NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
A ball falls freely from a height of 45 m . When the ball is at a height of 25 m , it explodes into two equal pieces. One of them acquires an additional horizontal component of velocity equal to 10 m s - 1 , while its vertical component remains the same. The distance between the two pieces when both strike the ground is
Options
- A10   m
- B20 m
- C15   m
- D30   m
Correct answer
B. 20 m
Step-by-step solution
Let at the time explosion velocity of one piece of mass is (10 i ^ ). If the velocity of other be v → 2 , then from conservation law of momentum (since there is no force in the horizontal direction), the horizontal component of v → 2 , must be - 1 0 i ^ . ∴ The relative velocity of two parts in the horizontal direction = 20  m  s - 1 Time taken by the ball to fall through 45 m, = t = 2 h g = 2 × 45 10 = 3   s and time taken by the ball to fall through first 20m, t ′ = 2 h