An oscillator of mass $M$ is at rest in its equilibrium position in a potential, $V = \dfrac12k(x - X)^2$. A particle of mass $m$ comes from the right with speed $u$ and collides completely inelastic 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: $(M = 10,\ m = 5,\ u = 1,\ k = 1)$
Answer: (B) $\dfrac{1}{\sqrt3}$
Collisions happen at the equilibrium position, where all the energy is kinetic.
- **1st collision:** momentum $mu$ (to the left) is shared, so the oscillator moves off with momentum $mu$.
- It returns through equilibrium moving to the right with the same momentum $mu$. The **2nd particle** brings $mu$ to the left, so the total momentum becomes zero and the system stops.
- The **3rd particle** sets it moving again with momentum $mu$, and so on.
After an odd number of collisions the momentum is $mu$. After 13 collisions the mass is $M + 13m = 10 + 65 = 75$, so
$$v = \frac{mu}{M + 13m} = \frac{5}{75} = \frac{1}{15}$$
Amplitude $A = \dfrac{v}{\omega}$ with $\omega = \sqrt{\dfrac{k}{M + 13m}} = \dfrac{1}{\sqrt{75}}$:
$$A = \frac{1}{15}\sqrt{75} = \frac{5\sqrt3}{15} = \frac{1}{\sqrt3}$$
Solution by Sreeraj P, M.Sc Physics