(b) The motion in a straight line is rectilinear motion.

(c) One to and fro motion of a clock pendulum takes time = 2 s .

(d) 36 km h^-1 = 10 m s^-1.

(e) Total distance travelled = average speed x total time taken.

(f) The weight of a girl is 36 kgf. Her mass will be 36 kg .

(g) The weight of a body is measured using spring balance .

(a) False
Correct Statement — Two trains going in opposite directions with the same speed are at motion relative to each other.

(b) False
Correct Statement — A ball is thrown vertically upwards. Its motion is non-uniform throughout.

(c) True

(d) True

(e) False
Correct Statement — A ball thrown by a boy from a roof-top has curvilinear motion.

(f) False
Correct Statement — Mass has only magnitude.

(g) True

(h) False
Correct Statement — Weight varies from place to place but mass does not.

An object can be in motion relative to one set of objects while at rest relative to some other set of objects. Thus, rest and motion are relative terms.

Example : If a boy is sitting on a bench in a bus stop and he sees a moving bus, for him the bus is in motion but the trees near by will appear to him at rest. But a boy who is sitting inside a bus, to him the trees and the boy outside sitting on the bench will appear to move in opposite direction. But the roof of the bus and driver will appear to him at rest.

From the above example it can be concluded that rest and motion are relative terms.

(a) A person walking in a compartment of a stationary train is in motion relative to the compartment and is in motion relative to the platform.

(b) A person sitting in a compartment of a moving train is at rest relative to the other person sitting by his side and is in motion relative to the platform.

The five different types of motion are:

1. Translatory motion
2. Circular motion
3. Oscillatory motion
4. Vibratory motion
5. Rotatory motion

A body is said to have translatory motion if every point on the moving body moves through the same distance in the same interval of time.

Example — Motion of a box when pushed from one corner of a room to the other.

A body is said to be in a rotatory motion if it moves about a fixed axis.
Examples — a spinning top, a merry-go-round.

The motion of a body along a circular path is called circular motion.

Example — Motion of a satellite around the earth.

The to and fro motion of a part of the body keeping its rest part fixed is called vibratory motion.

Example — Vibration of the membrane of tabla.

The motion of an object which takes path in no specific direction is called random motion.

Example — Motion of molecules of a gas.

(a) Rectilinear motion

(b) Rotatory motion

(c) Oscillatory motion and Periodic motion

(d) Non-periodic motion

(e) Circular motion

(f) Periodic motion, circular motion

(g) Rotatory motion

(a) Periodic motion

(b) Translatory and rotatory motion

(c) Translatory and rotatory motion

(d) Translatory and oscillatory motion

The weight of the body on earth is the force with which earth attracts the body. The S.I. unit of weight is newton (N).

1 kgf = 9.8 (nearly 10) N

Mass does not change by change of place.

The direction of weight is always directed vertically downwards.

A body is said to be at rest if it does not change its position with respect to its immediate surroundings.

A body is said to be in motion if it changes its position with respect to its immediate surroundings.

(i) The motion of a body in a straight line is called rectilinear or linear motion.

Example — A car moving on a straight road.

(ii) If the motion of a body is along a curved path, it is said to be a curvilinear motion.

Example — The motion of a cycle while taking a turn on the road.

The to and fro motion of a body from its rest position is called the oscillatory motion.

Example — Motion of a pendulum of a wall clock.

The bob of pendulum moves to side A from its rest position O, again comes back to position O and then moves to side B and then comes back to rest position O. This process is continuously repeated and this to and fro motion is called oscillatory motion.

The two examples that show a body can have two or more types of motion (mixed motion) simultaneously are:

1. The Earth rotates about its axis (rotatory motion) and at the same time, it revolves around the sun in a curved path (curvilinear or circular motion) in a fixed time interval (periodic motion).

2. A ball rolling on the ground has rotatory and translatory motion as it moves on the ground.

The average speed is calculated by finding the ratio of the total distance travelled by the body to the total time taken in the journey.

$\text {Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$

Given,

• Total distance travelled = 160 km
• Total time taken = 4 h

$$\begin{aligned}\text{Average Speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}} \\[1em] = \frac{\text{160}}{\text{4}} \\[1em] = 40 \text{ km h}^{-1}\end{aligned}$$

So, the average speed of the car is 40 km h^-1.

Given,

• Distance covered by the train = 300 km
• Average speed of the train = 60 km h^-1

Average speed of an object is given by,

$$\begin{aligned}\text{Average speed} = \frac{\text{Distance travelled}}{\text{Total time taken}} \\[1em] \Rightarrow \text {Total time taken} = \frac{\text{Distance travelled}}{\text{Average speed}} \\[1em] = \frac{300}{60} \\[1em] = 5 \text{ hours}\end{aligned}$$

So, time taken by the train to cover the distance is 5 hour.*

Given,

• Average speed of the boy = 10 m s^-1;
• Time taken = 20 min

Since

1 min = 60 sec ⟹ 20 min = 60 x 20 = 1200 sec

Average speed of an object is given by,

$$\begin{aligned}\text{Average speed} = \frac{\text{Total distance covered}}{\text{Time taken}} \\[1em] \Rightarrow \text{Total distance covered} = \text{Average speed} \times \text{Time taken} \\[1em] = 10 \times 1200 \\[1em] = 12000 \text{ m}\end{aligned}$$

So, distance covered by the boy is 12000 m or 12 km.*

Given,

• Initial distance covered by the boy = 30 m
• Time taken to cover 30 m = 1 minute
• Final distance covered by the boy = 30 m
• Time taken to cover 30 m = 1.5 minute

Since the boy covers equal distances in unequal intervals of time it means the motion of the boy is non-uniform.

Now

• Total distance travelled = 30 m + 30 m = 60 m
• Total time taken = 1 min + 1.5 min = 2.5 min

Since

1 min = 60 sec ⟹ 2.5 min = 2.5 x 60 = 150 sec
Average speed = ?

Average speed of an object is given by,

$$\begin{aligned}\text{Average Speed} = \frac{\text{Total distance travelled}}{\text{total time taken}} \\[1em] = \frac{\text{60}}{\text{150}} \\[1em] = 0.4 \text{ m s}^{-1}\end{aligned}$$

So, the average speed of the boy is 0.4 m s^-1.

Given,

• Distance covered in first hour = 1 km
• Distance covered in second hour = 0.5 km
• Distance covered in third hour = 0.3 km

(i) Now,

• Total distance travelled = 1 km + 0.5 km + 0.3 km = 1.8 km
• Total time taken = 1 h + 1 h + 1 h = 3 h

Average speed of an object is given by,

$$\begin{aligned}\text{Average Speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}} \\[1em] = \frac{\text{1.8}}{\text{3}} \\[1em] = 0.6 \text{ km h}^{-1}\end{aligned}$$

So, the average speed of the cyclist is 0.6 km h^-1.

(ii) As

• Total distance travelled = 1.8 km

Since

1 km = 1000 m ⟹ 1.8 km = 1.8 x 1000 = 1800 m

And

• Total time taken = 3 h

Since

1 h = 3600 sec ⟹ 3 h = 3 x 3600 = 10800 sec

$$\begin{aligned}\text{Average Speed} = \frac{\text{Total distance travelled}}{\text{total time taken}} \\[1em] = \frac{\text{1800}}{\text{10800}} \\[1em] = 0.167 \text{ m s}^{-1}\end{aligned}$$

So, the average speed of the cyclist is 0.167 m s^-1.

Given,

• Initial speed of the car = 30 km h^-1
• Initial time = 30 minutes = 0.5 hour
• Final speed of the car = 40 km h^-1
• Final time = 1 hour

(a) Case 1 : When the speed of the car is 30 km h^-1

$$\begin{aligned}\therefore \text{Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] \Rightarrow \text{Distance} = \text{Speed} \times \text{Time} \\[1em] = 30 \times 0.5 \\[1em] = 15 \text{ km}\end{aligned}$$

Similarly,

Case 2 : When the speed of the car is 40 km h^-1

$$\begin{aligned}\therefore \text{Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] \Rightarrow \text{Distance} = \text{Speed} \times \text{Time} \\[1em] = 40 \times 1 \\[1em] = 40 \text{ km}\end{aligned}$$

Now

Total distance travelled = 15 km + 40 km = 55 km

Hence, the total distance travelled by the car is 55 km.

(b) Total time taken = Initial time + Final time

= 1 h + 0.5 h

= 1.5 h

*So, the total time of travel is 1.5 hours.

(c) Average speed of an object is given by,

$$\begin{aligned}\text{Average Speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}} \\[1em] = \frac{\text{55}}{\text{1.5}} \\[1em] = 36.67 \text{ km h}^{-1}\end{aligned}$$

So, the average speed of the car is 36.67 km h^-1.

Given,

• Mass of the body on earth = 1.0 kg
• Weight of the body on earth = 10 N
• Mass of the boy = 37 kg

(a) Weight in kgf is equal to mass of the boy in kg,

So, the weight of the boy in kgf is 37 kgf.

(b) Since

1 kgf = 10 N ⟹ 37 kgf = 37 x 10 = 370 N

So, the weight of the boy in newton is 370 N.

Given,

• Mass of the body on moon = 6.0 kg
• Weight of the body on moon = 10 N
• Mass of the boy on earth = 30 kg

(a) Since mass of a body always remains constant then the mass of the boy still remains 30 kg on moon surface as mass of a body does not change with place.

So, the mass of the boy on moon surface is 30 kg.

(b) Weight of body of mass 30 kg on moon surface = $\frac{\text{10}}{\text{6.0}}\times 30$ = 50 N

So, the weight of the boy of 30 kg on moon surface is 50 N.

The solved crossword puzzle is given below:

Given,

• Mass of the body on earth = 100 kg
• Gravity on moon = 1/6th times the gravity of earth

Since the mass of a body does not change with location, so its mass on the Moon will remain 100 kg.

However, weight depends on gravitational pull.

So, on Earth, the weight of the body is given by,

Weight = m x g

• g = 9.8 N kg^-1 on earth

Then

Weight = 100 × 9.8 = 980 N

Since the Moon’s gravity is 1/6th of Earth’s, the weight on the Moon is given by,

$$\begin{aligned}\text {Weight on moon} = \frac{1}{6} \times \text {Weight on earth} \\[1em] = \frac{1}{6} \times 980 \\[1em] = 163.33\ \text N\\[1em] \approx 163\ \text N\end{aligned}$$

Thus, the mass remains the same on both Earth and Moon, but the weight decreases on the Moon due to weaker gravity.

Hence, weight of the body on earth is 980 N and on moon it is approximately 163 N.

(i) The student walking along the 200 m track shows uniform motion because he covers equal distances (every 20 m) in equal intervals of time. Therefore, his speed remains constant throughout.

(ii) Given,

• Distance covered by the student = 100 m
• Time taken = 20 s

Speed of an object is given by,

$$\begin{aligned}\text {Speed} = \frac {\text {Distance covered}}{\text {Time taken}} \\[1em] = \frac{100}{20} \\[1em] = 5 \text { m s}^{-1}\end{aligned}$$

So, the speed of the student is 5 m s^-1.

(iii) The student in the race shows non-uniform motion because his speed changes—he starts fast, slows down and then speeds up again.

Hence, he does not cover equal distances in equal intervals of time.

(iv) Average speed is the total distance travelled divided by the total time taken for the entire journey. Speed at a particular moment (instantaneous speed) is the speed of the object at a specific instant of time.