Question

A billiard ball of mass m = 0.25 kg strikes the cushion of a billiard table at θ₁ = 46° and a speed v₁ = 25 m/s. It bounces off at an angle of θ₂ = 67⁰ and a velocity of v₂ = 17 m/s. What is the magnitude of its change in momentum (in kg·m/s)?

Answer 1

Solution)

The horizontal component of initial momentum is 0.25*25 cos 46=4.34 Kg m/s
The vertical component of initial momentum is 0.25*25 sin46 = 4.49 Kgm/s

The horizontal component of final momentum is 0.25*17 cos 67= 1.66 Kg.m/s
The vertical component of final momentum is 0.25*17 sin 67= 3.9 Kgm/s

Change in momentum in the horizontal direction is 1.66-4.34 = -2.68 Kgm/s
Change in momentum in the vertical direction is 3.9-4.49 = -1 Kgm/s

Final momentum = √(-2.68)² + (-1) ²= 2.86 Kgm/s (Ans)

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