Question

Two particles A and B, of mass $m$ each, are joined by a rigid massless rod of length $l$. A particle P of mass $m$, moving with a speed $u$ normal to AB, strikes A and sticks to it. The centre of mass of 'A + B + P' system is C. Thus, immediately after the impact,

• A. $AC=\frac{l}{3}$.
• B. The angular momentum of the A+ B + P system about C is $\frac{1}{3}mul$.
• C. The moment of inertia of the A+ B + P system about C is $\frac{2}{3}m{l}^2$.
• D. The angular veloctiy of the A+ B + P system is $\frac{u}{2l}$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The angular momentum of the A+ B + P system about C is $\frac{1}{3}mul$.
B $AC=\frac{l}{3}$.
C The moment of inertia of the A+ B + P system about C is $\frac{2}{3}m{l}^2$.
D The angular veloctiy of the A+ B + P system is $\frac{u}{2l}$.

Since asked for immediately after impact, as there was no external force on the system, it follows LAW OF CONSERVATION OF ANGULAR MOMENTUM.

Given COM of system is at C, lets locate C and its distance from A is $x$.

$x=\frac{\sum \left(m_a{r}_a\right)}{\sum m_a}$

$=\frac{mL}{3m}$

$=L/3$

About COM,

$mv\frac{L}{3}=2m\left(\omega \frac{L}{3}\right)\frac{L}{3}+m\left(\omega \frac{2L}{3}\right)\frac{2L}{3}$ .{let "$\omega$" be the angular velocity}

$\Rightarrow \omega =u/2L$

The angular momentum of A+B+P about $C=muL/3$ {since, there is no change of angular momentum as discussed above i.e. since there is no external force on system}

Moment of inertia of A+B+P about $C=\sum \left(m_a{r}_a^2\right)$

$=m(\frac{L}{3}{)}^2+m(\frac{L}{3}{)}^2+m(\frac{2L}{3}{)}^2$

$=\frac{2}{3}m{L}^2$