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.
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
1 answer
1 answer