Question

A cue ball initially moving at 2.6 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball’s final speed is 1.3 m/s.Find the cue ball’s angle θ with respect to its original line of motion. Consider this to be an elastic collision (ignoring friction and rotational motion). Answer in units of ◦

Answer 1

initial x momentum = 2.6 m
initial y momentum = 0

final x momentum = 1.3 m cos T + V m cos A
final y momentum = 1.3 m sin T + V m sin A

T is theta, the cue ball angle from x and A is the 8 ball angle from x

now right off we know that the initial y momentum was zero so the final y momentum is zero
so
1.3 sin T = - V sin A

now what about conservation of energy to get V, the 8 ball speed
(1/2)m(2.6^2)=(1/2)m (1.3^2)+(1/2)m V^2
so
V^2 = 2.6^2 - 1.3^2
V = 2.25 m/s
so now we know
1.3 sin T = - 2.25 sin A

now the x momentum before and after
2.6 = 1.3 cos T + 2.25 cos A

well, we have two equations and two unknown but we would prefer to have all sines or all cosines so square that equation with the sines
1.69 sin^2 T = 5.06 sin^2 A remember A goes - y if T goes + y because we just lost the minus sign :)

so
1.69 (1 - cos^2 T) = 5.06(1 - cos^2 A)

5.06 cos^2 A = 3.37 + 1.69 cos^2 T
cos^2 A = .666 + .334 cos^2T
cos A = (.666 + .334 cos^2T)^.5
the x momentum equation was
2.6 = 1.3 cos T + 2.25 cos A
so
2.6=1.3cosT+2.25*(.666 +.334 cos^2T)^.5

(.666 +.334 cos^2T)^.5 =(2.6-1.3cosT)/2.25
so
.666+.334 cos^2T =(1/5.06)(6.76-6.76cosT +1.69 cos^2 T)
let cos T = p
.666 + .334 p^2 = (1/5.06)(6.76-6.76p +1.69 p^2

.666 +.334 p^2 = 1.34 -1.34 p +.334 p^2
-.674 = -1.34 p
p = .503 = cos T
so
T = 59.8 degrees