Question

A billiard ball moving at 4.0 m/s strikes a pair of billiard ball at rest. One of the ball moves off at 2.0 m/s in a direction 60 degrees from the line of motion of the first ball and the other moves off at 3.0 m/s in a direction 30 degrees from the line of the first ball on the other side. What is the velocity of the first ball after the collision?

Answer 1

M1*V1 + M2*V2 + M3*V3 = M1*V + M2*2[60o]
+ M3*3[30o]

M1 = M2 = M3?
M1*V1 + M1*V2 + M1*V3 = M1*V + M1*2[60o]
+ M1*3[30o]

M1*4 + M1*0 + M1*0 = M1*V + M1*2[60] +
M1*3[30]
Divide both sides by M1:
4 = V + 2[60] + 3[30]
4 = V + 2*Cos60 + 2*sin60 + 3*Cos30 + 3*sin30
4 = V + 1 + 1.73i + 2.60 + 1.5i
4 = V + 3.60 + 3.23i
-V = - 0.4 + 3.23i
V = 0.4 - 3.23i
V^2 = 0.4^2 + 3.23^2 = 10.59
V = 3.25 m/s.