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.
m=0.06 kg v=330 m/s t=0.15 s Average force is: Fa=δp/δt Impulse of the bullet is pb = m × δv = 0.06 kg × 330 m/s = 19.8 N×s Thus Fa = pb/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.
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