Glancing Elastic Collision Questionnaire

[Mysterio]

Homework Statement

In an elastic solution involving two masses A=0.5kg and B=0.3kg on a frictionless table. Mass A has an initial velocity of 4m/s in the positive x-direction and a final velocity of 2m/s in an unknown direction, while Mass B is initially at rest. Find the final speed of Mass B and the direction of masses after collision.

Relevant Equations

MAUi = M1V1Cosx + M2V2Cosy
0 = M1Vi1Sinx - M2V2Siny

I was able to solve for the velocity of MB and got my answer as 4.47m/s.
The main issue right now for me is how to get the angles. I'm really confused on what most people have been posting as we didn't get a groundwork on this topic and so most of the basics I had them self taught.
So far I stopped at:
(0.5*4) = (0.5*2*Cosx) + (0.3*4.47*Cosy)
(0.3*4.47)Siny = (O.5*2*Siny)

 

[vela]

Try squaring each equation and then adding them together. You should end up with an equation that will allow you to solve for $\cos(x+y)$.

 

[haruspex]

Try squaring each equation and then adding them together. You should end up with an equation that will allow you to solve for $\cos(x+y)$.

True, but that only finds x+y.
A more obvious approach, to me, is to get each equation so that on one side there is only the term involving, say, y. Now squaring and adding eliminates y.

 

[vela]

True, but that only finds x+y. A more obvious approach, to me, is to get each equation so that on one side there is only the term involving, say, y. Now squaring and adding eliminates y.

Yeah, that's a better approach.

Another approach is to work with the momentum vectors. Let $\vec p$ be the initial momentum of A, and $\vec p_A$ and $\vec p_B$ be the final momenta of A and B, respectively. Then $\vec p = \vec p_A + \vec p_B$. Squaring this equation was my approach above. But isolating $\vec p_B$ first and then squaring amounts to your suggestion.

 

[Steve4Physics]

Using the given data and conservation of kinetic energy (since it's an elastic collision), gives the final speed of B as 4.47m/s - agreed,

But for an elastic collision with a stationary object, I think the final speed of the initially stationary object cannot be greater than the impact speed. This would mean B’s final speed can’t more than 4m/s.

If so, there is a fault with the question Can someone correct me if this is wrong?

 

[haruspex]

the final speed of the initially stationary object cannot be greater than the impact speed.

Try a square on collision with the incoming object having the greater mass.

 

[Steve4Physics]

Try a square on collision with the incoming object having the greater mass.

Thank you @haruspex. I’m being silly. Too much food perhaps.

EDIT2. Got it. Apologies for diverting thread off-topic.