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 (figure 18-E15).

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 *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 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)

(b)

(c)

Mass of the system= M

Mass of the two blocks A and B= m

Radii of curvature of two concave mirror= R

At time t=0

Distance between two block A and B= 2R

Block B is moving at a speed v towards mirror.

Original position of whole system, x=0

(a) At time t= R/v

Distance moved by block B towards mirror= v x R/v= R

The distance of Block A from mirror=, u=-R

Focal length of the mirror, f= -R/2

By using mirror formula,

For the block A,

The object distance, u= -2R

By using mirror formula,

So, with respect to the given coordinate system, Position of A and B are respectively from origin.

(b) At time t= 3R/v

The block B after colliding with mirror must have come to rest because it is an 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,

The object distance, u= -R

By using mirror formula,

Therefore, the position of the image of block A is at -2R with respect to the given coordinate system. For block B, the image of 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. Positions of images of A and B are = –2R, 0 from origin.

(c) at time t= 5R/v

in the similar way, it can be proved that the position of the image of block A and B will be -3R and -4R/3 respectively

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