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Question: Answered & Verified by Expert
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 m s-1 with respect to the man. The speed of the man with respect to the surface is:
PhysicsCenter of Mass Momentum and CollisionJEE Main
Options:
  • A 0.20 m s-1
  • B 0.14 m s-1
  • C 0.47 m s-1
  • D 0.28 m s-1
Solution:
2026 Upvotes Verified Answer
The correct answer is: 0.20 m s-1

By conservation of linear momentum,

Pi=Pf

Where Pi and Pf are initial and final momentum of system (boy + man)

Now,  0=mbvb+mmvm......(i)

vb= velocity of boy w.r.t. ground

vm= velocity of man w.r.t. ground

vbm= velocity of boy w.r.t. man

So, vbm=vb-vm

or vb=vbm+vm

or vb=0.7+vm ........(ii)

From equation (i) and (ii),

0=200.7+vm+50 vm

vm=-1470=-0.20 m s-1

Negative sign shows velocity of man is opposite to boy.

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