Consider the situation shown in fig. 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 .
Text Solution
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Sol. Let us assume that the acceleration of blocks A and B is 'a' with respect to lift and a L is acceleration of lift.
Consider block B
System: Block
Frame of reference: Lift
Mg + ma L – T = ma … (1)
Now, consider block A
System: Block
Frame of reference: Lift
N = mg + ma L … (2)
T = ma … (3)
On adding eq. (1) and eq. (3), we get
a = 
= 
= 6 m/s 2 ∴ Distance fallen by block = 
at
2 = 
× 6 × (0.2)
2 = 0.12 m or 12 cm
Now, consider reflection at convex mirror
u = – (42 – 12) = – 40 cm
f = + 12 cm
∴ v = 
= 
= 8.75 cm
Therefore, the distance between the image of block and mirror is 8.57 cm.
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