Question

Jeanette is playing in a 9-ball pool tournament. She will win ifshe sinks the 9-ball from the final rack, so she needs to line upher shot precisely. Both the cue ball and the 9-ball have mass m,and the cue ball is hit at an initial speed of v_i. Jeanette carefully hits the cue ball into the9-ball off center, so that when the balls collide, they move awayfrom each other at the same angle θ from the direction in which the cue ball was originallytraveling (see figure). Furthermore, after the collision, the cueball moves away at speed v_f, while the 9-ball moves at speed v₉.

For the purposes of this problem, assume that the collision isperfectly elastic, neglect friction, and ignore the spinning of theballs.

Find the angle θ that the 9-ball travels away from the horizontal, asshown in the figure.

Express your answer in degrees tothree significant figures.

θ=

Answer 1

Concepts and reason

The concept used to solve this problem are perfectly elastic collision and the law conservation of energy.

Initially using the law of conservation of energy, establish the relationship between different velocity and finally calculate the magnitude of angle that the 9 ball travels away from the horizontal.

Fundamentals

For perfectly elastic collision,

The law of conservation of momentum is valid. And the law of conservation of kinetic energy is also valid.

From the law of conservation of kinetic energy,

$\frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{1}{2}mu_1^2 + \frac{1}{2}mu_2^2$

Here, ${m_1},{m_2}$ are the masses of the first and second objects, ${u_1},{u_2}$ is the initial velocities of the first and second object and ${v_1},{v_2}$ are the final velocities.

Consider diagram of the given system as,

From the law of conservation of kinetic energy,

$\frac{1}{2}{m_c}v_f^2 + \frac{1}{2}{m_9}v_9^2 = \frac{1}{2}{m_c}v_i^2 + \frac{1}{2}{m_9}u_i^2$ …… (1)

Here, ${m_c},{m_9}$ are the masses of the cue ball and 9 ball, ${v_f},{v_9}$ are the final velocities of the cue and 9 ball and ${v_i},{u_9}$ are the initial velocities of the cue and 9 ball.

The mass of all balls are equal.

Substitute $m$ for ${m_c}$ , $m$ for ${m_9}$ and $0$ for ${u_9}$ in the expression of the law of conservation of kinetic energy.

$\begin{array}{c}\\\frac{1}{2}{m_c}v_f^2 + \frac{1}{2}{m_9}v_9^2 = \frac{1}{2}{m_c}v_i^2 + \frac{1}{2}{m_9}\left( 0 \right)\\\\v_f^2 + v_9^2 = v_i^2\\\end{array}$

The velocities form a Pythagorean triplet.

$v_c^2 + v_9^2 = u_c^2$ …… (2)

This equation forms a Pythagorean triplet. It proves that the 9 ball and cue ball split at a right angle.

Then, the angle will be half of a right angle.

$\theta = {45^ \circ }$

Ans:

The angle, $\theta$ is ${45^ \circ }$ .