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 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 42.0cm. Find the distance between the image of the block B and the mirror at $t = 0.200s$. Take $g = 10m/s^2$.
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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),
$2ma - mg - 2m = 0$
$\Rightarrow2\text{ma}=\text{m}(\text{g}+2)$
$\Rightarrow\text{a}=\frac{10+2}{2}=\frac{12}{2}=6$
so, distance travelled by B in t = 0.2 sec is,
$\text{s}=\frac{1}{2}\text{at}^2=\frac{1}{2}\times6\times(0.2)^2=0.12\text{m}=12\text{cm}$
So, Distance from mirror, $\text{u}=-(42-12)=-30\text{cm};\text{f}=+12\text{cm}$
From mirror equation, $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}+\Big(-\frac{1}{30}\Big)=\frac{1}{12}$
$\Rightarrow\text{v}=8.57\text{cm}$
Distance between image of block B and mirror = 8.57cm.
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