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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.
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Solution
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)
`⇒ a = (10 + 2)/2`
`= 12/2 = 6 ms^-2`
Now, the distance travelled by block B of mass m in time t = 0.2 s is given by
`s = ut + 1/2at^2`
As `(u = 0)`
`s = 1/2 at^2`
On putting the respective values, we get:
`s = 1/2 xx 6 xx (0.2)^2`
`= 0.12 m = 12 cm`
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,
`1/v + 1/u = 1/f`
On putting the respective values, we get:
`1/v + 1/-30 = 1/12`
⇒ `1/v= 1/-30 = 1/ 12 + 1/30`
⇒ `v = 8.57 cm `
Hence, the distance between block B and mirror M is 8.57 cm.
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