On a frictionless horizontal air table, puck A (with mass 0.253 kg ) is moving toward puck B (with mass 0.366 kg ), which is initially at rest. After the collision, puck A has velocity 0.118 m/s to the left, and puck B has velocity 0.649 m/s to the right.
2 Answers

For A, initial x momentum = 0.03 * 7.79 = 0.2337
Final x momentum = 0.03 * vA * cos 65
Final y momentum = 0.03 * vA * sin 65
For B, final x momentum = 0.06 * vB * cos 37
Final y momentum = 0.06 * vB * sin 37
0.03 * vA * cos 65 + 0.06 * vB * cos 37 = 0.2337
0.03 * vA * sin 65 + 0.06 * vB * sin 37 = 0
0.03 * vA * sin 65 = 0.06 * vB * sin 37
vA = 2 * vB * sin 65 ÷ sin 37
0.03 * (2 * vB * sin 65 ÷ sin 37) * cos 65 + 0.06 * vB * cos 37 = 0.2337
vB * (0.6 * sin 65 ÷ sin 37 cos 65 + 0.06 * cos 37) = 0.2337
vB = 0.2337÷ [(0.6 * sin 65 ÷ sin 37 * cos 65 + 0.06 * cos 37)]
vB = 0.2337 ÷ 0.42948519
This is approximately 0.544 m/s.
vA = 2 * (0.2337 ÷ 0.42948519) * sin 65 ÷ sin 37
This is approximately 1.63 m/s.
Check:
0.03 * 1.63* cos 65 + 0.06 * 0.544 * cos 37 ≈ 0.2327
Since I rounded the velocities, I believe that this proves that they are correct.

Let v = velocity of A before collision to the right
Momentum before collision = 0.253v
Momentum after collision = (0.649)(0.366) – (0.253)(0.118)
= 0.20768
Therefore 0.253v = 0.20768
v = 0.20768/0.253
= 0.8208695652
= 0. 821 m/s
KE of A before collision = 1/2(0.253)(0.821)^2
= 0.0853 J
KE of A after collision = 1/2(0.253)(0.118)^2
= 0.00176 J
KE of B after collision = 1/2(0.366)(0.649)^2
= 0.0771 J
Total KE after collision = 0.0771 + 0.00176
= 0.07886
ΔK = 0.0853 – 0.07886
= 6.44 x 10^3 J ….[ = 0.00644 J]