Two equal blocks of mass M hang on either side of a massless, frictionless pulle

Question

Two equal blocks of mass M hang on either side of a massless, frictionless pulley. A small square rider of mass m is placed on one block. When the block is released, it accelerates for a distance H till the rider is caught by a ring that allows the block to pass. From then on the system moves at a constant speed which is measured by timing the fall through a distance D. Derive an expression for g in terms of M, m, D, H and t where t is the time that M moves at constant speed through the distance D. Present result in simplified factored form.

Explanation / Answer

writing the equation of motion for the blocks, we get,
T - Mg = Ma ------------(1)
(M + m)g - T = (M + m)a -----------(2)
here,
a --> acceleration of the blocks

T --> Tension in the string

Adding eqn (1) and (2), we get,

mg = (2M + m)a

=> a = mg/(2M + m) ----------(3)

Now, distance travelled by accelerating body = H

the velocity when the body has travelled distance H is given by

v2 = 2aH

=> v = (2aH)

now, v = D/t

=> D/t =  (2aH)

=> a = D2/(2t2H)

Substituting the value of a from eqn (3), we get,

=> [m/(2M + m)]g =   D2/(2t2H)

=> g = [ D2 ( 2M + m ) ]/(2 m t2 H)

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