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Answer to Question #3707 in Mechanics | Relativity for marisa

Question #3707
A bullet of mass 0.0600 kg is fired into a block of wood. <br>If the bullet is moving at 330 m/s as it enters the block and takes 0.15 s to stop, find the average force required to stop the bullet and the impulse exerted by the wood on the bullet.
Expert's answer
m=0.06 kg
v=330 m/s
t=0.15 s
Average force is:
Fa=&delta;p/&delta;t
Impulse of the bullet is
pb = m &times; &delta;v = 0.06 kg &times; 330 m/s = 19.8 N&times;s
Thus
Fa = pb/&delta;t = (19.8 N&times;s)/(0.15 s) = 132 N
The impulse exerted by the wood on the bullet is the exact initial impulse of the bullet.

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