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Step-by-Step Solution
Step 1: Understand the Physical Situation
We have an oscillator (mass M) initially at rest in its equilibrium position.
A particle (mass m) moving with speed u from the right collides completely
inelastically with the oscillator each time the system crosses the equilibrium position.
After the collision, the particle sticks to the oscillator, increasing the total mass.
We must find the amplitude of oscillation after 13 such collisions.
Step 2: List the Known Values
• Initial mass of oscillator, $M = 10$
• Each incoming particle’s mass, $m = 5$
• Speed of each incoming particle, $u = 1$
• Spring constant, $k = 1$
• Number of collisions to analyze, $n = 13$
Step 3: Use Momentum Conservation at the Equilibrium Crossings
Collisions occur exactly at the equilibrium position of the oscillator,
where the potential energy is zero and all the system’s energy is kinetic.
We use linear momentum conservation just before and right after each collision:
$$(\text{total incoming momentum}) = (\text{total final mass}) \times (\text{final velocity}).$$
Step 4: Pattern of Velocity After Each Collision
1. After the 1st collision:
• Before collision: mass $m$ moving at $u$; oscillator mass $M$ at rest.
• Total mass afterward: $M + m = 10 + 5 = 15.$
• Momentum conservation:
$$m u = (M + m)\,v_1 \quad \Longrightarrow \quad 5 \cdot 1 = 15 \,v_1 \quad \Longrightarrow \quad v_1 = \frac{1}{3}.$$
• The system then completes half an oscillation and returns to the equilibrium from the other side with velocity $-v_1.$
2. After the 2nd collision:
• The system of mass $15$ moves with speed $-v_1 = -\tfrac{1}{3}$ when it meets a new
incoming mass $m=5$ still moving from the right at speed $u = +1$.
• Total mass afterward: $M + 2m = 10 + 2 \times 5 = 20.$
• Momentum conservation:
$$(15)(-\tfrac{1}{3}) + (5)(1) = 20 \,v_2 \quad \Longrightarrow \quad -5 + 5 = 0 \quad \Longrightarrow \quad v_2=0.$$
• Hence, after every even-numbered collision, the combined system comes momentarily
to rest at the equilibrium position (velocity = 0).
3. General Pattern:
• After each even collision, the system is momentarily at rest.
• Then an odd-numbered collision occurs with a fresh mass $m$ arriving from the right at speed $u$.
• Because the system is at rest at that instant, the final velocity after each odd-numbered collision
can be found simply by $m u = (\text{total mass}) \times v_\text{final}.$
Step 5: Velocity After the 13th Collision
• Notice that all even-numbered collisions result in zero velocity. Thus,
just before the 13th collision (which is an odd-numbered collision), the system is at rest
with total mass $M + 12m = 10 + 12\times 5 = 70.$
• The incoming mass ($m=5$) has momentum $5 \cdot 1 = 5.$
• Total mass right after the 13th collision is $M + 13m = 75.$
Using momentum conservation for the 13th collision:
$$70 \times 0 + 5 \times 1 = (70 + 5)\,v_{13} \quad \Longrightarrow \quad 5 = 75\,v_{13} \quad \Longrightarrow \quad v_{13} = \frac{5}{75} = \frac{1}{15}.$$
Step 6: Convert Final Velocity to Amplitude of Oscillation
Right after the 13th collision, the entire system has mass $75$ and velocity $\tfrac{1}{15}.$
This kinetic energy is converted into the total energy of the simple harmonic motion:
$$\text{Kinetic Energy} = \frac{1}{2}\,(\text{total mass})\,v_{13}^2
= \frac{1}{2}\,(75)\,\Bigl(\!\frac{1}{15}\Bigr)^{2} = \frac{75}{2}\,\frac{1}{225} = \frac{75}{450} = \frac{1}{6}.$$
In a simple harmonic oscillator with spring constant $k=1,$
total energy $E = \frac{1}{2}\,k\,A^2.$
Therefore:
$$\frac{1}{2}\,A^2 = \frac{1}{6} \quad \Longrightarrow \quad A^2 = \frac{1}{3} \quad \Longrightarrow \quad A = \frac{1}{\sqrt{3}}.$$
Final Answer
The amplitude of oscillation after the 13th collision is
$$\boxed{\frac{1}{\sqrt{3}}}.$$
