Welcome to Chapter 8 on Force and Laws of Motion – a crucial topic in physics that lays the foundation for future learning. We have answered all in-text questions, exercise questions, additional exercises and in-text activities all in one place for your convenience. Our solutions are designed to make learning enjoyable and effective, ensuring you’re fully prepared not only for exams but also for the exciting journey ahead in higher classes. So, let’s dive in! Also don’t forget to practise those extra questions for that extra edge!
Solutions to In Text Questions of NCERT Class 9 Science Chapter 8 Force and Laws of Motion –
Page 91:
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 rupees coin and a one-rupee coin?
Answer:
Inertia of an object is measured by its mass. Heavier objects will have more inertia.
(a) The stone is heavier than the rubber ball, hence the stone has more inertia.
(b) The train is heavier than the bicycle, hence the train has more inertia.
(c) A five rupees coin is heavier a one-rupee coin, hence a five rupees coin has more inertia.
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.
Answer:
The velocity of the ball changes four times as follows:
(i) When the first player kicks the football to the second player. The agent is the first player.
(ii) When the second player kicks the football towards the goal. The agent is the second player.
(iii) When the goalkeeper collects the football. The agent is the goalkeeper.
(iv) When the goalkeeper kicks it towards a player of his own team. The agent is the goalkeeper.
3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
Shaking a tree branch causes it to move back and forth. However, the inertia of the leaves attached to the branch resists this motion. As a result, weakly attached leaves fall off due to inertia, while firmly attached leaves remain in place.
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?
Answer:
When a moving bus brakes to a stop, our body tends to continue in the same state of motion because of its inertia. Hence, we fall in the forward direction.
When the bus accelerates from rest, the lower portion of our body in contact with the bus moves along with the bus. But the rest of our body opposes this motion because of its inertia. Hence, we fall backwards.
Solutions to Exercises (Page No 97) of NCERT Class 9 Science Chapter 8 Force and Laws of Motion –
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.
Answer:
Yes, it is possible for the object to be travelling with a non-zero velocity. The object must move with constant velocity in the same direction. Such an object will continue to move with the same velocity in the same direction, even though the object experiences a net zero external unbalanced force.
2. When a carpet is beaten with a stick, dust comes out of it, Explain.
Answer:
When acarpet is beaten with a stick, dust comes out of it because of inertia of dust particles. When the carpet is beaten and set in motion, the dust particles tend to remain in a state of rest due to inertia. Hence, they get separated from the carpet and come out of it.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
When the bus accelerates from rest, the luggage tends to continue in the same state of rest because of inertia. This makes the luggage fall off backwards. On the other hand, when the bus is in motion and comes to a sudden halt, the luggage tends to continue in the state of motion because of inertia. This makes the luggage fall forwards. Hence, the luggage should be tied to the roof of the bus with a rope.
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.
Answer: (c) there is a force on the ball opposing the motion.
When the batsman hits a cricket ball, it comes to rest because of force of friction which opposes the motion of the ball.
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 its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Answer:
Initial velocity of truck (u) = 0.
Distance travelled (s) = 400 m.
Time taken (t) = 20 s.
Let the acceleration of the truck be a.
s = ut + (1/2)at2
or, 400 = 0 + (1/2)a(20)2
or, a = 2 ms-2
Mass (m) = 7 tonnes = 7 × 1000 kg = 7000 kg.
Using Newton’s second law of motion,
Force (F) = ma = 7000 × 2 ms-2 = 14000 N.
Summary:
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. The acceleration is 2 ms-2 and the force acting on it if its mass is 7 tonnes is 14000 N.
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?
Answer:
Initial velocity (u) = 20m/s
Since the stone comes to rest, final velocity (v) = 0 m/s
Distance travelled by the stone (s) = 50 m
Let acceleration be a.
We know,
v² = u² + 2as
Substituting the values in the above equation we get,
0² = (20)² + 2(a)(50)
-400 = 100a
a = -400/100 = -4m/s²
The negative value indicates that the acceleration is decreasing.
Mass of the stone (m) = 1kg
We know that.
Force (F) = ma (Newton’s second law of motion)
or, F = 1 × (-4) = – 4 N
The negative sign indicates that the force of friction is acting in a direction opposite to the motion of the stone.
Summary:
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. The force of friction between the stone and the ice is – 4 N.
7. A 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 and
(b) the acceleration of the train.
Answer:
(a) Force exerted by the engine = 40000 N.
Friction force exerted by the track = – 5000 N. The negative sign indicates that the force is acting in the opposite direction to motion.
Net force = 40000 N + (-5000 N) = 35000 N.
(b) Total mass (m) of the train = 8000 kg + 5 × 2000 kg = 18000 kg.
Let a be the acceleration.
We know according to Newton’s second law of motion,
Net force = ma
or, 35000 = 18000a
or, a = 35000/18000 = 1.94 m s-2
The acceleration of the train = 1.94 m s-2.
Summary:
A 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,(a) The net accelerating force is 35000 N and (b) the acceleration of the train is 1.94 m s-2.
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?
Answer:
Let mass of the automobile vehicle (m) = 1500 kg.
Acceleration of the vehicle (a) = – 1.7 m s-2.
Let force between vehicle and the road be F.
We know according to Newton’s second law of motion,
F = ma = 1500 (-1.7) = – 2550 N.
Therefore, the force between the vehicle and road is – 2550 N, opposite to the direction of motion.
Summary: An automobile vehicle has a mass of 1500 kg. The force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2 is – 2550 N.
9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv
Answer: (d) mv
Momentum = mass (m) × velocity (v)
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?
Answer:
Since the wooden cabinet moves across the floor at a constant velocity, the acceleration is 0. Therefore, the net force acting on the wooden cabinet is also 0. Hence, the magnitude of the friction force is equal and opposite to the horizontal force applied and is equal to – 200 N.
11. 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.
Answer:
The student’s logic is wrong because the two opposite and equal forces act on different bodies and hence they cannot cancel each other out. The truck does not move because it has a huge mass and hence it has negligible acceleration when pushed.
12. A hockey ball of mass 200 g travelling at 10 m s–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s–1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Mass of the ball (m) = 200 g
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = – 5m/s (The negative sign indicates that it is travelling in a direction opposite to its original path)
Initial momentum of the ball = mu = 200g × 10 ms-1 = 2000 g.m.s-1
Final momentum of the ball = mv = 200g × (–5 ms-1) = –1000 g.m.s-1
Therefore, the change in momentum = (mv – mu) = –1000 g.m.s-1 – 2000 g.m.s-1 = –3000 g.m.s-1
Magnitude in change in momentum = 3000 g.m.s-1.
Summary:
A hockey ball of mass 200 g travelling at 10 m s–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s–1. The magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick is 3000 g.m.s-1.
13. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–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.
Answer:
Mass of the bullet (m) = 10g = 0.01 kg.
Initial velocity of the bullet (u) = 150 ms-1
Final velocity of the bullet (v) = 0 ms-1
Time taken to come to rest (t) = 0.03 s
Let acceleration of the bullet while travelling through the block be a and distance of penetration be s
We know,
v = u + at
0 = 150 + a (0.03)
a = -5000 ms-2
We know,
v2 = u2 + 2as
0 = 1502 + 2 x (-5000)s
s = 2.25 m
According to Newton’s second law of motion,
F = ma
F = 0.01kg × (-5000 ms-2)
F = -50 N (The negative sign indicates that the force is acting opposite to direction of motion of the bullet)
Summary:
A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. The distance of penetration of the bullet into the block is 2.25 m and the magnitude of the force exerted by the wooden block on the bullet is 50 N.
14. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–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.
Answer:
Mass of object (m1) = 1 kg.
Initial velocity of object (u1) = 10 m s–1.
Mass of wooden block (m2) = 5 kg.
Initial velocity of wooden block (u2) = 0 (Since it is stationary).
Mass of the combined object = m1 + m2 = 1 kg + 5 kg = 6 kg.
Total momentum before impact = m1u1 + m2u2 = 1 × 10 + 0 = 10 kg ms-1.
According to law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Therefore, total momentum after impact = 10 kg ms-1.
Let velocity of combined object be v.
Therefore,
m1u1 + m2u2 = (m1 + m2)v
or, 1 × 10 + 0 = 6 × v
or, v = 1.67 m s-1
Summary:
An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–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. The total momentum just before the impact and just after the impact are both 10 kg ms-1. Also, the velocity of the combined object = 1.67 m s-1.
15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Mass of the object (m) = 100 kg
Initial velocity (u) = 5 ms-1
Final velocity (v) = 8 ms-1
Time (t) = 6s
Initial momentum = mu = 100kg × 5ms-1 = 500 kg.m.s-1
Final momentum = mv = 100kg × 8ms-1 = 800 kg.m.s-1
This change of momentum happens in 6 s.
Now,
Force = (mv – mu)/t = (800 – 500)/6 = 300/6 = 50 N.
Summary:
An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s. Initial momentum of the object = 500 kg.m.s-1 and final momentum = 800 kg.m.s-1. The magnitude of the force exerted on the object is 50 N.
16. Akhtar, Kiran and Rahul were riding in a motorcar 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.
Answer:
Kiran’s Statement:
Kiran’s statement is false. According the law of conservation of momentum, the total momentum before collision = total momentum after collision. It follows, that the change in momentum in one body = change in momentum in the second body. Hence, the statement that the insect suffered a greater than in momentum as compared to the change in momentum of the motorcar is false. The correct statement is that the changes in momentum of the insect and the motorcar are equal (but opposite in magnitude).
Akhtar’s Statement:
Akhtar’s statement is false. According to Newton’s third law, the force exerted by the motorcar on the insect = the force exerted by the insect on the motorcar.
Rahul’s Statement:
Rahul’s statement is correct. According to Newton’s third law, both the motorcar and the insect experienced the same force. It also follows from the law of conservation of momentum that the changes in momentum of the insect and the motorcar are equal.
17. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s–2.
Answer:
Mass of the dumb-bell (m) = 10 kg.
Height covered (s) = 80 cm = 0.8 m.
Initial velocity (u) = 0 (Since it is dropped from rest)
Acceleration (a) = 10 ms-2
Let final velocity be v.
We know,
v2 – u2 = 2as
or, v2 – 0 = 2 × 10 × 0.8 = 16 m2s-2
v = 4 ms-1
Momentum of the dumb-bell right when it hits the ground = mv = 10 kg × 4 ms-1 = 40 kg.m.s-1.
Summary: The momentum a dumb-bell of mass 10 kg will transfer to the floor if it falls from a height of 80 cm is 40 kg.m.s-1.
Solutions to Additional Exercises (Page No 99) of NCERT Class 9 Science Chapter 7 Force and Laws of Motion
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?
Answer:
(a) We can see from the table that with every passing time interval of 1 second, the distance covered is more. Therefore, we can say that the acceleration is increasing.
(b) From Newton’s second law of motion, Force = mass × acceleration. The mass of the object is constant and the acceleration is increasing. Therefore, the force is also increasing.
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 s-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Answer:
When two persons push, the car moves at uniform velocity.
When the third person pushes, the car accelerates. The acceleration is 0.2 m s-2.
Force exerted by the third person = mass × acceleration = 1200 kg × 0.2 m s-2 = 240 N.
All persons push the motorcar with the same muscular effort, hence the force exerted by the other two persons each also = 240 N.
Summary:
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 s-2. Each person pushes the motorcar with a force of 240 N.
A3. A hammer of mass 500 g, moving at 50 m s-1, 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?
Answer:
Mass of the hammer (m) = 500 g = 0.5 kg.
Initial velocity of the hammer (u) = 50 m s-1.
Final velocity of the hammer (v) = 0.
Time taken for change in momentum = 0.01 s.
Force exerted by the nail on the hammer = m(v – u)/t = 0.5 × (0 – 50)/0.01 = -2500 N.
Summary:
A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. The force of the nail on the hammer is 2500 N.
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.
Answer:
Mass of the car (m) = 1200 kg
Initial velocity (u) = 90 km/hour = 90000/3600 = 25 ms-1.
Final velocity (v) = 18 km/hour = 18000/3600 = 5 ms-1.
Time taken (t) = 4 seconds
Acceleration (a) = (v – u)/t = (5 – 25)/4 = – 5 ms-2
Change in momentum = mv – mu = 1200 kg × (5 ms-1) – 1200 kg × (25 ms-1) = -24000 kg ms-1.
External force required = ma = 1200 kg × (- 5 ms-2) = -6000 N.
Summary:
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. The acceleration is – 5 ms-2 and change in momentum = -24000 kg ms-1. The magnitude of the force required is -6000 N.
Solutions to All Activities of NCERT Class 9 Science Chapter 8 Force and Laws of Motion –
1. Complete Activity 8.1 (Page 90).
• Make a pile of similar carom coins on a table, as shown in Fig. 8.6.
• Attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker. If the hit is strong enough, the bottom coin moves out quickly. Once the lowest coin is removed, the inertia of the other coins makes them ‘fall’ vertically on the table.
Answer:
2. Complete Activity 8.2 (Page 91).
• Set a five-rupee coin on a stiff card covering an empty glass tumbler standing on a table as shown in Fig. 8.7. • Give the card a sharp horizontal flick with a finger. If we do it fast then the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia.
Answer:
3. Complete Activity 8.3 (Page 91).
• Place a water-filled tumbler on a tray.
• Hold the tray and turn around as fast as you can.
• We observe that the water spills. Why?
Answer:
4. Complete Activity 8.4 (Page 96).
• Request two children to stand on two separate carts as shown in Fig. 8.13.
• Give them a bag full of sand or some other heavy object. Ask them to play a game of catch with the bag.
• Does each of them experience an instantaneous force as a result of throwing the sand bag?
• You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.
Now, place two children on one cart and one on another cart. The second law of motion can be seen, as this arrangement would show different accelerations for the same force.
Answer:
Extra Questions to Complement Solutions to NCERT Class 9 Science Chapter 8 Force and Laws of Motion –
Very Short Answer Type Questions:
1. What is a frictionless surface called?
Answer:
Smooth surface.
2. An object moves in a straight line with acceleration. What kind of force acts on the object?
Answer:
Unbalanced force.
3. An object moves with uniform velocity in a circular path. What kind of force acts on the object?
Answer:
Unbalanced force.
4. A hammer exerts a force of 20 N when struck against a wall. Can you find the force exerted on the hammer?
Answer:
20 N (equal and opposite by Newton’s third law)
5. What is the SI unit of momentum.
Answer:
kg ms-1
6. How do you reduce the friction of a surface?
Answer:
By applying lubricant.
7. A car moving with constant velocity changes direction. Is there any force acting on it?
Answer:
Yes.
8. A brick is being pulled from two opposite sides with forces of 3 N and 2 N. What is the net unbalanced force acting on the body?
Answer:
1 N.
9. When a force is applied to an object, it tends to accelerate in the direction of the applied force. What property of the object resists this change in motion?
Answer:
Inertia.
10. Give the C.G.S unit of force.
Answer:
Dyne.
Multiple Choice Questions (MCQ):
1. When we press a balloon with both hands, it changes its shape. The force responsible for this is an example of
(A) Balanced forces
(B) Frictional forces
(C) Unbalanced forces
(D) Reaction forces
Answer: (A) Balanced forces
Balanced forces change shape and size of an object, but not its inertia.
2. If a body is released from the top of an inclined plane, it rolls down. The force responsible for this is an example of
(A) Balanced force
(B) Frictional force
(C) Unbalanced force
(D) Reaction force
Answer: (C) Unbalanced force
3. The velocity-time graph of a car is a straight line parallel to the time-axis. The net force acting on the particle is
(A) 10 N
(B) 0 N
(C) 5 N
(D) Need more information
Answer: (B) 0 N
The velocity does not change with time, that is the car is moving with constant velocity and has no acceleration. Therefore, the net force acting on the particle is 0 N.
4. A vehicle of mass 1000 kg decelerates at a constant rate of – 3 ms-2. Find the force of friction.
(A) 0 N
(B) – 2000 N
(C) – 3000 N
(D) 3000 N
Answer: (C) – 3000 N
Force of friction = mass × acceleration = 1000 kg × (-3ms-2) = -3000 N. The negative sign indicates that the force of friction is opposite to direction of motion.
5. An object rests on the ground, the forces involved are:
(A) A single force acting upwards
(B) A single force acting downwards
(C) Two equal and opposite forces
(D) No force at all
Answer: (C) Two equal and opposite forces
The answer follows from Newton’ third law.
Short Answer Type Questions:
1. A ball is rolling on a smooth horizontal place with uniform velocity. What will happen to the motion of the ball after five days?
Answer:
The ball will continue to move with the same uniform velocity. This is because the plane is smooth and there is no unbalanced frictional force acting on the body in the opposite direction. Due to conservation of energy principle, the body will maintain its energy by moving in the same direction forever.
2. If we jump from a certain height on a hard surface, we get hurt. However, if we jump on a surface covered with soft cushions, we do not get hurt. Explain why.
Answer:
The cushioned bed increases the time it takes for our feet to stop after landing. This decreases the rate of change of momentum and hence the force acting on our feet. Hence, we do not get hurt. In case of the hard surface, our feet come to a stop in a very short interval of time, which increases the force acting on our feet. Hence, we get hurt.
3. A bullet of mass 20 g fired from a gun of mass 5 kg attains an acceleration of 1000 ms-2. Find the acceleration of the gun.
Answer:
Mass of the bullet = 20 g. = 0.02 kg.
Acceleration of the bullet = 1000 ms-2.
Force exerted on the bullet = mass × acceleration = 0.02 kg × 1000 ms-2 = 20 N.
According to Newton’s third law, the force exerted by the bullet on the gun is equal and opposite to the force exerted by the gun on the bullet.
Therefore, the force exerted on the gun = -20 N.
Mass of the gun = 5 kg.
Therefore,
Mass × acceleration = -20 N
or, 5 kg × acceleration = -20 N
or, acceleration = -20 N/5kg = -4 ms-2
Therefore, acceleration of the gun = -4 ms-2. The negative sign indicates that the acceleration of the gun is in the opposite direction to the bullet.
4. A car of mass 1000 kg is traveling at a constant velocity of 20 m/s on a flat road. What is the net force acting on the car?
Answer:
Since the car travels with constant velocity and there is no acceleration, the net force acting on the car is 0 N.
5. An object of mass 5 kg travels with an acceleration of 10 ms-2. It is being pushed with a force of 100 N. what is the frictional force acting on the object?
Answer:
First, let us find the net force acting on the object using Newton’s second law:
Fnet = ma = 5 kg × 10 ms-2 = 50 N.
Now, frictional force acts opposite in direction to the force of push.
Therefore,
Net force = Force of push – Frictional force
or, 50 N = 100 N – Frictional force
or, Frictional force = 100 N – 50N = 50 N.
6. An inflated balloon when untied, flies away. Explain why.
Answer:
When the knot is undone, the pressurized air inside the balloon rushes out in one direction, creating an action force. This force is exerted by the air molecules as they escape from the balloon. According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Therefore, in response to the action force exerted by the escaping air, the balloon experiences an equal and opposite reaction force. Hence, it flies away.
7. A ping-pong ball of mass 10 g hits a wall at a velocity of 5 ms-1 and rebounds in the opposite direction with a velocity of 3 ms-1. What is the change in momentum of the ball?
Answer:
Initial momentum = mu = 0.01 kg × 5 ms-1 = 0.05 kg ms-1.
Final momentum = mv = 0.01 kg × (-3 ms-1) = – 0.03 kg ms-1.
Change of momentum = Final momentum – Initial momentum = (-0.03 – 0.05) kg ms-1 = -0.08 kg ms-1.
8. A child standing on a cart with wheel throws a luggage to the right of him. In which direction will he move?
Answer:
The child will move in the opposite direction, i.e. to the left. As the child exerts a force on the luggage to the right, the luggage exerts an equal and opposite force on the child towards the left according to Newton’s third law. Hence, the child accelerates towards the left.
9. A small bullet punctures our body while a tennis ball thrown at us does not. Explain.
Answer:
Although the mass of the bullet is smaller than a tennis ball, the bullet travels at a much higher velocity than a tennis ball. Since momentum = mass × velocity, the momentum of the bullet is much higher than a tennis ball. Hence, it can puncture our body.
10. Explain (i) walking and (ii) swimming in terms of Newton’s third of motion.
Answer:
(i) Walking: When you walk, your foot exerts a backward force on the ground, and the ground exerts an equal and opposite forward force on your foot according to Newton’s third law of motion. This reaction force from the ground propels you forward.
(ii) Swimming: In swimming, as you push against the water with your hands and feet, the water exerts an equal and opposite reaction force on you according to Newton’s third law of motion. This propels you forward.
Long Answer Type Questions:
1. An object of mass m moving in a straight line with velocity v strikes another object of mass 3m moving with velocity 2v and sticks to it. The combined object moves with what velocity?
Answer:
To find the velocity of the combined object after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant if no external forces act on it.
Before the collision, the total momentum of the system is the sum of the momenta of the two objects.
Initial momentum = m × v + 3m × 2v = 7mv.
Mass of the combined object = (m + 3m) = 4m.
Let velocity of the combined object is V.
Final momentum = initial momentum.
or, 4m × V = 7mv
or, V = 7mv/4m
or, V = 7v/4
Therefore, the combined object moves with velocity 7v/4.
Fill in the Blanks:
(a) Inertia is the tendency of undisturbed objects to stay at rest or to keep moving with the same __________.
(b) Newton’s __________ law states that force is proportional to acceleration produced in a body.
(c) A cart is being pulled along a horizontal road. The cart moves with uniform velocity. The force applied is equal to the _________ force.
(d) Even though action and reaction forces are always equal in magnitude, these forces may not produce accelerations of equal magnitudes because each force acts on a different object that may have a different __________.
(e) To every action there is an equal and opposite reaction and they act on __________ bodies.
Answers:
(a) Inertia is the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity.
(b) Newton’s second law states that force is proportional to acceleration produced in a body.
(c) A cart is being pulled along a horizontal road. The cart moves with uniform velocity. The force applied is equal to the frictional force.
(d) Even though action and reaction forces are always equal in magnitude, these forces may not produce accelerations of equal magnitudes because each force acts on a different object that may have a different mass.
(e) To every action there is an equal and opposite reaction and they act on different bodies.
Match and Pair:
| Column A | Column B |
| (i) Galileo | (a) Inertia |
| (ii) First law of motion | (b) kg ms-2 |
| (iii) Third law of motion | (c) High momentum |
| (iv) Unit of force | (d) Theory of motion |
| (v) Moving train | (e) Recoiling of a revolver |
Answer:
| Column A | Column B |
| (i) Galileo | (d) Theory of motion |
| (ii) First law of motion | (a) Inertia |
| (iii) Third law of motion | (e) Recoiling of a revolver |
| (iv) Unit of force | (b) kg ms-2 |
| (v) Moving train | (c) High momentum |
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Frequently Asked Questions (FAQs) on NCERT Solutions to Class 9 Science Chapter 8 Force and Laws of Motion
We have answered all in-text questions, exercises, additional exercises and activities thoroughly in this solutions material. With captivating illustrations to enhance your understanding, our material ensures a seamless grasp of every concept. Take your study sessions to the next level with our additional problem set, ideal for supercharging your exam preparation. And don’t forget, enjoy the convenience of free PDF downloads of all solutions, available anytime, anywhere!
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The following topics are covered:
8.1 Balanced and Unbalanced Forces
8.2 First Law of Motion
8.3 Inertia and Mass
8.4 Second Law of Motion
8.5 Third Law of Motion
7.6 Uniform Circular Motion
Here are the number of questions for the chapter:
(i) 6 Theoretical Questions (Questions 1 – 4, 12, 17)
(ii) 12 Numerical-based Questions (Questions 5 – 11, 13 – 16, 18)
Additional Exercises: 4 Numerical Questions
Learn the three laws of motion by Newton and practise numericals based on them. You are sure to see numericals based on them in your exams. Our solutions and extra problem set are geared towards helping you grasp the concepts and problem-solving techniques quickly, so make sure you study them well.
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