NTA Abhyas JEE Main2020PhysicsOscillationsPractice
A block of mass 1 kg is dropped on a spring-mass system as shown in the figure. The block travels 100 meters in the air before striking the 3 kg mass. Calculate maximum compression in the spring, if both the blocks move together after the collision. Spring constant of the string k = 1.25 × 10 6 .
Options
- A2   cm
- B4   cm
- C8   cm  
- D16   cm
Correct answer
A. 2   cm
Step-by-step solution
The velocity of 1   k g block just before striking V = 2 × 10 × 100 = 20 5   m   s - 1 Applying conservation of momentum, the blocks stick after the collision and move with velocity v Or 1 × 20 5 = 4 v Final velocity   v = 5 5 , Using energy conservation K E + P E = const 1 2 × 4 × 5 5 2 + 1 2 k 3 g k 2 = 0 + 1 2 k . x 2 ∴ x 2 = 4 × 25 × 5 1.25 × 10 6 + 30 × 30 1.25 × 10 6 2 x = 20 1000   m = 2   cm