Two concave mirrorsof equal radi of curvature R are fixed on a stand facing opposite directions. The whole system has a mass m nd is kept onn a frictionlesss horiztonal table figure

Two blocks A and B, each of mass m, are placed on the two side of the stnd. At t=0, teh separation between A and the mirrors is 2R and also the separation between B and the mirrors is 2R. The block B moves towards the mirror at a speed v. All collisions which take place are erastic. Taking the original position of the mirrors stand system to be x=0 and X -axis along AB, find the position of the image of A and B at t=
`a. R/(upsilon)`, b. `(3R)/(upsilon)` c. `(5R)/(upsilon)`

Options

A. At `t=(R)/(V),X_(A)=-(2r)/(3) and X_(B)=R`
B. At `t=(3R)/(V),X_(A)=-2R and X_(B)=0`
C. At `t=(5R)/(V),X_(A)=-3R and X_(B)=-(4R)/(3)`
D. At `t=(5R)/(V),X_(A)=-(4R)/(3) and X_(B)=-3R`

Answer

A, B, C

Explanation

At `t = (R)/(V)`
use mirror formula `- (2)/(R) = (1)/(V_(B)) - (1)/(R)`
`rArr V_(B) =-R`
Hence `X_(AB) = R`
similarly `X_(VA) = -(2R)/(3)`
(ii) At `t = (3R)/(3)`
As after `(2R)/(V)` time velocities are exchanged
`V_(A) = -R`
`:. X_(VA) = -2R`
and `V_(B) = -T`
`:. X_(VB) = 0`
(iii) at `t = (5R)/(V)`
As after `(2R)/(V)` time velocities are exchanged
Hence `V_(A) = -R` but position `X_(VA) = -3R`
and `V_(B) = -(2R)/(3)`
`:. X_(VB) = -(4R)/(3)`