Question

Consider the situation shown in figure. The elevator is going up with an acceleration of $2.00m/s^2$ and the focal length of the mirror is $12.0cm$. All the surfaces are smooth and 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 $42.0cm$. Find the distance between the image of the block $B$ and the mirror at $t=0.200s$. Take $g=10m/s^2$.

Answer 1

Let a = acceleration of the masses $A$ and $B$ (w.r.t.elevator). From the freebody diagrams,

$T-mg+ma-2m=0$...........(1)

similarly, $T-ma=0$............(2)

From (1) and (2), $2ma-mg-2m=0$

$\Rightarrow 2ma=m(g+2)$

$\Rightarrow a=\frac{10+2}{2}=\frac{12}{2}=6m{s}^{-2}$

so, distance travelled by B in $t=0.2$ sec is,

$s=\frac{1}{2}a{t}^2=\frac{1}{2}\times 6\times (0.2{)}^2=0.12m=12m$

So, distance from mirror $u=-(42-12)=-30cm;f=+12cm$

From mirror equation, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}+(-\frac{1}{30})=\frac{1}{12}$

$\Rightarrow v=8.57$ cm

Distance between image of block $B$ and mirror $=8.57cm$