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Asked by adjacentcalliber10 | 10 May, 2019, 11:41: AM
Expert Answer
Fig.(1) shows the movement of blocks when collar C attached with block B.
applying newton's law on block B and collar C , we have 2mg - T = 2ma ..................(1)
applying newton's law on block A , we have T = ma .........................(2)
where T is tension in the string and a is the acceleration of moving blocks
By eliminating T between eqn.(1) and (2) and solving for acceleration a we get ,
2mg - ma = 2ma or 3ma = 2mg or a = (2/3)g
if block-B and Collar-C have moved vertically from rest to a distance H/3 with acceleration (2/3)g,
then final speed v after falling through a distance H/3 is given by, v2 = 2×(2/3)g×(H/3) = (4/9)gH ....................(3)
After the distance (H/3), collar-C is detached. Movement of block-A and block-B is shown in figure-2
applying newton's law on block B , we have mg - T ' = m a' ..................(4)
applying newton's law on block A , we have T ' = m a' .........................(5)
By eliminating T ' between eqn.(4) and (5) and solving for acceleration a' we get ,
mg - ma' = ma' or mg = 2ma' or a' = g/2
if block-B has moved vertically from intial speed v as given by eqn.(3) to a distance H with acceleration g/2 ,
then final speed vf after falling through a distance H is given by, vf 2 = (4/9)gH + 2×(g/2)×H = (13/9)gH ....................(6)
hence final speed vf = 
Answered by Thiyagarajan K | 12 May, 2019, 10:25: AM
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