Question

For the given object (O) and mirror arrangement, the velocity of the image with respect to ground will be [All the velocities are given with respect to ground]

• A. $(64\hat{i}+25\hat{j}){\text{ms}}^{-1}$
• B. $(-64\hat{i}-24\hat{j}){\text{ms}}^{-1}$
• C. $(-64\hat{i}-25\hat{j}){\text{ms}}^{-1}$
• D. $(64\hat{i}+24\hat{j}){\text{ms}}^{-1}$

Answer 1

The correct option is B $(-64\hat{i}-24\hat{j}){\text{ms}}^{-1}$

For the given concave mirror.
$u=-40\text{cm},f=-30\text{cm}$

Relative velocity of object with respect mirror,
$\overrightarrow{V_{om}}={\overrightarrow{V}}_{og}-{\overrightarrow{V}}_{mg}$
$=(10\cos {53}^{\circ }\hat{i}+10\sin {53}^{\circ }\hat{j})-(-\hat{i})$
$=6\hat{i}+8\hat{j}+\hat{i}$
$=(7\hat{i}+8\hat{j})\text{m/s}$

Using mirror formula
$\frac{1}{v}-\frac{1}{40}=-\frac{1}{30}$
$\frac{1}{v}=\frac{1}{40}-\frac{1}{30}=\frac{-1}{120}$
$v=-120\text{cm}$

So, $m=\frac{-v}{u}=\frac{-(-120)}{-40}=-3$

Also, component of velocity of the image along the principal axis ($x$ - axis) is given by
$V_{im\left(x\right)}=-m^2{V}_{om\left(x\right)}$
$\Rightarrow V_{im\left(x\right)}=-(-3{)}^2\times 7=-63\text{m/s}$

Component of velocity of the image moving perpendicular to the principal axis ($y$ - axis) is given by
$V_{im\left(y\right)}=m{V}_{om\left(y\right)}$
$\Rightarrow V_{im\left(y\right)}=(-3)\times \left(8\right)=-24\text{m/s}$

So, net velocity of the image w.r.t to the mirror $=(-63\hat{i}-24\hat{j})\text{m/s}$

Now, the velocity of the image w.r.t the ground
${\overrightarrow{V}}_{ig}={\overrightarrow{V}}_{im}+{\overrightarrow{V}}_{mg}$
${\overrightarrow{V}}_{ig}=(-63\hat{i}-24\hat{j})+(-\hat{i})$
${\overrightarrow{V}}_{ig}=(-64\hat{i}-24\hat{j})\text{m/s}$

Why this Question? Note: Splitting velocity vector into components along principle axis and perpendicular to it is must for calculation of velocity of image.