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A trolley was moving horizontally on a smooth ground with velocity v w.r.t earth. Suddenly a man starts running from the rear end of the trolley with a velocity (3/2) v w.r.t trolley. After reaching the other end, the man turns back and continues running with a velocity (3/2) v w.r.t the trolley in opposite direction. If the length of the trolley is L, find the displacement of the man w.r.t earth when he reaches the starting point of the trolley. Mass of the trolley is equals to the mass of the man.
Dear Sanchit
initial velocity of trolly = v
let when start running then velocity of trolly become =v1
velocity of man with respect to earth = ( v1 + 3/2v)
conservation of momentum
(m+m)v = mv1 + m(v1 + 3/2v)
v1 =v/4
time taken by man to reach at the other end t1 = L /(3/2 v)
=2L/3v
when starts moving in backward direction let velocity of trolly =v2
then velocity of man w.r.t earth = v2 - 3/2 v
(m+m)v = mv2 + m(v2 - 3/2 v)
v2 = 5v/4
time taken by man to reach at the initial end t2 = L /(3/2 v)
so distance moved by trolly = V1 t1 + V2t2
=v/4 * 2L/3v + 5v/4 * 2L/3v
=L
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