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 speeds $2v$ and $v$ respectively strike the bar (as shown in figure) and stick to the bar after collision. By denoting angular velocity (about centre of mass), total energy and velocity of centre of mass after collision as $\omega ,E$ and $v_o$ respectively, we can say that

• A. $v_o=0$
• B. $\omega =\frac{6v}{5a}$
• C. $\omega =\frac{3v}{5a}$
• D. $E=\frac{3}{5}m{v}^2$

Answer 1

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

$E=\frac{3}{5}m{v}^2$

From conservation of momentum
$P_i=P_f$
$2mv-2mv=11m{v}_{cm}$
$v_{cm}=0$
Position of COM from O
$C.M\equiv \frac{2a\times 2m+8m\times 3a+m\times 5a}{11m}=3a$
Conserving angular momentum about COM
$L_{net}=I\omega$
$\Rightarrow 2mva+m\left(2v\right)\left(2a\right)=I\omega$
$\therefore I\omega =6mva$
$\Rightarrow (2m{a}^2+m\times 4a^2+\frac{1}{12}8m\times 36a^2)\times \omega =6mva$
$\Rightarrow (6a^2+24a^2)\omega =6va\therefore \omega =\frac{v}{5a}$
Energy of the system:
$E=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 30m{a}^2\times \frac{v^2}{25a^2}$
$=\frac{3}{5}{v}^2\times m$
$\therefore E=\frac{3}{5}m{v}^2$