# Frictional implications on non-elastic momentum

1. Mar 16, 2009

### anelmarx

1. The problem statement, all variables and given/known data
Consider a bullet of mass m fired at a speed of Vo into a wooden block of mass M. The bullet instantaneously comes to a rest in the block. The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic energy U.
How far (s) does the block slide before it comes to rest. Express your answer in terms of m,M,Vo,U and g.

2. Relevant equations
None given

3. The attempt at a solution
(m+M)Vf=mVo
thus
Vo=(m+M)/m*Vf
and Vf=root of (2gs)
So
(Vo*m/m+M)^2 = 2gs
s=(1/2g)(Vo*m/m+M)^2

Now my question is if this is correct. I know this is the equation used when the wood is on a pendulim. Where is the U that they give?

2. Mar 16, 2009

### LowlyPion

Is your u coefficient of kinetic energy not the coefficient of kinetic friction μk.

In which case your use of g in your equations should possibly be modified to reflect deceleration due to the force of kinetic friction?

F = m*a = μ*m*g

a = μ*g

3. Mar 16, 2009

### anelmarx

The U is definitely for kinetic energy. That's why I don't know where it is suppose to fit in.

4. Mar 16, 2009

### LowlyPion

Yes, but ... what is that?

From the law of the conservation of energy you can say that there is some energy that goes to friction related by some μ * Fn over the path of motion. That's energy robbed from the kinetic energy. But the μ is also commonly called μk - the coefficient of kinetic friction.

Energy to friction = Ef = μ*∫Fn(x) dx

It also relates to Force over that distance for a uniform Fn(x) over the distance as simply μ*Fn, or in your case with Fn = m*g, then ...

Ffr = μ*Fn = μ*m*g = m*a

As before then your deceleration is as stated a = μ*g