Question

A pendulum bob of mass $250\text{gm}$ is raised to a height of $4\text{cm}$ and then released. At the bottom-most position of its swing, an insect of mass $25\text{gm}$ sits on the bob. Find the maximum height to which the combined mass will rise. [Take $g=10{\text{m/s}}^2$]

• A. $5.6\text{cm}$
• B. $2.7\text{cm}$
• C. $3.3\text{cm}$
• D. $1.5\text{cm}$

Answer 1

The correct option is C $3.3\text{cm}$

The velocity of the bob at the bottom of its swing after falling through a vertical height $h=4\text{cm}$,
$v_1=\sqrt{2gh}=\sqrt{2\times 10\times 0.04}$
$\therefore v_1=\sqrt{0.8}\text{m/s}$
Initial velocity of the insect is $v_2=0\text{m/s}$
Let $v$ be the velocity of the combined mass of (insect+ bob) system.
$m_1\to$ mass of the bob
$m_2\to$ mass of the insect
Applying the conservation of momentum on system of (bob+insect) when insect sits on the bob
$P_i=P_f$
$\Rightarrow m_1{v}_1+m_2{v}_2=(m_1+m_2)v$
$\Rightarrow (0.25\times \sqrt{0.8})+(0.025\times 0)=(0.25+0.025)v$
$\therefore v=\frac{0.25\times \sqrt{0.8}}{0.275}=(0.909\times \sqrt{0.8})\text{m/s}$

Maximum height $(\text{final velocity}=0)$ of the combined mass is given by,
$v^2=2g{h}_{\text{max}}$
$\Rightarrow h_{\text{max}}=\frac{v^2}{2g}=\frac{(0.909\times \sqrt{0.8}{)}^2}{2\times 10}$
$\therefore h_{\text{max}}=\frac{0.66}{20}=0.033\text{m}=3.3\text{cm}$