Search any question & find its solution
Question:
Answered & Verified by Expert
A uniform rod of length $8 a$ and mass $6 m$ lies on a smooth horizontal surface. Two point masses $m$ and $2 m$ moving in the same plane with speed $2 v$ and $v$ respectively strike the rod perpendicularly at distances $a$ and $2 a$ from the mid point of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is :
Options:
Solution:
2929 Upvotes
Verified Answer
The correct answer is:
$\frac{6 v}{41 a}$
Applying conservation of angular momentum about point $O$,
$m(a)(2 v)+2 m(2 a)(v)=I \omega$
or $\omega=\frac{6 m a v}{I}$ ...(i)
Now, $\quad I=\frac{6 m(8 a)^2}{12}+m\left(a^2\right)+2 m(2 a)^2$
$=32 m a^2+m a^2+8 m a^2$
$=41 \mathrm{ma}^2$
Hence, from Eq. (i)
$\omega-=\frac{6 m a v}{41 m a^2}=\frac{6 v}{41 a}$
$m(a)(2 v)+2 m(2 a)(v)=I \omega$
or $\omega=\frac{6 m a v}{I}$ ...(i)
Now, $\quad I=\frac{6 m(8 a)^2}{12}+m\left(a^2\right)+2 m(2 a)^2$
$=32 m a^2+m a^2+8 m a^2$
$=41 \mathrm{ma}^2$
Hence, from Eq. (i)
$\omega-=\frac{6 m a v}{41 m a^2}=\frac{6 v}{41 a}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.