Question

An isolated smooth ring of mass M = 2m with two small beads each of mass m is as shown in the figure. Initially both the beads are at diametrically opposite points and have velocity $V_0$ (for each) in same direction. The speed of the beads just before they collide for the first time is (complete system is placed on a smooth horizontal surface and assume each point of ring is touching the surface)

• A. $V_0$
• B. $\frac{2}{\sqrt{3}}{V}_0$
• C. $\frac{V_0}{2}$
• D. $\frac{\sqrt{3}}{2}{V}_0$

Answer 1

The correct option is D $\frac{\sqrt{3}}{2}{V}_0$

Just before the collision, situation is as shown in figure.
Let the beads have velocity $V_1$ w.r.t. the ring and the ring has velocity $V_2$.
Then , by momentum conversion
$2m{V}_0=(M+2m)v_2$
$\Rightarrow V_2=\frac{V_0}{2}$........... (1)
If $V$ = velocity of beads w.r.t ground which is vector sum of $V_1$ and $V_2$
or $V$ = $\sqrt{(V_1{)}^2+(V_2{)}^2}$
Then by mechanical energy conservation
$\frac{1}{2}\left(2m\right)V_0^2=\frac{1}{2}2m(V{)}^2+\frac{1}{2}2m{\left(\frac{V_0}{2}\right)}^2$
So, $V=\frac{\sqrt{3}}{2}{V}_0$