Hockey puck momentum/ collision

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A 7 kg hockey puck moving at 3 m/s collides with a stationary identical puck on frictionless ice, resulting in two pucks moving at angles of 41.1647 degrees and 48.8353 degrees. The momentum conservation equation is applied, considering both x and y components of the collision. The initial calculations for the speed of the first puck after the collision yielded an incorrect result due to a calculator error. After identifying the mistake, the correct speed was determined to be 1.844 m/s. The discussion emphasizes the importance of accuracy in calculations during physics problems.
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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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