Q32P

Question

Two positive point charges, and $q_{B}$ (masses $m_{A}$ and $m_{B}$ )are at rest, held together by a massless string of length . Now the string is cut, and the particles fly off in opposite directions. How fast is each one going, when they are far apart? $q_{A}$

Step-by-Step Solution

Verified

Answer

The final speed of the charge $q_{A}$ is $v_{A}' = \sqrt{\frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})} .$

The final speed of the charge $q_{B}$ is $v_{B}' = \sqrt{\frac{2 k q_{A} q_{B}}{m_{B} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})} .$

1Step 1: Determine expression for energy .

Write the expression for the rule of energy conservation,

${kE}_{i} + {PE}_{i} = {KE}_{f} + {PE}_{f}$

Here, $K E_{i}$ is the initial kinetic energy of the system, $K E_{f}$ final kinetic energy of the system, $P E_{i}$ is the initial potential energy of the system, and $P E_{f}$ is the final potential energy of the system.

Write the expression for initial total energy system,

$T E_{i} = K E_{1} = K E_{2} + P E$

Here, ${K E}_{1}$ is the initial kinetic energy of the charge $q_{A}, K E_{2}$ is the initial kinetic energy of the charge $q_{B}$ , and PE is the potential energy of the system of charges.

Write the expression for the initial kinetic energy of the charge $q_{A}$ is given by,

$K E_{1} = \frac{1}{2} m_{A} v_{A}^{2}$

Here, $m_{A}$ is the mass of the charge $q_{A}$ and $v_{A}$ is the initial velocity of the charge .

Now, write the expression for the initial kinetic energy of the charge $q_{B}$ is given by,

$K E_{2} = \frac{1}{2} m_{B} v_{B}^{2}$

Here, $m_{B}$ is the mass of the charge $q_{B}$ and $v_{B}$ is the initial velocity of the charge $V_{B}$ .

Write the expression for the potential energy of a two-charge system:

$P E = \frac{k q_{A} q_{B}}{r}$

Here, K is the Coulomb’s constant and r is the distance between the two charges.

2Step 2: Determine the expression for the initial energy

Substitute $\frac{1}{2} m_{A} v_{A}^{2}$ for $K E_{1,} \frac{1}{2} m_{B} v_{B}^{2}$ for $K E_{2}$ , and $\frac{k q_{A} q_{B}}{r}$ for PE in equation

$T E_{i} = K E_{1} + K E_{2} + P E . T E_{i} = \frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} v_{B}^{2} + \frac{{Kq}_{A} q_{B}}{r}$

Now substitute 0 m/s for $v_{B}$ and a for b in above equation.

$T E_{i} = \frac{1}{2} m_{A} {(0 m / s)}^{2} + \frac{1}{2} m_{B} {(0 m / s)}^{2} + \frac{K q_{A} q_{B}}{a} = \frac{k q_{A} q_{B}}{a}$

Here $K E_{1}'$ is the final kinetic energy of the charge $q_{A}$ and $K E_{2}'$ is the final kinetic energy of the charge $q_{B}$ . $K E_{2}'$

3Step 3: Determine the expression for the final energy.

Write the expression for the final kinetic energy of the charge $q_{A}$ is given by,

$K E_{1}' = \frac{1}{2} m_{A} {(v_{A}')}^{2}$

Here, $v_{A}'$ is the final velocity of the charge $q_{A}$ .

Now, write the expression for the final kinetic energy of the charge $q_{A}$ is given by, $K E_{2}' = \frac{1}{2} m_{B} {(v_{B}')}^{2}$

Here, $v_{B}'$ is the final velocity of the charge $q_{B}$ .

Substitute $\frac{1}{2} m_{A} {(v_{A}')}^{2}$ for $K E_{1}', \frac{1}{2} m_{B} {(v_{B}')}^{2}$ for $K E_{2}'$ in equation $T E_{f} = K E_{1}' + K E_{2}' .$

$T E_{f} = \frac{1}{2} m_{A} {(v_{A}')}^{2} + \frac{1}{2} m_{B} {(v_{B}')}^{2}$

The initial total energy of the system of charges is equal to the final total energy of the system of charges, and is given by, according to the law of conservation of energy.

$T E_{i} = T E_{f} /$

Substitute $\frac{k q_{A} q_{B}}{a}$ for $T E_{i}$ and $\frac{1}{2} m_{A} {(v_{A}')}^{2} + \frac{1}{2} m_{B} {(v_{B}')}^{2}$ for $T E_{f}$ in above equation.

$\frac{K q_{A} q_{B}}{a} = \frac{1}{2} m_{A} {(v_{A}')}^{2} + \frac{1}{2} m_{B} {(v_{B}')}^{2}$

4Step 4: Determine final initial and the final speed of the charge.

From the conservation of the linear momentum, consider the expression.

$P_{i} = P_{f}$

Here, $P_{i}$ is the initial momentum of the system of two charges and $P_{f}$ is the final momentum of the system of two charges.

Write the expression for the linear momentum is given by,

P = mv

Here, m is the mass and v is the velocity of the object.

Consider that, the initial momentum of the system of charges is zero.

Now write the expression for the final momentum of the charge $q_{A}$ .

$P_{A} = m_{A} v_{A}'$

Here, the direction of the charge velocity is opposite the direction of the charge velocity, as indicated by the negative sign.

Consider the final momentum and it is given by,

Here, the direction of the charge $q_{B}$ velocity is opposite the direction of the charge $q_{A}$ velocity, as indicated by the negative sign.

Consider the final momentum and it is given by,

$P_{f} = P_{A} + P_{B}$

Substitute the $m_{A} v_{A}'$ for $P_{A,} - m_{A} v_{A}'$ for $P_{B}$ in above equation.

$P_{f} = m_{A} v_{A}' - m_{B} v_{B}'$

Substitute 0 for $P_{i}$ and $m_{A} v_{A}' - m_{B} v_{B}'$ for $P_{f}$ in the equation $P_{i} = P_{f}$ .

$0 = m_{A} m_{A}' - m_{B} m_{B}' m_{A} m_{A}' = m_{B} m_{B}' v_{B}' = (\begin{matrix} \frac{m_{A}}{m_{B}} \end{matrix}) v_{A}'$

Substitute $(\begin{matrix} \frac{m_{A}}{m_{B}} \end{matrix}) v_{A}'$ for $v_{B}'$ in equation (3).

$\frac{k q_{A} q_{B}}{a} = \frac{1}{2} m_{A} {(v_{A}')}^{2} + \frac{1}{2} m_{B} {(\begin{matrix} (\begin{matrix} \frac{m_{A}}{m_{B}} \end{matrix}) v_{A}' \end{matrix})}^{2} {(v_{A}')}^{2} (\begin{matrix} \frac{1}{2} \end{matrix} m_{A}) + (\begin{matrix} 1 + (\begin{matrix} \frac{m_{A}}{m_{B}} \end{matrix}) \end{matrix}) = \frac{k q_{A} q_{B}}{a} {(v_{A}')}^{2} = \frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix}) v_{A}' = \sqrt{\frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})}$

Hence, the final speed of the charge $q_{A}$ is $v_{A}' = \sqrt{\frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})}$

Substitute the $\sqrt{\frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})}$ for $v_{A}'$ in equation ${V_{B}}^{'} = [\frac{m_{A}}{m_{B}}] {V_{A}}^{'}$ $v_{B}' = (\begin{matrix} \frac{m_{A}}{m_{B}} \end{matrix}) v_{A}'$

$v_{B}' = (\begin{matrix} \sqrt{\frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})} \end{matrix}) = \sqrt{\frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})}$

Hence, the final speed of the charge $q_{B}$ is $v_{B}' = \begin{matrix} \sqrt{\frac{2 k q_{A} q_{B}}{m_{A} a} (\begin{matrix} \frac{m_{B}}{m_{A} + m_{B}} \end{matrix})} . \end{matrix}$

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