In a pool game the cue ball A must strike the eight all in t

In a pool game the cue ball A must strike the eight all in the position shown in order to send it to the pocket P with a velocity v2. The cue ball has a ve- locity v1 before impact and a velocity after im locity Di before impact and a velocity v,\' after im- pact. The coefficient of restitution is 0.9. Both balls have the same mass and diameter. Calculate the re- bound angle and the fraction n of the kinetic en- ergy which is lost during the impact. ergy which 1is lost during the impact

Solution

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Let the velocity of the 8 ball is v m/s and the velocity of the cue ball is u m/s at A* angle from the initial direction of the cue ball after the collision:-
By the law of momentum conservation:-
(i) In the initial direction(x-axis):-
=>m x 5 = m x v x cos30* + m x u x cosA*
=>5 = 0.87v + ucosA*
=>ucosA* = 5 - 0.87v ------------(i)
(ii) In the other direction(y-axis):-
=>vsin30* = usinA*
=>usinA* = 0.5v ------------------(ii)
By the law of energy conservation:- [as the collision is elastic]
=>1/2 x m x (5)^2 = 1/2 x m x u^2 + 1/2 x m x v^2
=>25 = u^2 + v^2 --------------------(iii)
By (i)^2 + (ii)^2:-
=>u^2 = [ 5 - 0.87v ]^2 + (0.5v)^2 {by putting the value of u^2 from (iii)}
=>25 - v^2 = 25 + 0.75v^2 -8.7v + 0.25v^2
=>2v^2 -8.7v = 0
=>v(2v-8.7) = 0
=>v = 4.33 m/s [as v=/=0 ] [answer for B]
By putting the value in (iii)
=>u = sqrt[6.25]
=>u = 2.5 m/s [answer for B]
By putting the value of u & v in (ii):-
=>2.5sinA* = 4.33/2
=>sinA* = 0.866 = sin60*
=>A* is 60* {opposite side of 30*} [answer for A]