Chapter 20, Problem 70P

(a)

To determine

The velocity of the particles.

Answer to Problem 70P

The velocity of the particles is $v_f=\left(m_1{v}_{1i}/m_1+m_2\right)$.

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.

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

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

Conclusion:

Substitute, $\left(m_1+m_2\right){\overrightarrow{v}}_f$ for ${\overrightarrow{P}}_f$, $\left(m_1{\overrightarrow{v}}_{1i}\right)+\left(m_2{\overrightarrow{v}}_{2i}\right)$ for ${\overrightarrow{P}}_i$, $0\text{m/s}$ for ${\overrightarrow{v}}_{2i}$ in Equation (I) to find ${\overrightarrow{v}}_f$.

  $\begin{array}{c}\left(m_1{\overrightarrow{v}}_{1i}\right)+\left(m_2{\overrightarrow{v}}_{2i}\right)=\left(m_1+m_2\right){\overrightarrow{v}}_f\\ {\overrightarrow{v}}_f=\frac{\left(m_1{\overrightarrow{v}}_{1i}\right)+\left(m_2\left(0\right)\right)}{\left(m_1+m_2\right)}\\ =\frac{\left(m_1{\overrightarrow{v}}_{1i}\right)}{\left(m_1+m_2\right)}\end{array}$

Thus, the velocity of the particles is $v_f=\left(m_1{v}_{1i}/m_1+m_2\right)$.

(b)

To determine

The closest distance.

Answer to Problem 70P

The closest distance is $\underset{\_}{r\left(2k{q}_1{q}_2\left(m_1+m_2\right)/m_1{m}_2{v}_{1i}^2\right)}$.

Explanation of Solution

Here, initial potential energy is zero.

Write the expression from conservation of energy.

$K_i=K_f+U_f$        (IV)

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.

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

Write the equation for final kinetic energy.

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

Write the equation for final potential energy.

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

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

Conclusion:

Substitute, $\frac{1}{2}\left(m_1{\left(v_{1i}\right)}^2\right)+\frac{1}{2}\left(m_2{\left(v_{2i}\right)}^2\right)$ for $E_i$, $\frac{1}{2}\left(m_1+m_2\right){\left(v_f\right)}^2$ for $E_f$, $\frac{k{q}_1{q}_2}{r}$ for $U_f$$q_2$, ,$0\text{m/s}$ for $v_{2i}$ in Equation (VII) to find $r$.

$\begin{array}{l}\frac{1}{2}\left(m_1{\left(v_{1i}\right)}^2\right)+\frac{1}{2}\left(m_2{\left(0\right)}^2\right)=\frac{1}{2}\left(m_1+m_2\right){\left(v_f\right)}^2+\left[\frac{k{q}_1{q}_2}{r}\right]\\ m_1{m}_2{v}_{1i}^2=\frac{2k{q}_1{q}_2\left(m_1+m_2\right)}{r}\\ r=\frac{2k{q}_1{q}_2\left(m_1+m_2\right)}{m_1{m}_2{v}_{1i}^2}\end{array}$

Thus, the closest distance is $\underset{\_}{r\left(2k{q}_1{q}_2\left(m_1+m_2\right)/m_1{m}_2{v}_{1i}^2\right)}$.

(c)

To determine

The velocity of first particle.

Answer to Problem 70P

The velocity of first particle is $\underset{\_}{v_{1f}=\left(\left(m_1-m_2\right)v_{1i}/m_1+m_2\right)}$.

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)

Conclusion:

Substitute, $v_{1i}+v_{1f}$ for $v_{2f}$, $\left(m_1+m_2\right){\overrightarrow{v}}_f$ for ${\overrightarrow{P}}_f$, $\left(m_1{\overrightarrow{v}}_{1i}\right)+\left(m_2{\overrightarrow{v}}_{2i}\right)$ for ${\overrightarrow{P}}_i$, $0\text{m/s}$ for ${\overrightarrow{v}}_{2i}$ in Equation (I) to find $v_{1f}$.

  $\begin{array}{l}{m}_1{v}_{1i}=m_1{v}_{1f}+m_2\left(v_{1i}+v_{1f}\right)\\ v_{1f}=\left(\frac{m_1-m_2}{m_1+m_2}\right)v_{1i}\end{array}$

Thus, the velocity of first particle is $\underset{\_}{v_{1f}=\left(\left(m_1-m_2\right)v_{1i}/m_1+m_2\right)}$.

(d)

To determine

The velocity of second particle.

Answer to Problem 70P

The velocity of second particle is $\underset{\_}{\left(2m_1{v}_{1i}/m_1+m_2\right)}$.

Explanation of Solution

Write the expression from relative velocity equation.

$v_{1i}=v_{2f}-v_{1f}$        (XI)

Conclusion:

Substitute, $\left(\left(m_1-m_2\right)v_{1i}/m_1+m_2\right)$ for $v_{1f}$ in Equation (XI) to find $v_{2f}$.

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

Thus, the velocity of second particle is $\underset{\_}{\left(2m_1{v}_{1i}/m_1+m_2\right)}$.