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.

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