Question

For the given object (O) and mirror arrangement, the velocity $($in $\text{m/s})$ of the image with respect to ground will be
$[$All the velocities are given with respect to ground$]$

• A. $+46\hat{i}-24\hat{j}$
• B. $+46\hat{i}+24\hat{j}$
• C. $-46\hat{i}-24\hat{j}$
• D. $-46\hat{i}+24\hat{j}$

Answer 1

The correct option is C $-46\hat{i}-24\hat{j}$

For the given concave mirror,
$u=-30\text{cm},f=-20\text{cm}$
Relative velocity of object with respect to mirror,
$v_{om}=v_{og}-v_{mg}=(15\cos {53}^{\circ }\hat{i}+15\sin {53}^{\circ }\hat{j})-(-2\hat{i})$
$\Rightarrow v_{om}=(9\hat{i}+12\hat{j})-(-2\hat{i})=(11\hat{i}+12\hat{j})\text{m/s}$
Using mirror formula,
$\frac{1}{v}-\frac{1}{30}=\frac{-1}{20}$
$\Rightarrow \frac{1}{v}=\frac{-1}{60}$
$\Rightarrow v=-60\text{cm}$
So, $m=\frac{-v}{u}=\frac{-(-60)}{-30}=-2$
Also, component of velocity of the image along the principal axis $(x-$axis$)$,
$v_{im}=-m^2{v}_{om}$
$\Rightarrow v_{im}=-(-2{)}^2\times 11=-44\text{m/s}$
Component of velocity of the image moving perpendicular to the principal axis $(y-$axis$)$,
$v_{im}=m{v}_{om}$
$\Rightarrow v_{im}=(-2)\times 12=-24\text{m/s}$
So, net velocity of the image with respect to the mirror,
$v_{im}=(-44\hat{i}-24\hat{j})\text{m/s}$
Now, the velocity of the image with respect to ground,
$v_{ig}=v_{im}+v_{mg}$
$\Rightarrow v_{ig}=(-44\hat{i}-24\hat{j})+(-2\hat{i})=(-46\hat{i}-24\hat{j})\text{m/s}$

$\text{Why this question?}$ $\text{Break the object velocity into components}\text{parallel and perpendicular to the principal axis,}\text{then apply the velocity formula for parallel and}\text{perpendicular direction respectively. The final image}\text{velocity is combination of obtained components.}$