Chapter 22, Problem 69E

(a)

To determine

The fractional transmitted energy.

Answer to Problem 69E

The fractional transmitted energy is $n_t=\frac{4m_1{m}_2}{{\left(m_1+m_2\right)}^2}$.

Explanation of Solution

Write the expression for conservation of momentum.

  $m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$

Here, $m_1$ is the mass of first body, $v_{1i}$ is the initial speed of first body, $v_{1f}$ is the final speed of first body, $m_2$ is the mass of second body, $v_{2i}$ is the initial speed of second body and $v_{2f}$ is the final speed of second body.

Substitute $0$ for $v_{2i}$ and rearrange the above equation for value $v_{2f}$.

  $v_{2f}=\frac{m_1}{m_2}\left(v_{1i}-v_{1f}\right)$

Write the expression for conservation of Kinetic energy.

  $\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$        (I)

Substitute $0$ for $v_{2i}$ in above expression and divide by $\frac{1}{2}{m}_1{v}_{1i}^2$.

  $1=\frac{v_{1f}^2}{v_{1i}^2}+\frac{\frac{1}{2}{m}_2{v}_{2f}^2}{\frac{1}{2}{m}_1{v}_{1i}^2}$

Substitute $n_t$ for  $\frac{\frac{1}{2}{m}_2{v}_{2f}^2}{\frac{1}{2}{m}_1{v}_{1i}^2}$ in above expression.

  $1=\frac{v_{1f}^2}{v_{1i}^2}+n_t$

Rearrange the above expression in terms of $n_t$.

  $n_t=1-\frac{v_{1f}^2}{v_{1i}^2}$        (II)

Here, $n_t$ is fractional transmitted energy.

Substitute $0$ for $v_{2i}$ and $\frac{m_1}{m_2}\left(v_{1i}-v_{1f}\right)$ for $v_{2f}$ in equation (I).

  $\begin{array}{c}\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{\left(0\right)}^2=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{\left[\frac{m_1}{m_2}\left(v_{1i}-v_{1f}\right)\right]}^2\\ m_1{v}_{1i}^2=m_1{v}_{1f}^2+\frac{m_1^2}{m_2}{\left(v_{1i}-v_{1f}\right)}^2\end{array}$

Simplify the above expression.

  $\begin{array}{c}{v}_{1i}^2-v_{1f}^2=\frac{m_1}{m_2}{\left(v_{1i}-v_{1f}\right)}^2\\ \left(v_{1i}+v_{1f}\right)\left(v_{1i}-v_{1f}\right)=\frac{m_1}{m_2}\left(v_{1i}-v_{1f}\right)\left(v_{1i}-v_{1f}\right)\\ v_{1i}+v_{1f}=\frac{m_1}{m_2}\left(v_{1i}-v_{1f}\right)\end{array}$

Rearrange the above expression in terms of $v_{1f}/v_{1i}$.

  $\frac{v_{1f}}{v_{1i}}=\frac{m_1-m_2}{m_1+m_2}$

Square both sides of above expression.

  ${\left(\frac{v_{1f}}{v_{1i}}\right)}^2={\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$

Substitute ${\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$ for ${\left(\frac{v_{1f}}{v_{1i}}\right)}^2$ in equation (II).

  $\begin{array}{c}{n}_t=1-{\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2\\ =\frac{{\left(m_1+m_2\right)}^2-{\left(m_1-m_2\right)}^2}{{\left(m_1+m_2\right)}^2}\end{array}$

Simplify the above expression.

  $n_t=\frac{4m_1{m}_2}{{\left(m_1+m_2\right)}^2}$        (III)

Conclusion:

Thus, the fractional transmitted energy is $n_t=\frac{4m_1{m}_2}{{\left(m_1+m_2\right)}^2}$.

(b)

To determine

The fractional reflected energy.

Answer to Problem 69E

The fractional reflected energy is $n_r={\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$.

Explanation of Solution

Write the expression for fractional reflected energy.

  $n_r=\frac{v_{1f}^2}{v_{1i}^2}$

Substitute ${\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$ for ${\left(\frac{v_{1f}}{v_{1i}}\right)}^2$ in above equation.

  $n_r={\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$        (IV)

Here, $n_r$ is fractional reflected energy.

Conclusion:

Thus, the fractional reflected energy is $n_r={\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$.

(c)

To determine

The fraction of energy transmitted and reflected.

Answer to Problem 69E

The fractional transmitted energy is $1$ and fractional reflected energy is $0$ for $m_1=m_2$; the fractional transmitted energy is $40/121$ and fractional reflected energy is $81/121$ for $m_1=10m_2$; the fractional transmitted energy is $0.04$ and fractional reflected energy is $0.96$ for $m_1=100m_2$.

Explanation of Solution

Conclusion:

For $m_1=m_2$:

Substitute $m_2$ for $m_1$ in equation (III).

  $\begin{array}{c}{n}_t=\frac{4m_2{m}_2}{{\left(m_2+m_2\right)}^2}\\ =\frac{4m_2^2}{4m_2^2}\\ =1\end{array}$

Substitute $m_2$ for $m_1$ in equation (IV).

  $\begin{array}{c}{n}_r={\left(\frac{m_2-m_2}{m_2+m_2}\right)}^2\\ ={\left(\frac{0}{2m_2}\right)}^2\\ =0\end{array}$

For $m_1=10m_2$:

Substitute $10m_2$ for $m_1$ in equation (III).

  $\begin{array}{c}{n}_t=\frac{4\left(10m_2\right)m_2}{{\left(10m_2+m_2\right)}^2}\\ =\frac{40m_2^2}{121m_2^2}\\ =\frac{40}{121}\end{array}$

Substitute $10m_2$ for $m_1$ in equation (IV).

  $\begin{array}{c}{n}_r={\left(\frac{10m_2-m_2}{10m_2+m_2}\right)}^2\\ ={\left(\frac{9m_2}{11m_2}\right)}^2\\ =\frac{81}{121}\end{array}$

For $m_1=100m_2$:

Substitute $100m_2$ for $m_1$ in equation (III).

  $\begin{array}{c}{n}_t=\frac{4\left(100m_2\right)m_2}{{\left(100m_2+m_2\right)}^2}\\ =\frac{400m_2^2}{10201m_2^2}\\ =0.04\end{array}$

Substitute $100m_2$ for $m_1$ in equation (IV).

  $\begin{array}{c}{n}_r={\left(\frac{100m_2-m_2}{10m_2+m_2}\right)}^2\\ ={\left(\frac{99m_2}{11m_2}\right)}^2\\ =\frac{9801}{10201}\\ =0.96\end{array}$

Thus, the fractional transmitted energy is $1$ and fractional reflected energy is $0$ for $m_1=m_2$; the fractional transmitted energy is $40/121$ and fractional reflected energy is $81/121$ for $m_1=10m_2$; the fractional transmitted energy is $0.04$ and fractional reflected energy is $0.96$ for $m_1=100m_2$.