A billiard ball of mass m = 0.250 kg hits the cushion of a billiard table at an angle of θ1 = 58.8° and a speed of v1 = 28.4 m/s. It bounces off at an angle of θ2 = 71.0° and a speed of v2 = 10.0 m/s.

Question

(a) What is the magnitude of the change in momentum of the billiard ball?
(b) In which direction does the change-of-momentum vector point? (Let up be the +y positive direction and to the right be the +x positive direction.)

Answer 1

a.m2*V2-m1*V1 = 0.250*10[71o]-0.250*28.4[58.8o] = 2.50[71o] - 7.1[58.8o]

X = 2.50*Cos71 - 7.1*Cos58.8 = -2.86
Y = 2.50*sin71 - 7.1*sin58.8 = -3.71

M^2=X^2 + Y^2 = (-2.86)^2 + (-3.71)^2 = 21.9

M = 4.68 = Magnitude of change in momentum.

b. Tan Ar = Y/X = -3.71/-2.86 = 1.29720
Ar = 52.4o = Reference angle.

A = 52.4 + 180 = 232.4o = Direction.