Theoretical Expression for %loss in Kinetic Energy

1. Nov 1, 2009

crazuiee

1. The problem statement, all variables and given/known data
I'm trying to derive an equation which should look like this in the end:
%loss in Kinetic energy= M/m+M*100%, I'm just not sure how to substitute everything, and cancel things out to get the expression. Can someone show me the steps?

2. Relevant equations
mVi=(M+m)Vf
%loss in Kinetic Energy= 1/2mVi^2-1/2(M+m)Vf^2/ 1/2mVi^2 *100%

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 1, 2009

rl.bhat

Pi = mVi
Pi^2 = (mVi)^2 = 2m*(1/2*m*Vi^2) = 2mEi.
So Ei = Pi^2/2m
Similarly find Ef
Then percent change = (Ei - Ef)/Ei*100

3. Nov 1, 2009

crazuiee

I'm sorry i really don't see how that's going to work out to M/M+m

4. Nov 1, 2009

rl.bhat

Substitute the values of Ei and Ef. While simplification substitute the value of Vf in terms of Vi using the first equation in relevant equations.

5. Nov 1, 2009

crazuiee

yea...that means nothing to me sorry..

6. Nov 2, 2009

rl.bhat

(Ei - Ef)/Ei = [1/2*m*vi^2 - 1/2*(m+M)*vf^2]/1/2*m*vi^2.
Cancel 1/2m
= vi^2 - [(m+M)/m*vf^2]/vi^2
Put vi^2 = [(m + M)/m*vf]^2 and simplify.