Momentum and collisions at an angle?

Question

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.039 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.059 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing.

Find the final speed of puck A.


Find the final speed of puck B.

Here is the drawing

http://www.webassign.net/CJ/07_32.gif

Answer 1

I'm using the notation 1 for A and 2 for B.
Total momentum P = m1v1 + m2v2 = const. = 5.5*m1 along +x axis
After collision,
m1v1cos65 = P-m2v2cos(-37) (eq. 1)
m1v1sin65 = -m2v2sin(-37) (eq. 2)
Dividing eq. 1 by eq. 2,
1/tan65 = -P/m2v2sin(-37) + 1/tan(-37)
P/m2v2sin(-37) = 1/tan(-37) - 1/tan65
Inverting,
m2v2sin(-37)/P = 1/(1/tan(-37) - 1/tan65)
v2 = P/((1/tan(-37) - 1/tan65)*m2sin(-37))
The solution is
v1 = 3.383930 m/s, v2 = 3.368578 m/s
This checks out with m1v1cos65 + m2v2cos(-37) = P, and
m1v1sin65 + m2v2sin(-37) = 0.