Question

Complete system shown in the figure is in horizontal plane. A rod A of mass 2 m and length 2L is hinged at O and is lying at rest. Another Identical rod B moving with velocity $V_0$ collides with rod A and sticks to it forming L-shape. Pick correct option(s) from the following.

• A. Angular velocity of system just after collision is $\frac{3V_0}{10L}$
• B. Velocity of centre of mass of system after collision is $\sqrt{\frac{2}{5}}V_0$
• C. Centre of mass of system is at a distance $\frac{\sqrt{5}}{2}L$ from O
• D. Kinetic energy of system just after collision is $\frac{3}{5}mV_0^2$

Answer 1

Answer: (A), (D)

Options 1 and 4.

Answer 2

Use angular momentum conservation about the hinge O.

Compute $I_A = (8/3)mL^2$ and $I_B = (32/3)mL^2$ giving $I_{total} = (40/3)mL^2$.

For rod B (CM at (2L, L)), angular momentum about O is $4mLV_0$.

Setting $4mLV_0 = I_{total} \omega$ gives $\omega = (3V_0)/(10L)$.

$KE = \frac{1}{2} I_{total} \omega^2 = (3/5)mV_0^2$.

The computed CM location does not match the given options for (2) and (3).