Question

A 164.7g cue ball (diameter = 58.1 mm) is struck with a cue stick over a period of 0.008 s with a force of 28 N. Assume no slipping or frictional forces, the cue ball elastically hits an identical billiard ball at rest. The cue ball deflects at 60degree and the target ball at an angle of 30°. a) The speed of the cue ball after it is struck by the cue stick is ____ m/s. b) The speed of the target ball after it is deflected from rest is ____ m/s c) The kinetic energy of the cue ball after collision is ____ J. d) The kinetic energy of the target ball after collision is ____ J.

Answer 1

here,

mass of ball = 164.7g = 0.1647 Kg

After Collision
v1'x = v1'Cos60
v1'y = v1'Sin60

v2'x = v2'Cos30
v2'y = v2'Sin30

A)
Force * time = mass * V1
V1 = F*t/m
v1 = 28*0.008/0.1647
v1 = 1.36 m/s

B)
From Conservation of momentum :
for X direction :
m1v1x = m1v1'x + m2v2'x

as m1 = m2 = m
v1x = v1'Cos60 + v2'Cos30
V2' = (v1x - v1'Cos60)/Cos30 ------------(1)

similarly for y direction
m1v1y = m1v1'y - m2v2'y
v1'y = v2'y
v1'Sin60 = v2'Sin30

v2' = v1'Sin60/Sin30 -------------------(2)

Therefore Eqn 1 becomes :

v1'*(Sin60/Sin30) = (1.36 - v1'Cos60)/Cos30
v1'*(Sin60*Cos30/Sin30) = (1.36 - v1'Cos60)
v1'*(0.86*0.86/0.5) = (1.36 - v1'0.86)
v1' = 0.581 m/s ------------------------------(3)

Therefore, eqn 1 become :
V2' = (1.36 - 0.581*0.86)/0.5
V2' = 1.720 m/s------------------------------(4)

C)

KE1 = 0.5* m * V1'^2
KE1 = 0.5 * 0.1647 *(0.581)^2
KE1 = 0.0277 J

D)

KE2 = 0.5* m * V2'^2
KE2 = 0.5 * 0.1647 *(1.720)^2
KE2 = 0.243 J