Conservation of Momentum with Friction

[derf709]

Homework Statement

A bullet of mass m = 6.32 g is fired into a block of mass M = 2.30 kg that is initially at rest on a rough surface. (The coefficient of friction is 0.28 between the block and the surface.) The bullet ends up stuck in the block, and together they slide across the surface and come to rest.

Homework Equations

KE=1/2 mv^2
p= mv

The Attempt at a Solution

(.00632)(vi)=(.00632+2.30)(Vf)

I was thinking I could set the total kinetic energy minus friction equal to zero, and then solve for the velocity, which could be used in the momentum equation. is this on the right track?

1/2(.00632+2.30)v^2-6.33=0
v=2.34 m/s

Plugging in final velocity, I got an initial velocity of 855 m/s. It sounds plausible, but the answer is incorrect. Any suggestions? :)

 

[PhanthomJay]

Your problem is missing information. You can't subtract friction force in Newtons from KE which is in in Newton-meters (joules). You need to know the distance traveled while coming to a stop.