Question

A block of mass m having positive charge q is
placed on smooth horizontal table as shown. A
uniform horizontal electric field E, is switched on
at t = 0. If collision of block with wall is inelastic
with e= 1/2 then total distance travelled by the
WIT
block before it finally comes to rest, is
m​

Answer 1

Answer: 5d/4

Explanation:

Let the initial separation between the block and the wall be 'd'.

According to the figure in answer,  

Initially, block moves towards right due to electric field given towards right .

After colliding with wall,  it changes velocity but not acceleration.

Hence,  block slows down and comes to rest.

1) So,  

In initial condition,

Initial velocity, u = 0

Force on block due to electric field,  F=qE

Hence ,a = qE/m

t = time taken to collide with  the wall.

Then,  

Using

$d=ut+\frac{at^2}{2}\\ \\=>d=0+\frac{at^2}{2}\\ \\=>t=\sqrt{\frac{2d}{a}}$

2)  Then,  

v = u+at

=>v=0+at=at

Since, e = 1/2=0.5

We get,  

$e=\frac{v_{separation}}{v_{approach}}\\ \\=>0.5=\frac{v_s}{v}\\ \\=>v_s=0.5v$

3)  Now, after collision with wall,  acceleration is opposite to velocity .

We have,  

Using,

$v^2=u^2+2as\\ \\=>0=v_s^2-2as\\ \\=>0=(0.5v)^2-2as\\ \\=>0=0.25v^2-2as\\ \\=>s=\frac{v^2}{8a}\\ \\=>s=\frac{a^2t^2}{8a}\\ \\=>s=\frac{at^2}{8}\\ \\=>s=\frac{a*2d}{8a}=\frac{d}{4}$

Hence, total distance covered by block is d+d/4 = 5d/4.