Homework Help: Physics

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.

Answer this Question

Related Questions

University Physics - A 8.3-g bullet, when fired from a gun into a 1.1-kg block ...
Momentum ~ Physics - A 25-g bullet is fired at 150 m/s into a 2.0-kg block of ...
Physics - A 15-g bullet is fired horizontally into a 3.000-kg block of wood ...
physics - A 25 g bullet is fired from a gun into a 1.35 kg block of wood ...
physics - a bullet of mass 0.06kg is fired into a block of wood. if the bullet ...
PHYSICS HELP!!!!! - A 7.0 g bullet is fired horizontally into a 1.5 kg block of ...
Physics - A bullet with a mass of 0.01kg is fired horizontally into a block of ...
Physics help please - A 13.2 gm bullet is fired into a 12.0 kg wood block that ...
Momentum - A 4.4 g bullet is fired into a block of wood with a mass of 21.3 g. ...
physics - A 5.88 g bullet is fired vertically into a 0.9 kg block of wood. The ...

For Further Reading

First Name:
School Subject:
Answer: