Question

Two particles A and B, of mass m each, are joined by a rigid massless rod of length l. The rod is kept in a horizontal plane. 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 immediately after impact, which of the following statement(s) hold true ?

• A. The velocity of A + P with respect to C is $\frac{u}{6}$, to the right.
• B. The angular velocity of A + P with respect to C is $\frac{u}{2l}$, clockwise.
• C. The velocity of B with respect to C is $\frac{u}{3}$, to the left.
• D. The angular velocity of B with respect to C is $\frac{u}{2l}$, clockwise.

Answer 1

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

The correct options are
A The velocity of A + P with respect to C is $\frac{u}{6}$, to the right.
B The angular velocity of A + P with respect to C is $\frac{u}{2l}$, clockwise.
C The angular velocity of B with respect to C is $\frac{u}{2l}$, clockwise.
D The velocity of B with respect to C is $\frac{u}{3}$, to the left.

Let $v_c$ be the velocity of the center of mass after collision. Thus using conservation of momentum we have
$mu=3m{v}_c$
or
$v_c=\frac{u}{3}$
Now using conservation of angular momentum we have
$mu\frac{l}{3}=I\omega$
or
$mu\frac{l}{3}=\left(2m\right(\frac{l}{3}{)}^2+m\left(\frac{2l}{3}{)}^2\right)\omega$(here l/3 is the distance from A+P to C and 2l/3 is the distance of B from C)
Thus solving it we get
$\omega =\frac{u}{2l}$
Thus option B is right.
Now velocity of A+P w.r.t center of mass is given as $v_c-r\omega -v_c$
or
$\frac{u}{3}-\frac{l}{3}\frac{u}{2l}-\frac{u}{3}=-\frac{u}{6}$(negative sign indicates opposite to the direction of velocity of center of mass)
Thus option A is right.
Similarly velocity of B w.r.t C is given as $v_c+r\omega -v_c$
or
$\frac{u}{3}+\frac{2l}{3}\frac{u}{2l}-\frac{u}{3}=\frac{u}{3}$
Thus option C is right.