Asked by hope
a 600 g pool ball moving at 2.0m/s hits a second pool of the same mass at rest. the first ball moves off at 1.5m/s at an angle of 30 degrees from the original direction. assume the surface of the pool table to b frictionless find the angle at which the second ball moves off and the speed of the second ball after impact.
Answers
Answered by
Damon
total y momentum = 0
m * 1.5 sin 30 = m v sin T
so
v sin T = 0.75
total x momentum = 2 m
2 m = 1.5 m cos 30 + m v cos T
v cos T = .7
v sin T/v cosT = tan T = .75/.7
T = 47 degrees
v = .7/cos 47
m * 1.5 sin 30 = m v sin T
so
v sin T = 0.75
total x momentum = 2 m
2 m = 1.5 m cos 30 + m v cos T
v cos T = .7
v sin T/v cosT = tan T = .75/.7
T = 47 degrees
v = .7/cos 47
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