Answer to Question #3707 in Mechanics | Relativity for marisa

Question

A bullet of mass 0.0600 kg is fired into a block of wood.
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.

Answer 1

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.