Question

Two balls of charge q₁ and q₂ 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 k₁ for the first ball. The electrostatic interaction between the balls should be neglected.

Answer 1

Let v₁ and v₂ be the velocities of the first and second balls after the removal of the uniform electric field. By hypothesis, the angle between the velocity v₁ and the initial velocity v is 60°. Therefore, the change in the momentum of the first ball is

Here we use the condition that v₁ = v/2, which implies that the change in the momentum Δp₁ of the first ball occurs in a direction perpendicular to the direction of its velocity v₁.

Since E || Δp₁ and the direction of variation of the second ball momentum is parallel to the direction of Δp₁, we obtain for the velocity of the second ball (it can easily be seen that the charges on the balls have the same sign)

The corresponding change in the momentum of the second ball is