Hi PhysicsForums. :P First time here. I would just like someone to verify whether I did this problem correctly.(adsbygoogle = window.adsbygoogle || []).push({});

A moving shuffleboard puck has a glancing collision with a stationary puck of the same mass. If friction is negligible, what are the speeds of the pucks after the collision?

Given variables: v1i = 0.95 m/s, after collision, m1 (puck 1) moves at a 50 degree angle to the horizontal, positive direction in the y (assuming a regular coordinate grid), while m2 (puck 2) moves at a 40 degree angle toe the horizontal, negative direction in the y.

Attempt:

Conservation of Momentum (X direction)

Pi = Pf

m1v1ix + m2v2ix = m1v1fx + m2v2fx

v1ix = v2fx-v1fx

0.95 = v2 (cos 40) - v1 (cos 50)

Conservation of Momentum (Y direction)

m1v1iy + m2v2iy = m1v1fy + m2v2fy

0 = v1fy + v2fy

-v1 (sin 50) = v2 (sin 40)

v1 is a vector, so since its moving in a positive direction (according to my grid), it's still positive anyway...

v2 = v1(sin50)/sin40

substitute this into the equation I got for C.o.M. in the X

0.95 m/s = v1 (cos50) + v1(sin50)(cos40)/sin(40)

0.95 m/s = v1 (cos 50 + sin50(cos 40)/sin(40)

v1 = 0.95m/s / (cos 50 + sin50cos40/sin40)

v1 = .6106482 m/s

substitute this into the equation I had for C.o.M. in the Y

v2 = .6106482 m/s (sin 50)/sin40

v2 = .7277422 m/s

Hopefully I got this right? >_>

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# Homework Help: Conservation of Momentum - Shuffleboard

Can you offer guidance or do you also need help?

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