Page No 128: - Chapter 9 Force & Laws Of Motion class 9 ncert solutions Science - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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:

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 a non-zero velocity. 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.

Answer:

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 resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.

Question 3:

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

Answer:

When the bus accelerates 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. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from the roof of the bus. To avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope.

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:

(c) 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 its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg).

Answer:

Initial velocity, u = 0 (since the truck is initially at rest)

Distance travelled, s = 400 m

Time taken, t = 20 s

According to the second equation of motion:

Where,

Acceleration = a

1 metric tonne = 1000 kg (Given)

∴ 7 metric tonnes = 7000 kg

Mass of truck, m = 7000 kg

From Newton’s second law of motion:

Force, F = Mass × Acceleration

F = ma = 7000 × 2 = 14000 N

Hence, the acceleration of the truck is 2 m/s² and the force acting on the truck is 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:

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,

Acceleration, a

(0)² = (20)² + 2 × a × 50

a = −4 m/s²

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 × Acceleration

F = ma

F = 1 × (− 4) = −4 N

Hence, the force of friction between the stone and the ice is −4 N.

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:

(a) 35000 N (b) 1.944 m/s² (c) 15552 N

(a)Force exerted by the engine, F = 40000 N

Frictional force offered by the track, Ff = 5000 N

Net accelerating force, Fa = F − Ff = 40000 − 5000 = 35000 N

Hence, the net accelerating force is 35000 N.

(b)Acceleration of the train = a

The engine exerts a force of 40000 N on all the five wagons.

Net accelerating force on the wagons, Fa = 35000 N

Mass of the wagons, m = Mass of a wagon × Number of wagons

Mass of a wagon = 2000 kg

Number of wagons = 5

∴ m = 2000 × 5 = 10000 kg

Total mass, M = m  = 10000 kg

From Newton’s second law of motion:

Fa = Ma

a=Fam   = 35000 10000    = 3.5 ms-2

Hence, the acceleration of the wagons and the train is 3.5 m/s².

(c)Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg

Acceleration of the wagons = 3.5 m/s²

Thus, force exerted on all the wagons except wagon 1

= 8000 × 3.5 = 28000 N

Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 â€‹ N.

Hence, the force exerted by wagon 1 on wagon 2 is 28000 â€‹ 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?

Answer:

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?

(d) mvAnswer:

Mass of the object = m

Velocity = v

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?

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 fictional 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 m s⁻¹ before the collision during which they stick together. What will be the velocity of the combined object after collision?

Answer:

Mass of one of the objects, m₁ = 1.5 kg

Mass of the other object, m₂ = 1.5 kg

Velocity of m₁ before collision, v₁ = 2.5 m/s

Velocity of m₂, moving in opposite direction 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:

Total momentum before collision = Total momentum after collision

m₁v₁ + 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.