Question

Two particles A and B, of mass $m$ each, are joined by a rigid massless rod of length $l$. A particle P of mass $m$, moving with a speed $u$ normal to AB, strikes A and sticks to it. The centre of mass of 'A + B + P' system is C. Then which of the following statement(s) is/are true ?

• A. The velocity of C before impact is $\frac{u}{3}$.
• B. The velocity of C after impact is $\frac{u}{3}$.
• C. The velocity of 'A + P' immediately after impact is $\frac{u}{2}$.
• D. The velocity of B immediately after impact is zero.

Answer 1

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

The correct options are
A The velocity of C before impact is $\frac{u}{3}$.
B The velocity of C after impact is $\frac{u}{3}$.
C The velocity of 'A + P' immediately after impact is $\frac{u}{2}$.
D The velocity of B immediately after impact is zero.

C is located on the line perpendicular to AB and intersecting it at a distance $\frac{L}{3}$ below A.
$M_C=\sum m=3m$
$M_C\times V_C=mu$
Since no external force acts on the system, the velocity of C remains unchanged before and after the collision.
$V_C=\frac{u}{3}$
=> option A and B are right
Also, angular momentum of the system about C is conserved.
$L=\frac{mul}{3}$
Let the rod acquire an angular velocity of $\omega$ about C.
Then the new velocities with respect C are:
$V_a=\frac{\omega l}{3}$ and $V_B=-\frac{2\omega l}{3}$
angular impulse at A = change in angular momentum of P and A
angular impulse at A = $\frac{2m{V}_A\times l}{3}-\frac{mul}{3}$ = $\frac{2\omega l^2}{9}-\frac{mul}{3}$
angular impulse at B = change in angular momentum of B
angular impulse at B = $-m\frac{2\omega l}{3}\frac{2l}{3}-0=\frac{-4\omega l^2}{9}$
angular impulse at A = angular impulse at B
=> $\omega =\frac{u}{2l}$
velocity of A $=V_A+V_C=\frac{u}{2}$
velocity of B $=V_B+V_C=0$
=> option C and D are also correct