Question

Two plane mirrors are placed as shown in the figure below:
A point object is approaching the intersection point of mirrors with a speed of 100 cm/s. The velocity of the image of object formed by $M_2$ w.r.to velocity of image of object formed by $M_1$, is

• A. $-128\hat{I}+48\hat{J}\left]\right(cm/s)$
• B. $-28\hat{I}+48\hat{J}\left]\right(cm/s)$
• C. $128\hat{I}-48\hat{J}\left]\right(cm/s)$
• D. $100\hat{I}+48\hat{J}\left]\right(cm/s)$

Answer 1

The correct option is A $-128\hat{I}+48\hat{J}\left]\right(cm/s)$

The components ofvarious velocities are as shown in the figure below:



${\overrightarrow{V}}_{I{M}_1}=\left(100\hat{I}\right)cm/s$
${\overrightarrow{V}}_{I{M}_2}$ is given by the vector sum of components of velocity of image w.r. to $M_2$ along the normal and perpendicular to the normal
${\overrightarrow{V}}_{I{M}_2}=\left[100si{n}^2{37}^0\hat{I}100sin{37}^{\circ }cos{37}^{\circ }j\right]+[-100co{s}^2{37}^{\circ }\hat{I}+100sin{37}^{\circ }cos{37}^{\circ }j]$
$\Rightarrow =(-28\hat{I}+48\hat{j})cm/s$
$\therefore {\overrightarrow{V}}_{I{M}_2}-{\overrightarrow{V}}_{I{M}_1}=[-128\hat{I}+48\hat{J}]cm/s$