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 ( $D_{1}$ )

$\begin{matrix} Column 1 & Column 2 & Column 3 (I) & t = \frac{R}{4 V_{0}} & (i) & V_{1} = \frac{8}{9} V_{0} & (P) & D_{1} = 3 R (I I) & t = \frac{3 R}{5 V_{0}} & (i i) & V_{1} = 24 V_{0} & (Q) & D_{1} = \frac{R}{2} (I I I) & t = \frac{R}{V_{0}} & (i i i) & V_{1} = 3 V_{0} & (R) & D_{1} = \frac{2}{3} R (I V) & t = \frac{2 R}{V_{0}} & (i v) & V_{1} = 0 & (S) & D_{1} = R \end{matrix}$

Which of the following is correctly matched?

(II) (iv) (S)

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(III) (iii) (P)

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(I) (ii) (P)

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(III) (ii) (Q)

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Solution

The correct option is D (III) (ii) (Q)

Before collision After collision

at t = $\frac{R}{4 v_{0}} u = - \frac{R}{4}$

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

$V_{1} = - {(m)}^{2} V_{0} m + V_{m} = - {(2)}^{2}$ V +V = - 3V

at t = $\frac{3 R}{5 V_{0}} u = - \frac{3 R}{5}$

V= $\frac{4 f}{u - f} V = - 3 R D_{1} = | V |$ = 3R

$V_{1} = - {(5)}^{2} V_{0} + V_{0} = - 24 V_{0}$

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 = $\frac{2 R}{V_{0}}$ u = -2R

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

$V_{1} = - {(\frac{1}{3})}^{2} V_{0} + V_{0} = \frac{8}{9} V_{0}$

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