Two-Body Relative Oscillation and Collision
For masses m1 and m2 connected by a spring k on a force-free line, the relative coordinate obeys mu r double-dot = -kr with reduced mass mu = m1m2/(m1+m2).
Why this shows up in the exam
A spring held between two free masses · Collision followed by spring compression · Three-particle collision passages with an oscillating pair
Learn the idea
When both ends move, the spring governs relative motion with reduced mass rather than either mass alone. Two free masses recoil in opposite directions; describing their separation turns the pair into one oscillator with a reduced inertia.
🧠 Memory hook: Free at both ends means reduced mass.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- mu = m1 m2/(m1 + m2) — reduced mass for relative motion
- omega_rel = sqrt(k/mu) — relative oscillation angular frequency
- V_cm = constant — centre-of-mass velocity without external force
How to approach it
- 1Separate centre-of-mass and relative coordinates
- 2Apply collision conservation laws at the event
- 3Evolve the relative coordinate with the reduced mass
Common slip-ups that cost marks
- •Treating one free mass as fixed
- •Using separation for centre-of-mass motion
- •Ignoring momentum conservation during a short collision
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A mass of 1 kg is attached to a spring of force constant 100 N/m. Find the angular frequency of small oscillations.
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