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 velocity by $\omega$, E and $v_c$ respectively after collision, then:

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

Answer 1

The correct option is A $v_c=0$

Total linear momentum$=m\left(2v\right)+2m(-v)=0$

so vel. of CoM after collision will also be zero.

using conservation of angular momentum to find angular velocity

$m\times 2v\times 2a+2m\times (-v)\times a+\frac{8m{a}^2}{12}\times 0=m\times 4a^2\times \omega +2m\times a^2\times \omega +\frac{8m{a}^2}{12}\times \omega \omega =\frac{24mva}{80ma\times a}=\frac{3v}{10a}$

total energy after collision.$=\frac{I{\omega }^2}{2}=\frac{1}{2}(\frac{8m{a}^2}{12}+4m{a}^2+2m{a}^2){\omega }^2=\frac{80m{a}^2}{24}\times (\frac{3v}{10a}{)}^2=\frac{3m{v}^2}{10}$

only option 'A' is correct