A billiard ball of mass m = 0.250 kg hits the cushion of a billiard table at an

Question

A billiard ball of mass m = 0.250 kg hits the cushion of a billiard table at an angle of ?1 = 58.0 A billiard ball of mass m = 0.250 kg hits the cushion of a billiard table at an angle of ?1 = 58.0 degree at a speed of v1 = 2.40 m/s. It bounces off at an angle of ?2 = 49.0 degree and a speed of v2 = 2.20 m/s. What is the magnitude of the change in momentum of the billiard ball? In which direction does the change of momentum vector point? (Take the x-axis along the cushion and specify your answer in degrees.)

Explanation / Answer

horizontal component of initial momentum is 0.25*2.4*cos(58 deg) = 0.3179 Kg.m/s
vertical component of initial momentum is 0.25*2.4*sin(58 deg) =0.5088 Kg.m/s

horizontal component of final momentum is 0.25*2.2*cos(49 deg)= 0.3608 Kg.m/s
vertical component of final momentum is 0.25*2.2*sin(49 deg)= 0.415 Kg.m/s

Change in momentum in the horizontal direction is 0.3608-0.3179 = 0.0429 Kg.m/s
Change in momentum in the vertical direction is 0.415-0.5088 = -0.0938 Kg.m/s

Resultant of the final momentums = sqrt((0.0429)^2 +(-0.0938)^2) = 0.1031 Kg.m/s
tan theta = vertical /horizontal = -0.0938 /0.0429
theta = -65.42 degrees

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