Consider the situation shown in figure (18-E16). The elevator is going up with an acceleration of 2.00 ms^{-2} and the focal length of the mirror is 12.) 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.) cm. Find the distance between the image of the block B and the mirror at t=0.200 s. Take g= 10 ms^{-2}

acceleration of the elevator, a= 2.00 ms^{-2}

focal length of the mirror M, f= 12 cm

mass of the blocks A and B = m

mass pulley system is released at time t=0

let the acceleration of the masses A and B w.r.t. the elevator = a

using free body diagram

using both the equations,

distance travelled by the block B of mass m in time t= 0.2 sec is given by

It is mentioned in the question that the distance of block B from the mirror is 42 cm.

The object distance u from the mirror = -(42-12) = -30cm

By using mirror formula,

Distance between image of block B and mirror = 8.57 cm.

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