Question

For the given object (O) and mirror arrangement, the velocity $($in $\text{cm/s})$ of the image will be

• A. $-2\hat{i}-3\hat{j}$
• B. $-2\hat{i}+3\hat{j}$
• C. $+2\hat{i}-3\hat{j}$
• D. $+2\hat{i}+3\hat{j}$

Answer 1

The correct option is A $-2\hat{i}-3\hat{j}$

For the given concave mirror,
$u=-60\text{cm},f=-20\text{cm}$
Velocity of object,
$v_o=10\cos {37}^{\circ }\hat{i}+10\sin {37}^{\circ }\hat{j}=(8\hat{i}+6\hat{j})\text{cm/s}$
Using mirror formula,
$\frac{1}{v}-\frac{1}{60}=\frac{-1}{20}$
$\Rightarrow \frac{1}{v}=\frac{-1}{30}$
$\Rightarrow v=-30\text{cm}$
So, $m=\frac{-v}{u}=\frac{-(-30)}{-60}=\frac{-1}{2}$
Also, component of velocity of the image along the principal axis $(x-$axis$)$,
$v_i=-m^2{v}_o$
$\Rightarrow v_i=-{\left(\frac{-1}{2}\right)}^2\times 8=-2\text{cm/s}$
Component of velocity of the image moving perpendicular to the principal axis $(y-$axis$)$,
$v_i=m{v}_o$
$\Rightarrow v_i=\left(\frac{-1}{2}\right)\times 6=-3\text{cm/s}$
So, net velocity of the image,
$v_i=(-2\hat{i}-3\hat{j})\text{cm/s}$

$\text{Why this question?}$ $\text{Tips: The continuous practice of resolving object velocity in parallel and perpendicular direction and applying velocity formula will give you an edge in such problems.}$