JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2019
- Medium
- 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 ........ \(ms^{-1}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(0.20\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\Rightarrow 0 = 50{V_1} - 20{V_2}\,and\,{V_1} + {V_2} = 0.7\\
\Rightarrow \,{V_1} = 0.2
\end{array}\)

\Rightarrow 0 = 50{V_1} - 20{V_2}\,and\,{V_1} + {V_2} = 0.7\\
\Rightarrow \,{V_1} = 0.2
\end{array}\)

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