Question

Two concave mirrors of equal radii of curvature R are fixed on a stand facing opposite directions. The whole system has a mass m and is kept on a frictionless horizontal table (figure). Two blocks A and B, each of mass m, are placed on the two sides of the stand. At t = 0, the separation between A and the mirrors is 2 R and also the separation between B and the mirrors is 2 R. The block B moves towards the mirror at a speed v. All collisions which take place are elastic. Taking the original position of the mirrors-stand system to be x = 0 and X-axis along AB, find the position of the images of A and B at t = (a) Rv (b) 3Rv (c) 5Rv.
Figure

Answer 1

The position of the images of A and B at t = (a) R/v (b) 3R/v (c) 5R/v has been derived.

Explanation:

(a) The mass B must have moved in time t = R / V :

        $\left(v \times \frac{R}{v}\right)=R$   Moving next to the mirror

So the block B :

$\begin{equation}\mathrm{u}=-\mathrm{R}\end$

$\begin{equation}\mathrm{f}=-\frac{\mathrm{R}}{2}\end$

$\begin{equation}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{-\frac{R}{2}}-\frac{1}{-R}\end$

$\begin{equation}\frac{1}{v}=\frac{2}{-R}-\frac{1}{-R}\end$

$\begin{equation}\frac{1}{v}=\frac{2-1}{-R}\end$

$\begin{equation}\frac{1}{v}=-\frac{1}{R}\end$

v = -R Same spot

To the Block A  :

u = -2R

$\begin{equation}f=-\frac{R}{2}\end$

$\begin{equation}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{-\frac{R}{2}}-\frac{1}{-2 R}\end$

$\begin{equation}\frac{1}{v}=\frac{2}{-R}-\frac{1}{-2 R}\end$

$\begin{equation}\frac{1}{v}=\frac{4-1}{-2 R}\end$

$\begin{equation}\frac{1}{v}=-\frac{3}{2 R}\end$

$\begin{equation}v=-\frac{2 R}{3}\end$

(b) When t = 3R/v :

Block B must have reached after colliding with the mirror stand rest (elastic collision) and the mirror established its initial position by moving a distance R to the left. So, right now, at this stage.

To Block A  :

u = -R

$\begin{equation}f=-\frac{R}{2}\end$      

Using the formula for lenses v = - R, from a mirror

So position XA = -2R, Source for managing framework

To block B  :

Picture at the same location as R distance from mirror.hence the picture position is zero

A and B Photo Positions are = -2R, 0 from origin  

(c) t = 3 R/v :

It can be proven similarly, that at the time  $\begin{equation}t=\frac{5 R}{V}\end$

The block position will be -3R and $\begin{equation}-\frac{4 R}{3}\end$ respectively.