Chapter 8, Problem 40TP

To determine

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

Answer to Problem 40TP

Center of mass velocity before collision is $-2\text{m}/\text{s}$.

Center of mass velocity before collision is $-2\text{m}/\text{s}$.

The change in the center of mass velocity of the system after collision will be zero.

Momentum is always conserved when no external force acts on the system.

Explanation of Solution

Given:

Mass of A is $m_{\text{a}}=1.0\text{kg}$.

Mass of B is $m_{\text{b}}=1.0\text{kg}$.

Initial velocity of mass A is $u_{\text{a}}=4\text{m}/\text{s}$.

Initial velocity of mass B is $u_{\text{b}}=-8\text{m}/\text{s}$

Formula used:

The conservation of momentum to find center of mass velocity of the bodies before collision is given by

  $\left(m_{\text{a}}+m_{\text{b}}\right)u_{\text{cm}}=m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}$

Final velocity after collision will be calculated as

  $m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}=\left(m_{\text{a}}+m_{\text{b}}\right)v$

The conservation of momentum to find center of mass velocity of the bodies after collision is given by

  $\left(m_{\text{a}}+m_{\text{b}}\right)v_{\text{cm}}=\left(m_{\text{a}}+m_{\text{b}}\right)v$

Change in the center of mass velocity of the system after collision is given as

  $\Delta v_{\text{cm}}=v_{\text{cm}}-u_{\text{cm}}$

Calculation:

The conservation of momentum to find center of mass velocity of the bodies before collision is given by

  $\begin{array}{c}\left(m_{\text{a}}+m_{\text{b}}\right)u_{\text{cm}}=m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}\\ \left(1.0+1.0\text{kg}\right)u_{\text{cm}}=\left(1.0\text{kg}\right)\left(4\text{m}/\text{s}\right)+\left(1.0\text{kg}\right)\left(-8\text{m}/\text{s}\right)\\ u_{\text{cm}}=\frac{-4}{2}\text{m}/\text{s}\\ u_{\text{cm}}=-2\text{m}/\text{s}\end{array}$

Final velocity after collision will be calculated as

  $\begin{array}{c}{m}_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}=\left(m_{\text{a}}+m_{\text{b}}\right)v\\ \left(1.0\text{kg}\right)\left(4\text{m}/\text{s}\right)+\left(1.0\text{kg}\right)\left(-8\text{m}/\text{s}\right)=\left(1.0+1.0\text{kg}\right)v\\ v=\frac{-4}{2}\text{m}/\text{s}\\ v=-2\text{m}/\text{s}\end{array}$

The conservation of momentum to find center of mass velocity of the bodies after collision is given by

  $\begin{array}{c}\left(m_{\text{a}}+m_{\text{b}}\right)v_{\text{cm}}=\left(m_{\text{a}}+m_{\text{b}}\right)v\\ \left(1.0+1.0\text{kg}\right)v_{\text{cm}}=\left(1.0+1.0\text{kg}\right)\left(-2\text{m}/\text{s}\right)\\ v_{\text{cm}}=\frac{-4}{2}\text{m}/\text{s}\\ v_{\text{cm}}=-2\text{m}/\text{s}\end{array}$

Change in the center of mass velocity of the system after collision is given as

  $\begin{array}{c}\Delta v_{\text{cm}}=v_{\text{cm}}-u_{\text{cm}}\\ =-2\text{m}/\text{s}-\left(-2\text{m}/\text{s}\right)\\ =0\end{array}$

Conclusion:

Center of mass velocity before collision is $-2\text{m}/\text{s}$.

Center of mass velocity before collision is $-2\text{m}/\text{s}$.

The change in the center of mass velocity of the system after collision is zero.

In an isolated system, where no external force acts on the system, the net momentum is always conserved irrespective of the nature of the collision.