Question

Three particles (A,B,C) each of mass $m$ are connected by three massless rods of length $l$. All three particles lie on smooth horizontal plane. A particle of mass $m$ moving along one of the rods with velocity $v_o$ strikes on a particle and stops (as shown in the diagram). Choose the correct option(s) among the following.

• A. Tension in any of the rod after collision is $\frac{m{v}_o^2}{36l}$
• B. Tension in any of the rod after collision is $\frac{m{v}_o^2}{28l}$
• C. Loss of energy due to collision is $\frac{7m{v}_o^2}{12}$
• D. Loss of energy due to collision is $\frac{7m{v}_o^2}{24}$

Answer 1

Answer: (A), (D)

The correct options are
A Tension in any of the rod after collision is $\frac{m{v}_o^2}{36l}$
D Loss of energy due to collision is $\frac{7m{v}_o^2}{24}$

Applying conservation of angular momentum with respect to centroid
Initial angular momentum $=$ Final angular momentum

Let r is the perpendicular distance between centroid and line of action.
$r=\frac{l}{2}\tan {30}^{\circ }=\frac{l}{2\sqrt{3}}$

After collision the system of masses start to translate in positive x direction and also start rotating about point P.
$AP=AD-PD$
Now,
$\tan {60}^0=\frac{AD}{l/2}$
$AD=\frac{\sqrt{3}}{2}l$
$AP=\frac{l}{\sqrt{3}}$

Now applying momentum conservation about point P
$m{v}_{o}r=I_{com}\omega$
$I_{com}=3m{R}^2$, where R is the distance of each mass from centroid so as there are three masses and each mass is at a distance $AP$. So, $AP=R$
$m{v}_o\frac{l}{2\sqrt{3}}=3m(\frac{l}{\sqrt{3}}{)}^2\omega ;$
$\Rightarrow \omega =\frac{v_o}{2\sqrt{3}l}$

For a single mass m the net force acting on it towards centroid will provide centripetal acceleration .

$2T\cos {30}^{\circ }=m{\omega }^2\frac{l}{\sqrt{3}}$
$\Rightarrow T=\frac{m{v}_o^2}{36l}$

Energy of system of three masses $=(\frac{1}{2}\left(3m\right)(\frac{v_o}{3}{)}^2+\frac{1}{2}(\frac{3m{l}^2}{3})\frac{v_o^2}{12l^2})$

Loss in energy
$=\frac{1}{2}m{v}_o^2-(\frac{1}{2}\left(3m\right)(\frac{v_o}{3}{)}^2+\frac{1}{2}(\frac{3m{l}^2}{3})\frac{v_o^2}{12l^2})$
$=\frac{7}{24}m{v}_o^2$