A small block of mass m and a concave mirror of radius R fitted with a stand lie on a smooth horizontal table with a separation d between them. The mirror together with its stand has a mass m. The block is pushed at t = 0 towards the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time t < d/V, (b) at a time t > d/V.

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Solution

Note :
(a) At time t = t,
Object distance, u = −(d − Vt)
Here d > Vt, $f = - \frac{R}{2}$

Mirror formula is given by:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{2}{R} + \frac{1}{d - V t}$
$= \frac{- 2 (d - V t) + R}{R (d - V t)}$
$v = \frac{- R (d - V t)}{R - 2 (d - V t)}$

Differentiating w.r.t. 't':

$\frac{d v}{d t} = \frac{- RV [R - 2 (d - V t)] - 2 V [R (d - V t)]}{{[R - 2 (d - V t)]}^{2}} = - \frac{R^{2} V}{{[2 (d - V t) - R]}^{2}}$
This is the required speed of mirror.
(b) When $t > \frac{d}{2}$ the collision between the mirror and mass will take place. As the collision is elastic, the object will come to rest and the mirror will start to move with the velocity V.

u = (d − Vt).
At any time, $t > \frac{d}{2}$
The distance of mirror from the mass will be:
$x = V (t - \frac{d}{V}) = V t - d$
Here,
Object distance, u = $- (V t - d)$ = $(d - V t)$
Focal length, f = $- \frac{R}{2}$
Now, mirror formula is given by:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ = $\frac{1}{- \frac{R}{2}} - \frac{1}{d - V t}$
$v \Rightarrow - [\frac{R (d - V t)}{R - 2 (d - V t)}]$

Velocity of image $(V_{i m a g e})$ is given by,
$V_{i m a g e} = \frac{d v}{dt} = \frac{d}{dt} [\frac{R (d - V t)}{R + 2 (d - V t)}]$
If y = d $-$ Vt
$\frac{d y}{dt}$ = $- V$

Velocity of image:

$V_{i m a g e} = \frac{d}{d t} [\frac{R y}{R + 2 y}] = - V_{r} [\frac{R + 2 y - 2 y}{{(R + 2 y)}^{2}}] = \frac{- V R^{2}}{{(R + 2 y)}^{2}}$
As the mirror is moving with velocity V,
Therefore,
V=V_image + V_mirror
Absolute velocity of image = $V [1 - \frac{R^{2}}{{(R + 2 y)}^{2}}]$
= $V [1 - \frac{R^{2}}{2 (V t - d) - R^{2}}]$

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