Chapter 10, Problem 30PQ

A billiard player sends the cue ball toward a group of three balls that are initially at rest and in contact with one another. After the cue ball strikes the group, the four balls scatter, each traveling in a different direction with different speeds as shown in Figure P10.30. If each ball has the same mass, 0.16 kg, determine the total momentum of the system consisting of the four balls immediately after the collision.

FIGURE P10.30

To determine

Total momentum of four balls after the collision.

Answer to Problem 30PQ

The total momentum of the four balls after collision is $\underset{\_}{\left(0.010\widehat{i}-0.028\widehat{j}\right)}$.

Explanation of Solution

The player sends a cue ball to hit on a group of four balls. All the four balls were initially at rest and after their collision with the cue ball each of them scattered to different directions.

Write the equation to find the total momentum of the four balls.

  $\overrightarrow{P_{tot}}=m_1\overrightarrow{v_1}+m_3\overrightarrow{v_3}+m_6\overrightarrow{v_6}$                                                                                   (I)

Here,$\overrightarrow{P_{tot}}$ is the total momentum of four balls,$m_1$ is the mass of first ball,$m_3$ is the mass of ball 3, $m_6$ is the mass of ball 6, $\overrightarrow{v_1}$ is the velocity of ball 1, $\overrightarrow{v_3}$ is the velocity of ball 3 and $\overrightarrow{v_6}$ is the velocity of ball 6.

Refer figure 10.30 provided in the problem. The figure gives the coordinate system of the system of four balls.

Consider each ball and write the component form of velocity vector for each ball.

Write the equation to find the velocity vector for ball 1.

  $\overrightarrow{v_1}=v_{1x}\cos \theta \widehat{i}+v_{1y}\sin \theta \widehat{j}$                                                                                   (II)

Here, $v_{1x}$ is the x component of velocity vector of ball 1,$v_{1y}$ is the y component of velocity vector of ball 1,$\theta$ is the angle made by vector and the x axis,$\widehat{i}$ is the unit vector along x axis, and $\widehat{j}$ is the unit vector along y axis.

Write the equation to find the velocity vector for ball 3.

  $\overrightarrow{v_3}=v_{3x}\cos \theta \widehat{i}+v_{3y}\sin \theta \widehat{j}$                                                                                (III)

Here, $v_{3x}$ is the x component of velocity vector of ball 3 and $v_{3y}$ is the y component of velocity vector of ball 3.

Write the equation to find the velocity vector for ball 6.

  $\overrightarrow{v_6}=v_{6x}\cos \theta \widehat{i}+v_{6y}\sin \theta \widehat{j}$                                                                                (IV)

Here, $v_{6x}$ is the x component of velocity vector of ball 6,$v_{6y}$ is the y component of velocity vector of ball 6

Write the equation to find the velocity vector for cue ball.

  $\overrightarrow{v_c}=v_x\cos \theta \widehat{i}+v_y\sin \theta \widehat{j}$                                                                                   (V)

Conclusion:

Substitute $0.35\text{m}/\text{s}$ for $v_x$,$0.35\text{m}/\text{s}$ for $v_y$ and $60°$ for $\theta$ in equation (II) to get $\overrightarrow{v_1}$.

  $\begin{array}{c}\overrightarrow{v_1}=\left(0.35\text{m}/\text{s}\right)\cos \left(60°\right)\widehat{i}+\left(0.35\text{m}/\text{s}\right)\sin \left(60°\right)\widehat{j}\\ =\left(0.175\widehat{i}+0.303\widehat{j}\right)\text{m}/\text{s}\end{array}$

Substitute $0.45\text{m}/\text{s}$ for $v_x$,$-0.45\text{m}/\text{s}$ for $v_y$ and $60°$ for $\theta$ in equation (III) to get $\overrightarrow{v_3}$.

  $\begin{array}{c}\overrightarrow{v_3}=\left(0.45\text{m}/\text{s}\right)\cos \left(60°\right)\widehat{i}-\left(0.45\text{m}/\text{s}\right)\sin \left(60°\right)\widehat{j}\\ =\left(0.390\widehat{i}-0.225\widehat{j}\right)\text{m}/\text{s}\end{array}$

Substitute $-0.25\text{m}/\text{s}$ for $v_y$ and $90°$ for $\theta$ in equation (IV) to get $\overrightarrow{v_6}$.

  $\begin{array}{c}\overrightarrow{v_6}=-\left(0.25\text{m}/\text{s}\right)\sin \left(90°\right)\widehat{j}\\ =\left(-0.25\widehat{j}\right)\text{m}/\text{s}\end{array}$

Substitute $-0.50\text{m}/\text{s}$ for $v_x$ and $0°$ for $\theta$ in equation (V) to get $\overrightarrow{v_c}$.

  $\begin{array}{c}\overrightarrow{v_c}=-\left(0.50\text{m}/\text{s}\right)\cos \left(0°\right)\widehat{i}\\ =\left(-0.50\widehat{i}\right)\text{m}/\text{s}\end{array}$

Substitute $0.16\text{kg}$ for $m_1,m_2,m_3\text{and}{m}_c$, $\left(0.175\widehat{i}+0.303\widehat{j}\right)\text{m}/\text{s}$ for $\overrightarrow{v_1}$, $\left(0.390\widehat{i}-0.225\widehat{j}\right)\text{m}/\text{s}$ for $\overrightarrow{v_3}$, $\left(-0.25\widehat{j}\right)\text{m}/\text{s}$ for $\overrightarrow{v_6}$ and $\left(-0.50\widehat{i}\right)\text{m}/\text{s}$ for $\overrightarrow{v_c}$ in equation (I) to get $\overrightarrow{P_{tot}}$.

  $\begin{array}{c}\overrightarrow{P_{tot}}=\left(0.16\text{kg}\right)\left[\left(0.175\widehat{i}+0.303\widehat{j}\right)\text{m}/\text{s}+\left(0.390\widehat{i}-0.225\widehat{j}\right)\text{m}/\text{s}+\left(-0.25\widehat{j}\right)\text{m}/\text{s}+\left(-0.50\widehat{i}\right)\text{m}/\text{s}\right]\\ =\left(0.028\widehat{i}+0.048\widehat{j}\right)\text{kg}\cdot \text{m}/\text{s}+\left(0.062\widehat{i}-0.036\widehat{j}\right)\text{kg}\cdot \text{m}/\text{s}-0.040\widehat{j}\text{kg}\cdot \text{m}/\text{s}-0.080\widehat{i}\text{kg}\cdot \text{m}/\text{s}\\ =\left(0.010\widehat{i}-0.028\widehat{j}\right)\end{array}$

Therefore, the total momentum of the four balls after collision is $\underset{\_}{\left(0.010\widehat{i}-0.028\widehat{j}\right)}$.