question_answer

Two plane mirrors are placed as shown in figure. A point object O is approaching the intersection point A of mirrors with a speed of 100 cm/s. The velocity of image of the object formed by \[{{M}_{2}}\], with respect to velocity of image of object formed by \[{{M}_{1}}\] is:

A) \[-128\,\hat{i}+96\hat{j}\,cm/s\]

B) \[-\,28\,\hat{i}+48\hat{j}\,cm/s\] 

C) \[128\,\hat{i}+48\hat{j}\,cm/s\]

D) \[100\,\hat{i}+48\hat{j}\,cm/s\]

Correct Answer: A

Solution :

The components of various velocities are as shown in figure

\[{{\overline{V}}_{I{{M}_{2}}}}\] is given by the vector sum of components of velocity of image with respect to \[{{M}_{2}}\] along the normal and perpendicular to the normal

\[{{\overline{V}}_{I{{M}_{2}}}}=[100{{\sin }^{2}}37{}^\circ \,\hat{i}+100\sin 37{}^\circ \cos 37{}^\circ \,\hat{j}]+\]

\[[-100{{\cos }^{2}}37{}^\circ \hat{i}+100\sin 37{}^\circ \cos 37{}^\circ \hat{j}]\]

\[=[-28\hat{i}+96\hat{j}]\,\,cm/s\]

\[\therefore \,\,\,\,{{\overline{V}}_{I{{M}_{2}},I{{M}_{1}}}}={{\overline{V}}_{I{{M}_{2}}}}-{{\overline{V}}_{I{{M}_{1}}}}\]\[=(-128\hat{i}+96\hat{j})cm/s\]