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Question

A man (mass=50 kg) and his son (mass=20 kg) are standing on a frictionless surface facing each other. The man pushes his son, so that he starts moving at a speed of 0.70 ms1 with respect to the man. The speed of the man with respect to the surface is

A
0.28 ms1
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B
0.20 ms1
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C
0.47 ms1
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D
0.14 ms1
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Solution

The correct option is B 0.20 ms1
The given situation can be shown as below

Using momentum conservation law,
(Total momentum)before collision
(m1×0)+(m2×0)=m1v1+m2v2
0=m1(v1)^i+m2v2^i
m1v1=m2v2
50v1=20v2
v2=2.5v1....(i)
Again, relative velocity 0.70 m/s
But from figure, relative velocity =v1+v2
v1+v2=0.7.....(ii)
From Eqs (i) and (ii), we get
v1+2.5v1=0.7
v1(3.5)=0.7
v1=0.73.5=0.20 m/s

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