How to solve an Elastic collision

For example for a elastic collision I have mass A that collides with mass B (initially at rest) with an given velocity. They both deflect by an angle from their original direction, theta (mass A) and phi (mass B). So only the initial velocity is given for mass A and only theta is given. If i have to find the final velocity of mass B how do I do this without knowing the final velocity of mass A and phi?


I am having a really hard time trying to solve this problem, so far I have these equations written down

MA = MB
VA0= initial velocity
VA = final velocity
VB = final velocity

Momentum x:

VA0 = VAcos(theta) + VBcos(phi)

Momentum y:

0 = VAsin(theta)-VBsin(phi)
VAsin(theta) = VBsin(phi)

Energy :

Ei = Ef
VA0^2 = VA^2 + VB^2


I tried to square the two momentum equations and adding them so that i can get rid of phi but i still ended up with VA, so i am not sure what to do, any help would be greatly appreciated. Thank you in advance.

Your momentum equations and energy equation are right. I'll copy and label them:

Mom x:
VA0 = VAcos(theta) + VBcos(phi) ------------ (1)

Mom y:
VAsin(theta) = VBsin(phi) --------------------------(2)

Energy:
VA0^2 = VA^2 + VB^2 -------------------------------(3)

====================================

We want to try to isolate the term involving VB (the term we would like to solve for).

From (1), we have
VB cos(phi) = VA0 - VA cos(theta)

Squaring both sides, we have
VB^2 cos^2(phi) =VA0^2 - 2 VA0 VA cos(theta) + VA^2 cos^2(theta)

The only other place where phi appears is in (2). This is as sin(phi). Fortunately, we can use an identity:

cos^2(phi) + sin^2(phi) = 1

So that we have, from (1) and the above working and identity:
VB^2 - VB^2 sin^2(phi) =VA0^2 - 2 VA0 VA cos(theta) + VA^2 cos^2(theta)

We now substitute for VB sin(phi) from (2) into this equation to get