Question

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

• A. $2\hat{i}$
• B. $4\hat{i}+\hat{j}+2\hat{k}$
• C. $3\hat{i}+2\hat{j}+3\hat{k}$
• D. $2\hat{i}-3\hat{k}$

Answer 1

The correct option is B $4\hat{i}+\hat{j}+2\hat{k}$

$x-$ axis is the direction of normal of the plane mirror lying in $yz-$plane.
Applying,
${\left({\overrightarrow{v}}_m\right)}_{\perp }=\frac{{\left({\overrightarrow{v}}_o\right)}_{\perp }+{\left({\overrightarrow{v}}_i\right)}_{\perp }}{2}$
$\Rightarrow {\left({\overrightarrow{v}}_m\right)}_x=\frac{{\left({\overrightarrow{v}}_o\right)}_x+{\left({\overrightarrow{v}}_i\right)}_x}{2}$
$\Rightarrow 3\hat{i}=\frac{2\hat{i}+{\left({\overrightarrow{v}}_i\right)}_x}{2}$
$\Rightarrow {\left({\overrightarrow{v}}_i\right)}_x=4\hat{i}$
Also the $y$ and $z$ component of image will be same as those of object. Hence
${\left({\overrightarrow{v}}_i\right)}_y=\hat{j},{\left({\overrightarrow{v}}_i\right)}_z=2\hat{k}$

$\Rightarrow {\overrightarrow{v}}_i=4\hat{i}+\hat{j}+2\hat{k}$