Hockey puck momentum, conservation of linear momentum

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SUMMARY

The discussion focuses on the conservation of linear momentum in a collision between two hockey pucks on an air-hockey table. Puck A, with a mass of 0.0480 kg and an initial velocity of +7.76 m/s, collides with puck B, which has a mass of 0.0960 kg and is initially at rest. The calculations provided indicate that after the collision, puck A has a speed of 4.77 m/s, while puck B has a speed of 3.595 m/s. These results are derived using momentum conservation equations for both the x and y directions.

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The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0480 kg and is moving along the x-axis with a velocity of +7.76 m/s. It makes a collision with puck B, which has a mass of 0.0960 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the speed of (a) puck A and (b) puck B.

There are no answers provided. Could someone please tell me if my answers are correct?

Momentum in the y direction
Piy = Pfy
0 = MaVasin65 + -MbVbsin37
MaVasin65 = MbVbsin37
Va = MbVbsin37 / Ma*sin65
Va = (0.096sin37Vb) / (0.048sin65)

Pix = Pfx
0.048(7.76) = 0.048cos65Va + 0.096cos37Vb

Plug Va in and you get Vb = 3.595 m/s and thus Va = 4.77 m/s ?
 
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