A particle of mass m is dropped from a height h above the ground. Simultaneously another particle of the same mass is thrown vertically upwards from the ground with a speed of 2 g h . If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground is

Step-by-step solution

The time for collision t 1 = h 2 g h After t 1 , v A = 0 - g t 1 = - g h / 2 V B = 2 g h - g t 1 = g h 2 - 1 2 At the time of collision P → i = P → f ⇒ m V → A + m V → B = 2 m V → f ⇒ - g h 2 + g h 2 - 1 2 = 2 V → f V f = 0 and height from ground = h - 1 2 g t 1 2 = h - h 4 = 3 h 4 so time = 2 × 3 h 4 g = 3 h 2 g