Chapter 11, Problem 41PQ

Initially, ball 1 rests on an incline of height h, and ball 2 rests on an incline of height h/2 as shown in Figure P11.40. They are released from rest simultaneously and collide elastically in the trough of the track. If m₂ = 4 m₁, m₁ = 0.045 kg, and h = 0.65 m, what is the velocity of each ball after the collision?

To determine

The velocity of each ball after collision.

Answer to Problem 41PQ

The velocity of ball 1 after collision is $-6.2\widehat{\text{i}}\text{ m/s}$ and that of ball 2 is $-8.7\times {10}^{-2}\widehat{\text{i}}\text{ m/s}$ .

Explanation of Solution

Apply conservation of energy for a ball falling from incline of height $y$.

  $mgy=\frac{1}{2}m{v}_i^2$

Here, $y$ is the height of inclination and $v_i$ is the speed of ball at lowest point, $m$ is the mass of ball and $g$ is the acceleration due to gravity.

Rearrange above equation to get $v_i$ .

  $\begin{array}{c}{v}_i^2=2gy\\ v_i=\sqrt{2gy}\end{array}$                                                                                                                 (I)

For ball 1 height $y=h$ and for ball 2 height is $y=\frac{h}{2}$.

Substitute $h$ for $y$ in equation (I) to find the magnitude of velocity of ball 1 just before collision.

  $v_{1i}=\sqrt{2gh}$

Here, $v_{1i}$ is the magnitude of velocity of ball1 just before collision.

From figure P11.40, it’s clear that ball 1 is moving along $x$ direction.

Write the expression for the velocity of ball 1 before collision.

  ${\overrightarrow{\text{v}}}_{1i}=\sqrt{2gh}\widehat{\text{i}}$                                                                                                             (II)

Here, ${\overrightarrow{\text{v}}}_{1i}$ is the velocity of ball 1 before collision.

Substitute $\frac{h}{2}$ for $y$ in equation (I) to find the velocity of ball 2 just before collision.

  $\begin{array}{c}{v}_{2i}=\sqrt{2g\frac{h}{2}}\\ =\frac{1}{\sqrt{2}}\sqrt{2gh}\end{array}$

Here, $v_{2i}$ is the magnitude of velocity of ball 2 just before collision.

From figure P11.40, it’s clear that ball 2 is moving along $-x$ direction.

Write the expression for the velocity of ball 2 before collision.

  ${\overrightarrow{\text{v}}}_{2i}=-\frac{1}{\sqrt{2}}\sqrt{2gh}\widehat{\text{i}}$

Here, ${\overrightarrow{\text{v}}}_{2i}$ is the velocity of ball 2 before collision.

Put equation (II) in the above equation.

  ${\overrightarrow{\text{v}}}_{2i}=-\frac{1}{\sqrt{2}}{\overrightarrow{\text{v}}}_{1i}$                                                                                                          (III)

It is given that the collision is elastic so that both kinetic as well as potential energy are conserved during the collision.

Write the expression for the velocity of the ball 1 after collision.

  ${\overrightarrow{\text{v}}}_{1f}=\left(\frac{m_1-m_2}{m_1+m_2}\right){\overrightarrow{\text{v}}}_{1i}+\frac{2m_2}{m_1+m_2}{\overrightarrow{\text{v}}}_{2i}$                                                                            (IV)

Here, ${\overrightarrow{\text{v}}}_{1f}$ is the velocity of the ball1 just after the collision, $m_1$ is the mass of ball 1 and $m_2$ is the mass of ball 2.

Write the expression for the velocity of the ball 2 after collision.

  ${\overrightarrow{\text{v}}}_{2f}=\left(\frac{m_2-m_1}{m_1+m_2}\right){\overrightarrow{\text{v}}}_{2i}+\frac{2m_1}{m_1+m_2}{\overrightarrow{\text{v}}}_{1i}$                                                                          (V)

It is given that $m_2=4m_1$ .

Put equation (III) in equation (IV) and replace $m_2$ by $4m_1$ .

  $\begin{array}{c}{\overrightarrow{\text{v}}}_{1f}=\left(\frac{m_1-4m_1}{m_1+4m_1}\right){\overrightarrow{\text{v}}}_{1i}+\left(\frac{2\times 4m_1}{m_1+4m_1}\right)\left(-\frac{1}{\sqrt{2}}{\overrightarrow{\text{v}}}_{1i}\right)\\ =\left(\frac{-3m_1}{5m_1}\right){\overrightarrow{\text{v}}}_{1i}-\frac{8m_1}{5\sqrt{2}{m}_1}{\overrightarrow{\text{v}}}_{1i}\\ =\left(-\frac{3}{5}-\frac{8}{5\sqrt{2}}\right){\overrightarrow{\text{v}}}_{1i}\\ =-\left(\frac{3\sqrt{2}+8}{5\sqrt{2}}\right){\overrightarrow{\text{v}}}_{1i}\end{array}$

Put equation (II) in the above equation to get final expression for ${\overrightarrow{\text{v}}}_{1f}$ .

  $\begin{array}{c}{\overrightarrow{\text{v}}}_{1f}=-\left(\frac{3\sqrt{2}+8}{5\sqrt{2}}\right)\sqrt{2gh}\widehat{\text{i}}\\ =-\left(\frac{3\sqrt{2}+8}{5}\right)\sqrt{gh}\widehat{\text{i}}\\ \cong -2.45\sqrt{gh}\widehat{\text{i}}\end{array}$                                                                                       (VI)

Put equation (III) in equation (V) and replace $m_2$ by $4m_1$ to get the expression for ${\overrightarrow{\text{v}}}_{2f}$ .

  $\begin{array}{c}{\overrightarrow{\text{v}}}_{2f}=\left(\frac{4m_1-m_1}{m_1+4m_1}\right)\left(-\frac{1}{\sqrt{2}}{\overrightarrow{\text{v}}}_{1i}\right)+\left(\frac{2m_1}{m_1+4m_1}\right){\overrightarrow{\text{v}}}_{1i}\\ =\left(-\frac{3m_1}{5\sqrt{2}{m}_1}\right){\overrightarrow{\text{v}}}_{1i}+\frac{2m_1}{5m_1}{\overrightarrow{\text{v}}}_{1i}\\ =\left(\frac{2}{5}-\frac{3}{5\sqrt{2}}\right){\overrightarrow{\text{v}}}_{1i}\\ =\left(\frac{2\sqrt{2}-3}{5\sqrt{2}}\right){\overrightarrow{\text{v}}}_{1i}\end{array}$

Put equation (II) in the above equation to get final expression for ${\overrightarrow{\text{v}}}_{2f}$ .

  $\begin{array}{c}{\overrightarrow{\text{v}}}_{2f}=\left(\frac{2\sqrt{2}-3}{5\sqrt{2}}\right)\sqrt{2gh}\widehat{\text{i}}\\ \cong -0.0343\sqrt{gh}\widehat{\text{i}}\end{array}$                                                                                      (VII)

Conclusion:

It is given that the value of $h$ is $0.65\text{ m}$ . The value of acceleration due to gravity is $9.81{\text{ m/s}}^2$ .

Substitute $9.81{\text{ m/s}}^2$ for $g$ and $0.65\text{ m}$ for $h$ in equation (VI) to find the value of ${\overrightarrow{\text{v}}}_{1f}$ .

  $\begin{array}{c}{\overrightarrow{\text{v}}}_{1f}=-2.45\sqrt{\left(9.81{\text{ m/s}}^2\right)\left(0.65\text{ m}\right)}\widehat{\text{i}}\\ \text{=}-6.2\widehat{\text{i}}\text{ m/s}\end{array}$

Substitute $9.81{\text{ m/s}}^2$ for $g$ and $0.65\text{ m}$ for $h$ in equation (VII) to find the value of ${\overrightarrow{\text{v}}}_{2f}$ .

  $\begin{array}{c}{\overrightarrow{\text{v}}}_{1f}=-0.0343\sqrt{\left(9.81{\text{ m/s}}^2\right)\left(0.65\text{ m}\right)}\widehat{\text{i}}\\ \text{=}-8.7\times {10}^{-2}\widehat{\text{i}}\text{ m/s}\end{array}$

Therefore, the velocity of ball 1 after collision is $-6.2\widehat{\text{i}}\text{ m/s}$ and that of ball 2 is $-8.7\times {10}^{-2}\widehat{\text{i}}\text{ m/s}$ .