Question

A billiard ball of mass m = 0.25 kg strikes the cushion of a
billiard table at θ_{1} = 46° and a speed v_{1} =
25 m/s. It bounces off at an angle of θ_{2} =
67^{0} and a velocity of v_{2} = 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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Good luck!:)

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