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# solve

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 Expert 12th May 2019, 10:25 AM
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