Question

In a pool game, a cue ball traveling at 0.75 m/s hits the stationary eight ball. The eight ball moves off with a velocity of 0.25 m/s at an angle of 37 degrees relative to the cue ball's initial direction. Assuming that the collision is inelastic, at what angle will the cue ball be deflected, and what will be its speed?

Answer 1

whether the collision is elastic or inelastic >> momentum conservation holds true.
We can take the initial direction of cue (going to hit) as (+x) and let collision takes place at origin.
let the cue ball deflect at angle θ with (+x) axis, whereas 8-ball made 37 deg angle with (+x) direction.

we will resolve momentum in x, y directions
X> before = after
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its presumed that masses of 2 balls are same
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0.75m + 0 = m vc cos θ + m (0.25) cos 37
0.75 = vc cos θ + 0.25 cos 37
4 vc cos θ + cos 37 = 3
4 vc cos θ = 3 - cos 37 ---(1)
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Y> before = after
0 + 0 = vc sin θ + 0.25 sin 37
4 vc sin θ = 4 - sin 37 ---(2)
squaring & adding
16 vc^2 = (3 - cos 37)^2 + (4 - sin 37)^2
16 vc^2 = [9+cos^2 37 - 6 cos 37 + 16+sin^2 37 - 8 sin 37]
16 vc^2 = [25+1 - 6 cos 37- 8 sin 37]
16 vc^2 = [16.394]
vc = 1.012 m/s
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divide (2) by (1)
tan θ = (4 - sin 37)/ (3 - cos 37) = 3.398/ 2.201 = 1.5438
θ = tan^-1[1.5438] = 57 degree
cue will exit at (57 deg) angle from its initial direction with speed = 1.012 m/s