Question

Determine the velocity of image with respect to ground.

• A. $8\hat{i}+6\hat{j}$
• B. $-8\hat{i}-6\hat{j}$
• C. $-8\hat{i}+6\hat{j}$
• D. $8\hat{i}-6\hat{j}$

Answer 1

The correct option is C $-8\hat{i}+6\hat{j}$



Velocity of object, ${\overrightarrow{v}}_0=10\cos 37{}^{\circ }\hat{i}+10\sin 37{}^{\circ }\hat{j}=8\hat{i}+6\hat{j}$

In the direction normal to the mirror,

|Relative velocity of image with respect to mirror| = |Relative velocity of object with respect to mirror|

$\text{(i.e)}{\overrightarrow{v}}_{ix}-{\overrightarrow{v}}_m=-({\overrightarrow{v}}_{ox}-{\overrightarrow{v}}_m)$

$\Rightarrow {\overrightarrow{v}}_{ix}=2{\overrightarrow{v}}_m-{\overrightarrow{v}}_{ox}$

From the data given in the question,

${\overrightarrow{v}}_m=0\hat{i}\text{m/s},{\overrightarrow{v}}_{ox}=8\hat{i}\text{m/s}$

$\Rightarrow {\overrightarrow{v}}_{ix}=-8\hat{i}\text{m/s}$

In the direction parallel to the surface of the mirror,

Along $Y-$ direction, ${\overrightarrow{v}}_{oy}={\overrightarrow{v}}_{iy}$

$\therefore {\overrightarrow{v}}_{iy}=6\hat{j}$

Therefore, velocity of image ${\overrightarrow{v}}_i={\overrightarrow{v}}_{ix}+{\overrightarrow{v}}_{iy}$

$\Rightarrow \overrightarrow{v}=-8\hat{i}+6\hat{j}$