Question

A block of mass $1\text{kg}$ containing a net positive charge ${10}^{-10}\text{C}$ is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure. The distance of the block from the wall is $2\text{m}$. A horizontal electric field ${10}^{10}\text{N/C}$ towards right is switched on. Assuming elastic collision (if any), find the time period of the resulting oscillatory motion.

• A. $1\text{s}$
• B. $2\text{s}$
• C. $3\text{s}$
• D. $4\text{s}$

Answer 1

The correct option is D $4\text{s}$

Force on the block, $F=qE$
$\Rightarrow F={10}^{-10}\times {10}^{10}=1\text{N}$
Acceleration toward left, $a=\frac{F}{m}$
$\Rightarrow a=\frac{1}{1}=1{\text{m/s}}^2$
Before collision, the electric force will accelerate the block and after collision (elastic), the electric force will retard the block.
Just before collision:
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 2=\frac{1}{2}\times 1\times t^2$
$\Rightarrow t=2\text{s}\dots \dots \left(1\right)$
Now,
$v=u+at$
$\Rightarrow v=0+1\times 2$
$\Rightarrow v=2\text{m/s}$
After collision:
As collision is elastic, the direction of velocity will reverse but magnitude remains the same.
Therefore, it will take same time to return to original position.
So, total time of oscillation,
$T=2+2=4\text{s}$ [from (1) and (2)]