Question

In figure shown rod A of mass 2 m and length 2t is kept on a horizontal frictional surface and hinged at point O such that it can rotate about vertical axis passing through O. Another identical rod B moving with a constant velocity v 0 ​ dollides with rod A at its end and stickes to it (both rods are perpencicular to each other)

• A. angular velocity of system just after collision is 3v 0 ​ 10ℓ ​
• B. Velocity of centre of mass at system just after collision is 9v 0 ​ 5 2 ​ ​
• C. Centre α mass of system is at a distance d 2 5 ​ ​ from 0
• D. Kinetic energy of system just alter collision is 5 3mv 0 2 ​ ​

Answer 1

Answer: (A), (D)

Options (A) and (D)

Answer 2

Solution:

We use conservation of angular momentum about the hinge point O. In the collision only rod B contributes initial angular momentum about O. By computing the rotational inertias of the two rods (with rod A of mass 2m and length 2L having

$I_A=\frac{1}{3}(2m)(2L)^2=\frac{8}{3}mL^2$

and rod B (identical in dimensions, mass 2m) having

$I_B=\frac{32}{3}mL^2$,

so that

$I_{\mathrm{total}}=\frac{8}{3}mL^2+\frac{32}{3}mL^2=\frac{40}{3}mL^2$,

and taking the effective lever‐arm for rod B (its centre of mass lies at (2L, L) relative to O) so that its angular momentum is

$L_B=4mLV_0$,

equating initial angular momentum $4mLV_0$ to $I_{\mathrm{total}}\omega$ we get

$\omega=\frac{3V_0}{10L}$.

The rotational kinetic energy after collision is then

$KE=\tfrac{1}{2}I_{\mathrm{total}}\omega^2 = \tfrac{1}{2}\left(\frac{40}{3}mL^2\right)\left(\frac{3V_0}{10L}\right)^2= \frac{3}{5}mV_0^2$.

Thus, the correct choices are the ones stating that the angular velocity is $\frac{3V_0}{10L}$ and the kinetic energy is $\frac{3}{5}mV_0^2$.