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4 June, 17:33

A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is deflected 38.00° to the right and moves off at 0.3400 m/s. Find the speed and direction of the second puck after the collision.

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  1. 4 June, 19:12
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    Speed = 0.283m / s

    Direction = 47.86°

    Explanation:

    Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

    MU1 = MU2cos38 + MV2cos y ... x plane

    0 = MU2sin38 - MV2sin y ... y plane

    Where M = mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 = 0.34, V2 = final velocity of puck 2, y = angular direction of puck2

    Substitute into equation above

    .46 =.34cos38 + V2cos y ... equ1

    .34sin38 = V2sin y ... equ2

    .19=V2cos Y ... x

    .21=V2sin Y ... y

    From x

    V2 = 0.19/cost

    Sub V2 into y

    0.21 = 0.19 (Sin y/cos y)

    1.1052 = tan y

    y = 47.86°

    Sub Y in to x plane equ

    .19 = V2 cos 47.86°

    V2=0.283m/s
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