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 following 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) `R/v`  (b)  `3R/v` (c) `5R/v`.

Answer 1

Given,
R is the radii of curvature of two concave mirrors and M is the mass of the whole system.
Mass of the two blocks A and B is m​.
As per the question,
At t = 0,
distance between block A and B is 2R
Block B is moving at a speed v towards the mirror.
Original position of the whole system at x = 0
(a)

At time t = `R/v`
The block B moved `( v xx R/v = R )` R distance towards the mirror.
For block A,
object distance, u = − 2R
focal length of the mirror, f = −`R/2`
Using the mirror formula:
`1/v + 1/u = 1/f`
`⇒ 1/v = 1/f - 1/u` 
`⇒ -2/R + 1/(2R)` 
Therefore, v = − 
Position of the image of block A is at  with respect to the given coordinate system.
 For block B,
 Object distance, u = − R
 Focal length of the mirror, f = −
Using the mirror formula: 
`1/v + 1/u = 1/f`

`⇒ 1/v = 1/f - 1/u`

`⇒ 1/v = -2/R + 1/R`

`⇒ 1/v = -1/R`
Therefore, v = − R
Position of the image of block B is at the same place.
Similarly,
(b)

At time `(3R)/v`
Block B, after colliding with the mirror must have come to rest because the collision is elastic. Due to this, the mirror has travelled a distance R towards the block A, i.e., towards left from its initial position.
So, at this time
For block A
Object distance, u = − R
Focal length of the mirror, f = −  `R/2`
Using the mirror formula: 
`1/v + 1/u = 1/f `
`⇒ 1/v = 1/f - 1/u`
`=-2/R + 1/R`
`=-1/R`

Therefore, v = − R 
Position of the image of block A is at − 2R with respect to the given coordinate system.

For Block B,
Image of the block B is at the same place as it is at a distance of R from the mirror.
Therefore, the image of the block B is zero with respect to the given coordinate system.
(c)

At time `( 5R )/v`
In a similar manner, we can prove that the position of the image of block A and B will be at − 3R and  `-(4R)/3` respectively.