Question

Two point masses m and 2m moving in the same horizontal plane with speeds 2v and v respectively, strike a bar as shown in fig. and stick to it after the collision. Denoting angular velocity (about the centre of mass), total energy and velocity of centre of mass by $\omega$, E and $v_c$ respectively, we have after collision :

• A. $v_c=0$
• B. $\omega =\frac{3v}{5a}$
• C. $\omega =\frac{v}{5a}$
• D. $E=\frac{3}{5}m{v}^2$

Answer 1

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

$v_c=0$
$\omega =\frac{v}{5a}$
$E=\frac{3}{5}m{v}^2$

Let the centre of mass of the system after collision lies at point C'.

Position of point C w.r.t point A is $3a$.

Position of C' from A, $x=\frac{m\left(a\right)+8m\left(3a\right)+2m\left(4a\right)}{m+8m+2m}=3a$

$⟹$ Points C and C' coincide.

For translational motion :

As no external force is acting on the system, thus linear momentum is conserved i.e $P_i=P_f$

$\therefore$ $2m\left(v\right)-m\left(2v\right)=11m\left(V_c\right)$

OR $0=11m\left(V_c\right)$ $⟹V_c=0$

For rotational motion :

Moment of inertia about C $I_c=m(2a{)}^2+\frac{\left(8m\right)(6a{)}^2}{12}+(2m)a^2=30m{a}^2$

Also, as no external torque acting on the system, thus angular momentum of the system about point C is conserved.

$\therefore$ $m\left(2v\right)\left(2a\right)+2m\left(v\right)\left(a\right)=I_{c}w$

OR $6mva=30m{a}^{2}w$ $⟹w=\frac{v}{5a}$

Total energy, $E=K.E_t+K.E_r$

$\therefore$ $E=0+\frac{1}{2}{I}_c{w}^2$

OR $E=0+\frac{1}{2}\times 30m{a}^2\times \frac{v^2}{25a^2}=\frac{3}{5}m{v}^2$