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 $v_0$ 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.

Answer 1

Two ball, A and B, each of mass m are joined rigidly to the ends of a light rod of length L. The system moves with a velocity $V_0$ 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.

(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 a velocity = $V_0$

If we consider the three bodies to be a system,

Applying L.C.L.M.

Therefore, $m{v}_0$ = $2m\times v$ : $\Rightarrow$ v = $\frac{v_0}{2}$

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

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

net external force = 0

Therefore,

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

= $\frac{m{v}_0+m{v}_0}{3m}$

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

(along the initial velocity as before collision.)

(c) The velocity of (A+P) w.r.t.

the center of mass = $\{\left(\frac{2v_0}{3}\right)-(v_0-2\left)\right\}$

(Only magnitude has been taken.)

Distance of the (A + P) from centre of mass

= $\frac{l}{3}$ and for B it is $\left(\frac{2l}{3}\right)$

$=\frac{v_0}{6}$

and the velocity of B w.r.t. the centre of mass $v_0-\frac{2v_0}{3}=\frac{v_0}{3}$

Therefore, $P_{cm}=l_{cm}\times \omega$

$\Rightarrow 2m\times \frac{V_0}{6}\times \frac{l}{m}+m\times \frac{V_0}{3}\times \frac{2l}{3}$

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

$\Rightarrow 6m\frac{V_{0}l}{18}=18\left(\frac{6ml}{9}\right)\omega$

$\Rightarrow \omega =\left(\frac{v_0}{2l}\right)$