- #1
Carlitos
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- 0
Homework Statement
Two blocks collide on a perfectly elastic collision, after sliding on a frictionless surface.
va = 1m/s
vb = 1m/s
ma = 0.05kg
mb = 0.03kb
I need to find the speed of both blocks after the crash.
Homework Equations
Conservation of momentum: mava+mbvb = mava' + mbvb'
Conservation of energy: 1/2mava^2 + 1/2mbvb^2 = 1/2mava'^2 + 1/2mbvb'^2
General solution for these two equations: -(vb−va)=vb'-va'
Refer to this thread for cool formatting of these equations (i don't really know how to do that):
https://www.physicsforums.com/threads/elastic-collision.753780/
I don't really understand where that last equation comes from or how to get there.
The Attempt at a Solution
I solved for the conservation of momentum:
mava+mbvb = mava' + mbvb'
0.05kg.1m/s + 0.03kg.1m/s = 0.05kg.va' + 0.03kg.vb'
0.08ms = 0.05va' + 0.03vb' (*)
Then on the third equation:
-(vb−va)=vb'-va'
-(-1m/s - 1m/s) = vb'-va'
2m/s = vb'-va'
2m/s +va' = vb'
Then i plug this on (*) and i get
0.08m/s = 0.05va' + 0.03(2m/s + va')
0.08m/s = 0.05va' + 0.06m/s + 0.03va'
0.02m/s = 0.08va'
0.25m/s = va'
then vb' = 2.25m/s
The expected solution is 0.5m/s and 2m/s .
So my two questions are, am i doing something wrong here? And how do i get to the third equation?