Two balls of charge $q_1$ and $q_2$ initially have a velocity of the same magnitude and direction. After a…
Two balls of charge $q_1$ and $q_2$ initially have a velocity of the same magnitude and direction. After a uniform electric field has been applied during a certain time, the direction of the velocity of the first ball changes by $60^{\circ}$, and the velocity magnitude is reduced by half. The direction of the velocity of the second ball changes thereby by $90^{\circ}$.
In what proportion will the velocity of the second ball change? Determine the magnitude of the charge-to-mass ratio for the second ball if it is equal to $k_1$ for the first ball. The electrostatic interaction between the balls should be neglected.
$\frac{k}{\sqrt{2}}$
$\frac{k}{\sqrt{3}}$
$\frac{k}{2}$
$$\frac{4}{3} k_1$$
Solution
Let $v_1$ and $v_2$ be the velocities of the first and second balls after the removal of the uniform electric field. By hypothesis, the angle between the velocity $v_1$ and the initial velocity $v$ is $60^{\circ}$. Therefore, the change in the momentum of the first ball is
$
\Delta p_1=q_1 E \Delta t=m_1 v \sin 60^{\circ}
$
Here we use the condition that $v_1=v / 2$, which implies that the change in the momentum $\Delta \mathrm{p}_1$ of the first ball occurs in a direction perpendicular to the direction of its velocity $v_1$.
Since $\mathrm{E} \| \Delta \mathrm{p}_1$ and the direction of variation of the second ball momentum is parallel to the direction of $\Delta p_1$, we obtain for the velocity of the second ball (it can easily be seen that the charges on the balls have the same sign)
$
v_2=v \tan 30^{\circ}=\frac{v}{\sqrt{3}}
$
The corresponding change in the momentum of the second ball is
$
\Delta p_2=q_2 E \Delta t=\frac{m_2 v}{\cos 30^{\circ}}
$
Hence we obtain
$
\begin{aligned}
& \frac{q_1}{q_2}=\frac{m_1 \sin 60^{\circ}}{m_2 / \cos 30^{\circ}}, \\
& \frac{q_2}{m_2}=\frac{4}{3} \frac{q_1}{m_1}=\frac{4}{3} k_1 .
\end{aligned}
$