Chapter 8, Problem 65P

(a)

To determine

The proof that $v_{\text{1f}}=v_{\text{2i}}$ and $v_{\text{2f}}=v_{\text{1i}}$ when two masses are equal.

Answer to Problem 65P

It is proved that $v_{\text{1f}}=v_{\text{2i}}$ and $v_{\text{2f}}=v_{\text{1i}}$ when two masses are equal.

Explanation of Solution

Formula used:

The expression for the final velocity of first object is given by,

  $v_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}$

The expression for the final velocity of second object is given by,

  $v_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}$

Calculation:

The expression for the final velocity of first object is calculated as,

  $\begin{array}{c}{v}_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}\\ =\frac{m_1-m_1}{m_1+m_1}{v}_{\text{1i}}+\frac{2m_1}{m_1+m_1}{v}_{\text{2i}}\\ =\frac{2m_1}{2m_1}{v}_{\text{2i}}\\ =v_{\text{2i}}\end{array}$

The expression for the final velocity of second object is given by,

  $\begin{array}{c}{v}_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}\\ =\frac{2m_1}{m_1+m_1}{v}_{\text{1i}}+\frac{m_1-m_1}{m_1+m_2}{v}_{\text{2i}}\\ =\frac{2m_1}{2m_1}{v}_{\text{1i}}\\ =v_{\text{1i}}\end{array}$

Conclusion:

Therefore, It is proved that $v_{\text{1f}}=v_{\text{2i}}$ and $v_{\text{2f}}=v_{\text{1i}}$ when two masses are equal.

(b)

To determine

The proof that $v_{\text{1f}}\approx -v_{\text{1i}}$ and $v_{\text{2f}}\approx 0$ when $m_2>>m_1$ and $v_{\text{2i}}=0$ .

Answer to Problem 65P

It is proved that $v_{\text{1f}}\approx -v_{\text{1i}}$ and $v_{\text{2f}}\approx 0$ when $m_2>>m_1$ and $v_{\text{2i}}=0$ .

Explanation of Solution

Calculation:

The expression for the final velocity of first object is calculated as,

  $\begin{array}{c}{v}_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}\\ =\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}\left(0\right)\\ =\frac{\frac{m_1}{m_2}-1}{\frac{m_1}{m_2}+1}{v}_{\text{1i}}\\ \approx \frac{0-1}{0+1}{v}_{\text{1i}}\end{array}$

Further simplify the above,

  $v_{\text{1f}}\approx -v_{\text{1i}}$

The expression for the final velocity of second object is given by,

  $\begin{array}{c}{v}_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}\\ =\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}\left(0\right)\\ =\frac{2\left(\frac{m_1}{m_2}\right)}{\left(\frac{m_1}{m_2}\right)+1}{v}_{\text{1i}}\\ \approx \frac{2\left(0\right)}{\left(0\right)+1}{v}_{\text{1i}}\end{array}$

Further simplify the above,

  $v_{\text{2f}}\approx 0$

Conclusion:

Therefore, it is proved that $v_{\text{1f}}\approx -v_{\text{1i}}$ and $v_{\text{2f}}\approx 0$ when $m_2>>m_1$ and $v_{\text{2i}}=0$ .

(c)

To determine

The proof that $v_{\text{1f}}\approx v_{\text{1i}}$ and $v_{\text{2f}}\approx 2v_{\text{1i}}$ when $m_1>>m_2$ and $v_{\text{2i}}=0$ .

Answer to Problem 65P

It is proved that $v_{\text{1f}}\approx v_{\text{1i}}$ and $v_{\text{2f}}\approx 2v_{\text{1i}}$ when $m_1>>m_2$ and $v_{\text{2i}}=0$ .

Explanation of Solution

Calculation:

The expression for the final velocity of first object is calculated as,

  $\begin{array}{c}{v}_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}\\ =\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}\left(0\right)\\ =\frac{1-\frac{m_2}{m_1}}{1+\frac{m_2}{m_1}}{v}_{\text{1i}}\\ \approx \frac{1-0}{1+0}{v}_{\text{1i}}\end{array}$

Further simplify the above,

  $v_{\text{1f}}\approx v_{\text{1i}}$

The expression for the final velocity of second object is given by,

  $\begin{array}{c}{v}_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}\\ =\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}\left(0\right)\\ =\frac{2\left(\frac{m_1}{m_1}\right)}{1+\left(\frac{m_2}{m_1}\right)}{v}_{\text{1i}}\\ \approx \frac{2}{1+0}{v}_{\text{1i}}\end{array}$

Further simplify the above,

  $v_{\text{2f}}\approx 2v_{\text{1i}}$

Conclusion:

Therefore, it is proved that $v_{\text{1f}}\approx v_{\text{1i}}$ and $v_{\text{2f}}\approx 2v_{\text{1i}}$ when $m_1>>m_2$ and $v_{\text{2i}}=0$ .