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 (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) $\frac{R}{v}$ (b) $\frac{3R}{v}$ (c) $\frac{5R}{v}$.
Figure

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 = $\frac{R}{v}$
The block B moved $\left(v\times \frac{R}{v}=R\right)$ R distance towards the mirror.
For block A,
object distance, u = − 2R
focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ =-\frac{2}{R}+\frac{1}{2R} \\ =-\frac{3}{2R} \end{gathered}$$
Therefore, v = − $\frac{2R}{3}$
Position of the image of block A is at $\frac{2R}{3}$ with respect to the given coordinate system.
For block B,
Object distance, u = − R
Focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \Rightarrow \frac{1}{v}=-\frac{2}{R}+\frac{1}{R} \\ \Rightarrow \frac{1}{v}=-\frac{1}{R} \end{gathered}$$
Therefore, v = − R
Position of the image of block B is at the same place.
Similarly,
(b)

At time $\frac{3R}{v}$
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 = − $\frac{R}{2}$
Using the mirror formula:
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ =-\frac{2}{R}+\frac{1}{R} \\ =-\frac{1}{R} \end{gathered}$$
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 $\frac{5R}{v}$
In a similar manner, we can prove that the position of the image of block A and B will be at − 3R and $\frac{-4R}{3}$ respectively.