Chapter 8, Problem 38TP

To determine

Initial and final momentum and kinetic energy, and to check whether momentum is conserved or not, and whether the collision is elastic or inelastic.

Answer to Problem 38TP

Initial momentum of body is $5.0\text{kg}\cdot \text{m}/\text{s}$.

Final momentum of body is $5.0\text{kg}\cdot \text{m}/\text{s}$.

Initial kinetic energy is $\text{4}\text{.165}\text{J}$.

Final kinetic energy is $\text{4}\text{.165}\text{J}$.

The momentum is conserved.

The collision is elastic.

Explanation of Solution

Given:

The positions of two objects before and after collision are depicted in the following figure.

  

  Figure (1)

Mass of object A, $m_1=1.0\text{kg}$.

Mass of object B, $m_2=3.0\text{kg}$.

Formula used:

Velocity of object from the figure is calculated as,

  $v=\frac{s_2-s_1}{t_2-t_1}$

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,

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

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 momentum is calculated as,

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

Change in kinetic energy is calculated as,

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

Calculation:

Initial velocity of object A, $u_{\text{a}}$ is calculated as,

  $\begin{array}{c}{u}_{\text{a}}=\frac{s_2-s_1}{t_2-t_1}\\ =\frac{1.5\text{m}-0\text{m}}{0.6\text{s}-0\text{s}}\\ =2.5\text{m}/\text{s}\end{array}$

Initial velocity of object B, $u_{\text{b}}$ is calculated as,

  $\begin{array}{c}{u}_{\text{b}}=\frac{s_2-s_1}{t_2-t_1}\\ =\frac{2.0-1.5\text{m}}{0.6-0\text{s}}\\ =0.83\text{m}/\text{s}\end{array}$

Final velocity of object A, $v_{\text{a}}$ is calculated as,

  $\begin{array}{c}{v}_{\text{a}}=\frac{s_2-s_1}{t_2-t_1}\\ =\frac{1.5-0\text{m}}{1.2-0.6\text{s}}\\ =2.5\text{m}/\text{s}\end{array}$

Final velocity of object B, $v_{\text{b}}$ is calculated as,

  $\begin{array}{c}{v}_{\text{b}}=\frac{s_2-s_1}{t_2-t_1}\\ =\frac{2.0\text{m}-1.5\text{m}}{1.2\text{s}-0.6\text{s}}\\ =0.83\text{m}/\text{s}\end{array}$

Initial momentum is calculated as,

  $\begin{array}{c}{p}_{\text{i}}=m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}\\ =\left(1.0\text{kg}\right)\left(2.5\text{m}/\text{s}\right)+\left(3.0\text{kg}\right)\left(0.83\text{m}/\text{s}\right)\\ =\left(2.5+2.5\right)\text{kg}\cdot \text{m}/\text{s}\\ =5.0\text{kg}\cdot \text{m}/\text{s}\end{array}$

Final Momentum is calculated as,

  $\begin{array}{c}{p}_{\text{f}}=m_{\text{a}}{v}_{\text{a}}+m_{\text{b}}{v}_{\text{b}}\\ =\left(1.0\text{kg}\right)\left(2.5\text{m}/\text{s}\right)+\left(3.0\text{kg}\right)\left(0.83\text{m}/\text{s}\right)\\ =\left(2.5+2.5\right)\text{kg}\cdot \text{m}/\text{s}\\ =5.0\text{kg}\cdot \text{m}/\text{s}\end{array}$

Initial kinetic energy is calculated as,

  $\begin{array}{c}K.E_{\text{i}}=\frac{1}{2}{m}_{\text{a}}{\left(u_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(u_{\text{b}}\right)}^2\\ =\frac{1}{2}\left(1.0\text{kg}\right){\left(2.5\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(3.0\text{kg}\right){\left(0.83\text{m}/\text{s}\right)}^2\\ =3.125+1.04\text{J}\\ =\text{4}\text{.165}\text{J}\end{array}$

Final kinetic energy is calculated as,

  $\begin{array}{c}K.E_{\text{f}}=\frac{1}{2}{m}_{\text{a}}{\left(v_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(v_{\text{b}}\right)}^2\\ =\frac{1}{2}\left(1.0\text{kg}\right){\left(2.5\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(3.0\text{kg}\right){\left(0.83\text{m}/\text{s}\right)}^2\\ =3.125+1.04\text{J}\\ =\text{4}\text{.165}\text{J}\end{array}$

Change in momentum is calculated as,

  $\begin{array}{c}\Delta p=p_{\text{f}}-p_{\text{i}}\\ =\left(5.0\text{kg}\cdot \text{m}/\text{s}\right)-\left(5.0\text{kg}\cdot \text{m}/\text{s}\right)\\ =0\end{array}$

Change in kinetic energy is calculated as,

  $\begin{array}{c}\Delta K.E=K.E_{\text{f}}-K.E_{\text{i}}\\ =\left(\text{4}\text{.165}\text{J}\right)-\left(\text{4}\text{.165}\text{J}\right)\\ =0\end{array}$

Conclusion:

Initial momentum of body is $5.0\text{kg}\cdot \text{m}/\text{s}$.

Final momentum of body is $5.0\text{kg}\cdot \text{m}/\text{s}$.

Initial kinetic energy is $\text{4}\text{.165}\text{J}$.

Final kinetic energy is $\text{4}\text{.165}\text{J}$.

The change in initial and final momentum is zero. Therefore, the momentum is conserved.

The change in initial and final kinetic energy is zero. Therefore, the collision is elastic in nature.