A bullet of mass 0.0018 kg moving at 472 m/s embeds itself in a large fixed...
A bullet of mass 0.0018 kg moving at 472 m/s embeds itself in a large fixed piece of wood and travels 0.61 m before coming to rest. Assume that the deceleration of the bullet is constant.
1. What force is exerted by the wood on the bullet?
F =
Homework Answers
Just apply kinematic equations 
U = 472 m/s, V =0, s=0.61 solving we get a = 182609 m/s^2
Force exerted = -ma = 0.0018*182609 =328.69 N
Force exerted is inertia force of bullet resisted by body
We can even use impulse momentum relation.
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