RBSE Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
When the tree’s branch is shaken vigorously the branch attain motion but the leaves stay at rest. Due to the inertia of rest, the leaves tend to remain in its position and hence detaches from the tree to fall down.

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?
Answer:
When a moving bus brakes to a stop : Our body is also in motion along with bus, but due to sudden brakes, the lower part of our body comes to rest as soon as the bus stops. But the upper part of our body continues to be in motion and hence we fall in forward direction due to inertia of motion.

When the bus accelerates from rest we fall backwards : When the bus is stationary our body is at rest but when the bus accelerates, the lower part of our body being in contact with the floor of the bus comes in motion, but the upper part of our body remains at rest due to inertia of rest. Hence, we fall in backward direction.

Page No. 126-127

Question 1.
If action is always equal to the reaction, explain how horses can pull a cart.
Answer:
The third law of motion states that action is always equal to the reaction but they act on two different bodies. In this case the horse exerts a force on the ground with its feet while walking, the ground exerts an equal and opposite force on the feet of the horse, which enables the horse to move forward and the cart is pulled by the horse.

Question 2.
Explain why is it difficult for a fireman to hold a hose, which ejects a large amount of water at a high velocity.
Answer:
The water that is ejected out from the hose in the forward direction comes out with a large momentum and equal amount of momentum is developed in the hose in the opposite direction and hence the hose is pushed backward. It becomes difficult for a fireman to hold a hose which experiences this large momentum.

Question 3.
From a rifle of mass 4 kg, a bullet of mass 50 g is Hired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.
Answer:
Mass of the bullet, m₁ = 50 g = \(\frac{50}{1000}\)kg = 0.05 kg
Mass of the rifle, m₂ = 4 kg
Initial velocity of bullet (u₁) = 0 ms^-1
Initial velocity of rifle, u₂ = 0 ms^-1
Final velocity of bullet, v₁ = 35 ms^-1
Final velocity of rifle = v₂
According to the law of conservation of momentum,
Total momentum before the fire = Total momentum after the fire
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0 = 0.05 kg × 0 ms^-1 + 4 kg × 0 ms^-1
0 = 0.05 kg × 35 ms^-1 + 4 kg × v₂
0 = 1.75 kg ms^-1 + 4 kg × v₂
v₂ = - 1.75 kg ms^-1/4 kg
= - 0.4375 ms^-1
The negative sign indicates that the rifle recoils backwards with a velocity of 0. 4375 m/s.

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 and 1 m/s respectively. They collide and after the collision the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.
Answer:
Mass of first object (m₁) = 100 g = 100/1000 kg = 0.1 kg
Mass of second object (m₂) = 200 g = 200/1000 kg = 0.2 kg
Initial velocity of the first object (u₁) = 2 ms^-1
Initial velocity of the second object (u₂) = 1 ms^-1
Final velocity of the first object (v₁) = 1.67 ms^-1
Final velocity of the second object (v₂) = ?
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.1 kg × 2 ms^-1 + 0.2 kg × 1 ms^-1 = 0.1 kg × 1.67 ms^-1 + 0.2 kg × v₂
0.4 kg ms^-1 = 0.167 kg ms^-1 + 0.2 kg v₂
0.4 kg ms^-1 - 0.167 kg ms^-1 = 0.2 kg v₂
0.233 kg ms^-1 = 0.2 kg v₂
v₂ = 0.233/0.2 = 1.165 ms^-1

RBSE Class 9 Science Chapter 9 Force and Laws of Motion Textbook Questions and Answers

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.
Answer:
When an object experiences a net zero external imbalanced force, in accordance with second law of motion its acceleration is zero. If the object was initially in a state of motion, then in accordance with the first law of motion, the object will continue to move in same direction with same speed. It means that the object may be travelling with a non-zero velocity but the magnitude as well as direction of velocity must remain unchanged or constant throughout.

Question 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.

Question 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 Aotion 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. It is tied with a rope to avoid the fall of the luggage.

Question 4.
A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the hall 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 hall.
(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.

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 its mass is 7 tonnes (Hint: 1 tone = 1000 kg).
Answer:
Initial velocity, (u) = 0 ms^-1
Distance (s) = 400 m
Time (t) = 20 s
Mass (m) = 7 metric tons = 7000 kg
By second equation of motion,
s = ut + \(\frac{1}{2}\)at²
400 = 0 ms^-1 × 20 s + \(\frac{1}{2}\) × a × (20 s)²
400 = \(\frac{1}{2}\) × a × 400 s²
a = 800 m/400 s² = 2 ms^-2
By Newton’s second law of motion :
F = m × a
= 7000 kg × 2 ms^-1 = 14000 N

Question 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 (m) = 1 kg
Initial velocity (u) = 20 ms^-1
Final velocity (v) = 0 ms^-1
Distance (s) = 50 m
By third equation of motion :
v₂ - u₂ = 2as
(0 ms^-1)² - (20 ms^-1)² = 2 × a × 50 m
- 400 m² s^-2 = 100 m × a
- 400 m²s^-2/100 m = a
- 4 ms^-2 = a
By Newton’s second law of motion :
F = m × a
= 1 kg × (- 4 ms^-2) = - 4N

Question 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,
(b) the acceleration of the train, and
(c) the force of wagon 1 on wagon 2.
Answer:
Mass of engine (m₁) = 8000 kg
Mass of each wagon (m₂) = 2000 kg
Number of wagons (n) = 5
Total mass of 5 wagons = 2000 kg × 5 = 10000 kg
Force exerted by engine, F₁ = 40,000 N
Frictional force offered by the track, F₂ = 5000 N
(a) Net accelerating force,
F = F₁ - F₂
= (40,000 - 5,000) N = 35,000 N
(b) Acceleration of the train, a = F/m
a = 35,000 N/10,000 kg = 3.5 ms^-2
(c) Force of wagon 1 on wagon 2
= (35,000 - 35,000/5) N
= (35,000 - 7000) N = 28,000 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 ms^-2?
Answer:
Mass of vehicle (m) = 1500 kg
Acceleration (a) = - 1.7 ms^-2
By second law of motion : F = m × a
= 1500 × - 1.7 = - 2550 N
Negative sign shows that the force is in a direction opposite to the direction of motion of the vehicle.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)²
(b) mv²
(c) 1/2 mv²
(d) mv
Answer:
(d) 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?
Answer:
A force of 200 N is applied in the forward direction. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the frictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet.

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 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
Initial velocity of the first object (u₁) = 2.5 ms^-1
Initial velocity of the second object (u₂) = - 2.5 ms^-1
Let the combined velocity = v ms^-1
According to the law of conservation of momentum,
Momentum before collision = Momentum after collision
m₁u₁ + m₂u₂ = (m₁ + m₂) v
1.5 (2.5) + 1.5 (- 2.5) = (1.5 + 1.5) v
3.75 - 3.75 = 3 v
0 = 3v
v = 0 ms^-1

Question 12.
According to the third law of motion when we push on an object, the object pushes hack 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:
According to the third law of motion, action and reaction forces are equal and opposite but they both act on different bodies. Hence, they cannot cancel each other. When we push, a massive truck, then the force applied on the truck is not sufficient to overcome the force of friction between the tyres of the truck and ground. Hence, the truck does not move. The truck will move only if the force applied on it is greater than the frictional force. Therefore, the justification of the student is not correct.

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.
Answer:
Mass of hockey ball (m) = 200 g = \(\frac{200}{1000}\) = 0.2 kg
Initial velocity of the ball (u) = 10 ms^-1
Final velocity of the ball (u) = - 5 ms^-1 (Hockey ball travels in opposite direction)
Initial momentum of the ball = m × u
= 0.2 kg × 10 ms^-1 = 2 kg ms^-1
Final momentum of the ball = m × v
= 0.2 kg × (- 5 ms^-1) = - 1 kg ms^-1
Change of momentum = mv - mu
= (- 1 - 2) kg ms^-1 = - 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.
Answer:
Mass (m) = 10 g = \(\frac{10}{1000}\) kg = 0.01 kg
Initial velocity (u) = 150 ms^-1
Final velocity (v) = 0 ms^-1
Time (t) = 0.03 s
According to first equation of motion, v = u + at
Acceleration (a) = \(\frac{v-u}{t}\)
= (0 - 150) ms^-1/0.03 s
= - 5000 ms^-2 (Negative sign indicates that the velocity of the bullet is decreasing.)
By second equation of motion :
s = ut + \(\frac{1}{2}\)at²
= 150 ms^-1 × 0.03 s + \(\frac{1}{2}\) × (5000 ms^-2) × (0.03 s)²
= 4.50 m - 2.25 m = 2 25 m
By second law of motion :
F = m × a
= 0.01 kg × (5000 ms^-2) = 50 N
Hence, the distance of penetration of the bullet into the block is 2.25 m and 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.
Answer:
Mass of object (m₁) = 1 kg
Initial velocity (u₁) = 10 ms^-1
Mass of stationary wooden block (m₂) = 5 kg
Velocity of wooden block before collision, (u₂) = 0 m/s
Let velocity of combined object = v
Total momentum just before impact = m₁ × u₁
= 1 kg × 10 ms^-1 = 10 kg ms^-1
By law of conservation of momentum,
Total momentum before impact = Total momentum after impact
m₁u₁ + m₂u₂ = (m₁ + m₂) v
1(10) + 5(0) = (1 + 5) v
v = \(\frac{10}{6}=\frac{5}{3}\)ms^-1 = 1.67 ms^-1

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.
Answer:
Mass of object (m) = 100 kg
Initial velocity (u) = 5 ms^-1
Final velocity (v) = 8 ms^-1
Time (t) = 6 s
Initial momentum = m × u
= 100 kg × 5 ms^-1 = 500 kg ms^-1
Final momentum = m × v
= 100 kg × 8 ms^-1 = 800 kg ms^-1
F = \(\frac{m v-m u}{t}\)
= \(\frac{(800-500) \mathrm{kg} \mathrm{ms}^{-1}}{6 \mathrm{~s}}\)
= \(\frac{300 \mathrm{~kg} \mathrm{~ms}^{-1}}{6 \mathrm{~s}}\)
= 50 N

Question 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 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 very small it will suffer greater change in velocity.

Question 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 (m) = 10 kg
Final velocity (u) = 0 ms^-1
Distance (s) = 80 cm = 0.8 m
Acceleration (a) = 10 ms^-1
By third equation of motion :
v₂ - u₂ = 2as
v₂ - (0 ms^-1) 2 = 2 × 10 ms^-2 × 0.8 m
v = 4 ms^-1
Momentum transferred to the floor :
= m × v
= 10 kg × 4 ms^-1
= 40 kg ms^-1

Additional Exercise from Textbook

Question 1.
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) There is an unequal change of distance in an equal interval of time. So, the given objects have a non-uniform motion. Thus, the acceleration is increasing as the velocity of the object increases with time.

(b) According to Newton’s second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. In the given case, the increasing acceleration of the given object indicates that the force acting on the object is also increasing.

Question 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 motorcar = 1200 kg
Only two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car, when it is pushed by the third person, a = 0.2 m/s²
Let the force applied by the third person be F.
From Newton’s second law of motion :
Force = Mass × acceleration
F = 1200 × 0.2 = 240 N
Thus, the third person applies a force of magnitude 240 N.
Hence, each person applies a force of 240 N to push the motorcar.

Question 3.
A hammer of mass 500 g, moving at 50 ms^-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
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion :
Force, F = m \(\frac{v-u}{t}\)
= 0.5 \(\frac{0-50}{0.01}\)
= - 2500 N
The hammer strikes the nail with a force of - 2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i. e. + 2500 N.

Question 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 motorcar, m = 1200 kg
Initial velocity of the motorcar, u = 90 km/h = 25 m/s
Final velocity of the motorcar, v = 18 km/h = 5 m/s
Time taken, t = 4 s
According to the first equation of motion :
v = u + at
⇒ 5 = 25 + a (4)
Accelaration, a = - 5 m/s²
Negative sign indicates that its a retarding motion, i.e. velocity is decreasing. Change in momentum = mv - mu = m (v - u)
= 1200 (5 - 25) = - 24000 kg ms^-1
Magnitude of the Force = Mass × acceleration
= 1200 × - 5 = - 6000 N
(Negative sign indicates retardation, decrease in momentum and retarding force)

Question 5.
A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to a halt after that. If the collision lasts for Is:
(a) Which vehicle experiences the greater force of impact?
(b) Which vehicle experiences the greater change in momentum?
(c) Which vehicle experiences the greater acceleration?
(d) Why is the car likely to suffer more damage than the truck?
Answer:
Let the mass of the truck be M and that of the car be m
Thus, M > m
Initial velocity of both vehicles, u
Final velocity of both vehicles, v' = 0 (since the vehicles come to rest after collision)
Time of impact, t = 1 s
From Newton’s second law of motion, the net force experienced by each vehicle is given by the relation :
F_car = m\(\frac{v^{\prime}-v}{t}\) = mv
F_truck = M\(\frac{v^{\prime}-v}{t}\) = Mv
Since, the mass of the truck is greater than that of the car, it will experiences a greater force of impact.

(b) Initial momentum of the car = mv
Final momentum of the car = 0
Change in momentum = mv
Initial momentum of the truck = Mv
Final momentum of the truck = 0
Change in momentum = Mv
Since, the mass of the truck is greater than that of the car, it will experience a greater change in momentum.

(c) From the first equation of motion, acceleration produced in a system is independent of the mass of the system, the initial velocity, the final velocity, and the time of impact remain the same in both the cases. Hence, both the car and the truck experience the same amount of acceleration.

According to Newton’s third law of motion, for every action there is an equal and opposite reaction that acts on different bodies. Since the truck experiences a greater force of impact (action), this larger impact force is also experienced by the car (reaction). Thus, the car is likely to suffer more damage than the truck.