An oscillator of mass M is at rest in its equilibrium position in a potential V=1/2k(x-X)² . A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is: (M=10, m=5, u=1, k=1).

Initial momentum of mass 'm' = mu = 5
Final momentum of system = (M+m)v = mu = 5
For second collision, mass (m=5, u = 1) coming from right strikes with system of mass 15, both momentum have opposite direction.
∴net momentum = zero
Similarly for 12th collision momentum is zero.
For 13th collision, total mass = 10 + 12×5 = 70
Using conservation of momentum
70×0 + 5×1 = (70+5)v'
v' = 1/15
Total mass = 10 + 13×5 = 75
Final KE of system = 1/2mv² = 1/2 × 75 × [1/15]²
1/2kA² = 1/2 × 75 × (1/15)² = 1/6
1/2 × 1 × A² = 1/6
A² = 1/3
A = 1/√3