Question

A uniform bar of length $6a$ and mass $8m$ lies on a smooth table. Two point masses $m$ and $2m$ moving in a horizontal plane with speeds $2V$ and $V$ strike the bar as shown and stick to the bar after collision.
lf $\omega$ is the angular Velocity about center of mass, $E$ is total energy and $v_c$ is velocity of the center of mass, after collision, then :

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

Answer 1

Answer: (A), (D)

The correct options are
A $v_c=0$
D $E=\frac{3m{v}^2}{5}$

The velocity of centre of mass of bar remains unchanged after the point masses strike and stick to it. The bar rotates about centre of mass as a result of collisions of point masses
or
$v_c=0$.

Thus option A is correct

By conversation of angular momentum, we have

$2m\left(va\right)+m\left(2v\right)\left(2a\right)=\left(2m{a}^2\right)\omega +m(2a{)}^2\omega +\frac{1}{12}(8m\left)\right(6a{)}^2\omega$

or

$6mva=2m{a}^2\omega +4m{a}^2\omega +24m{a}^2\omega$

$=30m{a}^2\omega$

or

$\omega =\frac{v}{5a}$

$\therefore E_k=\frac{1}{2}[2m{a}^2{\omega }^2+m(2a{)}^2{\omega }^2+\frac{1}{12}\left(8m\right)\left(6a{)}^2{\omega }^2\right]$

$=\frac{1}{2}[2m{a}^2{\omega }^2+4m{a}^2{\omega }^2+24m{a}^2{\omega }^2]$

$=\frac{1}{2}\left[30m{a}^2{\omega }^2\right]$

$=\frac{1}{2}\left[30m{a}^2{\left(\frac{v}{5a}\right)}^2\right]$

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

Thus option D is correct.