Question

Tom is playing billiards on a table set up as shown. He wants to hit the red ball into a corner pocket at an angle of 0 = 60 deg with the Y-axis. Both the red ball and the cue ball have mass m=0.16 kg and e = 0.9. Y e Iva 13. Where should he aim the cue ball? 14. First Tom hits the cue ball with the stick for a time period of 0.1 s, sending it rolling at 11 m/s along the Y-axis. 17.6 N is the force with which he hit the ball. 15. The cue ball hits the red ball, sending it toward the pocket. Mass and momentum is conserved during the collision. 16. What is the velocity vc2 of the cue ball after collision?

Answer 1

13. If the radius of the balls are r, then at the time of collision the distance between the center of masses of the balls is 2r. So if the x coordinate of the red ball is X, then Tom has to hit the cue ball from a point where the x coordinate of the center of mass of the cue ball is (X+
2r sin
) that is (X+ $r\sqrt{3}$ ) as $\sin 60=\sqrt{3}/2$ irrespective of y coordinate.[x coordinate increases from left to right]

14. We have to apply conservation of linear momentum in both x and y direction.The initial velocity and so is the initial momentum of the red ball is 0. Assuming there is no friction between the table and the balls, the cue ball hits the red ball at a velocity 11 m/s along the y axis. So from the conservation of initial momentum of the system along y axis, we get-

$m(V_{r2}\cos \theta +V_{c2}\cos \beta )=m\times 11$

  $\Rightarrow \frac{V_{r2}}{2}+V_{c2}\cos \beta =11$..................... (1)

From the conservation of initial momentum of the system along x axis, we get-

  $mV_{r2}\sin \theta =mV_{c2}\sin \beta$

  $\Rightarrow \frac{\sqrt{3}}{2}V_{r2}\sin \theta =V_{c2}\sin \beta$................. (2)

The coefficient of restitution is given by e=0.9. So -

$\sqrt{\frac{\frac{1}{2}mV_{r2}^{2}+\frac{1}{2}mV_{c2}^{2}}{\frac{1}{2}m\times (11)^{2}}}=0.9$

$\Rightarrow V_{r2}^{2}+V_{c2}^{2}=9.9^{2}$    ................... (3)

From (1) and (2), we can get the equation below through some simple algebraic steps -

  $V_{c2}^{2}-\frac{3}{4}V_{r2}^{2}=(11-\frac{V_{r2}}{2})^{2}$...................... (4)

Again from (3) and (4) we get -

  $2V_{r2}^{2}-11V_{r2}+22.99=0$

Upon solving the equation we get $V_{r2}$ as an imaginary number. Through a little algebra we can conclude that with hitting the cue ball with initial velocity V=11 m/s the red ball will never go into the pocket. The condition for the red ball to go into the pocket is that the initial velocity of the cue ball must be less than or equal to 10.58 m/s.

However, if we assume the initial velocity of the cue ball to be 10.58 m/s, then $\mathbf{V_{r2}=2.65\; m/s}$ and $\mathbf{V_{c2}=9.15\; m/s.}$