A plane mirror is placed along the vertical plane (yz-plane). If a particle is also projected in the vertical plane with speed of projection $10\text{m/s}$ and angle of projection as ${60}^{\circ }$ from the normal direction to mirror. The point of projection is at a distance of $5\text{m}$ from the mirror, and the particle is directed towards the mirror. The magnitude of velocity of approach of particle and image, just before the particle touches the mirror is

The correct option is A $10\text{m/s}$
The normal for plane mirror will be along x-direction. ($\therefore$ mirror is in yz plane)


For object, resolving the velocity and applying
$(v_{o,m}{)}_x=(v_{i,m}{)}_x$


The y-component of velocity of object and image will be same at all time.
Thus, relative velocity, $(v_{o,i}{)}_{xy}=0$
Velocity of approach for image and object will be,
$|v_{approach}|=|(v_{o,i}{)}_x|=|10\cos {60}^{\circ }-(-10\cos {60}^{\circ })|$
$\Rightarrow |v_{approach}|=10\cos {60}^{\circ }+10\cos {60}^{\circ }$
$\therefore |v_{approach}|=20\cos {60}^{\circ }=20\times \frac{1}{2}=10\text{m/s}$