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Asked by adjacentcalliber10 10th May 2019, 11:41 AM
Answered by 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 = begin mathsize 12px style square root of 13 over 9 g H end root end style
Answered by Expert 12th May 2019, 10:25 AM
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