Chapter 9.2, Problem 29PP

To determine

The velocity of each ball after collision.

Answer to Problem 29PP

Final velocity of the stationary ball is $3.46\text{ m/s, 30}\text{.0}°$ to the right.

Final velocity of the moving ball is $2.0\text{ m/s, 60}\text{.0}°$ to the left.

Explanation of Solution

Given:

Let $"A"$ be the stationary billiard ball and $"B"$ be the billiard ball which is motion.

Both the balls are identical and their mass is $m_A=m_B=m=0.17\text{ kg}$

Since the ball $"A"$ is at stationary, its initial velocity is, $v_{Ai}=0\text{ m/s}$

Initial velocity of the ball $"B"$ is $v_{Bi}=4.0\text{ m/s}$

After the collision the ball $"B"$ moves $\text{60}\text{.0}°$ to the left of its original direction and the ball $"A"$ moves ${30.0}^0$ to the right of the moving ball’s original direction.

This can be diagrammatically represented as shown in the figure $1$ .

  

Figure 1.

Formula used:

Since the momentum is conserved, initial momentum of the system is equal to the final momentum of the system. That is,

  $p_i=p_f$$\left(1\right)$

In this problem,

  $p_i=p_{Ai}+p_{Bi}$ , and

  $p_f=p_{Af}+p_{Bf}$

Therefore,

  $p_{Ai}+p_{Bi}=p_{Af}+p_{Bf}$

From the vector diagram, the equation for final momentum of the ball $"A"$ can be written as,

  $p_{Af}=p_{Bi}\sin \text{60}\text{.0}°$$\left(2\right)$

And, the equation for final momentum of the ball $"B"$ can be written as,

  $p_{Bf}=p_{Bi}\cos \text{60}\text{.0}°$$\left(3\right)$

But the momentum $\left(p\right)$ is the product of mass $\left(m\right)$ and velocity $\left(v\right)$ of a moving body. That is,

  $p=mv$

Applying this relation, equations $\left(2\right)$ and $\left(3\right)$ can be written as,

  $m{v}_{Af}=m{v}_{Bi}\sin \text{60}\text{.0}°$

  $\Rightarrow v_{Af}=v_{Bi}\sin \text{60}\text{.0}°$$\left(4\right)$

And,

  $m{v}_{Bf}=m{v}_{Bi}\cos \text{60}\text{.0}°$

  $\Rightarrow v_{Bf}=v_{Bi}\cos \text{60}\text{.0}°$$\left(5\right)$

Where, $v_{Af}$ and $v_{Bf}$ are the final velocity of the balls $"A"$ and $"B"$ respectively.

Calculation:

Final velocity of the ball $"A"$ is calculated by substituting the numerical values in equation $\left(4\right)$ ,

  $v_{Af}=\left(4.0\text{ m/s}\right)\sin \text{60}\text{.0}°$

  $=3.46\text{ m/s, 30}{\text{.0}}^0$ to the right

Final velocity of the ball $"B"$ is calculated by substituting the numerical values in equation $\left(5\right)$ ,

  $v_{Bf}=\left(4.0\text{ m/s}\right)\cos \text{60}\text{.0}°$

  $=2.0\text{ m/s, 60}\text{.0}°$ to the left

Conclusion:

Final velocity of the stationary ball is $3.46\text{ m/s, 30}\text{.0}°$ to the right.

Final velocity of the moving ball is $2.0\text{ m/s, 60}\text{.0}°$ to the left.