Question

The figure gives an overhead view of the path taken by a 0.163 kg cue ball as it bounces from a rail of a pool table. The ball's initial speed is 1.82 m/s, and the angle θ₁ is 58.2°. The bounce reverses the y component of the ball's velocity but does not alter the x component. What are (a) angle θ₂ and (b) the magnitude of the change in the ball's linear momentum? (The fact that the ball rolls is irrelevant to the problem.)

Answer 1

rail po V2 Given that, a cree ball bounces from the core of a table. m-0.163kg Speed of ball, Vo = 1.82 m/s Vi 02 0.- 58.2° Before bouncing, x-component of velocity = Visina, = 1.82 sin58120 = 1.55 m/s Y-component of velocity = V. Coso, -1.82 COS 58.2° = 0.96 m/s Also given that, after bouncing of the ball, the x-component remains same but the Y- component gets reversed. So after bouncing, ne a-component of veloutyova 1.55 m/s Y-component of velocitytag - 0.96m/s Now, velocity of ball after bouncing, 2 : [ Van jaf (Way) 2 + 2 van. Vay cos ao - V2 = 1.552 (+0.96) 2 V2 2 V 2. 40a5 +0.9216 » V2 = 13.3241 & V2 = 1.82 m/s

(a) Now, according to the of velocity after bouncing figure, the x-component is = V2 Sinda But we have x-component = 1.55m/s. so Vasinda = 1.55 m/s → Sinoa 1.55 * Da = 58.49 (5) Initial momentum of ball, P = my, cosa = 0.163 x 0.96 = 0.156 kgm/s final momentum of ball, P2 = ml-racosoa) = 0.163 xl-4.82 COS 58.4°) = -0. 155 kg m/s change in momentum = -0.155 -0.156 = – 0.311 kgm/s Magnitude of change in momentum is 0.311 kgm/s.