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In terms of m1, m2 and g, find the acceleration of both the blocks as shown in the figure. Neglect all friction and the masses of the pulleys.

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From the free body diagram of the block of mass m1:

m1a1 = T     ....(i)

From the free body diagram of the block of mass m2:

m2a2 = m2g - 2T     ....(ii)

Substituting for T in the equation (ii), we have

m2a2 = m2g - 2m1a1    ...(iii)

When the blocks move, the total length of the string remains unchanged. Therefore, if the block of mass m1 moves towards right through a distance x, the block of mass m2 moves down through a distance x/2. Hence, the acceleration of the block of mass m2, must be one half as that of the block mass m2

i.e., a2 = 1/2 a1, or a1 = 2a2

Therefore, the equation (iii) becomes

m2a2 = m2g - 2m2 x 2a2     ....(iv)

or, 4m1a2 + m2a2 = m2g

or, a2 = {m2g}/{4m1 + m2}

Also, a1 = 2a2 = {2m2g}/{4m1 + m2}

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