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. 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 42.0 cm. Find the distance between the image of the block B and the mirror at t = 0.200 s. Take g = 10 m s⁻².
Figure

Answer 1

Given,
Acceleration of the elevator, a = 2.00 m/s²
Focal length of the mirror M, f = 12.00 cm
Acceleration due to gravity, g = 10 m/s²
Mass of blocks A and B = m​
As per the question, the mass–pulley system is released at time t = 0.
Let the acceleration of the masses A and B with respect to the elevator be a.


Using the free body diagram,
T − mg + ma − 2m = 0 ...(i)
Also,
T − ma = 0 ...(ii)
From (i) and (ii), we get:
2ma = m(g + 2)
$$\begin{gathered} \Rightarrow a=\frac{10+2}{2} \\ =\frac{12}{2}=6{\mathrm{ms}}^{-2} \end{gathered}$$
Now, the distance travelled by block B of mass m in time t = 0.2 s is given by
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ \mathrm{As}(u=0) \\ s=\frac{1}{2}a{t}^2 \end{gathered}$$
On putting the respective values, we get:
$$\begin{gathered} s=\frac{1}{2}\times 6\times {\left(0.2\right)}^2 \\ =0.12\mathrm{m}=12\mathrm{cm} \end{gathered}$$

As given in the question, the distance of block B from the mirror is 42 cm.
Object distance u from the mirror = − (42 − 12) = − 30 cm
Using the mirror equation,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
On putting the respective values, we get:
$$\begin{gathered} \frac{1}{v}+\frac{1}{-30}=\frac{1}{12} \\ \Rightarrow \frac{1}{v}=\frac{1}{-30}=\frac{1}{12}+\frac{1}{30} \\ \Rightarrow v=8.57\mathrm{cm} \end{gathered}$$
Hence, the distance between block B and mirror M is 8.57 cm.