Question

A concave mirror of radius of curvature R is moving towards object of mass m with constant velocity $V_0$. Mirror make elastic collision with object at pole at t = 0. Mass of mirror is very large compare to mass of object. Column-1 Shown time, Column-2 shows speed of image of object, column-3 shows distance of image from pole of mirror at given time $\left(D_1\right)$



$\begin{array}{cccccc}& \text{Column 1}& & \text{Column 2}& & \text{Column 3}\\ \text{(I)}& t=\frac{R}{4V_0}& \text{(i)}& V_1=\frac{8}{9}{V}_0& \text{(P)}& D_1=3R\\ \text{(II)}& t=\frac{3R}{5V_0}& \text{(ii)}& V_1=24V_0& \text{(Q)}& D_1=\frac{R}{2}\\ \text{(III)}& t=\frac{R}{V_0}& \text{(iii)}& V_1=3V_0& \text{(R)}& D_1=\frac{2}{3}R\\ \text{(IV)}& t=\frac{2R}{V_0}& \text{(iv)}& V_1=0& \text{(S)}& D_1=R\end{array}$

Which of the following correctly matched

• A. (II) (iv) (S)
• B. (III) (iii) (P)
• C. (I) (ii) (P)
• D. (I) (iii) (Q)

Answer 1

The correct option is D (I) (iii) (Q)


To perceive this we can go on the reference of the mirror, then it will be analogous to the situation where a ball hits a wall elastically.
at $t=\frac{R}{4V_0}u=-\frac{R}{4}$

$V=\frac{uf}{u-f}\frac{(-\frac{R}{4})(-\frac{R}{2})}{-\frac{R}{4}+\frac{R}{2}}=\frac{R}{2}$

$V_1=-(m{)}^2{V}_0+V_0=-(2{)}^{2}V+V=-3V$ Here m is velocity magnification i.e $\frac{v^2}{u^2}$

Now, at $t=\frac{3R}{5V_0}u=-\frac{3R}{5}$

$V=\frac{4f}{u-f}V=-3R{D}_1=|V|=3R$

$V_1=-(5{)}^2{V}_0+V_0=-24V_0$

at $t=\frac{R}{V_0}u=-R$

$V=-R{D}_1=|V|=R$

$V_1=-(1{)}^2{V}_0+V_0=0$

at $t=\frac{2R}{V_0}u=-2R$

$V=\frac{uf}{u-f}=\frac{2}{3}RM=\frac{-V}{u}=\frac{1}{3}$

$V_1=-{\left(\frac{1}{3}\right)}^2{V}_0+V_0=\frac{8}{9}{V}_0$