Chapter 9, Problem 75A

To determine

The velocity of the cue ball and the 8 balls after collision.

Answer to Problem 75A

The velocity of the cue ball after collision is $v_{\text{Cf}}=2.8\text{m}/\text{s}$ .

The velocity of the eight balls after collision is $v_{\text{8f}}=2.8\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the cue ball is $m_{\text{C}}=0.16\text{ kg}$ .

The velocity of the cue ball is $v_{\text{Ci}}=\text{4}\text{.0 }\text{m}/\text{s}$ .

The angle of collision is $\theta =45°$ .

The mass of the eight balls is equal to the mass of the cue ball $m_8=m_{\text{C}}$ .

Formula used:

The expression for the momentum is

  $p=mv$

Here, $m$ is the mass of the system and $v$ is the velocity of the system.

Calculation:

Consider the initial momentum of the cue ball is $p_{\text{Ci}}$ and the momentum of the stationary balls to be zero.

The x and y components of the balls after collision are shown below.

  

The initial velocity of the cue ball is $v_{\text{Ci}}$ .

The final velocity of the cue ball is $v_{\text{Cf}}$ .

The final velocity of the eight balls is $v_{\text{8f}}$ .

The initial momentum of the cue ball is,

  $\begin{array}{l}{p}_{\text{Ci}}=m_C{v}_{\text{Ci}}\\ p_{\text{Ci}}=\left(\text{0}\text{.16 kg}\right)\times \left(\text{4}\text{.0 }\text{m}/\text{s}\right)\\ p_{\text{Ci}}=0.64\text{kg}\cdot \text{m}/\text{s}\end{array}$

By the law of conservation of momentum, the initial momentum and final momentum of the system are equal.

  $p_{\text{i}}=p_{\text{f}}$

The final velocity of the cue ball using the relation of the final momentum of the cue ball is,

  $\begin{array}{l}{p}_{\text{Cf}}=p_{\text{Ci}}\sin 45°\\ m_{\text{C}}{v}_{\text{Cf}}=m_{\text{C}}{v}_{\text{Ci}}\times \sin 45°\\ v_{\text{Cf}}=\left(\text{4}\text{.0}\text{m}/\text{s}\right)\times \sin 45°\\ v_{\text{Cf}}=2.8\text{m}/\text{s}\end{array}$

The final velocity of the eight balls using the relation of the final momentum of the eight balls is,

  $\begin{array}{l}{p}_{\text{8f}}=p_{\text{Ci}}\cos 45°\\ m_{\text{8}}{v}_{\text{8f}}=m_{\text{C}}{v}_{\text{Ci}}\times \cos 45°\\ v_{\text{8f}}=\left(\text{4}\text{.0}\text{m}/\text{s}\right)\times \cos 45°\\ v_{\text{8f}}=2.8\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of the cue ball after collision is $v_{\text{Cf}}=2.8\text{m}/\text{s}$ and the velocity of the eight balls after collision is $v_{\text{8f}}=2.8\text{m}/\text{s}$ .