Kerala SSLC Class 9 Solutions for Physical Sciences – Motion
(English Medium) Part1
Page No.46:
Question 1:
A car travels with a uniform speed. Figure shows the positions reached by the car at different time intervals. Observe the figure and complete table.
Time  0s  2s  —–  —–  —–  —– 
Position  0m  10m  —–  —–  —–  —– 
Let’s see how to mark on the graph the data given in the table. You can mark time on the Xaxis and position on the Yaxis.
 What is the highest value given in the tables?
 How can this be marked on the graph paper?
 Why can’t we take a scale, 1cm → 10 m.
 If so what scale can we consider for marking the time?
Solution:
Time  0s  2s  4s  6s  8s  10s 
Position  0m  10m  20m  30m  40m  50m 
 The highest value in the given table is 50 m.
 We can mark it on the graph paper by using a suitable scale along the X and Yaxis. If we take a scale 1 cm → 10 m then the portion of the graph covered will be very less.
 We can consider that 1 cm = 2 s on the graph.
Page No.47:
Question 1:
Mark the scale on the right hand top corner of the graph.
Scale
X axis : 1 cm → ______
Y axis : 1 cm → 10 m
Observe the table:
Time  0s  2s  4s  6s  8s  10s 
Position  0m  10m  20m  30m  40m  50m 
When the time is zero, where is the position of the car?
 What is the shape of the graph that you get?
 Using this graph how can you find out the distance travelled by the car at a given time?
Solution:
Scale: X axis : 1 cm = 2 sY axis : 1 cm = 10 m
When the time is 0 s the position of the car is also at 0 m.
 We get a straight line graph.
 We can do this by drawing a line perpendicular to the Xaxis from a point at which the distance has to be measured. From the point where it touches the straightline graph, draw a line perpendicular to the Yaxis. The corresponding point on the Yaxis gives the distance travelled by the car.
Example: To find the distance travelled by the car in 9 s. Draw a line perpendicular from the point 9 s on the Xaxis such that it meets the straightline graph.
 From this point draw a line perpendicular to the Yaxis.
 The measurement corresponding to this on the Yaxis, i.e. 45 m is the distance travelled by the car. Thus, the car travels a distance of 45 m in 9 s.
Page No.48:
Question 1:
In the table below, the time taken by a car to reach different positions is given.
Using this data draw a positiontime graph
Time (s)  Position (m) 
0  0 
2  1 
4  4 
6  9 
8  16 
10  25 
Solution:
Question 2:
Given below is the graph showing how the brothers Hari and Prasad started from their home and reached the school 2.5 km away travelling along the same path.
 Who started first?
 Who reached the school first?
 How many minutes after Prasad started from home did he meet Hari?
 How much time, in minutes, did each of them take to reach the school?
Solution:
 Prasad started first.
 Hari reached the school first.
 Prasad met Hari after 15 minutes he started from his home.
 Prasad – 25 minutes
Hari – 21.5 minutes
Question 3:
Draw a graph showing time and velocity with the data given in Table just as you have drawn the position – time graph.
Time (s)  0  5  10  15  20  25  30  35  40 
Velocity (m/s)  0

10  20  20  20  15  10  5  0 
Solution:
Page No.49:
Question 1:
What will be a suitable scale to mark time on the X axis and velocity on the Y axis?
Scale
X axis: 1cm →………..
Y – axis: 1cm → …….
Find the following details from the graph.
 What will be the velocity of the car at the 8^{th} second?
 When did the vehicle have a velocity of 12m/s?
Solution:
 Scale: Xaxis: 1 cm = 5 sYaxis: 1 cm = 5 m/s
 The velocity of the car at 8 s will be 16 m/s.
 The vehicle had velocity of 12 m/s at 6s and 28 s.
Question 2:
Look at the graph of a vehicle travelling with a velocity of 10 m/s.
Consider the graph from the 10^{th} to the 20^{th} second. Here the line AB shows the initial velocity (u) of the vehicle.
 If so which is the line showing the period of time (t)?
 In the square ABCD, AB and BC are two adjacent sides. How can you find the area of this square according to the scale in the graph?
 What is the relationship between the area of this square and the displacement of the vehicle?
Solution:
 The line BC shows the time period of the vehicle.
 Area of the square ABCD = AB × BC
= 10× 10 = 100 m
We know that Displacement = Velocity × Time
From the graph,
Length of line AB = Velocity of the vehicle = 10 m
Length of line BC = Time taken by the vehicle = 10 s
Displacement = 10×10 = 100 m
Thus, the area of the square is equal to the displacement of the vehicle.
Question 3:
Given below is the velocity of a stone at each second as it falls from a height.
 When the object is at A what is its acceleration?
 When the object is at C?
 What is the peculiarity of the acceleration of this object?
Solution:
 The stone is dropped from point O. We have to find the acceleration a at point A. So, from the figure we get, Initial velocity, u (at point O) = 0 m/s Final velocity v (at point A) = 9.8 m/s Time, t = 1 s
 To find the acceleration at point C, consider points O and C. So, in this case, Initial velocity, u (at point O) = 0 m/s Final velocity, v (at point C) = 29.4 m/s Time, t = 3 s
 When the object is dropped, it moves in the downward direction due to the gravitational force of the Earth. A uniform acceleration is produced by the gravitational pull. Thus, the acceleration at each point is 9.8 m/s^{2}.
Page No.51:
Question 1:
A train starting from rest attained a velocity of 30 m/s in 30 minutes. What is its acceleration? What is the distance travelled during this time?
Solution:
Initial velocity, u=0 m/s
Final velocity, v=30 m/s
Time, t=30 minutes=30×60 s=1800 s
v=u+at
Acceleration of the train = 0.016 m/s^{2
}The distance travelled by the train within this time
Question 2:
A ball is dropped from a height of 5 m and another from a height of 20 m. What is the velocity with which each hits the ground? (acceleration due to gravity, g = 10 m/s^{2})
Which one has greater velocity when it touches the ground? Does the velocity with which an object hits the ground change with its height from the ground?
Solution:
Let the height of the ball dropped from height 5 m be ‘s_{1}‘ and that dropped from a height of 20 m be ‘s_{2}‘.
Let ‘u_{1}‘ and ‘v_{1}‘ be the initial and final velocity of one ball. Similarly, let the initial and final velocity of the other ball be ‘u_{2}‘ and ‘v_{2}‘ respectively.
We know that
The ball dropped from a height of 20 m has greater velocity.
We can see that the final velocity of a freely falling body changes according to the height from which it is dropped. .
The ball dropped from a height of 20 m has greater velocity.
We can see that the final velocity of a freely falling body changes according to the height from which it is dropped. .
Page No.52:
Question 1:
A plastic ball and an iron ball are dropped from different positions in a tray filled with wet sand. Observe the depth of the pits formed in each case.
 Which are the occasions on which the depth of the pits increased?
a. When the mass increased / decreased.
b. When the velocity increased / decreased.  What were the factors, the increase of which enhanced the depth of the pit or the impact made by the ball?
Solution:
 The depth of the pit or the impact caused by the ball increased due to increase in the height and the mass of the ball. when the mass of the ball increased, the depth of the pit increased. Also, the depth of the pit increased with an increase in the height of the ball because of an increase in its velocity.
 The depth of the pit or the impact caused by the ball increased due to increase in the height and the mass of the ball.
Question 2:
Let a bullet of mass 0.005 kg leave a pistol with a velocity of 150 m/s. Another bullet of the same mass is thrown with a velocity of 10 m/s.
 What is the momentum of the bullet as it leaves the pistol?
 What is the momentum of the bullet when it is thrown?
Solution:
 Momentum of the bullet when it leaves the piston:
Mass, m = 0.005 kg
Velocity, v = 150 m/s
Momentum, P =?  Momentum of the bullet when it is thrown is:
Mass, m = 0.005 kg
Velocity, v = 10 m/s
Momentum, P =?
Thus we can conclude that if an object has greater momentum, the impact produced by it is also more.
Question 3:
What is the advantage in drawing your hand back slightly as you take a catch in cricket?
Imagine that a cricket ball of mass 0.6 kg comes towards you with a velocity of 20m/s.
 What is the momentum of the cricket ball?Imagine that a player brings the ball to rest drawing back his hand for one second after the ball comes into contact with his hand.
 Now the velocity has become zero. What is its momentum?
 What is the difference in the momentum of the ball just before the hand touches the ball and the ball has come to rest in the hand?
 So much change of momentum has occurred in one second. If so what is the rate of change of momentum?
 If the ball comes to rest three seconds after it has touched the hand, what will be the rate of change of momentum?
 What happens to the rate of change of momentum if the time interval for the change of momentum is increased? What is the effect of the impact when the rate of change of momentum is decreased?
Solution:
 The momentum of the ball p
 When the velocity becomes zero, the momentum also becomes zero.
 The difference in the momentum of the ball just before the hand touches the ball and the ball has come to rest in the hand is given as: Change in momentum = 12 – 0 = 12 kg m/s
 Rate of change of momentum
 Rate of change of momentum
 When the time interval for the change of momentum is increased, the rate of change of momentum is decreased. As a result of this decrease in rate change of momentum the force exerted on the hand is decreased and there by the impact is also decreased.
Page No.53:
Question 1:
Why is a cushion bed/sand bed arranged for the athletes of high jump / pole vault to land on?
Solution:
A cushion bed/sand bed arranged for the athletes of high jump / pole vault to land to prevent them from injury. Cushion bed / sand bed allow to slowly sink into them.
According to newton’s second law of motion,
Force = mass × acceleration
When an athlete jumps with a larger force, it gives greater acceleration. As they fall on a cushion bed, the rate of change of momentum decreases and hence body reduces the impact.
Question 2:
Haven’t you studied Newton’s first law of motion? Try writing
it.
Suppose a body of mass m moves with a velocity change to v in t seconds. Then
 What is the initial momentum of the body?
 The final momentum?
 What is the change in momentum of the body?
 What is the rate of change of momentum?
 What is to be done to increase the rate of change of momentum of a moving body?
Solution:
Newton’s first law states that a body remains at rest or in motion in a straight line with uniform speed, unless it is acted upon by an external force. Thus, all objects resist the change in their state of motion. This resistance is known as the inertia of the body.
 Initial momentum of the body, P_{1} = mu
 Final momentum of the body, P_{2} = mv
 Change in the momentum of the body is given as P_{2 }– P_{1} = mv – mu = m(v – u)
 The rate of change of momentum of the body is
 A large force is to be applied on the moving body in order to increase its rate of change of momentum.
Page No.54:
Question 1:
Examine the following two situations:
i. An auto rickshaw at rest is pushed for t seconds. Then its velocity increased to 2 m/s.
ii. When the same auto rickshaw was pushed by two persons for the same time its velocity increased to 3 m/s.
a. Of the above situations which is the case of greater change in momentum?
b. Which is the case in which rate of change of momentum is greater?
Solution:
a. In the second case, the change in momentum is greater.
b. In the second case, the rate of change of momentum is greater.
Question 2:
A constant force is applied for 2 s on a body of mass of 5 kg. As a result the velocity of the body increases from 3 m/s to 7 m/s. What is the magnitude of the force applied?
Solution:
Given that
Mass of the body, m = 5 kg
Time, t = 2 s
Initial velocity, u = 3 m/s
Final velocity, v = 7 m/s
Thus, the magnitude of the force applied is 10 N.
Question 3:
A ball of mass 6 kg rolls on a floor. Its velocitytime graph is given in figure below:
What can be the force exerted by the floor on the ball to bring it to rest?
Solution:
From the graph we get
Initial velocity, u = 4 m/s
Final velocity, v =0 m/s
Time taken, t = 8 s
Thus, the force exerted by the floor on the ball to bring it to rest is 3 N. The negative sign indicates that the direction of the force exerted by the floor is in the opposite direction to the motion of the ball.
Page No.55:
Question 1:
Don’t you launch little rockets during festivals like Vishu, Deepavali, etc.
a. When ignited what happens to the gun powder inside these little rockets?
b. Are the directions of the effluent gases produced on firing the rocket and direction of motion of the rocket the same? If not, how are they?
Solution:
a. When the rocket is ignited, the gun powder inside the rockets burns and releases gas in a great force because of which the rocket rises up.
b. No, the directions of the effluent gases produced on firing the rocket and direction of motion of the rocket are not in the same direction.
They are in the opposite direction.
Question 2:
A person jumps to the shore from a boat floating on water.
a. When the person jumped to the shore why did the boat move back?
b. What is the action involved when the person jumps from the boat? Which is the object on which the action is exerted?
c. What about reaction?
d. Do action and reaction act on the same object?
e. Is the action in the same direction as reaction?
Solution:
a. According to Newton’s third law of motion, for every action there is equal and opposite reaction. When the person jumps forward, he applies a backward force on boat. Because of this force of action, the boat moves in the backward direction.
b. When the person jumps out of the boat he exerts force on the boat in the backward direction. The boat is the object on which action is exerted.
c. The reaction is the opposite force exerted by the boat on the person. It is because of this force that the man jumps out of the boat and reaches the ground.
d. No, action and reaction force does not act on the same object.
e. No, the action and reaction are not in the same direction.
Page No.56:
Question 1:
Write down the action and reaction in the following situations:
 When shot is fired from a pistol, shot moves forward and the pistol backwards.
 The swimmer moves forward.
Solution:
 When a bullet is fired from a gun, the force exerted by the gun on the bullet because of which the bullet moves in the forward direction, is the action. An equal force, because of which the gun moves in the backward direction, is the reaction.
 The swimmer pushes the water backward using his hands and feet to move in the forward direction. The water exerts an equal force which pushes the swimmer in the forward direction. The force exerted by the swimmer is the action and the force exerted by the water is the reaction.
Question 2:
Why do people walking on smooth wet surface slip down?
Solution:
People walking on a smooth wet surface slip down because the friction force on the surface is very less. Because of which we cannot exert a backward action force on it that would produce a forward reaction force on people.
Question 3:
Imagine that the mass of each ball is 15g. Let 5 m/s be the velocity with which the first ball hits others. Let the velocity of the pushed off ball also be 5m/s. Calculate the total momentum of the ball before and after collision Just before collision:
a. What will be the momentum of the first ball?
b. What will be the total momentum of the remaining 4 balls at rest?
c. What will be the total momentum?
Solution:
a. The momentum of the first ball
p = mv = 0.015 kg × 5 m/s
= 0.075 kg m/s
b. The total momentum of the remaining 4 balls at rest is: p = mv = 0.015 kg × 0 = 0
c. The total momentum is 0.075 kg m/s.
Question 4:
Imagine that the mass of each ball is 15g. Let 5 m/s be the velocity with which the first ball hits others. Let the velocity of the pushed off ball also be 5m/s. Calculate the total momentum of the ball before and after collision
After the collision:
a. What will be the momentum of the pushed off ball?
b. What will be the momentum of the remaining 4 undeflected balls?
c. What will be the total momentum?
Solution:
a. The momentum of the pushed off ball is p = mv = 0.015 kg × 5 m/s
= 0.075 kg m/s
b. The momentum of the remaining 4 undeflected balls is: p = mv = 0.015 kg × 0 = 0
c. Total momentum is 0.075 kg m/s.