Q. When two blocks $A$ and $B$ coupled by a spring on a frictionless table are stretched and the released, then
A) Kinetic energy of body at any instant after releasing is inversely proportional to their masses
B) Kinetic energy of body at any instant may or may not be inversely proportional to their masses
C) $\frac{KE\ of\ B}{KE\ of\ A} = \frac{mass\ of\ B}{mass\ of\ A},$ when spring is massless
D) Both (b) and (c) is correct
Explanation
(c) When spring is massless then according to momentum conservation principle
${\overrightarrow{p}}_{i} = {\overrightarrow{p}}_{f}$ or $0 = m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{1}$
$\therefore m_{1}{\overrightarrow{v}}_{1} = - m_{2}{\overrightarrow{v}}_{1}$
$\therefore m_{1}v_{1} = m_{2}v_{2}$ or $p_{1} = p_{2}$
$\therefore K_{1} = \frac{p_{1}^{2}}{2m_{1}},\ \ \ \ \ K_{2} = \frac{p_{2}^{2}}{2m_{2}}$
$\therefore\ \frac{K_{1}}{K_{2}} = \frac{m_{2}}{m_{1}}\ \ \ \ \ \ \ (\because\ \ p_{1} = p_{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.