Posted by **Help Please** on Saturday, July 14, 2012 at 2:12am.

1. A bullet of 0.0500 kg is fired into a block of wood. Knowing that the bullet left the gun

with a muzzle velocity of 350. m/s, and the bullet penetrates .15 m into the block of

wood, determine:

a) The average force required to stop the bullet.

b) The impulse exerted by the wood on the bullet.

c) The change in momentum of the bullet

- Physics -
**Elena**, Saturday, July 14, 2012 at 7:20am
If the block doesn’t move with bullet inside, the magnitude of acceleration (deceleration) of the bullet is

a =v²/2s=350²/2•0.15=408333 m/s².

F=ma= 0.05•408333= 20417 N.

The change in momentum of the bullet

Δp=p2-p1=0-m•v=-0.05•350=-1.4•10^-4 kg•m/s.

The impulse exerted by the wood on the bullet = Δp=1.4•10^-4 kg•m/s.

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