NCERT Solutions Class 9 Science Chapter 9 Force and Laws Of Motion

NCERT Solutions Class 9 Science 

The NCERT Solutions in English Language for Class 9 Science Chapter – 9 (Force and Laws Of Motion) has been provided here to help the students in solving the questions from this exercise.

Chapter – 9 (Force and Laws Of Motion) 

Questions 

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?
Answer –
(a) A stone of the same size
(b) a train
(c) a five-rupees coin
As the mass of an object is a measure of its inertia, objects with more mass have 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 football changes four times.
First, when a football player kicks to another player.
Second when that player kicks the football to the goalkeeper.
Third when the goalkeeper stops the football.
Fourth when the goalkeeper kicks the football towards a player of his own team.
Agent supplying the force:

• First case – First player
• Second case – Second player
• Third case – Goalkeeper
• Fourth case – Goalkeeper

3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer – Some leaves of a tree get detached when we shake its branches vigorously because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

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 –
(a) In a moving bus, passengers are in motion along with bus. When brakes are applied to stop a moving bus, bus comes in the position of rest. But because of tendency to be in the motion a person falls in forward direction.

(b) Similarly, when a bus accelerates from rest, we tend to fall backwards because when a bus is accelerated from rest, the tendency to be in rest, a person in the bus falls backwards.

Questions

1. If action is always equal to the reaction, explain how a horse can pull a cart.
Answer – A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.

2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer – When large amount of water is ejected from a hose at a high velocity, according to Newton’s Third Law of Motion, water pushes the hose in backward direction with the same force. Therefore, it is difficult for a fireman to hold a hose in which ejects large amount of water at a high velocity.

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.
Answer –
the Bullet’s mass (m₁) = 50 g
The rifle’s mass (m₂) = 4kg = 4000 g
Initial velocity of the fired bullet (v₁) = 35 m/s
Let the recoil velocity be v₂.
Since the rifle was initially at rest, the initial momentum of the rifle = 0
The total momentum of the rifle and bullet after firing = m₁v₁ + m₂v₂
As per the law of conservation of momentum, the total momentum of the rifle and the bullet after firing = 0 (same as initial momentum)
Therefore, m₁v₁ + m₂v₂ = 0
This implies that v₂ = –
= –
= – 0.4375 m/s
The negative sign indicates that recoil velocity is opposite to the bullet’s motion.

4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms^–1 and 1 ms^–1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^–1. Determine the velocity of the second object.
Answer –
Assuming that the first object is object A and the second one is object B, it is given that:
Mass of A (m₁) = 100g
Mass of B (m₂) = 200g
Initial velocity of A (u₁) = 2 m/s
Initial velocity of B (u₂) = 1 m/s
Final velocity of A (v₁) = 1.67 m/s
Final velocity of B (v₂) =?
Total initial momentum = Initial momentum of A + initial momentum of B
= m₁u₁ + m₂u₂
= (100g) × (2m/s) + (200g) × (1m/s) = 400 g.m.sec^-1
As per the law of conservation of momentum, the total momentum before collision must be equal to the total momentum post collision.
therefore,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ = 400 g.m.sec^-1
100g × 1.67 m/s + 200g × v₂ = 400 g.m.sec^-1
v₂ = m. s^-1

v₂ = 1.165 m. s^-1
Therefore, the velocity of object B after the collision is 1.165 meters per second.

Exercises

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, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.

2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer – The carpet with dust is in state of rest. When it is beaten with a stick the carpet is set in motion, but the dust particles remain at rest. Due to inertia of rest the dust particles retain their position of rest and falls down due to gravity.

3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer – In moving vehicle like bus, the motion is not uniform, the speed of vehicle varies and it may apply brake suddenly or takes sudden turn. The luggage will resist any change in its state of rest or motion, due to inertia and this luggage has the tendency to fall sideways, forward or backward. To avoid the fall of the luggage, it is tied with the 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.

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.)
Answer –
Distance covered by the truck (s) = 400 meters
Time taken to cover the distance (t) = 20 seconds
The initial velocity of the truck (u) = 0 (since it starts from a state of rest)
We know, s = ut + ½ at²
400 = 0 + ½ a (20)²
a = 2 ms^–2
m = 7 metric tonne = 7000 kg,
a = 2 ms^–2
F = ma = 7000 × 2
= 14000 N Ans.
Therefore, the force acting on the truck is of 14000 N

6. A stone of 1 kg is thrown with a velocity of 20 ms^-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 –
Mass of the stone (m) = 1kg
Initial velocity (u) = 20m/s
Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)
Distance travelled by the stone (s) = 50 m
As per the third equation of motion
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 (retardation)
We know that
F = m×a
Substituting above obtained value of a = -4 in F = m x a
We get,
F = 1 × (-4) = -4N
Here the negative sign indicates the opposing force which is Friction

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 and
(b) the acceleration of the train
Answer –
(a) The force exerted by the train (F) = 40,000 N
Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N

(b) Total mass of the train = mass of engine + mass of each wagon
= 8000kg + 5 × 2000kg
The total mass of the train is 18000 kg.
As per the second law of motion, F = ma (or: a = F/m)
Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)
= 35,000/18,000 = 1.94 ms^-2
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 ms^-2?
Answer –
Mass of the vehicle (m) = 1500 kg
Acceleration (a) = -1.7 ms^-2
As per the second law of motion, F = ma
F = 1500kg × (-1.7 ms^-2)
= -2550 N
Hence, the force between the automobile and the road is -2550 N, in the opposite direction of the automobile’s motion.

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 – The momentum of an object is defined as the product of its mass m and velocity v
Momentum = mass x velocity
Hence, the correct answer is mv i.e option (d)

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, a horizontal force of 200N is used to move a wooden cabinet, thus a friction force of 200N will be exerted on the cabinet. Because according to third law of motion, an equal magnitude of force will be applied in the opposite direction.

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 ms^-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer –
Mass of first object, m₁ = 1.5 kg
Mass of second object, m₂ = 1.5 kg
Velocity of first object before collision, v₁ = 2.5 m/s
The velocity of the second object which is moving in the opposite direction, v₂ = -2.5 m/s
We know that,
Total momentum before collision = Total momentum after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)v
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5)v
3.75 – 3.75 = 3v
v = 0
Therefore, the velocity of the combined object after the collision is 0 m/s

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.
Answer – The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.

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 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) = 200g
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = – 5m/s
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
This implies that the momentum of the ball reduces by 1000 g.m.s^-1 after being struck by the hockey stick.

14. 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 (or 0.01 kg)
Initial velocity of the bullet (u) = 150 m/s
Terminal velocity of the bullet (v) = 0 m/s
Time period (t) = 0.03 s
To find the distance of penetration, the acceleration of the bullet must be calculated
Let the distance of penetration be s
As per the first law of motion
v = u + at
0 = 150 + a (0.03)
a = -5000 ms^-2
v² = u² + 2as
0 = 150² + 2 × (-5000)s
s = 2.25 m
As per the second law of motion, F = ma
F = 0.01kg × (-5000 ms^-2)
F = -50 N

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.
Answer –
Mass of the object (m₁) = 1kg
Mass of the block (m₂) = 5kg
Initial velocity of the object (u₁) = 10 m/s
Initial velocity of the block (u₂) = 0
Mass of the resulting object = m₁ + m₂ = 6kg
Velocity of the resulting object (v) =?
Total momentum before the collision = m₁u₁ + m₂u₂ = (1kg) × (10m/s) + 0 = 10 kg.m.s^-1
As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision. Therefore, the total momentum post the collision is also 10 kg.m.s^-1
Now, (m₁ + m₂) × v = 10kg.m.s^-1

v =
= 1.66 ms^-1
The resulting object moves with a velocity of 1.66 meters per second.

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.
Answer –
Mass of the object (m) = 100kg
Initial velocity (u) = 5 m/s
Terminal velocity (v) = 8 m/s
Time period (t) = 6s
Now, initial momentum (m × u) = 100kg × 5m/s = 500 kg.m.s^-1
Final momentum (m × v) = 100kg × 8m/s = 800 kg.m.s^-1
Acceleration of the object (a) =
= ms^-2
Therefore, the object accelerates at 0.5 ms^-2. This implies that the force acting on the object (F = ma) is equal to:
F = (100kg) × (0.5 ms^-2) = 50 N
Therefore, a force of 50 N is applied on the 100kg object, which accelerates it by 0.5 ms^-2.

17. 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 – Rahul gave the correct reasoning and explanation that both the motorcar and the insect experienced the same force and a change in their momentum. As per the law of conservation of momentum.
When 2 bodies collide:
Initial momentum before collision = Final momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
The equal force is exerted on both the bodies but, because the mass of insect is every small it will suffer greater change in velocity.

18. 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 ms^–2.
Answer –
Mass of the dumb-bell (m) = 10kg
Distance covered (s) = 80cm = 0.8m
Initial velocity (u) = 0 (it is dropped from a position of rest)
Acceleration (a) = 10ms^-2
Terminal velocity (v) =?
Momentum of the dumb-bell when it hits the ground = mv
As per the third law of motion
v² – u² = 2as
Therefore, v² – 0 = 2 (10 ms^-2) (0.8m) = 16 m²s^-2
v = 4 m/s
The momentum transferred by the dumb-bell to the floor = (10kg) × (4 m/s) = 40 kg.m.s^-1

Additional Exercises

1. The following is the distance-time table of an object in motion:

(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) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.

(b) As per the second law of motion, force = mass × acceleration. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well

2. 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 ms^-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)
Answer –
Mass of the car (m) = 1200kg
When the third person starts pushing the car, the acceleration (a) is 0.2 ms^-2. Therefore, the force applied by the third person (F = ma) is given by:
F = 1200kg × 0.2 ms^-2 = 240N
The force applied by the third person on the car is 240 N. Since all 3 people push with the same muscular effort, the force applied by each person on the car is 240 N.

3. 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) = 500g = 0.5kg
Initial velocity of the hammer (u) = 50 m/s
Terminal velocity of the hammer (v) = 0 (the hammer is stopped and reaches a position of rest).
Time period (t) = 0.01s
Therefore, the acceleration of the hammer is given by :
a =
=
= -5000ms^-2
Therefore, the force exerted by the hammer on the nail (F = ma) can be calculated as:
F = (0.5kg) × (-5000 ms^-2)
= -2500 N
As per the third law of motion, the nail exerts an equal and opposite force on the hammer. Since the force exerted on the nail by the hammer is -2500 N, the force exerted on the hammer by the nail will be +2500 N.

4. 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) = 1200kg
Initial velocity (u) = 90 km/hour = 25 meters/sec
Terminal velocity (v) = 18 km/hour = 5 meters/sec
Time period (t) = 4 seconds
The acceleration of the car be calculated with the help of the formula
a =

= ms^-2

= -5ms^-2
Therefore, the acceleration of the car is -5 ms^-2.
Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg.m.s^-1
Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg.m.s^-1
Therefore, change in momentum (final momentum – initial momentum)
= (6,000 – 30,000) kg.m.s^-1
= -24,000 kg.m.s^-1
External force applied = mass of car × acceleration = (1200kg) × (-5 ms^-2) = -6000N
Therefore, the magnitude of force required to slow down the vehicle to 18 km/hour is 6000 N

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