Blocks of masses m, 2m, 4m and 8m are arranged in a line on a frictionless floor. Another block of mass m, moving with speed v along the same line (see figure) collides with mass m in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass 8m starts moving the total energy loss is p% of the original energy. Value of ‘p’ is close to :
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a
37
b
87
c
94
d
77
answer is C.
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Detailed Solution
Conservation of momentum, mv=16m v' ⇒v'=v16 Percentage loss in kinetic energy =Ki−KfKi×100 =12mv2−12(16m)(v16)212mv2×100=1516×100=93.75%