Chapter 8, Problem 39TP

To determine

The change in the center-of-mass velocity.

Answer to Problem 39TP

Option (a) There will be no change in the centre-of-mass velocity.

Explanation of Solution

Given:

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

Mass of B is $m_{\text{b}}=3.0\text{kg}$.

Initial velocity of mass A is $u_{\text{a}}=8\text{m}/\text{s}$.

Initial velocity of mass B is $u_{\text{b}}=0\text{m}/\text{s}$

Formula used:

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

  $\begin{array}{c}\left(m_{\text{a}}+m_{\text{b}}\right)u_{\text{cm}}=m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}\\ \left(1.0+3.0\text{kg}\right)u_{\text{cm}}=\left(1.0\text{kg}\right)\left(8\text{m}/\text{s}\right)+\left(3.0\text{kg}\right)\left(0\text{m}/\text{s}\right)\\ u_{\text{cm}}=\frac{8}{4}\text{m}/\text{s}\\ u_{\text{cm}}=2\text{m}/\text{s}\end{array}$

Final velocity after collision will be calculated as

  $\begin{array}{c}{m}_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}=\left(m_{\text{a}}+m_{\text{b}}\right)v\\ \left(1.0\text{kg}\right)\left(8\text{m}/\text{s}\right)+\left(3.0\text{kg}\right)\left(0\text{m}/\text{s}\right)=\left(1.0+3.0\text{kg}\right)v\\ v=\frac{8}{4}\text{m}/\text{s}\\ v=2\text{m}/\text{s}\end{array}$

The conservation of momentum to find center of mass velocity of the bodiesaftercollision is given by

  $\begin{array}{c}\left(m_{\text{a}}+m_{\text{b}}\right)v_{\text{cm}}=\left(m_{\text{a}}+m_{\text{b}}\right)v\\ \left(1.0+3.0\text{kg}\right)v_{\text{cm}}=\left(1.0+3.0\text{kg}\right)\left(2\text{m}/\text{s}\right)\\ v_{\text{cm}}=\frac{8}{4}\text{m}/\text{s}\\ v_{\text{cm}}=2\text{m}/\text{s}\end{array}$

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

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

Conclusion:

The change in the center of mass velocity of the system after collision is zero. Thus, the correct option is (a).