# 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:

F

Impulse of the bullet is

p

Thus

F

The impulse exerted by the wood on the bullet is the exact initial impulse of the bullet.

v=330 m/s

t=0.15 s

Average force is:

F

_{a}=δp/δtImpulse of the bullet is

p

_{b}= m × δv = 0.06 kg × 330 m/s = 19.8 N×sThus

F

_{a}= p_{b}/δt = (19.8 N×s)/(0.15 s) = 132 NThe impulse exerted by the wood on the bullet is the exact initial impulse of the bullet.

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