Question

Jeanette is playing in a 9-ball pool tournament. She will win if she sinks the 9-ball from the final rack, so she needs to line up her shot precisely. Both the cue ball and the 9-ball have mass m and the cue ball is hit at an initial speed of v1 Jeanette carefully hits the cue ball into the 9-ball off center, so that when the balls collide, they move away from each other at the same angle theta from the direction in which the cue ball was originally traveling (see figure). Furthermore, after the collision, the cue ball moves away at speed vt, while the 9-ball moves at speed v9. (Figure 1) For the purposes of this problem, assume that the collision is perfectly elastic, neglect friction, and ignore the spinning of the balls. Find the angle theta through which the 9-ball travels away from the horizontal, as shown in the figure Perhaps surprisingly, you should be able to obtain an expression that is independent of any of the given variables! Express your answer in degrees to three significant figures.

Answer 1

45 degree

The original momentum of the cue ball has to be conserved. If you define the original direction of the cue ball as moving along the y-axis, then at the start there is zero momentum along the x-axis and all of it is along the y-axis. After the collision the vector sum of the momentums in the x-direction must still equal zero, meaning the x-component of the velocity of both balls must be equal in magnitude and opposite. Their y-components of velocity must add up to v_i. In addition you know that the angles Theta are equal, so if the x components are equal, Thetas are equal, then the y-components of the velocity must be also equal. You can prove this using v_f Sin(Theta) + v_9 Sin(Theta) = v_i and v_f Cos(Theta) = v_9 Cos(Theta) and conservation of energy (v_f)^2 + (v_9)^2 = (v_i)^2.

Answer 2

solution:

m is the mass of each ball
m*vi=m*(vf*cos?+v9cos?)
and
0=m*(vf*sin?-v9sin?)

and
.5*m*vi^2=.5*m*vf^2+.5*m*v9^2

solve for ?

vi=vf*cos?+v9cos?
0=vf*sin?-v9sin?
vf=v9

vi^2=2*vf^2
vi=2*vf*cos?

vi^2=4*vf^2*cos^2?

sqrt(.5)=cos?
?=45 degrees