Chapter 8, Problem 24TP

To determine

The velocity of the center of mass of the system before and after the collision,the momentum of the center of mass of the system before and after the collision, and the change in momentum of the system.

Answer to Problem 24TP

The velocity of the center of mass of the system before and after the collision is $1.6\text{m}/\text{s}$.

The momentum of the center of mass of the system before and after the collision is $1.6\text{kg}\cdot \text{m}/\text{s}$.

The change in momentum of system is $0\text{kg}\cdot \text{m}/\text{s}$.

Explanation of Solution

Given:

The mass of puck A is $m_{\text{A}}=200\text{g}$.

The mass of puck B is $m_{\text{B}}=800\text{g}$.

Formula used:

The initial center of mass velocity of system is given by

  $v_{\text{cm}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

The conservation of momentum for combined bodies is given by

  $\left(m_1+m_2\right)v_{\text{cm}}=\left(m_1+m_2\right){v^{\prime }}_{\text{cm}}$

Here, $v_{\text{cm}}$ is the initial velocity of the center of mass of the system and ${v^{\prime }}_{\text{cm}}$ is the final velocity of the center of mass of the system.

The initial momentum of center of mass of system is given by

  $p_{\text{i}}=\left(m_1+m_2\right)v_{\text{cm}}$

The final momentum of center of mass of system is given by

  $p_{\text{f}}=\left(m_1+m_2\right){v^{\prime }}_{\text{cm}}$

The change in momentum is given by

  $\Delta p=p_{\text{f}}-p_{\text{i}}$

Calculation:

The initial center of mass velocity of system is calculated as

  $\begin{array}{c}{v}_{\text{cm}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}\\ =\frac{\left(200\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(8\text{m}/\text{s}\right)+\left(800\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(0\text{m}/\text{s}\right)}{\left(200\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)+\left(800\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}\\ =\frac{1.6\text{kg}\cdot \text{m}/\text{s}}{1\text{kg}}\\ =1.6\text{m}/\text{s}\end{array}$

The final center of mass velocity of system is calculated as

  $\begin{array}{c}{v^{\prime }}_{\text{cm}}=\frac{\left(m_1+m_2\right)v_{\text{cm}}}{\left(m_1+m_2\right)}\\ =v_{\text{cm}}\\ =1.6\text{m}/\text{s}\end{array}$

The initial momentum of center of mass of system is calculated as

  $\begin{array}{c}{p}_{\text{i}}=\left(m_1+m_2\right)v_{\text{cm}}\\ =\left[\left(200\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)+\left(800\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right]\left(1.6\text{m}/\text{s}\right)\\ =\left(1\text{kg}\right)\left(1.6\text{m}/\text{s}\right)\\ =1.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

The final momentum of center of mass of system is calculated as

  $\begin{array}{c}{p}_{\text{f}}=\left(m_1+m_2\right){v^{\prime }}_{\text{cm}}\\ =\left[\left(200\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)+\left(800\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right]\left(1.6\text{m}/\text{s}\right)\\ =\left(1\text{kg}\right)\left(1.6\text{m}/\text{s}\right)\\ =1.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

The change in momentum is calculated as

  $\begin{array}{c}\Delta p=p_{\text{f}}-p_{\text{i}}\\ =\left(1.6\text{kg}\cdot \text{m}/\text{s}\right)-\left(1.6\text{kg}\cdot \text{m}/\text{s}\right)\\ =0\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

The velocity of the center of mass of the system before and after the collision is equal to $1.6\text{m}/\text{s}$.

The momentum of the center of mass of the system before and after the collision is equal to $1.6\text{kg}\cdot \text{m}/\text{s}$.

The change in momentum of system is equal to $0\text{kg}\cdot \text{m}/\text{s}$.