Hockey puck momentum/ collision

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SUMMARY

The discussion centers on a physics problem involving two identical hockey pucks, each with a mass of 7 kg. The first puck, moving at 3 m/s, collides with a stationary puck on frictionless ice. After the collision, the first puck moves at an angle of 41.1647 degrees with a speed of 1.844 m/s, while the second puck moves at an angle of 48.8353 degrees. The participant initially miscalculated the speed due to a calculator error but ultimately resolved the issue.

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Homework Statement


A puck of mass 7 kg moving at 3 m/s strikes an identical puck that is stationery on frictionless ice. After the collision, the first puck leaves with a speed of v1 at an angle of 41.1647 degrees with respect to its original line of motion and the second puck leaves with a speed of v2 at 48.8353 degrees.


Homework Equations


MaVa + MbVb = MaVa' + MbVb'

The Attempt at a Solution


Puck along x-axis:
MVa = MVa' * Cos 41.1647 + MVb' * Cos -48.8353

y-axis:
0 = MVa' * Sin 41.1647 + MVb' * Sin -48.8353


I solved for Va' and got 1.844 (rounded to 3 decimal places), but that answer is wrong. What did I do wrong?
 
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Figured it out. Calculator error.
 

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