A block of mass 200 g is suspended from a vertical spring. At equilibrium, the spring is stretched by 1.0 cm . A particle of mass 120 g is dropped onto the block from a height of 45 cm . Following the impact, the particle sticks to the block. Determine the maximum extension of the spring. Take g = 10 m/s ^2 .

Step-by-step solution

Mass of the block, M = 200 g = 0.2 kg Initial extension, x₀ = 1.0 cm = 0.01 m Spring constant, k = Mg x₀ = 0.2 10 0.01 = 200 N/m Velocity of the particle of mass m = 120 g = 0.12 kg just before impact: v = 2gh = 2 10 0.45 = 3 m/s By conservation of linear momentum during the inelastic collision: mv = (M + m)V 0.12 3 = (0.2 + 0.12)V V = 0.36 0.32 = 9 8 m/s The new equilibrium position of the spring is: x_ eq = (M + m)g k = 0.32 10 200 = 0.016 m = 1.6 cm The angular frequency of the oscillation is: = k M + m = 200 0.