PDA

View Full Version : Theoretical Expression for %loss in Kinetic Energy


crazuiee
Nov1-09, 11:54 PM
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

rl.bhat
Nov2-09, 12:08 AM
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

crazuiee
Nov2-09, 12:23 AM
I'm sorry i really don't see how that's going to work out to M/M+m

rl.bhat
Nov2-09, 12:44 AM
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.

crazuiee
Nov2-09, 12:53 AM
yea...that means nothing to me sorry..

rl.bhat
Nov2-09, 01:45 AM
(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.