Question

A uniform bar of length 6a and mass 8 m 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 and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by $\omega$, $E$ and $v_c$ respectively, we have after collision

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

Answer 1

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

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

Using momentum conservation, $2m\times v-m\times 2v=(8m+2m+m)v_c\Rightarrow v_c=0$
Now, using conservation of angular momentum, $2m\times v\times a+m\times 2v\times 2a=I\omega$ where $I=8m\frac{(6a{)}^2}{12}+2m{a}^2+m(2a{)}^2=30m{a}^2$
Solving we get, $\omega =\frac{v}{5a}$
Now, energy $\frac{1}{2}I{\omega }^2=\frac{3}{5}m{v}^2$