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)`

Explanation

At `t=(R)/(v)`, the position of blocks and mirror is as shown
`A:(1)/(v)+(1)/(-2R)=(1)/((-R)/(2))`
`(1)/(v)=-(2)/(R)+(1)/(2R)=(-4+1)/(2R)`
`v=(-2R)/(3)`
B: image of B is formed at B itself.
Position of images of A and B w.r.t. O
`x_A=(-2R)/(3),x_B=R`
(b) During collision of mirrors and block B, blocks B comes to rest, mirrors moves with speed v.
At `t=(3R)/(v)`:
Images of A and B are formed at the same position.
`x_A=-2R,x_B=0`
(c) During collision of mirrors and block A, mirrors comes to rest and block A moves with speed v.
At `t=(5R)/(v)`,
Image of A is formed on A itself i.e., `x_A=-3R`
B: `(1)/(v)+(1)/(-2R)=(1)/((-R)/(2))impliesv=(-2R)/(3)`
`x_B=-((2R-2R)/(3))=(-4R)/(3)`
`x_A:` distance of image of A from O
`x_B:` distance of image of B from O