Chapter 11, Problem 51PQ

In Figure P11.51, a cue ball is shot toward the eight-ball on a pool table. The cue ball is shot at the eight-ball with a speed of 8.00 m/s in a direction 30.0° from the y axis. Both balls have the same mass of 0.170 kg. After the balls undergo an elastic collision, the eight-ball travels in the negative x direction into the side pocket. What is the velocity of the cue ball after this collision?

FIGURE P11.51

To determine

The velocity of the cue ball after the collision.

Answer to Problem 51PQ

The velocity of the cue ball after the collision is $\underset{\_}{6.93\widehat{\text{j}}\text{m/s}}$.

Explanation of Solution

Given that the speed of the cue ball is $8.00\text{m/s}$ in a direction $30.0°$ from the $y$ axis, and the mass of both the balls are $0.170\text{kg}$. The diagrammatic representation of the situation is shown in Figure P11.51.

From the given Figure, deduce the $x$ and $y$ components of the initial velocity of the cue ball.

  $v_{1iy}=v_1\cos \theta$                                                                                                         (I)

  $v_{1ix}=-v_1\sin \theta$                                                                                                       (II)

Here, $v_{1iy}$ is the $y$ component of the initial velocity of the cue ball, $v_{1ix}$ is the $x$ component of the initial velocity of the cue ball, and $\theta$ is the angle made by the cue ball’s velocity vector with the $x$ axis, and $v_1$ is the speed of the cue ball.

Write the expression for the momentum conservation along $x$ and $y$ directions in the given collision.

  $m_1{v}_{1ix}+m_2{v}_{2ix}=m_1{v}_{1fx}+m_2{v}_{2fx}$                                                                              (III)

  $m_1{v}_{1iy}+m_2{v}_{2iy}=m_1{v}_{1fy}+m_2{v}_{2fy}$                                                                            (IV)

Here, $m_1$ is the mass of the cue ball, $m_2$ is the mass of the eight ball, $v_{2ix}$ is the $x$ component of the initial velocity of the eight ball, $v_{2iy}$ is the $y$ component of the initial velocity of the eight ball, $v_{1fx}$ is the $x$ component of the final velocity of the cue ball, $v_{2fx}$ is the $x$ component of the final velocity of the eight ball, $v_{2fx}$ is the $x$ component of the final velocity of the eight ball, and $v_{2fy}$ is the $y$ component of the final velocity of the eight ball

Since the eight ball has no final $y$ velocity, the $y$ momentum (and hence velocity) of the cue ball remains unchanged in the collision. Moreover, the masses of the balls are equal. Thus, equation (III) and (IV) can be reduced to,

  $v_{1ix}=v_{1fx}+v_{2fx}$                                                                                                      (V)

  $v_{1iy}=v_{1fy}$                                                                                                               (VI)

Apply the conservation of kinetic energy to the collision process.

  $\frac{1}{2}m{v}_{1i}^2=\frac{1}{2}m{v}_{2f}^2+\frac{1}{2}m{v}_{1f}^2$                                                                                    (VII)

Reduce equation (VII).

  $\begin{array}{c}\frac{1}{2}{v}_{1i}^2=\frac{1}{2}m{v}_{2fx}^2+\frac{1}{2}m\left(v_{1fx}^2+v_{1fy}^2\right)\\ v_{1i}^2=v_{2fx}^2+v_{1fx}^2+v_{1fy}^2\end{array}$                                                                         (VIII)

Conclusion:

Substitute $8.00\text{m/s}$ for $v_1$, $30.0°$ for $\theta$ in equation (I) to find $v_{1iy}$.

  $\begin{array}{c}{v}_{1iy}=\left(8.00\text{m/s}\right)\cos 30.0°\\ =6.93\text{m/s}\end{array}$

Substitute $8.00\text{m/s}$ for $v_1$, $30.0°$ for $\theta$ in equation (II) to find $v_{1ix}$.

  $\begin{array}{c}{v}_{1ix}=-\left(8.00\text{m/s}\right)\sin 30.0°\\ =-4.00\text{m/s}\end{array}$

Substitute $-4.00\text{m/s}$ for $v_{1ix}$ in equation (V) and solve for $v_{1fx}$

  $\begin{array}{c}-4.00\text{m/s}=v_{1fx}+v_{2fx}\\ v_{1fx}=-4.00\text{m/s}-v_{2fx}\end{array}$                                                                                (IX)

Substitute $6.93\text{m/s}$ for $v_{1iy}$ in equation (VI).

  $6.93\text{m/s}=v_{1fy}$                                                                                                      (X)

Substitute $8.00\text{m/s}$ for $v_{1i}$, and $6.93\text{m/s}$ for $v_{1iy}$ in equation (VIII) and reduce.

  $\begin{array}{c}{\left(8.00\text{m/s}\right)}^2=v_{2fx}^2+v_{1fx}^2+{\left(6.93\text{m/s}\right)}^2\\ 16.0{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}=v_{2fx}^2+v_{1fx}^2\end{array}$                                                                  (XI)

Use equation (IX) in (XI) and solve for $v_{2fx}$.

  $\begin{array}{c}16.0{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}=v_{2fx}^2+{\left(-4.00\text{m/s}-v_{2fx}\right)}^2\\ 2v_{2fx}^2+\left(8.00\text{m/s}\right)v_{2fx}=0\\ v_{2fx}=-4.00\text{m/s}\text{or}{v}_{2fx}=0\end{array}$

The solution for $v_{2fx}$ that is consistent with the problem is $v_{2fx}=-4.00\text{m/s}$.

Substitute $-4.00\text{m/s}$ for $v_{2fx}$ in equation (XI) and solve for $v_{1fx}$.

  $\begin{array}{c}16.0{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}={\left(-4.00\text{m/s}\right)}^2+v_{1fx}^2\\ v_{1fx}=\sqrt{\left(16.0{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)-{\left(-4.00\text{m/s}\right)}^2}\\ =0.00\text{m/s}\end{array}$

Write the expression for the velocity of the cue ball after the collision.

  ${\overrightarrow{\text{v}}}_{1f}=v_{1fx}\widehat{\text{i}}+v_{1fy}\widehat{\text{j}}$                                                                                                 (XII)

Substitute $0.00\text{m/s}$ for $v_{1fx}$, and $6.93\text{m/s}$ for $v_{1fy}$ in equation (XII) to find ${\overrightarrow{\text{v}}}_{1f}$.

  $\begin{array}{c}{\overrightarrow{\text{v}}}_{1f}=\left(0.00\text{m/s}\right)\widehat{\text{i}}+\left(6.93\text{m/s}\right)\widehat{\text{j}}\\ \text{=}6.93\widehat{\text{j}}\text{m/s}\end{array}$

Therefore, the velocity of the cue ball after the collision is $\underset{\_}{6.93\widehat{\text{j}}\text{m/s}}$.