Test SeriesFree PYQsExamsPricing

JEE Main Physics question

PhysicsSystems of Particles and Rotational MotionJEE Main

Two blocks A and B of mass 3 kg and 6 kg are connected by a massless spring of force constant 2×10⁻³ N/m. They are placed on a smooth horizontal plane. The spring is stretched by an amount X and then released. The relative velocity of the blocks when the spring comes to its natural length, v = NX m/s where N = × 10⁻². (Take √10 = 3.16)

Answer: 3.16

Solution

Reduced mass μ = m₁m₂/(m₁+m₂) = (3×6)/(3+6) = 18/9 = 2 kg. By conservation of mechanical energy (in the centre-of-mass

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-154230