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## Homework Statement

http://img403.imageshack.us/img403/1751/problem1r.jpg [Broken]

## Homework Equations

[itex]v_f = \frac{m v_1 + M v_2}{m+M}[/itex]

## The Attempt at a Solution

Let "A" be the configuration immediately before the collision and "B" the configuration immediately after the collision, and "C" the very final state.

I have used the above equation and solved for the velocity when the bullet enters the system (immidiently after the collision):

[itex]v_B = \frac{m v_{1A}}{m+M}[/itex]

The expression for the total kinetic energy of the system right after the collision is

[itex]K_B = 1/2 (m + M) v_B^2[/itex]

Substituting v

_{B}we get

[itex]K_B = \frac{m^2v_{1A}^2}{2(m+M)}[/itex]

We then apply the conservation of mechanical energy principal to the system to get:

[itex]K_B+U_B = K_C + U_C[/itex]

[itex]\frac{m^2v_{1A}^2}{2(m+M)} + 0 = 0+ (m+M) gh[/itex]

[itex]v_{1A} = \left( \frac{M+m}{m} \right) \sqrt{2gh}[/itex]

This is very close to the answer, but how can I bring

**μ**(coefficient of friction) into my equation? The potential energy is given by mgh, so how do I write it in terms of μ

_{k}_{k}and d?

Any help is appreciated.

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