Chapter 20, Problem 69P

(a)

To determine

The velocity of the particles.

Answer to Problem 69P

The velocity of the particles is $6.00\widehat{i}\text{m/s}$.

Explanation of Solution

Write the expression from conservation of momentum.

${\overrightarrow{P}}_i={\overrightarrow{P}}_f$        (I)

Here, ${\overrightarrow{P}}_i$ is the initial momentum and ${\overrightarrow{P}}_f$ is the final momentum.

Write the equation for initial momentum.

  ${\overrightarrow{P}}_i=\left(m_1{\overrightarrow{v}}_{1i}\right)+\left(m_2{\overrightarrow{v}}_{2i}\right)$        (II)

Here, $m_1,m_2$ are the masses, ${\overrightarrow{v}}_{1i},{\overrightarrow{v}}_{2i}$ are the initial velocities of the objects.

Write the equation for final momentum at the distance of closest approach.

  ${\overrightarrow{P}}_f=\left(m_1+m_2\right){\overrightarrow{v}}_f$        (III)

Here, ${\overrightarrow{v}}_f$ are the final velocities of the objects.

Rewrite the expression from equation (I) by using (II), (III) in terms of final velocity.

$\begin{array}{l}\left(m_1{\overrightarrow{v}}_{1i}\right)+\left(m_2{\overrightarrow{v}}_{2i}\right)=\left(m_1+m_2\right){\overrightarrow{v}}_f\\ {\overrightarrow{v}}_f=\frac{\left(m_1{\overrightarrow{v}}_{1i}\right)+\left(m_2{\overrightarrow{v}}_{2i}\right)}{\left(m_1+m_2\right)}\end{array}$        (IV)

Conclusion:

Substitute, $2.00\text{g}$ for $m_1$, $5.00\text{g}$ for $m_2$, $21.0\widehat{i}\text{m/s}$ for ${\overrightarrow{v}}_{1i}$, $0\text{m/s}$ for ${\overrightarrow{v}}_{2i}$ in Equation (IV) to find ${\overrightarrow{v}}_f$.

Thus, the velocity of the particles is $6.00\widehat{i}\text{m/s}$.

(b)

To determine

The closest distance.

Answer to Problem 69P

The closest distance is $\underset{\_}{3.64\text{m}}$.

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$K_i=K_f+U_f$        (V)

Here, $K_i,K_f$ are the initial and final kinetic energy and $U_f$ is the final potential energy.

Write the equation for initial kinetic energy.

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

Write the equation for final kinetic energy.

  $K_f=\frac{1}{2}\left(m_1+m_2\right){\left(v_f\right)}^2$        (VII)

Write the equation for final potential energy.

  $U_f=\frac{k{q}_1{q}_2}{r}$        (VIII)

Here, $q_1,q_2$ are the charges, $k$ is Coulomb’s constant, $r$ is distance.

Rewrite the expression from equation (V) by using (VI), (VII) and (VIII) in terms of distance.

$r=\frac{2k{q}_1{q}_2\left(m_1+m_2\right)}{m_1{m}_2{v}_{1i}^2}$        (IX)

Conclusion:

Substitute, $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $15.0\text{μC}$ for $q_1$, $8.50\text{μC}$ for $q_2$, $2.00\text{g}$ for $m_1$, $5.00\text{g}$ for $m_2$, $21.0\text{m/s}$ for $v_{1i}$ in Equation (IX) to find $r$.

$\begin{array}{c}r=\frac{2\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(15.0\text{μC}\right)\left(8.50\text{μC}\right){\left(\frac{1\times {10}^{-6}\text{C}}{1\text{μC}}\right)}^2\right]\left[\left\{\begin{array}{l}\left(2.00\text{g}\right)+\\ \left(5.00\text{g}\right)\end{array}\right\}\left(\frac{1\times {10}^{-3}\text{kg}}{1\text{g}}\right)\right]}{\left[\left\{\left(2.00\text{g}\right)\left(5.00\text{g}\right)\right\}{\left(\frac{1\times {10}^{-3}\text{kg}}{1\text{g}}\right)}^2\right]{\left(21.0\text{m/s}\right)}^2}\\ =3.64\text{m}\end{array}$

Thus, the closest distance is $\underset{\_}{3.64\text{m}}$.

(c)

To determine

The velocity of first particle.

Answer to Problem 69P

The velocity of first particle is $\underset{\_}{-9.00\widehat{\text{i}}\text{m/s}}$.

Explanation of Solution

The initial velocity of second particle is zero.

Write the expression from relative velocity equation.

$v_{1i}=v_{2f}-v_{1f}$        (X)

Rewrite the expression from equation (IV) by using (X) in terms of final velocity of first particle.

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

Conclusion:

Substitute, $2.00\text{g}$ for $m_1$, $5.00\text{g}$ for $m_2$, $21.0\text{m/s}$ for $v_{1i}$ in Equation (XI) to find $v_{1f}$.

  $\begin{array}{c}{v}_{1f}=\left(\frac{2.00\text{g}-5.00\text{g}}{2.00\text{g}+5.00\text{g}}\right)\left(21.0\text{m/s}\right)\\ =-9.00\text{m/s}\end{array}$

Thus, the velocity of first particle is $\underset{\_}{-9.00\widehat{\text{i}}\text{m/s}}$.

(d)

To determine

The velocity of second particle.

Answer to Problem 69P

The velocity of second particle is $\underset{\_}{12.0\widehat{\text{i}}\text{m/s}}$.

Explanation of Solution

Write the expression from relative velocity equation.

$v_{1i}=v_{2f}-v_{1f}$        (XII)

Conclusion:

Substitute, $-9.00\text{m/s}$ for $v_{1f}$, $21.0\text{m/s}$ for $v_{1i}$ in Equation (XII) to find $v_{2f}$.

  $\begin{array}{l}\left(21.0\text{m/s}\right)=v_{2f}-\left(-9.00\text{m/s}\right)\\ v_{2f}=12.0\text{m/s}\end{array}$

Thus, the velocity of second particle is $\underset{\_}{12.0\widehat{\text{i}}\text{m/s}}$.