I was able to get the conservation of momentum equations for x and y (I think) t

Question

I was able to get the conservation of momentum equations for x and y (I think) then I got stuck.

(x) mv_A + mv_B = mv'_Acos(theta'_A)+ mv'_B
(y) mv_Asin(theta_A) + mv_B = mv'_Asin(theta'_A) + mv'_Bsin(theta'_B)

For x
A is moving in the y direction only before the collision so mv_A=0
B is moving in the y direction only after the collision so mv'_B=0
For y
Before the collision A is moving in the positive y direction and after the collision B is moving in the positive y direction so sin(theta_A)=0 and sin(theta'_B)=0
B is moving in the x direction only before the collision so mv_B=0

and m_A = m_B (=m)

so...
v_B = v'_Acos(theta'_A)
v_A = v'_Asin(theta'_A) + v'_B

This is where I got stuck because I have 3 unknowns: v'_A ; theta'_A ; v'_B . I can only figure out how to get rid of one of the variables through substitution.

Explanation / Answer

Think of this one as a conceptual question: If these balls are equal mass and there is a collision there will be a vector conservation of momentum. Rephrase the question as: "Ball A is going north at a speed of 'x'. Ball B is going east at a speed of 'y'. A and B have the same mass."

We are told that after the collision Ball B is now going north.

So, since they are the same mass, and since Ball B is now moving in the exact same direction that Ball A originally was, we can think of Ball B as absorbing all of the vector momentum that was originally in ball A. We know that ball B is now moving at A's original speed ('x').

The remaining system momentum (the original momentum in ball B) is now transferred in ball A. So ball A now has all of ball B's old momentum (that is, heading east at speed 'y').

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So, back to the question you asked, Ball A is now going east (at an angle of zero). Ball A has a speed of 3.7 and Ball B has a speed of 2.

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