Chapter 8, Problem 115P

(a)

To determine

The speed at which the astronauts move.

Answer to Problem 115P

The speed of the firs astronaut is $\left(\frac{-m_{\text{b}}}{{\left(m_1+m_{\text{b}}\right)}^2}\right)\left(\frac{m_2\left(2m_1+m_{\text{b}}\right)+m_1}{m_{\text{b}}+m_2}\right)v$ and the speed of the second astronaut is $\frac{m_{\text{b}}}{\left(m_{\text{b}}+m_2\right)}\left(1+\frac{m_1}{\left(m_1+m_{\text{b}}\right)}\right)v$ .

Explanation of Solution

Given:

The mass of the first astronaut is $m_1$

The mass of second astronaut is $m_2$ .

The mass of the ball is $m_{\text{b}}$

Calculation:

Let the speed of the ball be $v_{\text{b}}$ and the speed of the first astronaut be $v_1$ .

The expression for conservation of linear momentum related to the speed of astronaut and the ball after the first throw is given by,

  $m_1{v}_1+m_{\text{b}}{v}_{\text{b}}=0$ ....... (I)

The expression for the velocity of the ball in the laboratory frame relative to the first astronaut is given by,

  $\begin{array}{c}v=v_{\text{b}}-v_1\\ v_{\text{b}}=v+v_1\end{array}$

Equation (I) can be rewritten as,

  $\begin{array}{c}{m}_1{v}_1+m_{\text{b}}{v}_{\text{b}}=0\\ m_1{v}_1+m_{\text{b}}\left(v+v_1\right)=0\\ \left(m_1+m_{\text{b}}\right)v_1=-m_{\text{b}}v\\ v_1=\frac{-m_{\text{b}}v}{m_1+m_{\text{b}}}\end{array}$

The expression for the speed of the ball is given by,

  $\begin{array}{c}{v}_b=v-v_1\\ =v-\left(\frac{-m_{1}v}{m_1+m_{\text{b}}}\right)\\ =\frac{m_{\text{b}}{v}_{\text{b}}}{m_1+m_{\text{b}}}\end{array}$

The expression for conservation of momentum for the second astronaut is given by,

  $\begin{array}{c}{m}_{\text{b}}{v}_{\text{b}}=\left(m_2+m_{\text{b}}\right)v_2\\ v_2=\frac{m_{\text{b}}{v}_{\text{b}}}{\left(m_2+m_{\text{b}}\right)}\\ =\frac{m_{\text{b}}}{\left(m_2+m_{\text{b}}\right)}\left(\frac{m_{1}v}{m_1+m_b}\right)\\ =\left[\frac{m_{\text{b}}{m}_1}{\left(m_2+m_{\text{b}}\right)\left(m_1+m_{\text{b}}\right)}\right]v\end{array}$

Let $v_{\text{bf}}$ be the speed of the ball thrown by the second astronaut and $v_{\text{2f}}$ is the speed of the second astronaut when ball is thrown.

The expression for the conservation of momentum is given,

  $\left(m_2+m_{\text{b}}\right)v_2=m_{\text{b}}{v}_{\text{bf}}+m_2{v}_{\text{2f}}$ ....... (II)

The expression for the speed of the ball in laboratory frame relative to the second astronaut is given by,

  $\begin{array}{c}v=v_{2\text{f}}-v_{\text{bf}}\\ v_{\text{bf}}=v_{\text{2f}}-v\end{array}$

Equation (II) can be rewritten as,

  $\begin{array}{c}\left(m_2+m_{\text{b}}\right)v_2=m_{\text{b}}\left(v_{2\text{f}}-v\right)+m_2{v}_{2\text{f}}\\ \left(m_2+m_{\text{b}}\right)v_2+m_{\text{b}}v=\left(m_{\text{b}}+m_2\right)v_{2\text{f}}\\ v_{2\text{f}}=\frac{\left(m_2+m_{\text{b}}\right)v_2+m_{\text{b}}v}{\left(m_{\text{b}}+m_2\right)}\\ =v_2+\frac{m_{\text{b}}}{\left(m_{\text{b}}+m_2\right)}v\end{array}$

Solve further,

  $\begin{array}{c}{v}_{2\text{f}}=\left(\frac{m_{\text{b}}{m}_1}{\left(m_{\text{b}}+m_2\right)\left(m_1+m_{\text{b}}\right)}\right)v+\frac{m_{\text{b}}}{\left(m_{\text{b}}+m_2\right)}v\\ =\frac{m_{\text{b}}}{\left(m_{\text{b}}+m_2\right)}\left(1+\frac{m_1}{\left(m_1+m_{\text{b}}\right)}\right)v\end{array}$

The expression for the velocity of the ball in laboratory frame is given by,

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

The expression for the speed of the astronaut when the catch is taken as,

  $\begin{array}{c}{v}_{1\text{f}}=\frac{m_{\text{b}}{v}_{\text{bf}}+m_1{v}_1}{m_1+m_{\text{b}}}\\ =\left(\frac{m_{\text{b}}}{m_1+m_{\text{b}}}\right)\left(\left(\frac{m_{\text{b}}}{m_{\text{b}}+m_2}-1\right)\left(1+\frac{m_1}{m_1+m_{\text{b}}}\right)v\right)+\left(\frac{m_1}{m_1+m_{\text{b}}}\right)\left(\frac{-m_{\text{b}}v}{m_1+m_{\text{b}}}\right)\\ =\left(\frac{m_{\text{b}}}{m_1+m_{\text{b}}}\right)\left(\left(\left(\frac{-m_2}{m_{\text{b}}+m_2}\right)\left(\frac{2m_1+m_{\text{b}}}{m_1+m_{\text{b}}}\right)v\right)-\left(\frac{m_{1}v}{m_1+m_{\text{b}}}\right)\right)\\ =\left(\frac{-m_{\text{b}}}{{\left(m_1+m_{\text{b}}\right)}^2}\right)\left(\frac{m_2\left(2m_1+m_{\text{b}}\right)+m_1}{m_{\text{b}}+m_2}\right)v\end{array}$

Conclusion:

Therefore, the speed of the firs astronaut is $\left(\frac{-m_{\text{b}}}{{\left(m_1+m_{\text{b}}\right)}^2}\right)\left(\frac{m_2\left(2m_1+m_{\text{b}}\right)+m_1}{m_{\text{b}}+m_2}\right)v$ and the speed of the second astronaut is $\frac{m_{\text{b}}}{\left(m_{\text{b}}+m_2\right)}\left(1+\frac{m_1}{\left(m_1+m_{\text{b}}\right)}\right)v$ .

(b)

To determine

The change in system’s kinetic energy and from where this change came from.

Answer to Problem 115P

The change in system’s kinetic energy is $\frac{1}{2}\frac{m_{\text{b}}^2}{{\left(m_1+m_{\text{b}}\right)}^2{\left(m_{\text{b}}+m_2\right)}^2}\left(\frac{m_1{\left(m_2\left(\left(2m_1+m_{\text{b}}\right)+m_1\right)\right)}^2}{{\left(m_1+m_b\right)}^2}+m_2{\left(2m_1+m_{\text{b}}\right)}^2\right)v^2$ and this change came from the chemical energy stored in the astronaut’s bodies.

Explanation of Solution

Given:

The mass of the first astronaut is $m_1$

The mass of second astronaut is $m_2$ .

The mass of the ball is $m_{\text{b}}$

Formula Used:

Let $v_{\text{bf}}$ be the speed of the ball thrown by the second astronaut and $v_{\text{2f}}$ is the speed of the second astronaut when ball is thrown.

The expression for change in system’s kinetic energy is given by

  $\Delta K=\frac{1}{2}{m}_1{v}_{1\text{f}}^2+\frac{1}{2}{m}_2{v}_{2\text{f}}^2$

Calculation:

The change in systems kinetic energy is calculated as,

  $\begin{array}{l}\Delta K=\left[\begin{array}{l}\frac{1}{2}{m}_1{\left(\left(\frac{-m_{\text{b}}}{{\left(m_1+m_{\text{b}}\right)}^2}\right)\left(\frac{m_2\left(2m_1+m_{\text{b}}\right)+m_1}{m_{\text{b}}+m_2}\right)v\right)}^2+\\ \frac{1}{2}{m}_2{\left(\frac{m_{\text{b}}}{\left(m_{\text{b}}+m_2\right)}\left(1+\frac{m_1}{\left(m_1+m_{\text{b}}\right)}\right)v\right)}^2\end{array}\right]\\ =\frac{1}{2}\frac{m_{\text{b}}^2}{{\left(m_1+m_{\text{b}}\right)}^2{\left(m_{\text{b}}+m_2\right)}^2}\left(\frac{m_1{\left(m_2\left(\left(2m_1+m_{\text{b}}\right)+m_1\right)\right)}^2}{{\left(m_1+m_b\right)}^2}+m_2{\left(2m_1+m_{\text{b}}\right)}^2\right)v^2\end{array}$

This energy generally came from the chemical energy in the astronaut’s bodies.

Conclusion:

Therefore, change in system’s kinetic energy is $\frac{1}{2}\frac{m_{\text{b}}^2}{{\left(m_1+m_{\text{b}}\right)}^2{\left(m_{\text{b}}+m_2\right)}^2}\left(\frac{m_1{\left(m_2\left(\left(2m_1+m_{\text{b}}\right)+m_1\right)\right)}^2}{{\left(m_1+m_b\right)}^2}+m_2{\left(2m_1+m_{\text{b}}\right)}^2\right)v^2$ and this change came from the chemical energy stored in the astronaut’s bodies.