Goa Board Class 9 Solutions for Physics – Force And Laws Of Motion (English Medium)
Page No. 118:
Question 1:
Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five-rupee coin and a one-rupee coin?
Solution:
Mass is a measure of the inertia of a body. The greater the mass of a body; the greater is its inertia.
(a) Mass of a stone is more than the mass of a rubber ball of same size. Hence, inertia of a stone is greater than that of a rubber ball of same size.
(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of a train is greater than that of a bicycle.
(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of a five rupee coin is greater than that of a one-rupee coin.
Concept Insight: More mass means more inertia.
Question 2:
In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.
Solution:
The velocity of the ball changes four times.
(i) As a football player kicks the football, its speed changes from zero to a certain value. As a result, the velocity of the ball gets changed. In this case, the force is applied by the kick of the player.
(ii) When the ball reaches another player, he kicks the ball towards the goal post. As a result, the direction of the ball and its speed both get changed. Therefore, its velocity also changes. In this case, the force is applied by the kick of the player.
(iii) When the goalkeeper collects the ball, the ball comes to rest, i.e. its speed reduces to zero from a certain value. Thus, the velocity of the ball changes. In this case, the force is applied by the hands of the goalkeeper.
(iv) The goalkeeper then kicks the stationary ball towards his team player, i.e., the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again. In this case, the force is applied by the kick of the goalkeeper.
Concept Insight:– Velocity is a vector quantity. It has both magnitude and direction.
Question 3:
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Solution:
Some leaves of a tree may get detached when we shake its branch vigorously. This is because when the branch of the tree is shaken, it moves to and fro, but due to inertia its leaves tend to remain at rest. Due to this reason, the leaves fall down from the tree.
Concept Insight:- Inertia resists change in state of motion.
Question 4:
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Solution:
When a moving bus brakes to a stop, we fall in the forward direction because though the lower part of our body comes to a stop when the bus stops but the upper part of the body continues to be in motion in the forward direction due to its inertia, thus making us fall in the forward direction.
When a bus accelerates from rest, we fall backwards because though the lower part of our body starts moving with the bus but the upper part of the body tries to remain at rest due to its inertia, thus making us fall in the backward direction.
Concept Insight:- Inertia resists any change in the state of motion of a body.
Page No. 126:
Question 1:
If action is always equal to the reaction, explain how a horse can pull a cart.
Solution:
A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the ground on the horse in the forward direction. As a result, the horse moves forward along with the cart.
Concept Insight:- Action and reaction forces act on two different bodies and that too in opposite directions.
Question 2:
Explain why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Solution:
Due to the backward reaction of the water ejecting from the hose pipe.
When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the hose pipe tends to go backward and slips from the hands of fireman which makes it difficult for the fireman to hold the hose pipe.
Question 3:
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s-1. Calculate the initial recoil velocity of the rifle.
Solution:
Mass of the bullet, m2 = 50 g = 0.05 kg
Recoil velocity of the rifle = v1
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum (before firing) of the rifle and bullet system = (m1 + m2)v =0
Total final momentum (after firing) of the rifle and bullet system
= 4v1 + 1.75
According to the law of conservation of momentum:
Concept Insight:- Total momentum after the firing = Total momentum before the firing
4v1 + 1.75 = 0
4v1 = − 1.75
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
Page No. 127:
Question 4:
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s-1 and 1 m s-1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s-1. Determine the velocity of the second object.
Solution:
Mass of first object, m1 = 100 g = 0.1 kg
Mass of second object, m2 = 200 g = 0.2 kg
Velocity of first object before collision, v1 = 2 m/s
Velocity of second object before collision, v2 = 1 m/s
Velocity of first object after collision, v3 = 1.67 m/s
Velocity of second object after collision = v4
According to the law of conservation of momentum:
Concept Insight: Total momentum before collision = Total momentum after collision
Hence, the velocity of the second object becomes 1.165 m/s after the collision.
Page No. 128:
Question 1:
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Solution:
Yes. Even when an object experiences a net zero external unbalanced force, it is possible that the object is travelling with a non-zero velocity. This is possible only when the object has been moving with a constant velocity in a particular direction. Then, there is no net unbalanced force applied on the body. The object will keep moving with the same non-zero velocity.
Concept Insight: To change the state of motion, a net non-zero external unbalanced force must be applied on the object.
Question 2:
When a carpet is beaten with a stick, dust comes out of it. Explain
Solution:
Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to retain their state of rest. Hence, the dust particles come out of the carpet.
Concept Insight: Inertia resists change in state of motion.
Question 3:
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Solution:
When the bus suddenly accelerates from rest and moves forward, it acquires a state of motion. However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of rest and hence may fall down from the roof of the bus.
Similarly, when the moving bus stops suddenly, then due to its inertia of motion, the luggage kept on the roof of the bus tends to remain in motion and hence may fall down from the roof of the bus.
Hence, it is advised to tie the luggage kept on the roof of a bus with a rope so that it does not fall down when the bus starts or stops suddenly.
Concept Insight: Inertia resists change in state of motion
Question 4:
A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) The batsman did not hit the ball hard enough.
(b) Velocity is proportional to the force exerted on the ball.
(c) There is a force on the ball opposing the motion.
(d) There is no unbalanced force on the ball, so the ball would want to come to rest.
Solution:
(c) There is a force on the ball opposing the motion.
A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion.
Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is responsible for stopping the cricket ball.
Question 5:
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if it’s mass is 7 tonnes (Hint: 1 tonne = 1000 kg).
Solution:
Initial velocity, u = 0 (since the truck is initially at rest)
Distance travelled, s = 400 m
Time taken, t = 20 s
Acceleration, a=?
According to the second equation of motion:
Question 6:
A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 (finally the stone comes to rest)
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v2 = u2 + 2as
where, a = acceleration
Concept Insight: The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force, F = Mass x Acceleration
F = ma
F = 1 × (- 4) = -4 N
Hence, the force of friction between the stone and the ice is -4 N.
Question 7:
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Solution:
(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Hence, net accelerating force, Fa = F – Ff = 40000 – 5000 = 35000 N
(b) Let acceleration of the train be a.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, M = Mass of a wagon × Number of wagons = 2000 x 5 = 10000 kg
From Newton’s second law of motion:
Concept Insight:- Force = Mass x Acceleration
Fa = Ma
Hence, the acceleration of the train is 3.5 m/s2.
(c) Mass of the four wagons behind the first wagon = 4 × 2000 = 8000 kg
Acceleration of the wagons = 3.5 m/s2
Thus, force of wagon 1 on remaining four wagons behind it = 8000 x 3.5 = 28000 N
Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.
Question 8:
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s-2?
Solution:
Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 (finally the automobile stops)
Acceleration of the automobile, a = -1.7 ms-2
From Newton’s second law of motion:
Force = Mass × Acceleration = 1500 × (-1.7) = -2550 N
Hence, the force between the automobile and the road is -2550 N, in the direction opposite to the motion of the automobile.
Question 9:
What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) mv2 (d) mv
Solution:
(d) mv
Mass of the object = m
Velocity = v
Concept Insight:- Momentum = Mass × Velocity
Momentum = mv
Question 10:
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution:
Force applied, P = 200 N
Force of friction, F = ?
As the wooden cabinet is to move across the floor with a constant velocity, no force (f) is spent in accelerating the cabinet, i.e.,
f = P-F = 0
or, F = P = 200 N
Concept Insight:- For a non-accelerated motion, no net force is required.
Question 11:
Two objects, each of mass 1.5 kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Solution:
Mass of first object, m1 = 1.5 kg
Mass of second object, m2 = 1.5 kg
Velocity first object before collision, v1 = 2.5 m/s
Velocity of second object before collision, v2 = -2.5 m/s
(Negative sign arises because mass m2 is moving in an opposite direction)
After collision, the two objects stick together.
Total mass of the combined object = m1 + m2
Velocity of the combined object = v
According to the law of conservation of momentum:
Concept Insight:- Total momentum before collision = Total momentum after collision
m1 v1 + m2 v2 = (m1 + m2) v
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5) v
3.75 – 3.75 = 3 v
v = 0
Hence, the velocity of the combined object after collision is 0 m/s.
Page No. 129:
Question 12:
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Solution:
When we push a massive truck parked along the roadside, it does not move. The justification given by the student that the two opposite and equal forces cancel each other is totally wrong. This is because force of action and reaction never act on one body. There is no question of their cancellation. The truck does not move because the push applied is far less than the force of friction between the truck and the road.
Concept Insight:- Action and reaction forces act on different objects.
Question 13:
A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution:
Mass of the hockey ball, m = 200 g = 0.2 kg
Hockey ball travels with velocity, v1 = 10 m/s
Initial momentum = mv1
After being struck by the stick, the hockey ball travels in the opposite direction with velocity, v2 = -5 m/s
Final momentum = mv2
Concept Insight:- Change in momentum = Final momentum – Initial momentum
Change in momentum = mv2 – mv1 = m (v2 – v1) = 0.2 [-5-10] = 0.2 (-15) = -3 kg ms-1
Hence, the change in momentum of the hockey ball is -3 kg ms-1.
Question 14:
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution:
Mass of the bullet, m = 10 g = 0.01 kg
It is given that the bullet is travelling with a velocity of 150 m/s.
Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t = 0.03 s
According to the first equation of motion,
v = u + at
where , a is the acceleration of the bullet
0 = 150 + (a × 0.03)
According to the third equation of motion:
v2 = u2 + 2 as
0 = (150)2 + 2×(-5000)×s
0 = 22500 + 2×(-5000)×s
0=22500 – 10000 s
10000 s = 22500
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Concept Insight: Force, F = Mass Acceleration
Mass of the bullet, m = 0.01 kg
Acceleration of the bullet, a = -5000 m/s2
F = ma = 0.01×(-5000) = -50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.
Question 15:
An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Solution:
Mass of the object, m1 = 1 kg
Velocity of the object before collision, v1 = 10 m/s
Mass of the stationary wooden block, m2 = 5 kg
Velocity of the wooden block before collision, v2 = 0 m/s
Total momentum before collision = m1v1+m2v2 = (1×10)+(5×0) = 10 kg m s-1
According to the law of conservation of momentum, the total momentum just after the impact will be the same as the total momentum just before the impact.
i.e., the total momentum just after the impact will be 10 kg m s-1.
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m1 + m2
Velocity of the combined system = v
According to the law of conservation of momentum:
Concept Insight:- Total momentum before collision = Total momentum after collision
m1v1+m2v2 = (m1+m2)v
(1×10)+(5×0) = (1+5)v
10+0 = 6v
10 = 6v
Hence, velocity of the combined object after collision will be 1.67 m/s.
Question 16:
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Solution:
Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time taken by the object to accelerate, t = 6 s
Initial momentum of the object = mu = 100 � 5 = 500 kg ms-1
Final momentum of the object = mv = 100 � 8 = 800 kg ms-1
Question 17:
Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Solution:
The suggestion made by Kiran that the insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong. The suggestion made by Akhtar that the motor car exerted a larger force on the insect because of large velocity of motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insect with motor car, both experience the same force as action and reaction are always equal and opposite. Further, changes in their momenta are also the same. Only the signs of changes in momenta are opposite, i.e., change in momenta of the two occur in opposite directions, though magnitude of change in momentum of each is the same.
Question 18:
How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms-2 .
Solution:
Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
Concept Insight Choose the equation of motion wisely out of the three, to minimize the number of steps in calculations.
According to the third equation of motion:
v2 = u2 + 2as
v2 = 0 + 2 (10) 0.8
v = 4 m/s
Hence, the momentum with which the dumbbell hits the floor = mv = 10×4 = 40 kg m s-1
Momentum transferred to the floor is 40 kg m/s.
Page No. 130:
Question A1:
The following is the distance time table of an object in motion.
Time in seconds | Distance in metres |
0 | 0 |
1 | 1 |
2 | 8 |
3 | 27 |
4 | 64 |
5 | 125 |
6 | 216 |
7 | 343 |
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing or zero?
(b) What do you infer about the forces acting on the object?
Solution:
(a) A careful observation of the distance-time table shows that
s ∝ t3
It is known that
(i) for motion with uniform velocity (zero acceleration)
s=ut
i.e; s ∝ t
(ii) for motion with uniform acceleration
s=ut + (½ ) at3
In the present case, s ∝ t3 . Therefore, we conclude in this case that acceleration must be increasing uniformly with time.
(b) As F = ma, therefore, F ∝ a. Hence, the force must also be increasing uniformly with time.
Question A2:
Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m/s2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Solution:
Here, mass of motorcar, m = 1200 kg
Let each person exert a push F on the motorcar.
Total push of two persons = F + F = 2F
As this push gives a uniform velocity to the motorcar along a level road, it must be a measure of the force of friction (f) between the motorcar and the road,
i.e., f = 2F.
When three person push, total force applied = F + F + F = 3F
Force that produces acceleration (a=0.2 m/s2),
i.e., ma = 3F-f = 3F-2F = F
or, F = ma = 1200 × 0.2 = 240 N
Question A3:
A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Solution:
Thus, the force of the nail on the hammer is 2500 N. Negative sign indicates the opposing force.
Question A4:
A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution:
Change in momentum = mv-mu = m(v-u) = 1200×(5-25)= -24000 kg m/s
Force required , F = ma = 1200×(-5) = -6000 N
Magnitude of force required = 6000 N
Negative sign shows that force is opposing the motion.