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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 msâˆ’1 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 ms−1The 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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