Four equal masses, m each are placed at the comers of a square of length (l) as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be :

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, body 'Q' will have a maximum speed equal to '2u' after collision.

Reason R : During elastic collision, the momentum and kinetic energy are both conserved.

In the light of the above statements, choose the most appropriate answer from the options given below :

Particle A of mass m₁ moving with velocity $$\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}}$$ collides with another particle B of mass m₂ which is at rest initially. Let $$\overrightarrow {{V_1}} $$ and $$\overrightarrow {{V_2}} $$ be the velocities of particles A and B after collision respectively. If m₁ = 2m₂ and after
collision $$\overrightarrow {{V_1}} = $$$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ , the angle between $$\overrightarrow {{V_1}} $$ and $$\overrightarrow {{V_2}} $$ is :

A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle $$\theta $$ before momentarily coming to rest. If the rod has mass M = 2 kg, and length $$l$$ = 1 m, the value of $$\theta $$ is approximately :
(take g = 10 m/s²)