Question

A small disc and a thin uniform rod of length L, whose mass is n times greater than the mass of the disc, lie on a smooth horizontal plane. The disc is set in motion, in horizontal direction and perpendicular to the rod, with velocity v, after which it elastically collides with the end of the rod. Then determine the value of n for which the velocity of the disc after the collision be equal to zero is?

Answer 1

Here angular momentum converted into linear momentum and angular momentum.

I = moment of inertia of disc about C

$I_1$ = moment of inertia of disc about C

angular momentum conservation.

$I\omega =I^{\prime }{\omega }^{\prime }mv\frac{l}{2}=nm\frac{l^2}{12}{\omega }^{\prime }\to \left(1\right)$

linear momentum conservation

$mv=nm{v}_0\to \left(2\right)$

equation of coefficient of resistution

$e=1=\frac{({\omega }^{\prime }\frac{l}{2}+v_0)}{v}v={\omega }^{\prime }\frac{l}{2}+v_0\to \left(3\right)$

from equation (1) ${\omega }^{\prime }=\frac{6v}{nl}$

$v=\frac{6vl}{nl\times 2}+v_0=\frac{3v}{n}+v_{0}v(1-\frac{3}{n}v)=v_0$

putting in equation (2)

$\frac{m{v}_0}{\left(\frac{1-3}{n}\right)}=nm{v}_{0}1=n(1-\frac{3}{n})1=n-3n=4$