Chapter 8, Problem 91P

To determine

The total kinetic energy before collision in terms of $m_1,m_2\text{ and }{p}_1$ , the total kinetic energy after collision in terms of $m_1,m_2\text{ and }{p^{\prime }}_1$ and the situation for ${p^{\prime }}_1=+p_1$ .

Answer to Problem 91P

The total kinetic energy before collision in terms of $m_1,m_2\text{ and }{p}_1$ is ${\left(K.E\right)}_I=\frac{p_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)$ , The total kinetic energy after collision in terms of $m_1,m_2\text{ and }{p^{\prime }}_1$ is ${\left(K.E\right)}_F=\frac{{p^{\prime }}_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)$ . If ${p^{\prime }}_1=-p_1$ , then particle collides with each other and rebounds back and have same speed. The situation ${p^{\prime }}_1=+p_1$ can only occur if particle does not collide, but they pass by each other.

Explanation of Solution

Given:

The mass of particle 1 is $m_1$ .

The mass of particle 2 is $m_2$ .

Formula Used:

The expression for conservation of momentum is given by,

  ${p^{\prime }}_1+{p^{\prime }}_2=p_{cm}$

The expression for kinetic energy before collision is given by,

  $$

  ${\left(K.E\right)}_I=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2$

The expression for kinetic energy after collision is given by,

  $$

  ${\left(K.E\right)}_F=\frac{1}{2}{m}_1{v^{\prime }}_1{}^2+\frac{1}{2}{m}_2{v^{\prime }}_2{}^2$

The expression for conservation of energy is,

  ${\left(K.E\right)}_I={\left(K.E\right)}_F$

Calculation:

The expression for conservation of momentum is calculated as,

  $\begin{array}{c}{p^{\prime }}_1+{p^{\prime }}_2=p_{cm}\\ {p^{\prime }}_1+{p^{\prime }}_2=0\\ {p^{\prime }}_2=-{p^{\prime }}_1\end{array}$

The expression for kinetic energy before collision is calculated as,

  $\begin{array}{c}{\left(K.E\right)}_I=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2\\ =\frac{1}{2}{m}_1{v}_1{}^2\cdot \frac{m_1}{m_1}+\frac{1}{2}{m}_2{v}_2{}^2\cdot \frac{m_2}{m_2}\\ =\frac{m_1^2{v}_1{}^2}{2m_1}+\frac{m_1^2{v}_2{}^2}{2m_2}\\ =\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}\end{array}$

Further simplify the above,

  $\begin{array}{c}{\left(K.E\right)}_I=\frac{p_1^2}{2m_1}+\frac{{\left(-p_1\right)}^2}{2m_2}\\ =\frac{p_1^2}{2m_1}+\frac{p_1{}^2}{2m_2}\\ {\left(K.E\right)}_I=\frac{p_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\end{array}$

The expression for kinetic energy after collision is calculated as,

  $\begin{array}{c}{\left(K.E\right)}_F=\frac{1}{2}{m}_1{v^{\prime }}_1{}^2+\frac{1}{2}{m}_2{v^{\prime }}_2{}^2\\ =\frac{1}{2}{m}_1{v^{\prime }}_1{}^2\cdot \frac{m_1}{m_1}+\frac{1}{2}{m}_2{v^{\prime }}_2{}^2\cdot \frac{m_2}{m_2}\\ =\frac{m_1^2{v^{\prime }}_1{}^2}{2m_1}+\frac{m_1^2{v^{\prime }}_2{}^2}{2m_2}\\ =\frac{{p^{\prime }}_1^2}{2m_1}+\frac{{p^{\prime }}_2^2}{2m_2}\end{array}$

Further simplify the above,

  $\begin{array}{c}{\left(K.E\right)}_F=\frac{{p^{\prime }}_1^2}{2m_1}+\frac{{\left(-{p^{\prime }}_1\right)}^2}{2m_2}\\ =\frac{{p^{\prime }}_1^2}{2m_1}+\frac{{p^{\prime }}_1{}^2}{2m_2}\\ {\left(K.E\right)}_F=\frac{{p^{\prime }}_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\end{array}$

The expression for conservation of energy is calculated as,

  $\begin{array}{c}{\left(K.E\right)}_F={\left(K.E\right)}_I\\ \frac{{p^{\prime }}_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=\frac{p_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\\ {p^{\prime }}_1^2=p_1^2\\ {p^{\prime }}_1=\pm p_1\end{array}$

Conclusion:

Therefore, the total kinetic energy before collision in terms of $m_1,m_2\text{ and }{p}_1$ is. ${\left(K.E\right)}_I=\frac{p_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)$ , The total kinetic energy after collision in terms of $m_1,m_2\text{ and }{p^{\prime }}_1$ is ${\left(K.E\right)}_F=\frac{{p^{\prime }}_1^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)$ . If ${p^{\prime }}_1=-p_1$ , then particle collides with each other and rebounds back and have same speed. The situation ${p^{\prime }}_1=+p_1$ can only occur if particle does not collide, but they pass by each other.