Class 9 Science Chapter 8 question answer Force and laws of motion

Ncert Solutions for Class 9 Science Chapter 8: Force and laws of motion class 9 questions with answers

Textbook	Ncert
Class	Class 9
Subject	Science
Chapter	Chapter 8
Chapter Name	Force and laws of motion class 9 ncert solutions
Category	Ncert Solutions
Medium	English

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In Text Questions Page No: 91

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 fiverupees coin and a one-rupee coin?

Answer 1: (a) Stone (more mass) Even though they are the same size, the stone has more mass, so it has more inertia.
(b) Train (more mass) A train is much more massive than a bicycle, so it has much greater inertia.
(c) Five-rupees coin (more mass) The five-rupees coin is heavier than the one-rupee coin, so it has 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.

Answer 2: Scenario Description:

1. Player A kicks the football to Player B.
2. Player B kicks the football towards the goal.
3. The goalkeeper stops the football.
4. The goalkeeper kicks it to his own team.

Number of times velocity changes:

• 1st kick (Player A) → velocity changes
• 2nd kick (Player B) → velocity changes
• Ball is stopped by goalkeeper → velocity changes
• Goalkeeper kicks the ball → velocity changes

Total = 4 times the velocity changes.

Agents applying force:

1. Player A
2. Player B
3. Goalkeeper (to stop the ball)
4. Goalkeeper (to kick the ball)

Question 3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer 3: 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 4: In a moving bus, a passenger moves with the bus due to inertia of motion. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain to inertia of motion. As a result, a forward force is exerted on him.

Similarly, the passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.

Exercises

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.

Answer 1: Yes, it is possible. An object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.

Question 2: When a carpet is beaten with a stick, dust comes out of it, Explain.

Answer 2: Using a stick to beat a carpet; causing the carpet to move quickly, while dust particles trapped in the carpet’s pores prefer to stay still, since inertia of an item resists any change in its state of rest or motion. The dust particles, according to Newton’s first rule of motion, remain at rest as the carpet moves. As a result, dust particles emerge from the carpet.

Question 3: Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer 3: If luggage is not tied with a rope to the body of the bus, it will move backward due to inertia of rest when the bus starts moving and will move forward due to inertia of motion when the bus stops.

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.

Answer 4: Correct answer: (c) there is a force on the ball opposing the motion.

Explanation: The ball stops because of friction between the ball and the ground — a force that opposes motion. It’s not about how hard the ball was hit; without friction, it would keep rolling.

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 tonne = 1000 kg.)

Answer 5:

• Distance covered by the truck (s) = 400 m
• Time (t) = 20 s
• Initial velocity (u) = 0
• Mass of the truck: m = 7 tons = 7000 kg (1 metric ton = 1000 kg)
• According to the second equation of motion
• S = ut + 12at²
• 400 = 0 × 20 + 12 × a × (20)²
• a = 2m/s²
• Hence, acceleration of the truck = 2m/s²
• Since force (F) = mass × acceleration (a),
• F = 7000 × 2
• F = 14000N
• Hence, force applied on the truck = 14000 N

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?

Answer 6: Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance covered by the stone, s = 50 m
Since, v² – u² = 2as,
Or, 0 – 20² = 2a × 50,
Or, a = –4 ms^-2
Force of friction, F = ma = – 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 and
(b) the acceleration of the train.

Answer 7:

• (a) Given, 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.

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?

Answer 8: m = 1,500 kg; a = – 1.7 ms^-2
F = m a = (1,500) (-1.7) = – 2,550 N
Backward force = 2,550 N (in a direction opposite to the motion of vehicle.)

Question 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 9: (d) mv
Mass of the object = m
Velocity = v
Momentum = Mass x 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?

Answer 10: If the net external force on the cabinet is zero, it will move at a constant velocity. Because a horizontal force of 200 N acts on the cabinet in the forward direction, a frictional force of 200 N will act on it, resulting in a net external force of zero. Thus, a frictional force of 200 N will be applied to the cabinet.

Question 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 11: 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.

Question 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 12: Given, 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.

Question 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 13: Given:

• Mass of bullet, m = 10g = 0.01kg
• Initial velocity, u = 150m/s
• Final velocity, v = 0m/s
• Time, t = 0.03s

Find distance of penetration ( s )

We use the equation:
\(s = \frac{(u + v)}{2} \cdot t\)

\(s = \frac{(150 + 0)}{2} \cdot 0.03 = 75 \cdot 0.03 = \boxed{2.25\,m}\)

Find the force exerted

First, calculate acceleration using:
\(a = \frac{v – u}{t} = \frac{0 – 150}{0.03} = -5000\,m/s^2\)

Now apply Newton’s second law:
\(F = m \cdot a = 0.01 \cdot (-5000) = \boxed{-50\,N}\)

(The negative sign shows the force is in the opposite direction of motion.)

Answer:

• Distance of penetration = 2.25 m
• Force by block on bullet = 50 N (opposite to motion)

Question 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 14: Given:

• Mass of object = m₁ ​= 1kg, velocity = u₁ = 10 m/s
• Mass of block = m₂ ​= 5kg, velocity = u₂ = 0 m/s
• After collision, both stick together → combined mass = m₁+m₂ = 6 kg

1. Total momentum before the impact:

Total momentum = \(m_1 \cdot u_1 + m_2 \cdot u_2 = 1 \cdot 10 + 5 \cdot 0 = \boxed{10\,kg·m/s}\)

2. Total momentum after the impact:

Since momentum is conserved:
Total momentum after = 10 kg·m/s

3. Velocity after collision (combined object):

\(v = \frac{\text{Total momentum}}{\text{Total mass}} = \frac{10}{6} = \boxed{1.67\,m/s}\)

Final Answers:

• Momentum before impact: 10 kg·m/s
• Momentum after impact: 10 kg·m/s
• Velocity after collision: 1.67 m/s

Question 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 15: Given:

• Mass m = 100 kg
• Initial velocity u = 5 m/s
• Final velocity v = 8m/s
• Time t = 6s

Initial momentum

p_1​ = m⋅u = 100⋅5 = 500kg⋅m/s

Final momentum

p₂​ = m⋅v = 100⋅8 = 800kg⋅m/s​

Force (from rate of change of momentum)

\(F = \frac{p_2 – p_1}{t} = \frac{800 – 500}{6} = \boxed{50\,N}\)

Answer:

• Initial momentum = 500 kg·m/s
• Final momentum = 800 kg·m/s
• Force = 50 N

Question 16: 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.

Answer 16: 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 17: 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 m s⁻².

Answer 17: Given, 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