Question

2 balls of charges Q1 and Q2 initially have same velocity are subjected to uniform electric field for same time as a result the velocity of the first ball is reduced to half of its initial value and its direction changes by 60 degree that the direction of the velocity of the second ball is found to change by 90 degree if the new velocity of second charged particle has a magnitude x times the initial velocity then find xPLEASE GIVE ME EXPLAIN ATION FOR IT

Answer 1

The magnitude of change in velocity of the second ball is $\frac{1}{\sqrt{3}}$ times the initial velocity "v".

Explanation:

Given data

• Charges of two balls = $q_1 \text {and} q_2$
• Two balls are subjected to same uniform electric field for same time
• Initial velocity is same for both the balls "v"
• Change in position for first ball = 60 degree
• Change in position for second ball = 90 degree
• Find the magnitude of the velocity 'x

In x- axis, the acceleration of the first ball is

$\left(a_{1}\right)_{x}=\frac{3 V}{4 t} \hat{i}$

In y-axis, $\left(a_{1}\right)_{y}=\frac{\sqrt{3} V}{4 t}(-\hat{j})$

$\begin{aligned}&\left(a_{2}\right)_{x}=\frac{-V}{t}(\hat{i})\\&\left(a_{2}\right)_{y}=\frac{-V^{\prime}}{t}(\hat{j})\end{aligned}$

$\overrightarrow{\mathrm{a}}_{1}=\frac{\overrightarrow{\mathrm{E}} \cdot \mathrm{q}_{1}}{\mathrm{m}}=-\frac{3 \mathrm{v}}{4 \mathrm{t}} \hat{\mathrm{i}}-\frac{\sqrt{3} \mathrm{v}}{4 \mathrm{t}} \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{E}}=\frac{-3 \mathrm{mv}}{4 \mathrm{q}_{1} \mathrm{t}} \hat{\mathrm{i}}-\frac{\sqrt{3} \mathrm{mv}}{4 \mathrm{q}_{1} \mathrm{t}}$

$\overrightarrow{\mathrm{a}}_{2}=\frac{\overrightarrow{\mathrm{E}} \mathrm{q}_{2}}{\mathrm{m}}=\frac{-3 \mathrm{v} \mathrm{q}_{2} \hat{\mathrm{i}}}{4 \mathrm{q}_{1} \mathrm{t}}-\frac{\sqrt{3} \mathrm{v} \mathrm{q}_{2}}{4 \mathrm{q}_{1} \mathrm{t}} \hat{\mathrm{j}}$

                 $=\frac{-v}{t} \hat{i}-\frac{v^{1}}{t} \hat{j}$

$\Rightarrow \frac{\mathrm{q}_{2}}{\mathrm{q}_{1}}=\frac{4}{3} \text { and } \mathrm{V}^{1}$

      $=\frac{\sqrt{3} v}{4}\left(\frac{4}{3}\right)$

$=\frac{v}{\sqrt{3}}$

Therefore the velocity of the second ball changed in magnitude is $\frac{1}{\sqrt{3}}$ times the initial velocity 'v', when the position of the first ball changed to 60 degree and second ball as 90 degree and have the same initial velocity.

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