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_{a}=δp/δt

Impulse of the bullet is

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

Thus

F_{a} = p_{b}/δt = (19.8 N×s)/(0.15 s) = 132 N

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

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.

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