Question

A billiard ball of mass m = 0.25 kg strikes the cushion of a billiard table at θ θ 1 = 54° and a speed v1 = 22 m/s. It bounces off at an angle of θ θ 2 = 590 and a velocity of v2 = 15 m/s. What is the magnitude of its change in momentum (in kg·m/s)?

Answer 1

The horizontal component of initial momentum   is 0.25*22 cos 54=3.233Kg.m/s

The vertical component of initial momentum   is 0.25*22 sin 54 = 4.450 Kg.m/s

The horizontal component of final momentum   is 0.25*15 cos 59= 1.931 Kg.m/s

The vertical component of final momentum   is 0.25*15 sin59= 3.214 Kg.m/s

Change in momentum in the horizontal direction is 1.931 -3.233 = -1.302 Kg.m/s

Change in momentum in the vertical direction is 3.214 – 4.450 = -1.236 Kg.m/s

Thus, Resultant of the final momentums = √ {(-1.302)² + (-1.236) ²} = 1.80 Kg.m/s

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