Question

A uniform bar of length $6a$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speed $2v$ and $v$ respectively, strike the bar (as shown in figure) and stick to the bar after collision. C represents centre of mass of bar. Denoting angular velocity (about the centre of mass), total energy and centre of mass by $\omega$, E and $v_c$ respectively, then after collision we have

• A. $v_c=0$
• B. $\omega =\frac{3v}{5a}$
• C. $\omega =\frac{v}{5a}$
• D. $E=\frac{3m{v}^2}{5}$

Answer 1

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

As $F_{ext}=0$, so linear momentum of the system is conserved, Thus
$-2mv+m\times 2v+0=(2m+m+8m)v_c$
or
$v_c=0$
Now as ${\tau }_{ext}=0$, angular momentum of the system is also conserved, Thus
$m_1{v}_1{r}_1+m_2{v}_2{r}_2=(I_1+I_2+I_3)\omega$
or
$2mva+m\left(2v\right)\left(2a\right)=(2m{a}^2+m(2a{)}^2+\frac{8m\times (6a{)}^2}{12})\omega$
or
$6mva=30m\omega a^2$
or
$\omega =\frac{v}{5a}$
Now as the system has no translatory motion but only rotatory motion we have
$E=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(30m{a}^2\right)(\frac{v}{2a}{)}^2=\frac{3}{5}m{v}^2$