A hockey puck B rests on a smooth surface of ice and is struck by a second puck

Question

A hockey puck B rests on a smooth surface of ice and is struck by a second puck A , which was originally traveling at v = 40.0m/s and which is deflected ?1 = 33.0? from its original direction. (See the figure below (Figure 1) .) Puck Bacquires a velocity at a ?2 = 43.0? to the original direction of A . The pucks have the same mass
Compute the speed of puck A after the collision.
Compute the speed of puck B after the collision.
What fraction of the original kinetic energy of puck A
  dissipates during the collision
I get stuck doing the simultaneous equations.

Explanation / Answer

Momentum conservation in perpendicular to A initial velo
M. Va sin 33= M. Vb sin 43
Va = 1.25 Vb------------------------------------------------1

Momentum conservation in A direction
M.*40 = M. Va cos 33 + M. Vb cos 43------------------2

40 = 1.25 Vb * cos 33 + Vb * cos 43

Vb(0.7313 +1.048) = 40

Vb =22.47 m/s
Va = 28.1 m/s

Initial KE = 0.5 * m * 40^ 2= 800m
Final ke = 0.5mvf^2 +0.5 m vi^2 = 647.255 m
So fraction dissipation = 800-647.255/800 = 19.1 %

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