Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

21 of 130

Question 21 of 130

A man (mass =50kg=50 kg ) and his son (mass =20kg=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 0.70ms−10.70 m{s}^{-1} with respect to the man. The speed of the man with respect to the surface is:

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Answer: (A) 0.20ms−10.20m{s}^{-1}

By conservation of linear momentum,
P⃗i=P⃗f{\vec{P}}_{i}={\vec{P}}_{f}
Where P⃗i{\vec{P}}_{i} and P⃗f{\vec{P}}_{f} are initial and final momentum of system (boy + man)
Now, 0=mbv⃗b+mmv⃗m0={m}_{b}{\vec{v}}_{b}+{m}_{m}{\vec{v}}_{m}......(i)
vb={v}_{b}= velocity of boy w.r.t. ground
vm={v}_{m}= velocity of man w.r.t. ground
v⃗bm={\vec{v}}_{bm}= velocity of boy w.r.t. man
So, v⃗bm=v⃗b−v⃗m{\vec{v}}_{bm}={\vec{v}}_{b}-{\vec{v}}_{m}
or v⃗b=v⃗bm+v⃗m{\vec{v}}_{b}={\vec{v}}_{bm}+{\vec{v}}_{m}
or v⃗b=0.7+v⃗m{\vec{v}}_{b}=0.7+{\vec{v}}_{m} ........(ii)
From equation (i) and (ii),
0=20(0.7+v⃗m)+50v⃗m0=20\left(0.7+{\vec{v}}_{m}\right)+50{\vec{v}}_{m}
v⃗m=−1470=−0.20ms−1{\vec{v}}_{m}=-\dfrac{14}{70}=-0.20m{s}^{-1}
Negative sign shows velocity of man is opposite to boy.

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