JEE PYQ: Laws Of Motion Question 27
Question 27 - 2019 | 12 Apr Shift 1] (MCQ)
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 :
(1) 0.28 ms$^{-1}$
(2) 0.20 ms$^{-1}$ [Correct]
(3) 0.47 ms$^{-1}$
(4) 0.14 ms$^{-1}$
Show Answer
Answer: (2)
Solution
By momentum conservation: $0 = 20(0.7 - v) - 50v$. $v = 0.2$ m/s.
BETA
