First slide

Conservation of mechanical energy

Question

A particle of mass m moving with velocity V₀ strikes a simple pendulum of mass m and sticks to it. The maximum height attained by the pendulum will be

Moderate

A

$h = \frac{V_{0}^{2}}{8 g}$
B

$\sqrt{V_{0} g}$
C

$2 \sqrt{\frac{V_{0}}{g}}$
D

$\frac{V_{0}^{2}}{4 g}$

Solution

Initial momentum of particle = mV₀
Final momentum of system (particle + pendulum) = 2mv
By the law of conservation of momentum
$\Rightarrow {mV}_{0} = 2 mv \Rightarrow$ Initial velocity of system $v = \frac{V_{0}}{2}$
$∴$ Initial K.E. of the system = $\frac{1}{2} (2 m) v^{2} = \frac{1}{2} (2 m) {(\frac{V_{0}}{2})}^{2}$
If the system rises up to height h then P.E. = 2mgh
By the law of conservation of energy
$\frac{1}{2} (2 m) {(\frac{V_{0}}{2})}^{2} = 2 mgh \Rightarrow h = \frac{V_{0}^{2}}{8 g}$

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