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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.

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 =
  1. $\frac{\text{R}}{\text{v}}$
  2. $\frac{3\text{R}}{\text{v}}$
  3. $\frac{5\text{R}}{\text{v}}$
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  1. In time, $\text{t}=\frac{\text{R}}{\text{V}}$ the mass B must have moved $\Big(\text{v}\times\frac{\text{R}}{\text{v}}\Big)=\text{R}$ closer to the mirror stand
So,

For the block B:

$\text{u}=-\text{R}, \ \text{f}=-\frac{\text{R}}{2}$

$\therefore \ \frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{2}{\text{R}}+\frac{1}{\text{R}}=-\frac{1}{\text{R}}$

$\Rightarrow\text{v}=-\text{R}$ at the same place

For the block A:

$\text{u}=-2\text{R}, \ \text{f}=-\frac{\text{R}}{2}$

$\therefore \ \frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{2}{\text{R}}+\frac{1}{2\text{R}}=\frac{-3}{2\text{R}}$

$\Rightarrow\text{v}=\frac{-2\text{R}}{3}$ image of A at $\frac{2\text{R}}{3}$ from PQ in the x-direction.

So, with respect to the given coordinate system,

$\therefore$ Position of A and B are $\frac{-2\text{R}}{3},$ R respectively from origin.
  1. When $\text{t}=\frac{3\text{R}}{\text{v}},$ the block B after colliding with mirror stand must have come to rest (elastic collision) and the mirror have travelled a distance R towards left form its initial position.
So, at this point of time,

For block A:

$\text{u}=-\text{R}, \ \text{f}=-\frac{\text{R}}{2}$

Using lens formula, v = -R (from the mirror),

So, position xA = -2R (from origin of coordinate system)

For block B:

Image is at the same place as it is R distance from mirror.

Hence, position of image is ‘0’.

Distance from PQ (coordinate system)

$\therefore$ positions of images of A and B are = -2R, 0 from origin.
  1. Similarly, it can be proved that at time $\text{t}=\frac{5\text{R}}{\text{v}},$
The position of the blocks will be -3R and $-\frac{4\text{R}}{3}$ respectively.
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