Question

9. A cue ball traveling at 0.65 m/s hits the stationary 8-ball, which moves off with a speed of 0.22 m/s at an angle of 40° relative to the cue ball's initial direction. Assuming that the balls have equal masses and the collision is inelastic, what will be the speed of the cue ball?

Answer 1

Given that the collision is inelastic.

Now, in inelastic collision also, momentum shall be conserved.

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 ? with (+x) axis, whereas 8-ball made 40 deg angle with (+x) direction.

we will resolve momentum in x, y directions
X> before = after

Also, given that the balls have equal masses.

So -

0.65m + 0 = m vc cos ? + m (0.22) cos 40

=> 0.65 = vc cos ? + 0.22 cos 40

=> vc cos ? + 0.17 = 0.65

=> vc cos ? = 0.48 ----------------------------------(i)

Again -

Y> before = after

0 + 0 = vc sin ? + 0.22 sin 40
=> vc sin ? = - 0.14 ---------------------------------------(ii)

squaring & adding

vc^2 = 0.48^2 + (-0.14)^2 = 0.25

=> vc = 0.50 m/s

Now, divide (ii) by (i) -

tan ? = -0.14 / 0.48 = -0.292

=> ? = tan^-1[-0.292] = -16.3 degree = 180 - 16.3 = 163.7 deg. from the positive direction of x - axis.