Question

A particle of mass m moving with velocity $V_0$ strikes a simple pendulum of mass $m$ and sticks 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. $\frac{V_0^2}{4g}$

Answer 1

Answer: (A)

Initial momentum of particle $=m{V}_0$
Final momentum of system (particle + pendulum)
$=2mv$
By the law of conservation of momentum
$\Rightarrow m{V}_0=2mv$
$\Rightarrow$ Initial velocity of system $v=\frac{V_0}{2}$
$\therefore$ Initial $K.E.$ of the system
$=\frac{1}{2}\left(2m\right)v^2$
$=\frac{1}{2}\left(2m\right){\left(\frac{V_0}{2}\right)}^2$
If the system rises up to height h then $P.E.=2mgh$
By the law of conservation of energy
$=\frac{1}{2}\left(2m\right){\left(\frac{V_0}{2}\right)}^2=2mgh$
$\Rightarrow h=\frac{V_0^2}{8g}$