Question
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
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
Answers
Elena
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.
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.