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…
JEE Main 2019 — Physics Mechanics
2019mcqeasy
A man (mass =50kg ) and his son (mass =20kg ) 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−1 with respect to the man. The speed of the man with respect to the surface is:
Official previous-year question
Held on 12 Apr 2019 · Verified 6 Jul 2026.
Options
A
0.20ms−1
B
0.14ms−1
C
0.47ms−1
D
0.28ms−1
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Solution
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=20(0.7+vm)+50vm
vm=−7014=−0.20ms−1
Negative sign shows velocity of man is opposite to boy.
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