Question

Consider the situation shown in figure. The elevator is going up with an acceleration of $2.00 \,m / s ^{2}$ and the focal length of the mirror is $12.0 \,cm$. The entire surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at $t=0$ when the distance of $B$ from the mirror is $42.0\, cm$. Find the distance (in cm) between the image of the block $B$ and the mirror at $t=0.200 \,s$. Take $g=10 \,m / s ^{2}$.

Answer 1

Consider observer in elevator.
Consider free body diagram. of blocks $A$ and $B$,

For motion of $A , T = ma _{\text {rel }}$...(1)
For motion of $B, m g+ m a_{0}-T=m a_{\text {rel }}$..(2)
$(1)+(2) \Rightarrow mg + ma _{0}=2 ma _{\text {rel }}$
$a _{\text{ rel }}=\frac{ g + a _{0}}{2}$
Using, $S=u t+\frac{1}{2} a t^{2}$
$S=(0)(6)+\frac{1}{2}(6)(0.2)^{2}=12\, cm$
At $t=0.2 \,\sec , u =30\, cm , f =-12\, cm$
Using $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{30}=\frac{1}{-12} $
$\Rightarrow v=-\frac{30 \times 12}{42}=-8.57\, cm$
$\therefore $ Required distance $=8.57 \,cm$