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 37.00° to the right and moves off at 0.3700 m/s. Find the speed and direction of the second puck after the collision.

Answer 1

.342m/s at 43 degrees to the right

Answer 2

initial x momentum = .46 m
initial y momentum = 0

final x momentum = .37 m cos 37 + v m cos T
final y momentum = .37 m sin 37 - v m sin T

so
.46 = .37 cos37 + v cos T
and
.37 sin 37 = v sin T

Answer 3

Given:
M1 = M kg, V1 = 0.46 m/s.
M2 = M kg, V2 = 0.
V3 = 0.37m/s[37o] = Velocity of M1 after collision.
V4 = ? = Velocity of M2 after collision.

Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + + M2*V4.
M*0.46 + M*0 = M*0.37[37] + M*V4,
Divide both sides by M:
0.46 + 0 = 0.37[37o] + V4,
0.46 = 0.295+0.223i + V4,
V4 = 0.165 -.223i = 0.277[-53.5] = 53.5o S. of E. = 0.277m/s[306.5o] CCW.

Answer 4

Ahockey