Two blocks of masses m₁ and m₂ are connected by a spring of spring constant k . The block of mass m₂ is given a sharp empulse so that it acquires a velocity v₀ towards right. Find the maximum elongation that the spring will suffer.

Step-by-step solution

The velocity of centre of mass v_ cm = m₁ v₁+m₂ v₂ m₁+m₂ When v₁=0 and v₂=v₀ , then v_ c m = m₂ v₀ m₁+m₂ Now, let x be the elongation in the spring. Change in potential energy = potential energy stored in spring gathered 1 2 m₂ v₀^2- 1 2 (m₁+m₂ ) ( m₂ V₀ m₁+m₂ )^2= 1 2 k x^2 gathered This gives, x= ( m₁ m₂ m₁+m₂ )^ 1 2 v₀