Question

Two concave mirrors 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 mirror is 2R and also the separation between B and the mirror is 2R. 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 standard system to be x = 0 and x-axis along AB, find the position of the image of A and B at; 

 (a) t = (R/v) 

(b) t = (3R/v) 

(c) t = 5R/v

Answer 1

 (a) At t = R/v 

For block A,

u = -2R 

∴ (1/v) + (1/-2R) = 2/-R 

or v = -2R/3 

For block B: The distance travels by block B in time (R/v) is R 

Thus                  u = -R 

(1/v) + (1/-R) = (2/-R) 

or                         v= -R

The x- coordinate of the image of the block with respect to the mirror will be +R 

(b) At                          t = 3R/v 

The block B will collide with the stand after time 2R/v. 

After collision block B becomes at rest and mirror starts moving with the same velocity v. In the remaining time R/v, the distance moved by the mirror is R. The position of blocks and mirror are shown in fig. (b)  

At this time the blocks lie at the centre of curvature of the respective mirrors. Their images will form at the centres of curvature. So their coordinates are:

 For block A,                               x = -R 

For block B,                                x = +R 

(c) At                                           t = 5R/v 

The block B will collide to the mirror after a time (2R/v), thereafter mirror starts moving towards block A with velocity v, At t = 4R/v, the mirror will collide with block A and stops after collision. The positions of blocks and mirror are shown in fig. (c)  

For block A: Its image will form on the same place. 

Therefore the positions of the blocks are xA = -3R 

For block B: u = -2R (1/v) + (1/-2R) = (2/-R) 

v = -2R/3 

The co-ordinates of B : – (2R-(2R/3)) = -4R/3