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Question

# A man (mass=70 kg) and his son (mass=20 kg) are standing on a frictionless surface facing in same direction. The man pushes his son, so that he starts moving with a speed of 0.90 ms−1 with respect to man. The speed of man with respect to the surface is

A
0.20 ms1
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B
0.5 ms1
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C
0.67 ms1
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D
0.85 ms1
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Solution

## The correct option is A 0.20 ms−1There is no external force in horizontal direction on the system of (man+son), so momentum of system remains conserved in horizontal or x− direction. ⇒ Considering man to be moving w.r.t surface with velocity (vm,S)=v′ along −ve x−direction, so that final momentum of system could become zero, velocity of son w.r.t man is (vs,m)=0.90 ms−1 along +ve− x direction. ⇒velocity of son w.r.t surface is given as: →vs,S=→vs,m+→vm,S vs,S=+(0.90)+(−v′)=0.90−v′ Applying momentum conservation: Pi=Pf=0 ⇒Pf=(ms×vs,S)+(mm×vm,S)=0 ⇒20×(0.90−v′)+70×(−v′)=0 18−20v′−70v′=0 ∴v′=1890=0.2 ms−1

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