How to solve an Elastic collision
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How to solve an Elastic collision

[From: ] [author: ] [Date: 12-11-12] [Hit: ]
but we can solve this by using (3) (energy conservation). We have,......
VB^2 - VA^2 sin^2(theta) = VA0^2 - 2 VA0 VA cos(theta) + VA^2 cos^2(theta).

Now we move VA^2 sin^2(theta) to the right hand side
VB = VA0^2 - 2 VA0 VA cos(theta) + VA^2 cos^2(theta) + VA^2 sin^2(theta) .

The trig identity
cos^2(phi) + sin^2(phi) = 1

is valid for any choice of phi. We can also say
cos^2(theta) + sin^2(theta) = 1

So we have
VB = VA0^2 - 2 VA0 VA cos(theta) + VA^2 .

This still involves VA, but we can solve this by using (3) (energy conservation). We have, from (3):
VA = sqrt(VA0^2 + VB^2)

So substituting this into the preceeding equation gives
VB^2 = VA0^2 - 2 VA0 sqrt(VA0^2 + VB^2) cos(theta) + VA0^2 + VB^2

Cancelling and collecting terms:
0 = 2VA0^2 - 2VA0 sqrt(VA0^2 + VB^2) cos(theta)

Rearranging:
sqrt(VA0^2 + VB^2) cos(theta) = VA0

Squaring:
(VA0^2 + VB^2)cos^2(theta) = VA0^2

So
VB^2 cos^2(theta) = VA0^2(1-cos^2(theta) ) = VA0^2 sin^2(theta)
[from trig identity]

Hence
VB = VA0 tan(theta)
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