Question

As shown in figure, two particles $A$ and $B$ of same mass $2\text{kg}$ are joined together by a rigid massless rod of length $2\text{m}$. A particle $P$ of mass $2\text{kg}$, moving normal to $AB$ with a speed $u=5\text{m/s}$ , strikes at end $A$ and sticks to it. The centre of mass of the system ${}^{\prime }A+B+P^{\prime }$ is $C$, then which of the following statement(s) holds true?

• A. The velocity of $C$ before impact is $\frac{5}{3}\text{m/s}$
• B. The velocity of $C$ after impact is $\frac{5}{3}\text{m/s}$
• C. The velocity of $(A+P)$ immediately after impact is $\frac{5}{2}\text{m/s}$
• D. None of the above

Answer 1

Answer: (A), (B), (C)

The correct options are
A The velocity of $C$ before impact is $\frac{5}{3}\text{m/s}$
B The velocity of $C$ after impact is $\frac{5}{3}\text{m/s}$
C The velocity of $(A+P)$ immediately after impact is $\frac{5}{2}\text{m/s}$

$C$ is the COM of the system.So velocity of $C$ before collision,
$v_{\text{COM}}=\frac{m_1{v}_1+m_2{v}_2+m_3{v}_3}{m_1+m_2+m_3}$
$v_{\text{COM}}=\frac{mu+m\left(0\right)+m\left(0\right)}{m+m+m}$
$\therefore v_{\text{COM}}=\frac{u}{3}=\frac{5}{3}\text{m/s}$

$F_{\text{ext}}=0$, hence velocity of COM will not change before and after the collision.
$v_{\text{COM}}=\frac{5}{3}\text{m/s}=\text{constant}$

Just after collision:


$\text{COM}$ 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, 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 $\left(i\right)$,
$mu\left(\frac{l}{3}\right)=\frac{2m{l}^2}{3}\omega$
$\Rightarrow \omega =\frac{u}{2l}=\frac{5}{2\times 2}=\frac{5}{4}\text{rad/s}$
$\therefore$Velocity of $(A+P)$ just after impact:
$v=v^{\prime }+\omega r...\left(iii\right)$
where $r=\frac{l}{3}$, is the radius of circular path for $(A+P)$ about $\text{COM}$ of the system, and $v^{\prime }=v_{COM}$ is translational velocity of $\text{COM}$ of the system.

$\therefore v=\frac{5}{3}+(\frac{5}{4}\times \frac{2}{3})=\frac{5}{2}\text{m/s}$
Options A, B, C are correct.