JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Conservation of Momentum

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

ms^{- 1}

with respect to the man. The speed of the man with respect to the surface is :

(A) $0.47 ms^{- 1}$
(B) $0.28 ms^{- 1}$
(C) $0.14 ms^{- 1}$
(D) $0.20 ms^{- 1}$ ✓ Correct

Answer: (D)

Solution

On a frictionless floor the two push off from rest, so the total momentum of the man-and-son system must stay zero. Let the man drift backwards with ground speed

v_{1}

and the son move forwards with ground speed

v_{2}

. Equating the magnitudes of their momenta,

50 v_{1} = 20 v_{2}

, which gives

v_{2} = 2.5 v_{1}

. Because they travel in opposite directions, the son's speed as seen by the man is the sum

v_{1} + v_{2}

, and we are told this equals

0.70 m/s

. Substituting,

v_{1} + 2.5 v_{1} = 0.70

, so

3.5 v_{1} = 0.70

and

v_{1} = 0.20 m/s

, which is the man's speed relative to the ground.

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Conservation of Momentum

Solution

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