Question

A hockey puck B rests on frictionless, level ice and is struck by a second puck A, which was
originally traveling at 40.0 m/s and which is deflected 30.0° from its original direction. Puck B
acquires a velocity at a 45.0° angle to the original direction of A. The pucks have the same mass.
Determine the speed of each puck after the collision.

I'm kind of stuck on this question. I know that its an elastic collision, and that B is initially at rest. However, I have no idea what to do from here.

Answer 1

call the original direction the x-direction

all of the original momentum is in the x-direction ... carried by A

after the collision, there are momenta in the ± y-direction
... they are equal and opposite (cancel each other)

the sum of the two x-momenta is equal to the original momentum

40.0 m/s = vA cos(30º) + vB cos(45º)

vA sin(30º) = vB sin(45º)

Answer 2

Oh okay, I think you helped me figure it out.
Since the y-momenta cancel, piy = pif.
initially, piy = 0
pif = vA sin 30 + vB sin 45
0 = vA sin 30 + vB sin 45
vA = vB sin45 / sin 30
substitute that into your equation to find VB and then solve the rest
thanks alot!