Chapter 8: Force and Laws of Motion

Q11Written answer

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?

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Worked solution

Given

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

Q12Written answer

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.

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Worked solution

Since the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. This implies that the two forces are equal in magnitude but opposite in direction (since the person pushing the truck is not displaced when the truck doesn’t move). Therefore, the student’s logic is correct.

Q13Written answer

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.

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Worked solution

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.

Q14Written answer

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.

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Worked solution

Given, 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 x (-5000)s

s = 2.25 m

As per the second law of motion, F = ma

F = 0.01kg × (-5000 ms^-2)

F = -50 N

Q15Written answer

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.

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Worked solution

Given, 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

NCERT Solutions For Class 9 Science Chapter 9 Image 6 `

The resulting object moves with a velocity of 1.66 meters per second.

Q16Written answer

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.

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Worked solution

Given, 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

NCERT Solutions For Class 9 Science Chapter 9 Image 7

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.

Q17Written answer

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.

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Worked solution

As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car (since force is proportional to mass).

Akhtar’s assumption is partially right. Since the mass of the car is very high, the force exerted on the insect during the collision is also very high.

Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum. The velocity of insect changes accordingly due to its mass as it is very small compared to the motorcar. Similarly, the velocity of motorcar is very insignificant because its mass is very large compared to the insect.

Rahul’s statement is completely right. As per the third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. However, Rahul’s suggestion that the change in the momentum is the same contradicts the law of conservation of momentum.

Q18Written answer

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.

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Worked solution

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

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Exercises · Page 128,129