Two hockey pucks of equal mass are involved in a perfectly elastic, glancing collision, as shown in the figure below. The orange puck is initially moving to the right at
voi = 3.95 m/s, when it strikes the initially stationary blue puck, and moves off in a direction that makes an angle of
θ = 37.0° with the horizontal axis while the blue puck makes an angle of
ϕ = 53.0° with this axis. Note that for an elastic collision of two equal masses, the separation angle θ + ϕ = 90.0°. Determine the speed of each puck after the collision.
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1 answer

first, note momentum in the perpendicular direction is zero.
So M*v1*sin37=M*v2sin53
that will give you a ratio for v1/v2
now momentum in the horizontal direction...
M*3.95=Mcos37*V1 + M cos53*V2
from the ratio you found earlier, you can solve that for V1, then V2
finally, energy: You have one more equation, not that I see you need it except to verify it was an elastic collision
1/2 m *3.95^2=1/2 m v1^2 + 1/2 m v2^2 👍
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answered by bobpursley
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