An apple with mass 0.4 kg is hanging at rest from the lower end of a light vertical rope. A dart of mass 0.1 kg is shot vertically upward, strikes the bottom of the apple, and remains embedded in it. a) If the speed of the dart is 5m/just before it strikes the apple, what is the speed of the apple just after the collision

The speed of the apple just after the collision is 1m/s

Explanation:

Since the two bodies collided, we will apply the law of conservation of momentum to find their common velocity which states that the sum of momentum of the bodies before collision is equal to the sum of momentum of the bodies after collision. After collision both bodies will move with a common velocity v.

Given momentum = mass × velocity

Initial momentum of the 0.4kg apple hanging at rest will be;

Momentum = 0.4kg×0m/s = 0kgm/s (the velocity of apple is 0m/s since it is hanging at rest)

Initial momentum of the 0.1kg dart moving with speed of 5m/s will be;

Momentum of dart = 0.1kg × 5m/s

Momentum of dart = 0.5kgm/s

Momentum of the bodies after collision will give;

Momentum after collision = (0.4kg+0.1kg) × v (v is their common velocity)

Momentum after collision = 0.5v

Using the law of conservation of momentum to calculate their common velocity, we will have;

0+0.5 = 0.5v

0.5 = 0.5v

v = 0.5/0.5

v = 1m/s

Therefore the speed of both bodies after collision is 1m/s which will also serve as the speed of the apple after collision.

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