Question

A uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $J$ is applied to the end $B$ perpendicular to the rod in horizontal direction. Speed of particle $P$ at a distance $\frac{l}{6}$ from the centre towards $A$ of the rod after time $\text{t}=\frac{\pi \text{ml}}{12 \text{J}}$ is

• A. $2 \frac{\text{J}}{\text{m}}$
• B. $\frac{\text{J}}{\sqrt{2} \text{m}}$
• C. $\frac{\text{J}}{\text{m}}$
• D. $\sqrt{2} \frac{\text{J}}{\text{m}}$

Answer 1

Answer: (D)

Let v and ω be the linear and angular speeds of the rod
after applying an impulse J at B. Then from :
impulse = change in momentum
we have
$\text{mv} = \text{J or} \text{v} = \frac{\text{J}}{\text{m}}$ ...(1)
$\text{I} \omega = \text{J} \cdot \frac{\textit{l}}{2}$
or $\frac{\text{m} \textit{l}^{2}}{1 2} \omega = \text{J} \cdot \frac{\textit{l}}{2} \text{or} \omega = \frac{6 \text{J}}{\text{m} \textit{l}}$ ...(2)
After the given time $\text{t} = \frac{\pi \text{m} \textit{l}}{1 2 \text{J}} \text{,}$ the rod will rotate an angle
$\theta = \omega \text{t} = \left(\frac{6 \text{J}}{\text{m} \textit{l}}\right) \left(\frac{\pi \text{m} \textit{l}}{1 2 \text{J}}\right) = \frac{\pi }{2}$
$\frac{\textit{l}}{6} \text{.} \omega = \left(\frac{\textit{l}}{6}\right) \left(\frac{6 \text{J}}{\text{m} \textit{l}}\right) = \frac{\text{J}}{\text{m}} = \text{v}$
∴ $\left|\text{v}_{\text{P}}^{\overset{ \rightarrow }{}}\right| = \sqrt{2} \text{v} = \sqrt{2} \frac{\text{J}}{\text{m}}$