Chapter 7, Problem 25P

To determine

The speed and direction of each tennis ball after collision.

Answer to Problem 25P

  $v_1=0.88\text{ m/s}$ , in the original direction

  $v_2=2.23\text{ m/s}$ , in the original direction.

Explanation of Solution

Given:

Mass of first tennis ball is $m_1=0.06\text{ kg}$

Speed of the first tennis ball before collision is $u_1=2.50\text{ m/s}$

Mass of second tennis ball is $m_2=0.09\text{ kg}$

Speed of the second tennis ball before collision is $u_2=1.15\text{ m/s}$

Formula used:

Let $v_1$ and $v_2$ be the final speed of the first and second tennis ball respectively. Since the collision between the balls are perfectly elastic, momentum is conserved. That is total initial momentum of the balls is equal to the total final momentum of the balls.

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$$\left(1\right)$

Also, for a perfectly elastic one dimensional collision,

  $u_1-u_2=v_2-v_1$

  $\Rightarrow v_2=v_1+u_1-u_2$$\left(2\right)$

Substituting this in equation $\left(1\right)$ ,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2\left(v_1+u_1-u_2\right)$

  $\Rightarrow m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_1+m_2{u}_1-m_2{u}_2$

  $\Rightarrow m_1{v}_1+m_2{v}_1=m_1{u}_1+m_2{u}_2-m_2{u}_1+m_2{u}_2$

  $\Rightarrow v_1\left(m_1+m_2\right)=-m_2{u}_1+2m_2{u}_2+m_1{u}_1$

  $\Rightarrow v_1=\frac{\left(-m_2{u}_1+2m_2{u}_2+m_1{u}_1\right)}{\left(m_1+m_2\right)}$$\left(3\right)$

Calculation:

Substituting the numerical values in equation $\left(3\right)$

  $v_1=\frac{\left[-\left(0.09\text{ kg}\right)\left(2.50\text{ m/s}\right)+2\left(0.09\text{ kg}\right)\left(1.15\text{ m/s}\right)+\left(0.06\text{ kg}\right)\left(2.50\text{ m/s}\right)\right]}{\left(0.06\text{ kg}+0.09\text{ kg}\right)}$

  $=\frac{\left[-0.225\text{ kg}\text{.m/s+0}\text{.207 kg}\text{.m/s+}0.15\text{ kg}\text{.m/s}\right]}{0.15\text{ kg}}$

  $=\frac{0.132\text{ kg}\text{.m/s}}{0.15\text{ kg}}$

  $=0.88\text{ m/s}$ , in the original direction

Substituting the value of $v_1$ and other numerical values in equation $\left(2\right)$ ,

  $v_2=0.88\text{ m/s + }2.50\text{ m/s - }1.15\text{ m/s}$

  $=2.23\text{ m/s}$ , in the original direction.

Conclusion:

The speed and direction of each tennis ball after collision is,

  $v_1=0.88\text{ m/s}$ , in the original direction

  $v_2=2.23\text{ m/s}$ , in the original direction.