Question

Two small balls A and B, each of mass m, are joined rigidly tot he ends of a light rod of length L. The system translates on a frictionless horizontal surface with a velocity $v_0$ in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface stickes 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 od the system A+B+P and
c) the angular speed of the system about C after the collision

Answer 1

Two balls $A$ and $B$ each, of mass $m$ are joined rigidly to the ends of light of rod of length $L$.

The system movies in a velocity $v_0$ in a direction $\perp$ 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.

$a)$ The light rod will exert a force on the ball $B$ only along its length.

So collision will not affect its velocity.

$B$ has velocity $=v_0$

If we consider the three bodies to be a system

Applying Law of conservation of linear momentum,

$m{v}_0=2m{v}^{\prime }\Rightarrow v^{\prime }=\frac{v_0}{2}$

Therefore $A$ has velocity $=\frac{v_0}{2}$

$b)$ If we consider the three bodies to be a system

Therefore, net external force $=0$

Therefore, the velocity of the center of mass is given as:

$V_{cm}=\frac{m\times v_0+2m\left(\frac{v_0}{2}\right)}{m+2m}$

$=\frac{m{v}_0+m{v}_0}{3m}=\frac{2v_0}{3}$ (along the initial velocity as before collision)

$c)$ The velocity of $(A+P)$ w.r.t. the centre of mass

$=\frac{2v_0}{3}-\frac{v_0}{2}=\frac{v_0}{6}$

The velocity of $B$ w.r.t. the centre of mass

$v_0-\frac{2v_0}{3}=\frac{v_0}{3}$ [Only magnitude has been taken]

Distance of the $(A+P)$ from centre of mass $=l/3$ & for $B$ it is $2l/3$.

Therefore, using conservation of angular momentum:-

$P_{cm}=I_{cm}\times \omega$

$\Rightarrow 2m\times \frac{v_0}{6}\times \frac{l}{3}+m\times \frac{v_0}{3}\times \frac{2l}{3}=2m{\left(\frac{l}{3}\right)}^2+m{\left(\frac{2l}{3}\right)}^2\times \omega$

$\Rightarrow \frac{6m{v}_{0}l}{18}=\frac{6ml}{9}\times \omega$

$\Rightarrow \omega =\frac{v_0}{2l}$