# A man weighing 80Kg is standing on a trolley weighing 320Kg. The trolley is resting on a frictionless surface. If the man starts walking on the trolley along its length at speed 1 m/s, then his relative displacement with respect to ground after 4s is_______

148 Points
14 years ago

Dear sindhu

let initial c.o.m of trolley and man  from y axis is a and b repectivily

Xc.m=(320*a +80*b)/400

at final position after 4 sec =Xc.m ={320*(a-x1) +80*(b+4-x1)}/400

so equate

{320*(a-x1) +80*(b+4-x1)}/400  =(320*a +80*b)/400

-400x1 +320 =0

x1=4/5

so 4-x1=4-4/5 =16/5

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Ramandeep Kaur
13 Points
6 years ago
In order to conserve the momentum, the trolley starts to move backwards. The backward momentum will be shared by both trolley and man
Hence,
60*1=(240+60)v
V=0.20m/s
Resultant speed of man wrt ground=0.80m/s
Displacement =0.8*4=3.2m
Dhruv
19 Points
5 years ago
According to conservation og linear momentum,

1×80=(80+320)×V
V = 80÷400
V = 0.2m/s
Trolley travels 4×0.2 = 0.8 metres.
Man travels 4 ×1 =4 metres in opposite direction of trolley. Net displacement of men = 4 - 0.8 = 3.2 meters.