Question

Two small balls A and B, each of mass m, are joined rigidly to the ends of a light rod of length L (figure 10-E10). The system translates on a frictionless horizontal surface with a velocity ${\nu }_0$ in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. Find
(a) the linear speeds of the balls A and B after the collision, (b) the velocity of the centre of mass C of the system A + B + P and (c) the angular speed of the system about C after the collision.

Figure

Answer 1

(a) Collision will not affect the velocity of ball B because the light rod will exert a force on it only along its length.
Therefore, we have:
Velocity of B = v₀

For ball A,

On applying the law of conservation of linear momentum, we get:
$m{\nu }_0=2m\times \nu$
$\Rightarrow v=\frac{v_0}{2}$
$\therefore$ Velocity of A$=\frac{{\nu }_0}{2}$

(b) If we consider the three bodies to be a system, we have:
Net external force = 0
$$\begin{gathered} v_{\mathrm{CM}}=\frac{m\times v_0+2m\times \left(v_0/2\right)}{n+2m} \\ =\frac{m{v}_0+m{v}_0}{3m} \\ =\frac{2v_0}{3} \end{gathered}$$
(Direction will be same as the initial velocity before collision.)

(c) Velocity of (A + P) w.r.t. the centre of mass$=\left\{\frac{2v_0}{3}-\frac{v_0}{2}\right\}=\frac{v_0}{6}$
Velocity of B w.r.t. the centre of mass$=v_0-\frac{2v_0}{3}=\frac{v_0}{3}$
Distance of the (A + P) from the centre of mass $=\frac{l}{3}$
Distance of the B from the centre of mass =$\frac{2l}{3}$
$$\begin{gathered} \mathrm{Applying}m{v}_{com}r=l_{cm}\times \omega ,\mathrm{we}\mathrm{have}: \\ \left(2m\times \frac{v_o}{6}\times \frac{l}{3}\right)+\left(m\times \frac{v_0}{3}\times \frac{2l}{3}\right)=\left\{2m{\left(\frac{l}{3}\right)}^2+{\left(\frac{2l}{3}\right)}^{2}m\right\}\times \omega \\ \Rightarrow \frac{6m{\mathrm{v}}_{0}l}{18}=\frac{6ml}{9}\omega \\ \Rightarrow \omega =\frac{v_0}{2l} \end{gathered}$$