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 ...
- physics - a 100-g bullet is fired horizontally into a 14.9kg block of wood ...
- physics - A gun is fired vertically into a 1.55 kg block of wood at rest ...
- physics - A 25 g bullet is fired from a gun into a 1.35 kg block of wood ...
- physics - A 12.0 g bullet is fired into a 13.9kg wooden block that is hanging ...
- College Physics - A 7.00 g bullet, when fired from a gun into a 1.00 kg block of...
- Physics - a point 22 caliber rifle fires a 1.8 gram bullet at a 400m/s into a ...
- Physics - A 15-g bullet is fired horizontally into a 3.000-kg block of wood ...
- pHysics !! - A 20-g bullet is fired horizontally into a 980 g block of wood ...
- physics - a bullet of mass 7g is fired into a block of wood weight 7kg the ...

More Related Questions