Question

A plane mirror is moving with velocity $(4\hat{i}+5\hat{j}+7\hat{k})\text{m/s}$. A point object in front of the mirror moves with a velocity $(3\hat{i}+4\hat{j}+5\hat{k})\text{m/s}$. Here $\hat{k}$ is along the normal to plane mirror and facing towards the object. The velocity of image w.r.t ground is (in SI units)

• A. $3\hat{i}+4\hat{j}+9\hat{k}$
• B. $3\hat{i}+4\hat{j}+11\hat{k}$
• C. $-3\hat{i}-4\hat{j}+11\hat{k}$
• D. $3\hat{i}-4\hat{j}+9\hat{k}$

Answer 1

The correct option is A $3\hat{i}+4\hat{j}+9\hat{k}$

We know that for fixed plane mirror,
$\left(1\right)$ Perpendicular to the mirror ($x$ - axis), $v_o=-v_i$
$\left(2\right)$ Parallel to the mirror ($y$ - axis), $v_o=v_i$
Since $\hat{k}$ is along the normal to mirror.
Applying $({\overrightarrow{v}}_{o,m}{)}_{\perp }=-({\overrightarrow{v}}_{i,m}{)}_{\perp }$
$\Rightarrow (v_{o,m}{)}_z=-(v_{i,m}{)}_z$
$\Rightarrow (v_o{)}_z-(v_m{)}_z=-\left[\right(v_i{)}_z-\left(v_m{)}_z\right]$
$\Rightarrow 5\hat{k}-7\hat{k}=-(v_i{)}_z+(v_m{)}_z$
or $-2\hat{k}=-(v_i{)}_z+7\hat{k}$
$\therefore (v_i{)}_z=9\hat{k}\text{m/s}$
Normal direction for mirror is $\hat{k}$ (z - axis), thus mirror is in (xy - plane). Hence, $x$ and $y$ component of velocity of the image will be same as that of the object.
$\Rightarrow (v_i{)}_y=4\hat{j}\&(v_i{)}_x=3\hat{i}$
$\therefore \overrightarrow{v_i}=(3\hat{i}+4\hat{j}+9\hat{k})\text{m/s}$