Question

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.

Answer 1

The velocity of the image,

(a) at a time t < d/V, is given by $\mathrm{V}_{\text {image }}=\frac{\mathrm{V} \mathrm{R}^{2}}{\left[2 \times(\mathrm{d}-\mathrm{v} \mathrm{t})-\mathrm{R}^{2}\right]^{2}}$.

(b) at a time t > d/V, is given by $V\left[1-\frac{R^{2}}{2(v t-d)-R^{2}}\right]\end$.

Explanation:

(a) when t<d/v

The target ia approaches the reflector

$\mathrm{V}_{\text {image }}=\frac{\text { velocity of the object } \times \mathrm{R}^{2}}{2 \times \text { distance between them }-\mathrm{R}^{2}}$

$\mathrm{V}_{\text {image }}=\frac{\mathrm{V} \mathrm{R}^{2}}{\left[2 \times(\mathrm{d}-\mathrm{v} \mathrm{t})-\mathrm{R}^{2}\right]^{2}}$  

At any time , x = d- Vt  

b) After a time t > d / V

A conflict between the mirror and the mass will occur. The body (mass) must come to rest as the collision is completely elastic and the mirror begins to move away with the same velocity V. At any point t > d / V, the mirror's distance from the mass is

$\mathrm{x}=\mathrm{v}\left(\mathrm{t}-\frac{\mathrm{d}}{\mathrm{V}}\right)=\mathrm{vt}-\mathrm{d}$

Here,

u = - (vt-d) = d-vt

$\mathrm{f}=-\frac{\mathrm{R}}{2}$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}=-\frac{R}{2}-\frac{1}{-(v t-d)}$

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

$\mathrm{v}=-\left[\frac{\mathrm{R}(\mathrm{d}-\mathrm{V} \mathrm{t})}{\mathrm{R}+2(\mathrm{d}-\mathrm{V} \mathrm{t})}\right]=\text { image distance }$

And picture velocity will be,

$\mathrm{V}_{\text {image }}=\frac{\mathrm{d}}{\mathrm{dt}}(\text { image distance })$

$\mathrm{V}_{\text {image }}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\mathrm{R}(\mathrm{d}-\mathrm{Vt})}{\mathrm{R}+2(\mathrm{d}-\mathrm{Vt})}\right]$

Let y = v-dt

   $\frac{d y}{d t}=-v$

$\mathrm{V}_{\text {image }}=\frac{\mathrm{d}}{\mathrm{dt}} \frac{\mathrm{Ry}}{\mathrm{R}+2 \mathrm{y}}$

$\mathrm{V}_{\text {image }}=\frac{(\mathrm{R}+2 \mathrm{y})(\mathrm{R}(-\mathrm{V})-\mathrm{Ry}(+2)(-\mathrm{V})}{(\mathrm{R}+2 \mathrm{y})^{2}}$

           $\begin{equation}=V r\left[\frac{(R+2 y)-2 y}{(R+2 y)^{2}}\right]\end$    

           $\begin{equation} =\frac{-V R^{2}}{(R+2 y)^{2}}\end$

Since, the mirror it passes with velocity V

Imagine velocity absolute  $\begin{equation}=V\left[1-\frac{V R^{2}}{(R+2 y)^{2}}\right]\end$

$\begin{equation}\text { since } \mathrm{V}=\mathrm{V}_{\text {image }}+\mathrm{V}_{\text {mirrer }}\end$

           $\begin{equation}=V\left[1-\frac{R^{2}}{2(v t-d)-R^{2}}\right]\end$

Thus, the velocity of the image has been determined.