Question

A mass of 1g carrying charge q falls through a potential difference V. The kinetic energy acquired by it is E. When a mass of 2g carrying the charge q falls through a potential difference V. What will be the kinetic energy acquired by it?
A). 0.25 E
B). 0.50 E
C). 0.75 E
D). E

Answer 1

Hint: To solve this question, we need to apply conservation of momentum theorem. Conservation of momentum states that for a collision occurring between 2 bodies in an isolated system, the total momentum of those bodies before collision will be equal to the momentum of those bodies after collision. With this approach we will solve the given question.

Complete step by step answer:
We must consider the basic formula of electrostatic force,
$\text{E = q }\times \text{ }{{\text{V}}_{\text{p}}}\text{ }\ldots \text{(i)}$
Where,
E = energy
q = charge
${{\text{V}}_{\text{p}}}$ = potential difference
In this,
Momentum is constant for both bodies.
Therefore, we can say,
$\text{Momentum (M) = mass (m) }\times \text{ velocity (v)}$
$\therefore \text{ m }\times \text{ v = constant}$
This implies that,
$\text{m }\; \alpha \; \text{ }\dfrac{1}{\text{v}}\text{ }\ldots \text{(ii)}$
We know, Kinetic Energy,
$\text{KE = }\dfrac{1}{2}\text{m}{{\text{v}}^{2}}$
According to equation (ii),
$\text{KE }\; \alpha\; \text{ m}{{\left( \dfrac{1}{\text{m}} \right)}^{2}}$
$\therefore \text{KE }\; \alpha\\!\\!\text{ }\left( \dfrac{1}{\text{m}} \right)$
We can consider ${{\text{m}}_{\text{1}}}\text{ = 1g}$ and ${{\text{m}}_{\text{2}}}\text{ = 2g}$.
With this into consideration,
Let Kinetic energy acquired by ${{\text{m}}_{\text{2}}}$ be ${{\text{E}}_{\text{2}}}$.
Let Kinetic energy acquired by ${{\text{m}}_{1}}$be $\text{E}$.
Hence, we can determine that,
$\dfrac{{{\text{E}}_{\text{2}}}}{\text{E}}\text{ = }\dfrac{1}{2}$
$\therefore \text{ }{{\text{E}}_{\text{2}}}\text{ = 0}\text{.5E}$
Hence, the correct option is Option B.

Note: The conservation of momentum, as we can see from the answer is simply the third law of Newton. During the collision, the forces acting on the colliding body are always equal and act opposite at every instant taken into consideration.