Force and Laws of Motion NCERT Solution, study material, CBSE Notes

NCERT Solution: Force and Laws of Motion

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?

The cabinet will move with constant velocity only when the net force on it is zero.
Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.

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?

Mass of one of the objects, m₁ = 1.5 kg
Mass of the other object, m₂ = 1.5 kg
Velocity of m1 before collision, u₁ = 2.5 m/s
Velocity of m2, moving in opposite direction before collision, u₂ = −2.5 m/s
Let v be the velocity of the combined object after collision. By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m₁ + m₂) v = m₁u₁ + m₂u₂
Or, (1.5 + 1.5) v = 1.5 × 2.5 +1.5 × (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms^–1

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.

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.

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 change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Mass of the hockey ball, m = 200 g = 0.2 kg
Hockey ball travels with velocity, v₁ = 10 m/s
Initial momentum = mv₁
Hockey ball travels in the opposite direction with velocity, v₂ = −5 m/s
Final momentum = mv₂
Change in momentum = mv₁ − mv₂ = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s⁻¹
Hence, the change in momentum of the hockey ball is 3 kg m s⁻¹.

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.

Initial velocity, u= 150 m/s
Final velocity, v= 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v= u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)a = -150 / 0.03 = -5000 m/s²

(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:
v²= u2²2as
0 = (150)²+ 2 (-5000)
= 22500 / 10000
= 2.25 m

Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s²
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

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.

Mass of the object, m₁ = 1 kg
Velocity of the object before collision, v₁ = 10 m/s
Mass of the stationary wooden block, m₂ = 5 kg
Velocity of the wooden block before collision, v₂ = 0 m/s
∴ Total momentum before collision = m₁ v₁+ m₂ v₂
= 1 (10) + 5 (0) = 10 kg m s⁻¹

It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = 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 (10) + 5 (0) = (1 + 5) v
v = 10 / 6
= 5 / 3

The total momentum after collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg m s−1
Total momentum just after the impact = (m1 + m2) v = 6 × 5 / 3 = 10 kg ms-1
Hence, velocity of the combined object after collision = 5 / 3 ms-1

An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s−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.

Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg m s⁻¹
Final momentum = mv = 100 × 8 = 800 kg m s⁻¹
Force exerted on the object, F = mv - mu / t
= m (v-u) / t
= 800 - 500
= 300 / 6
= 50 N

Initial momentum of the object is 500 kg m s⁻¹.
Final momentum of the object is 800 kg m s.⁻¹
Force exerted on the object is 50 N.

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, 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

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.