Question

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 $60 °$ , and the velocity magnitude is reduced by half. The direction of the velocity of the second ball changes thereby by $90 °$ .

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.

Accepted Answer

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 $60 °$ . Therefore, the change in the momentum of the first ball is, $Δ p_{1} = F ∆ t = q_{1} E Δ t = m_{1} v \sin 60 °$ .

Here we use the condition that $v_{1} = \frac{v}{2}$ , which implies that the change in the momentum $Δ p_{1}$ of the first ball occurs in a direction perpendicular to the direction of its velocity $v_{1}$ .

Since $E ‖ Δ p_{1}$ and the direction of variation of the second ball momentum is parallel to the direction of $Δ p_{1}$ , we obtain for the velocity of the second ball (it can easily be seen that the charges on the balls have the same sign)

$v_{2} = v \tan 30 ° = \frac{v}{\sqrt{3}}$
The corresponding change in the momentum of the second ball is

$Δ p_{2} = q_{2} E Δ t = \frac{m_{2} v}{\cos 30 °}$

Hence we obtain,

$\frac{q_{1}}{q_{2}} = \frac{m_{1} \sin 60 °}{\frac{m_{2}}{\cos 30 °}}, \frac{q_{2}}{m_{2}} = \frac{4}{3} \frac{q_{1}}{m_{1}} = \frac{4}{3} k_{1}$ .

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