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 :

Blocks of masses m, 2m, 4m and 8m are arranged in a line on a frictionless floor. Another block of mass m, moving with speed v along the same line (see figure) collides with mass m in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass 8m starts moving the total energy loss is p% of the original energy. Value of 'p' is close to :

A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m/s² . Assume there is no rotational motion and loss of energy after the collision is negligable.]

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)