Question

Two balls $A$ and $B$ of mass $m=1\text{kg}$ each are joined together by a rigid massless rod of length $l=2\text{m}$. A ball $P$ of mass $m=1\text{kg}$ moving with speed $(u=12\text{m/s})$ normal to $AB$, strikes end $A$ of the rod and sticks to it. The centre of mass of the system $"A+B+P"$ is ${}^{\prime }{C}^{\prime }$. Then immediately after impact, the velocity of $A+P$ in $\text{m/s}$ with respect to $C$ will be

Answer 1

$\text{COM}$ of the system from point $A\left(\text{origin}\right)$ is,
$y_{\text{COM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$y_{\text{COM}}=\frac{m\left(0\right)+m\left(0\right)+m\left(l\right)}{m+m+m}$
$\therefore y_{\text{COM}}=\frac{l}{3}=\frac{2}{3}\text{m}$
Just after collision, the system starts moving with velocity $v_{\text{COM}}=v^{\prime }$ (translational velocity) and with angular velocity $\omega$ about COM.
$\because {\tau }_{\text{ext}}=0$, applying angular momentum conservation for system about $\text{COM}$,
$L_i=L_f$
$\Rightarrow mu\left(y_{\text{COM}}\right)=L_{\text{trans}}+L_{\text{rot}}$
[taking antilockwise sense of rotation as $+ve$]
$\Rightarrow -mu\left(\frac{l}{3}\right)=0+(-I\omega )$
$\Rightarrow mu\left(\frac{l}{3}\right)=I\omega ...\left(i\right)$
[$L_{\text{trans}}=0$,because velocity vector passes through COM]

$\text{MOI}$ of system about centre of mass is,
$I=2m{\left(\frac{l}{3}\right)}^2+m{\left(\frac{2l}{3}\right)}^2$
$I=\frac{2m{l}^2}{3}...\left(ii\right)$
Substituting in Eq $\left(i\right)$,
$mu\left(\frac{l}{3}\right)=\frac{2m{l}^2}{3}\omega$
$\Rightarrow \omega =\frac{u}{2l}=\frac{12}{2\times 2}=3\text{rad/s}$

$\therefore$Velocity of $(A+P)$ just after impact with respect to centre of mass of system $C$ is the tangential velocity given as:
$v=\omega r...\left(iii\right)$
where $r=\frac{l}{3}$, is the radius of circular path for $(A+P)$ about $\text{COM}$ of the system

$\therefore v=(3\times \frac{2}{3})=2\text{m/s}$