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 $t < \frac{d}{V}$ ,

(b) at a time $t > \frac{d}{V}$ .

Open in App

Solution

(a) At time t = t,

u = -(d - Vt)

Here d > Vt, $t = - \frac{R}{2}$

By mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

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

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

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

Differentiating w.r.t 't'

$\frac{d v}{d t} = \frac{- R V [R - 2 (d - V t)] - 2 V [R (d - V t)]}{[R - 2 (d - V t)]^{2}}$

$= - \frac{R^{2} V}{[R - 2 (d - V t)]^{2}}$

This is the required speed of mirror.

(b) Similar as above, using u = (d - Vt).

Suggest Corrections

Similar questions

\frac{R}{v}

\frac{3 R}{v}

\frac{5 R}{v}

M

m

v

f

V_{0}

(D_{1})

\begin{matrix} Column 1 & Column 2 & Column 3 (I) & t = \frac{R}{4 V_{0}} & (i) & V_{1} = \frac{8}{9} V_{0} & (P) & D_{1} = 3 R (II) & t = \frac{3 R}{5 V_{0}} & (ii) & V_{1} = 24 V_{0} & (Q) & D_{1} = \frac{R}{2} (III) & t = \frac{R}{V_{0}} & (iii) & V_{1} = 3 V_{0} & (R) & D_{1} = \frac{2}{3} R (IV) & t = \frac{2 R}{V_{0}} & (iv) & V_{1} = 0 & (S) & D_{1} = R \end{matrix}

V_{0}

(D_{1})

\begin{matrix} Column 1 & Column 2 & Column 3 (I) & t = \frac{R}{4 V_{0}} & (i) & V_{1} = \frac{8}{9} V_{0} & (P) & D_{1} = 3 R (II) & t = \frac{3 R}{5 V_{0}} & (ii) & V_{1} = 24 V_{0} & (Q) & D_{1} = \frac{R}{2} (III) & t = \frac{R}{V_{0}} & (iii) & V_{1} = 3 V_{0} & (R) & D_{1} = \frac{2}{3} R (IV) & t = \frac{2 R}{V_{0}} & (iv) & V_{1} = 0 & (S) & D_{1} = R \end{matrix}

m

\to u

m

Join BYJU'S Learning Program