Chapter 25, Problem 55AP

(a)

To determine

The velocity of the particles at the instant of closest approach.

Answer to Problem 55AP

The velocity of the particles at the instant of the closest approach is $6.00\widehat{i}\text{m}/\text{sec}$.

Explanation of Solution

Write the expression for the velocity of the closest approach using the law of conservation of energy.

    $v_{\text{c}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$                              (I)

Here, $v_{\text{c}}$ is the velocity at the closest approach, $m_1$ is the mass of the particle, $m_2$ is the mass of the second particle, $v_1$ is the velocity of the first particle AND $v_2$ is the velocity of the second particle.

Conclusion:

Substitute $2.00\text{gm}$ for $m_1$, $5.00\text{gm}$ for $m_2$, $21.0\widehat{i}\text{m}/\text{sec}$ for $v_1$ and $0.00\text{m}/\text{sec}$ for $v_2$ in equation (I) to calculate the value of $v_{\text{c}}$.

    $\begin{array}{c}{v}_{\text{c}}=\frac{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)\left(21.0\widehat{i}\text{m}/\text{sec}\right)+\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)\left(0\text{m}/\text{sec}\right)}{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)+\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}\\ =\frac{\left(42.0\widehat{i}\text{m}/\text{sec}\right)}{7}\\ =6.00\widehat{i}\text{m}/\text{sec}\end{array}$

Therefore, the velocity of the particles at the instant of the closest approach is $6.00\widehat{i}\text{m}/\text{sec}$.

(b)

To determine

The distance of the closest approach.

Answer to Problem 55AP

The distance of the closest approach is $3.64\text{m}$.

Explanation of Solution

Write the expression for the law of conservation of energy.

    $\left(K{E}_{\text{f}}+P{E}_{\text{f}}\right)=\left(K{E}_{\text{i}}+P{E}_{\text{i}}\right)$                          (II)

Here, $K{E}_{\text{f}}$ is the final kinetic energy, $K{E}_{\text{i}}$ is the initial kinetic energy, $P{E}_{\text{f}}$ is the final potential energy and $P{E}_{\text{i}}$ is the initial potential energy.

Substitute $\frac{1}{2}\left(m_1+m_2\right)v_{\text{c}}{}^2$ for $K{E}_{\text{f}}$, $\frac{k_{\text{e}}{q}_1{q}_2}{r}$ for $P{E}_{\text{f}}$, $\frac{1}{2}\left(m_1{v}_1{}^2+m_2{v}_2{}^2\right)$ for $K{E}_{\text{i}}$ and $0$ for $P{E}_{\text{i}}$ in equation (II) to get the expression for $r$.

    $\begin{array}{c}\left(\frac{1}{2}\left(m_1+m_2\right)v_{\text{c}}{}^2+\frac{k_{\text{e}}{q}_1{q}_2}{r}\right)=\left(\frac{1}{2}\left(m_1{v}_1{}^2+m_2{v}_2{}^2\right)+0\right)\\ \frac{1}{2}\left(m_1+m_2\right)v_{\text{c}}{}^2+\frac{k_{\text{e}}{q}_1{q}_2}{r}=\frac{1}{2}\left(m_1{v}_1{}^2+m_2{v}_2{}^2\right)\end{array}$                           (III)

Here, $q_1$ is the charge on the first particle, $q_2$ is the charge on the second particle, $k_{\text{e}}$ is the Coulomb’s constant and $r$ is the distance of closest approach.

Conclusion:

Substitute $2.00\text{gm}$ for $m_1$, $5.00\text{gm}$ for $m_2$, $21.0\widehat{i}\text{m}/\text{sec}$ for $v_1$, $0.00\text{m}/\text{sec}$ for $v_2$, $6.0\widehat{i}\text{m}/\text{sec}$ for $v_{\text{c}}$, $15.0\text{μC}$ for $q_1$, $8.50\text{μC}$ for $q_2$ and $8.9876\times {10}^9\text{N}$ for $k_{\text{e}}$ in equation (III) to calculate the value of $r$.

    $\begin{array}{c}\left[\begin{array}{l}\frac{1}{2}\left(\begin{array}{l}\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)+\\ \left(5.00\text{gm}\frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)\end{array}\right){\left(6.0\widehat{i}\text{m}/\text{sec}\right)}^2\\ +\left(\frac{\begin{array}{l}\left(8.9876\times {10}^9\text{N}\right)\left(15.0\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\\ \left(8.50\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\end{array}}{r}\right)\end{array}\right]=\left[\frac{1}{2}\left(\begin{array}{l}\left(\begin{array}{l}\left(2.00\text{gm}\frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)\\ {\left(21.0\widehat{i}\text{m}/\text{sec}\right)}^2\end{array}\right)\\ +\left(\begin{array}{l}\left(5.00\text{gm}\frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)\\ {\left(0.00\text{m}/\text{sec}\right)}^2\end{array}\right)\end{array}\right)\right]\\ 0.126{{\text{kg-m}}^2/\text{sec}}^2+\frac{1.13475{\text{N-C}}^{\text{2}}}{r}=0.441{{\text{kg-m}}^2/\text{sec}}^2\\ \frac{1.145919{\text{N-C}}^{\text{2}}}{r}=0.441{{\text{kg-m}}^2/\text{sec}}^2-0.126{{\text{kg-m}}^2/\text{sec}}^2\\ r=\frac{1.145919{\text{N-C}}^{\text{2}}}{0.315{{\text{kg-m}}^2/\text{sec}}^2}\end{array}$

Solving further,

    $r=3.64\text{m}$

Therefore, the distance of the closest approach is $3.64\text{m}$.

(c)

To determine

The velocity of $2.00\text{gm}$ particle.

Answer to Problem 55AP

The velocity of $2.00\text{gm}$ particle is $-9.00\widehat{i}\text{m}/\text{s}$.

Explanation of Solution

Since the collision between the charged particle is elastic. Hence, the expression for the final velocity of the first particle is,

    $v_{\text{1f}}=\left[\left(\frac{m_1-m_2}{m_1+m_2}\right)v_1+\left(\frac{2m_2}{m_1+m_2}\right)v_2\right]$                                 (IV)

Here, $v_{\text{1f}}$ is the final velocity of the initial particle.

Conclusion:

Substitute $2.00\text{gm}$ for $m_1$, $5.00\text{gm}$ for $m_2$, $21.0\widehat{i}\text{m}/\text{sec}$ for $v_1$ and $0.00\text{m}/\text{sec}$ for $v_2$ in equation (IV) to calculate the value of $r$.

    $\begin{array}{c}{v}_{\text{1f}}=\left[\begin{array}{l}\left(\frac{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)-\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)+\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}\right)21.0\widehat{i}\text{m}/\text{sec}+\\ \left(\frac{2\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)+\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}\right)0.0\text{m}/\text{sec}\end{array}\right]\\ =\frac{-3}{7}\left(21.0\widehat{i}\text{m}/\text{sec}\right)\\ =-9.00\widehat{i}\text{m}/\text{s}\end{array}$

Therefore, the velocity of the $2.00\text{gm}$ particle is $-9.00\widehat{i}\text{m}/\text{s}$.

(d)

To determine

The velocity of $5.00\text{gm}$ particle.

Answer to Problem 55AP

The velocity of $5.00\text{gm}$ particle is $12.0\widehat{i}\text{m}/\text{sec}$.

Explanation of Solution

Since the collision between the charged particle is elastic. Hence, the expression for the final velocity of the second particle is,

    $v_{\text{2f}}=\left[\left(\frac{m_1-m_2}{m_1+m_2}\right)v_2+\left(\frac{2m_1}{m_1+m_2}\right)v_1\right]$                                 (V)

Here, $v_{\text{2f}}$ is the final velocity of the second particle.

Conclusion:

Substitute $2.00\text{gm}$ for $m_1$, $5.00\text{gm}$ for $m_2$, $21.0\widehat{i}\text{m}/\text{sec}$ for $v_1$ and $0.00\text{m}/\text{sec}$ for $v_2$ in equation (V) to calculate the value of $r$.

    $\begin{array}{c}{v}_{\text{2f}}=\left[\begin{array}{l}\left(\frac{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)-\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)+\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}\right)0.0\text{m}/\text{sec}+\\ \left(\frac{2\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}{\left(2.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)+\left(5.00\text{gm}\times \frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)}\right)21.0\widehat{i}\text{m}/\text{sec}\end{array}\right]\\ =\frac{4}{7}\left(21.0\widehat{i}\text{m}/\text{sec}\right)\\ =12.0\widehat{i}\text{m}/\text{sec}\end{array}$

Therefore, the velocity of the $5.00\text{gm}$ particle is $12.0\widehat{i}\text{m}/\text{sec}$.