NTA Abhyas JEE Main2020PhysicsOscillationsPractice
Two blocks P and Q of masses 0.3 kg and 0.4 kg respectively are stuck to each other by some weak glue as shown in the figure. They hang together at the end of a spring with a spring constant k = 200 N m - 1 . The block Q suddenly falls free due to failure of glue, then find the maximum kinetic energy of the block P during subsequent motion (in mJ ).
Correct answer
40
Step-by-step solution
Let A be the amplitude, then we have kA = mg       ⇒ A = mg k A = 0.4 × g 200 = 2   cm Motion starts from the lower extreme. KE max = 1 2 mω 2 A 2 = 1 2 × 0.3 × k m × A 2 = 1 2 × 0.3 × 200 0.3 × 2 100 2 = 0.04   J = 40   mJ