Chapter 8, Problem 30TP

To determine

The velocities of the both the masses using conservation of momentum and conservation of kinetic energy.

Answer to Problem 30TP

The velocity of mass A after the collision is $0\text{m}/\text{s}$.

The velocity of mass B after the collision is $24\text{m}/\text{s}$.

The system is consistent with both conservation of momentum and conservation of kinetic energy as,

  $\begin{array}{c}\Delta p=\Delta K.E\\ =0\end{array}$

Explanation of Solution

Given:

Mass of A, $m_{\text{a}}=3m$.

Mass of B, $m_{\text{b}}=m$.

Velocity of mass A before collision, $u_{\text{a}}=12\text{m}/\text{s}$.

Velocity of mass B before collision, $u_{\text{b}}=-12\text{m}/\text{s}$.

Formula used:

Final velocity of mass A using conservation of momentum is calculated as,

  $v_{\text{a}}=\left(\frac{m_{\text{a}}-m_{\text{b}}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{a}}+2\left(\frac{m{\text{b}}_{}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{b}}$

Final velocity of mass B using conservation of momentum is calculated as,

  $v_{\text{b}}=\left(\frac{m_{\text{b}}-m_{\text{a}}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{b}}+2\left(\frac{m{\text{a}}_{}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{a}}$

Initial momentum is calculated as,

  $p_{\text{i}}=m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}$

Final momentum is calculated as,

  $p_{\text{f}}=m_{\text{a}}{v}_{\text{a}}+m_{\text{b}}{v}_{\text{b}}$

Change in momentum is calculated as,

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

Initial kinetic energy is calculated as,

  $K.E_{\text{i}}=\frac{1}{2}{m}_{\text{a}}{\left(u_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(u_{\text{b}}\right)}^2$

Final kinetic energy is calculated as,

  $K.E_{\text{f}}=\frac{1}{2}{m}_{\text{a}}{\left(v_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(v_{\text{b}}\right)}^2$

Change in kinetic energy is calculated as,

  $\Delta K.E=K.E_{\text{f}}-K.E_{\text{i}}$

Calculation:

Final velocity of mass A using conservation of momentum is calculated as,

  $\begin{array}{c}{v}_{\text{a}}=\left(\frac{m_{\text{a}}-m_{\text{b}}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{a}}+2\left(\frac{m{\text{b}}_{}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{b}}\\ =\left(\frac{3m-m}{3m+m}\right)\left(12\text{m}/\text{s}\right)+2\left(\frac{m}{3m+m}\right)\left(-12\text{m}/\text{s}\right)\\ =\left(\frac{2}{4}\right)\left(12\text{m}/\text{s}\right)+2\left(\frac{1}{4}\right)\left(-12\text{m}/\text{s}\right)\\ =0\text{m}/\text{s}\end{array}$

Final velocity of mass B using conservation of momentum is calculated as,

  $\begin{array}{c}{v}_{\text{b}}=\left(\frac{m_{\text{b}}-m_{\text{a}}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{b}}+2\left(\frac{m{\text{a}}_{}}{m_{\text{a}}+m_{\text{b}}}\right)u_{\text{a}}\\ =\left(\frac{m-3m}{3m+m}\right)\left(-12\text{m}/\text{s}\right)+2\left(\frac{3m}{3m+m}\right)\left(12\text{m}/\text{s}\right)\\ =\left(-\frac{2}{4}\right)\left(-12\text{m}/\text{s}\right)+2\left(\frac{3}{4}\right)\left(12\text{m}/\text{s}\right)\\ =24\text{m}/\text{s}\end{array}$

Initial momentum is calculated as,

  $\begin{array}{c}{p}_{\text{i}}=m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}\\ =3m\left(12\text{m}/\text{s}\right)+m\left(-12\text{m}/\text{s}\right)\\ =\left(36m-12m\right)\text{kg}\cdot \text{m}/\text{s}\\ =\left(24m\right)\text{kg}\cdot \text{m}/\text{s}\end{array}$

Final Momentum is calculated as,

  $\begin{array}{c}{p}_{\text{f}}=m_{\text{a}}{v}_{\text{a}}+m_{\text{b}}{v}_{\text{b}}\\ =m\left(0\text{m}/\text{s}\right)+m\left(24\text{m}/\text{s}\right)\\ =\left(24m\right)\text{kg}\cdot \text{m}/\text{s}\end{array}$

Change in momentum is calculated as,

  $\begin{array}{c}\Delta p=p_{\text{f}}-p_{\text{i}}\\ =\left(24m\right)\text{kg}\cdot \text{m}/\text{s}-\left(24m\right)\text{kg}\cdot \text{m}/\text{s}\\ =0\end{array}$

Initial kinetic energy is calculated as,

  $\begin{array}{c}K.E_{\text{i}}=\frac{1}{2}{m}_{\text{a}}{\left(u_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(u_{\text{b}}\right)}^2\\ =\frac{1}{2}\left(3m\right){\left(12\text{m}/\text{s}\right)}^2+\frac{1}{2}m{\left(-12\text{m}/\text{s}\right)}^2\\ =\left(216{{\text{m}}^2/\text{s}}^2\right)m+\left(72{{\text{m}}^2/\text{s}}^2\right)m\\ =\left(288{{\text{m}}^2/\text{s}}^2\right)m\end{array}$

Final kinetic energy is calculated as,

  $\begin{array}{c}K.E_{\text{f}}=\frac{1}{2}{m}_{\text{a}}{\left(v_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(v_{\text{b}}\right)}^2\\ =\frac{1}{2}\left(3m\right){\left(0\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(m\right){\left(24\text{m}/\text{s}\right)}^2\\ =0+\left(288{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)m\\ =\left(288{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)m\end{array}$

Change in kinetic energy is calculated as,

  $\begin{array}{c}\Delta K.E=K.E_{\text{f}}-K.E_{\text{i}}\\ =\left(288{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)m-\left(288{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)m\\ =0\end{array}$

Conclusion:

The velocity of mass A after the collision is $0\text{m}/\text{s}$.

The velocity of mass B after the collision is $24\text{m}/\text{s}$.

The system is consistent with both conservation of momentum and conservation of kinetic energy as,

  $\begin{array}{c}\Delta p=\Delta K.E\\ =0\end{array}$