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Billiard Ball?
A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an angle of ?1 = 46.0° at a speed of v1 = 27.0 m/s. It bounces off at an angle of ?2 = 43.0° and a speed of v2 = 13.0 m/s. What is the magnitude of the change in momentum of the billiard ball?
I don’t know which equation to use. I tried using:
? = cos^-1 (p || / p)
(The || means parallel)
Thanks.
For some reason, that is the wrong answer. I don’t see how it is, because what you did made sense.
Change in momentum means final momentum – intial momentum.
You can’t directly add or subtract vectors, so they need to be resolved into x and y components that can be subtracted.
I’m making the assumption here that the ball hits a side rail and the angles are measured from the rail, so that the ball was going somewhat up and right before impact and somewhat up and left afterwards.
final x momentum = mvcos43 = .15 x 13 x cos43 = -1.43
initial x momentum = mvcos46 = .15 x 27 x cos46 = 2.81
change in x momentum = -4.24 (subtract above)
final y momentum = mvsin43 = .15 x 13 x sin43 = 1.33
initial y momentum = mvsin46 = .15 x 27 x sin46 = 2.91
change in y momentum = -1.58 (subtract above)
Finally the magnitude of the change will be the combined x and y changes using the pythagorean theorem. I dropped the negative signs since they’ll go away when you square the numbers anyway.
change = sqroot(4.24^2 + 1.58^2) = 4.5
