Chapter 25, Problem 25.56AP

Review. From a large distance away, a particle of mass m₁, and positive charge q₁ is fired at speed υ in the positive x direction straight toward a second particle, originally stationary but free to move, with mass m₂, and positive charge q₂. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity, (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the particle of mass m₁, and (d) the particle of mass m₂.

To determine

The velocity at the instant when the both particle moves with same velocity.

Answer to Problem 25.56AP

The velocity at the instant of closest approach when the both particle moves with same velocity is $\left(\frac{m_{1}v}{m_1+m_2}\right)$.

Explanation of Solution

The initial velocity of the second particle is $0$.

The momentum is conserved of an isolated system.

$m_1{\overrightarrow{v}}_{1\text{i}}+m_2{\overrightarrow{v}}_{\text{2i}}=m_1{\overrightarrow{v}}_{1\text{f}}+m_2{\overrightarrow{v}}_{\text{2f}}$

Here,

$m_1$ is the mass of the first particle.

$v_{1\text{i}}$ is the initial velocity of the first particle.

$m_2$ is the mass of the second particle.

$v_{\text{2i}}$ is the initial velocity of the second particle.

Substitute $0$ for ${\overrightarrow{v}}_{2\text{i}}$ , $v$ for ${\overrightarrow{v}}_{\text{1i}}$, $v_{\text{c}}$ for ${\overrightarrow{v}}_{1\text{f}}$ and $v_{\text{c}}$ ${\overrightarrow{v}}_{\text{2f}}$ for in above equation.

$\begin{array}{c}{m}_{1}v\widehat{\text{i}}+0=m_1{v}_{\text{c}}\widehat{\text{i}}+m_2{v}_{\text{c}}\widehat{\text{i}}\\ v_{\text{c}}=\left(\frac{m_{1}v}{m_1+m_2}\right)\end{array}$

Conclusion:

Therefore, the velocity at the instant of closest approach when the both particle moves with same velocity is $\left(\frac{m_{1}v}{m_1+m_2}\right)$.

To determine

The closet distance.

Answer to Problem 25.56AP

The closet distance is $\frac{2k_{\text{e}}{q}_1{q}_2\left(m_1+m_2\right)}{m_1{m}_2{v}^2}$.

Explanation of Solution

From part (a) the value of $v_{\text{c}}$ is $\left(\frac{m_{1}v}{m_1+m_2}\right)$.

Write the expression for initial the kinetic energy of first particle.

$k_{\text{1i}}=\frac{1}{2}{m}_1{v}_{\text{1i}}{}^2$

Here,

$m_1$ is the mass of the first particle.

$v_{1\text{i}}$ is the initial velocity of the first particle.

$k_{\text{1i}}$ is the initial kinetic energy of the first particle.

Write the expression for final the kinetic energy of first particle.

$k_{\text{1f}}=\frac{1}{2}{m}_1{v}_{\text{c}}{}^2$

Here,

$v_{\text{c}}$ is the final velocity of the first particle.

$k_{\text{1f}}$ is the final kinetic energy of the first particle.

Write the expression for initial the kinetic energy of second particle.

$k_{\text{2i}}=\frac{1}{2}{m}_2{v}_{\text{2i}}{}^2$

Here,

$m_2$ is the mass of the second particle.

$v_{\text{2i}}$ is the initial velocity of the second particle.

$k_{\text{2i}}$ is the initial kinetic energy of the second particle.

Write the expression for final the kinetic energy of second particle.

$k_{\text{2f}}=\frac{1}{2}{m}_2{v}_{\text{c}}{}^2$

Here,

$v_{\text{c}}$ is the final velocity of the second particle.

$k_{\text{2f}}$ is the final kinetic energy of the second particle.

Total initial kinetic energy is given by,

$k_{\text{i}}=k_{\text{1i}}+k_{\text{2i}}$

Substitute $\frac{1}{2}{m}_1{v}_{1\text{i}}{}^2$ for $k_{1\text{i}}$ and $\frac{1}{2}{m}_2{v}_{\text{2i}}{}^2$ for $k_{\text{2i}}$ in above equation.

$k_{\text{i}}=\frac{1}{2}{m}_1{v}_{1\text{i}}{}^2+\frac{1}{2}{m}_2{v}_{\text{2i}}{}^2$

Substitute $0$ for $v_{2\text{i}}$ and $v$ for $v_{1\text{i}}$ in above equation.

$\begin{array}{c}{k}_{\text{i}}=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{\left(0\right)}^2\\ =\frac{1}{2}{m}_1{v}^2\end{array}$

Total final kinetic energy is given by,

$k_{\text{f}}=k_{\text{1f}}+k_{\text{2f}}$

Substitute $\frac{1}{2}{m}_1{v}_{\text{c}}{}^2$ for $k_{1\text{i}}$ and $\frac{1}{2}{m}_2{v}_{\text{c}}{}^2$ for $k_{\text{2i}}$ in above equation.

$\begin{array}{c}{k}_{\text{f}}=\frac{1}{2}{m}_1{v}_{\text{c}}{}^2+\frac{1}{2}{m}_2{v}_{\text{c}}{}^2\\ =\frac{1}{2}\left(m_1+m_2\right)v_{\text{c}}{}^2\end{array}$

The initial electric potential energy is $0$.

$U_{\text{i}}=0$

Here,

$U_{\text{i}}$ is the initial electric potential energy.

The final electric potential energy is expressed as,

$U_{\text{f}}=\frac{k_{\text{e}}{q}_1{q}_2}{r_{\text{c}}}$

Here,

$k_{\text{e}}$ is the coulomb’s constant.

$r_{\text{c}}$ is the closest distance.

$q_1$ is first charge.

$q_2$ is the second charge.

$U_{\text{f}}$ is the final electric potential energy.

The energy is conserved within the isolated system.

$k_{\text{i}}+U_{\text{i}}=k_{\text{f}}+U_{\text{f}}$

Substitute $\frac{1}{2}{m}_1{v}^2$ for $k_{\text{i}}$, $0$ for $U_{\text{i}}$, $\frac{1}{2}\left(m_1+m_2\right)v_{\text{c}}{}^2$ for $k_{\text{f}}$ and $\frac{k_{\text{e}}{q}_1{q}_2}{r_{\text{c}}}$ for $U_{\text{f}}$ in above equation.

$\frac{1}{2}{m}_1{v}^2+0=\frac{1}{2}\left(m_1+m_2\right)v_{\text{c}}{}^2+\frac{k_{\text{e}}{q}_1{q}_2}{r_{\text{c}}}$

Substitute $\left(\frac{m_{1}v}{m_1+m_2}\right)$ for $v_{\text{c}}$ in above equation.

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

Conclusion:

Therefore, the closet distance is $\frac{2k_{\text{e}}{q}_1{q}_2\left(m_1+m_2\right)}{m_1{m}_2{v}^2}$.

To determine

The velocity of the particle of mass $m_1$.

Answer to Problem 25.56AP

The velocity of the particle of mass $m_1$ is $\left(\frac{m_1-m_2}{m_1+m_2}\right)v$.

Explanation of Solution

The expression for the relative velocity is,

$v_{\text{1i}}-v_{\text{2i}}=v_{\text{2f}}-v_{\text{1f}}$

Substitute $0$ for $v_{\text{2i}}$ and $v$ for $v_{\text{1i}}$ in above equation.

$\begin{array}{c}v-0=v_{\text{2f}}-v_{\text{1f}}\\ v_{\text{2f}}=v+v_{\text{1f}}\end{array}$

The overall elastic collision is described by the conservation of the momentum.

$m_1{\overrightarrow{v}}_{1\text{i}}+m_2{\overrightarrow{v}}_{\text{2i}}=m_1{\overrightarrow{v}}_{1\text{f}}+m_2{\overrightarrow{v}}_{\text{2f}}$

Substitute $0$ for $v_{2\text{i}}$ and $v$ for $v_{1\text{i}}$ in above equation.

$\begin{array}{c}{m}_{1}v\widehat{\text{i}}+0=m_1{v}_{1\text{f}}\widehat{\text{i}}+m_2{v}_{\text{2f}}\widehat{\text{i}}\\ m_{1}v=m_1{v}_{1\text{f}}+m_2{v}_{\text{2f}}\end{array}$

Substitute $v+v_{\text{1f}}$ for $v_{\text{2f}}$ in above equation.

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

Conclusion:

Therefore, the velocity of the particle of mass $m_1$ is $\left(\frac{m_1-m_2}{m_1+m_2}\right)v$.

To determine

The velocity of the particle of mass $m_2$.

Answer to Problem 25.56AP

The velocity of the particle of mass $m_2$ is $\left(\frac{2m_1}{m_1+m_2}\right)v$.

Explanation of Solution

From part (c) the value of $v_{\text{1f}}$ is $\left(\frac{m_1-m_2}{m_1+m_2}\right)v$.

From part (c) the expression for $v_{\text{2f}}$ is,

$v_{\text{2f}}=v+v_{\text{1f}}$

Substitute $\left(\frac{m_1-m_2}{m_1+m_2}\right)v$ for $v_{\text{1f}}$ in above equation.

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

Conclusion:

Therefore, the velocity of the particle of mass $m_2$ is $\left(\frac{2m_1}{m_1+m_2}\right)v$.