Two blocks having masses m₁ and m₂ are connected by a spring with spring constant k . A sharp impulse is administered to the block of mass m₂ , causing it to gain a velocity v₀ towards the right. Calculate the maximum elongation that the spring will undergo.

Step-by-step solution

Let v be the common velocity of the two blocks at the instant of maximum elongation. By conservation of linear momentum: m₂ v₀ = (m₁ + m₂)v v = m₂ v₀ m₁ + m₂ By conservation of mechanical energy: 1 2 m₂ v₀^2 = 1 2 (m₁ + m₂)v^2 + 1 2 k x_ max ^2 Substituting the value of v : 1 2 m₂ v₀^2 = 1 2 (m₁ + m₂) ( m₂ v₀ m₁ + m₂ )^2 + 1 2 k x_ max ^2 m₂ v₀^2 = m₂^2 v₀^2 m₁ + m₂ + k x_ max ^2 k x_ max ^2 = m₂ v₀^2 ( 1 - m₂ m₁ + m₂ ) k x_ max ^2 = m₁ m₂ m₁ + m₂ v₀^2 x_ max = v₀ [ m₁ m₂ (m₁ + m₂)k ]^ 1/2