Name: Total Internal Reflection From Prism
Uploaded: 2019-07-19
Duration: 1 min 56 s
Description: You must have seen reflectors on a bicycle. Why are they attached to the mudguards? What use do they have? Let’s watch this video on reflectors to find out.

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 $g = 10 m s^{- 2}$ .

Accepted Answer

Given,

acceleration of elevator, $a = 2.00 m s^{- 2} (upward)$ ,

the focal length of the convex mirror, $B$ .

Elevator is accelerating, therefore, in elevator frame pseudo force needs to be applied.

Let, acceleration of blocks $B$ in the elevator frame.

Apply Newton's Second Law,

In the vertical direction for $B$ ,

$m g + m a - T = m a_{0} \dots (i)$

In the horizontal direction for $A$ ,

$T = m a_{0} \dots (ii)$ .

Add equations $(i)$ and $(ii)$ ,

$m g + m a = 2 m a_{0}$

$\Rightarrow \frac{g + a}{2} = a_{0}$

$\Rightarrow \frac{10 + 2}{2} = a_{0}$

$\Rightarrow a_{0} = 6.00 m s^{- 2}$ .

In the elevator frame,

displacement of $B$ ,

apply kinematic equation of motion,

$S = u t + \frac{1}{2} a t^{2}$

$\Rightarrow S = 0 \times 0.200 + \frac{1}{2} \times 6.00 \times (. 200)^{2}$

$\Rightarrow S = 0.120 m = 12.0 cm$ .

The distance of $B$ from the mirror, $u = - 30 cm$ .

Mirror formula,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{v} + \frac{1}{- 30} = \frac{1}{12}$

$\Rightarrow v = \frac{60}{7} cm = 8.57 cm$ .

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