Q. A particle of mass $m$ moving with velocity $V_{0}$ strikes a simple pendulum of mass $m$ and strikes to it. The maximum height attained by the pendulum will be
A) $h = \frac{v_{0}^{2}}{8g}$
B) $\sqrt{V_{0}g}$
C) $2\sqrt{\frac{V_{0}}{g}}$
D) $\frac{V_{0}^{2}}{4g}$
Explanation
(a) Initial momentum of particle $= mV_{0}$
Final momentum of system (particle + pendulum) $= 2mv$
By the law of conservation of momentum
$\Rightarrow mV_{0} = 2mv \Rightarrow$ Initial velocity of system $v = \frac{V_{0}}{2}$
$\therefore$ Initial K.E. of the system $= \frac{1}{2}(2m)v^{2} = \frac{1}{2}(2m)\left( \frac{V_{0}}{2} \right)^{2}$
If the system rises up to height $h$ then P.E$. = 2mgh$
By the law of conservation of energy
$\frac{1}{2}(2m)\left( \frac{V_{0}}{2} \right)^{2} = 2mgh \Rightarrow h = \frac{V_{0}^{2}}{8g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.