1.

A man weighing  80kg is standing  in a trolley 320kg  .the trolley is resting on frictional  horizontal rails. If the man starts walking on the trolley with a speed of 1m/s then after 4sec his displacement relative to the ground will be

Answer»

Let Vmg = velocity of man with respect to ground
Vtg = velocity of trolley with respect to ground
Vmt = velocity of man with respect to trolley = 1 m/s
Let M = mass of person = 80 kg
Let M' = the mass of trolley = 320kg

from relative motion 
Vtg = Vmg - Vmt --------------------(1)

Initially trolley and man were at rest initial momentum = 0
According to law of conservation of momentum

VmgM + VtgM' = 0
Substituting value of Vtg from equation (1) 
VmgM + (Vmg - Vmt)M' = 0  Vmg(M + M') - VmtM'
Vmg(320 + 80) = 1(320)  Vmg = 320/400 = 0.8 m/s
Thus displacement of man with respect to observer = Vmg × time = 0.8 × 4 = 3.2 m



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