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Question
A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest but free to move. At what speed does the target move off ?
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Solution
Here,
Mass of the bullet, m = 10 g = 0.01 kg
Velocity of the bullet, v = 200 m/s
Mass of the target, M = 2 kg
Mass of the (target+bullet) after hitting = V (say)
Momentum of the system before hitting = (0.01)(200) + (2)(0) = 2
Momentum of the system after hitting = (2 + 0.01)V
(after hitting the bullet sticks to the target)
Now,
By law of conservation of momentum,
Momentum of the system before hitting = Momentum of the system after hitting
⇒ 2 = (2 + 0.01)V
⇒ V = 0.995 m/s
This is the velocity with which the bullet and the target moves after hitting.
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