Question

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.​

Answer 1

The speed of the second ball after impact is 1.43 m/s at 60.8⁰.

Speed of the second ball

Apply the principle of conservation of linear momentum;

in y direction

0.6(2) = 0.6(1.5 sin30) + 0.6(vy)

1.2 = 0.45 + 0.6vy

0.6vy = 0.75

vy = 0.75/0.6

vy = 1.25 m/s

in x direction

0.6(2) = 0.6(1.5 cos30) + 0.6(vx)

1.2 = 0.78 + 0.6vx

0.6vx = 0.42

vx = 0.42/0.6

vx = 0.7 m/s

Resultant speed

v = √(vy² + vx²)

v = √(1.25² + 0.7²)

v = 1.43 m/s

Direction of second mass

θ = arc tan(vy/vx)

θ = arc tan (1.25/0.7)

θ = 60.8⁰

Thus, the speed of the second ball after impact is 1.43 m/s at 60.8⁰.

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