Projectile Collisions, Explosions, and Centre of Mass
For a short internal event, total momentum is conserved. During subsequent flight, the centre of mass obeys M a_CM = sum F_external and therefore follows the gravity-driven path fixed by its state at the event.
Why this shows up in the exam
Projectile breakup at the highest point · Mid-air inelastic collisions · Locating unseen fragments from centre-of-mass motion
Learn the idea
Internal collisions or explosions redistribute relative motion while the centre of mass continues under external gravity. Fragments can follow very different paths, yet their mass-weighted average position behaves like the original projectile because internal impulses cancel in the total momentum balance.
🧠 Memory hook: Fragments may scatter, but their centre of mass keeps the external-force story.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- sum p_before = sum p_after — momentum conservation during a short internal event
- r_CM = (sum m_i r_i)/(sum m_i) — centre-of-mass position
- M a_CM = sum F_external — centre-of-mass motion under external forces
How to approach it
- 1Isolate the instant of collision or explosion
- 2Conserve vector momentum across that instant
- 3Propagate fragments or centre of mass under gravity afterward
Common slip-ups that cost marks
- •Conserving kinetic energy in a completely inelastic collision
- •Applying momentum conservation over a long interval with gravity impulse
- •Following one fragment and forgetting the centre of mass
🌟 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 particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?
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