Question

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


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

Which of the following is correctly matched?

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

Answer 1

The correct option is B (IV) (i) (R)


Before collision After collision

at t = $\frac{R}{4v_0}u=-\frac{R}{4}$

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

$V_1=-{\left(m\right)}^2{V}_{0}m+V_m=-{\left(2\right)}^{2}V+V=-3V$

at t = $\frac{3R}{5V_0}u=-\frac{3R}{5}$ V= $\frac{4f}{u-f}V=-3R{D}_1=|V|=3R$

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

at t = $\frac{R}{v_0}$ u = - R V = - R $D_1=|V|$ = R

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

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

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

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