A particle of mass m is projected with a speed u from the ground at an angle $$\theta = {\pi \over 3}$$ w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $$u\widehat i$$ . The horizontal distance covered by the combined mass before reaching the ground is:

Two particles of equal mass m have respective
initial velocities $$u\widehat i$$ and $$u\left( {{{\widehat i + \widehat j} \over 2}} \right)$$.
They collide completely inelastically. The energy lost in the process is :

As shown in figure, when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centered at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e, on the surface of the cavity. R can be detemined by the equation :

A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $$\sqrt {2gh} $$. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of $$\sqrt {{h \over g}} $$ is :