A man standing on a trolley pushes another identical trolley (initially both trolleys are at rest on a rough surface), so that they are set in motion and stop after sometime. If the ratio of mass of first trolley + man to mass of second trolley is 3, then find the ratio of the stopping distances of the second trolley to that of the first trolley. (Assume coefficient of friction to be same for both the trolleys).
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mAvA + mBvB = 0
3vA +vB = 0 ... (1)
coefficient of friction µ is same
FrA = µN
= µmg = mA aB
frB = µmBg = mBaB
aA = µg & aB = µg
2aAxA = vA2 – 02
2aBxB = vB2 – 02
xB : xA = 9