Question

In the figure shown, two plane mirrors $M_1$ and $M_2$ are fixed on the two blocks $A$ and $B$ as shown in the figure. An object is placed midway between the two mirrors . Find the acceleration of the image formed by mirror-2 w.r.t the image formed by mirror-1.

• A. $\frac{22g}{15}$
• B. $\frac{-22g}{15}$
• C. $-\frac{44g}{15}$
• D. $\frac{44g}{15}$

Answer 1

The correct option is D $\frac{44g}{15}$


Acceleration of the mirror $M_1$ is given by
$a_1=\frac{-4mg}{5m}=\frac{-4g}{5}$
Acceleration of the block $M_2$ is given by
$a_2=\frac{2mg}{3m}=\frac{2g}{3}$

Let $I_1$ and $I_2$ be the images formed by $M_1$ and $M_2$ respectively.
Acceleration of $I_1$ w.r.t $O$ is
${\overrightarrow{a}}_{I_1}=2\overrightarrow{a_1}=\frac{-8g}{5}$

[since $O$ is stationary and $M_1$ is moving away with acceleration $a_1$]

Acceleration of $I_2$ w.r.t $O$ is
$\overrightarrow{a_{I_2}}=2\overrightarrow{a_2}=\frac{4g}{3}$

Thus, Acceleration of image formed by mirror $2$ w.r.t to image formed by mirror $1$ is given by
$\overrightarrow{a_{I_2{I}_1}}=\overrightarrow{a_{I_2}}-\overrightarrow{a_{I_1}}$
$=\frac{4g}{3}-(-\frac{8g}{5})$

or $\overrightarrow{a_{I_2{I}_1}}=\frac{44g}{15}$ (towards right - +ve)