Question

An object $O$ is moving with velocity of $(\hat{i}+2\hat{j}+3\hat{k})\text{m/s}$ and a plane mirror in $yz$ plane facing the object is moving with velocity of $\left(2\hat{i}\right)\text{m/s}$. The velocity of the image with respect to ground (in SI units) will be

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

Answer 1

The correct option is C $3\hat{i}+2\hat{j}+3\hat{k}$


$x$ - axis is the direction of normal of the plane mirror lying in $yz$ - plane.
Applying,
$({\overrightarrow{v}}_m{)}_{\perp }=\frac{(\overrightarrow{v_o}{)}_{\perp }+({\overrightarrow{v}}_i{)}_{\perp }}{2}$
$\Rightarrow ({\overrightarrow{v}}_m{)}_x=\frac{({\overrightarrow{v}}_o{)}_x+({\overrightarrow{v}}_i{)}_x}{2}$
$\Rightarrow 2\hat{i}=\frac{1\hat{i}+({\overrightarrow{v}}_i{)}_x}{2}$
$\therefore ({\overrightarrow{v}}_i{)}_x=3\hat{i}\text{m/s}$
Since, the mirror is lying along $yz$ - plane, the $y$ and $z$ components of the velocity of the image will be the same as that of the object.
$\Rightarrow ({\overrightarrow{v}}_i{)}_y=2\hat{j}\text{m/s}$
$\Rightarrow ({\overrightarrow{v}}_i{)}_z=3\hat{k}\text{m/s}$
Hence, the velocity of image $w.r.t$ ground will be,
${\overrightarrow{v}}_i=({\overrightarrow{v}}_i{)}_x+({\overrightarrow{v}}_i{)}_y+({\overrightarrow{v}}_i{)}_z$
$\Rightarrow {\overrightarrow{v}}_i=(3\hat{i}+2\hat{j}+3\hat{k})\text{m/s}$