Chapter 7, Problem 44P

To determine

The final direction of ball A and the final velocities of two balls.

Answer to Problem 44P

The final direction of ball A is $0°$ .

The final velocity of ball A is $3.7\text{m/s}$ .

The final velocity of the ball B is $2\text{ m/s}$ .

Explanation of Solution

Given:

The mass of balls is $m_1=m_2=m$ .

The initial velocity of the ball A is $u_1=2\text{m}/\text{s}$ .

The initial velocity of the ball B is $u_2=3.7\text{m}/\text{s}$ .

Formula used:

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the ball A, $m_2$ is the mass of ball B, $u_1$ is the initial velocity of ball A, $u_2$ is the initial velocity of the ball B, $v_1$ is the final velocity of ball A, and $v_2$ is the final velocity of the ball B.

The expression for conservation of kinetic energy is,

  $\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

Here, $m_1$ is the mass of the ball A, $m_2$ is the mass of ball B, $u_1$ is the initial velocity of ball A, $u_2$ is the initial velocity of the ball B, $v_1$ is the final velocity of ball A, and $v_2$ is the final velocity of the ball B.

Calculation:

Sketch the representation of the billiard ball before and after collision as shown in Figure 1.

  

Apply the conservation of momentum along x direction.

  $\begin{array}{l}m{u}_2=m{v}_1\cos \theta \\ u_2=v_1\cos \theta \end{array}$     .......(1)

Apply the conservation of momentum along y direction.

  $\begin{array}{l}m{u}_1=m{v}_1\sin \theta +m{v}_2\\ u_1=v_1\sin \theta +v_2\\ u_1-v_2=v_1\sin \theta \end{array}$     .....(2)

Squaring both sides of the equations (1) and (2) and adding them,

  $\begin{array}{l}{\left(v_1\cos \theta \right)}^2+{\left(v_1\sin \theta \right)}^2=u_2^2+{\left(u_1-v_2\right)}^2\\ v_1^2\left({\cos }^2\theta +{\sin }^2\theta \right)=u_2^2+u_1^2-2u_1{v}_2+v_2^2\\ v_1^2=u_2^2+u_1^2-2u_1{v}_2+v_2^2\end{array}$     ......(3)

Apply the conservation of energy.

  $\begin{array}{l}\frac{1}{2}m{u}_1^2+\frac{1}{2}m{u}_2^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2\\ u_1^2+u_2^2=\left(u_2^2+u_1^2-2u_1{v}_2+v_2^2\right)+v_2^2\\ 2v_2^2=2u_1{v}_2\\ v_2=u_1\\ v_2=2\text{ m/s}\end{array}$

Find the final velocity of ball A using Equation (3).

  $\begin{array}{l}{v}_1^2=u_2^2+u_1^2-2u_1{v}_2+v_2^2\\ v_1^2={\left(3.7\text{m/s}\right)}^2+{\left(2\text{m/s}\right)}^2-2\times \left(2\text{m/s}\right)\left(2\text{m/s}\right)+{\left(2\text{ m/s}\right)}^2\\ v_1=\sqrt{13.69}\\ v_1=3.7\text{m/s}\end{array}$

Find the final direction of ball A using Equation (1).

  $\begin{array}{l}{u}_2=v_1\cos \theta \\ \left(3.7\text{ m/s}\right)=\left(3.7\text{ m/s}\right)\cos \theta \\ \cos \theta =1\\ \theta =0°\end{array}$

Conclusion:

Thus, the final direction of ball A is $\theta =0$ .

Thus, the final velocity of ball A is $v_1=3.7\text{m/s}$ .

Thus, the final velocity of the ball B is $v_2=2\text{ m/s}$ .