# Question is in the description below...

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

A. 5 m

B. 4.8 m

C. 3.2 m

D. 3.0 m

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Explanation:

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