Noesis
1.4.11Momentum & Collisions

Motion of centre of mass — external force determines a_CM

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🇮🇳 Yeh note Hinglish mein padho →

WHAT is the centre of mass?

WHY mass-weighted? Because a heavy particle "anchors" the system more than a light one. If you just averaged positions, a feather and an anvil would balance at their midpoint — clearly wrong. Weighting by mass puts the balance point near the anvil.

HOW the external force controls aCMa_{CM} — full derivation

We derive F⃗ext=Ma⃗CM\vec{F}_{ext} = M\vec{a}_{CM} from scratch.

Step 1 — Start from the definition and multiply by MM: MR⃗CM=∑imir⃗iM\vec{R}_{CM} = \sum_i m_i \vec{r}_i Why this step? It clears the fraction so differentiation is clean.

Step 2 — Differentiate once (velocity): MV⃗CM=∑imiv⃗i=∑ip⃗i=P⃗totalM\vec{V}_{CM} = \sum_i m_i \vec{v}_i = \sum_i \vec{p}_i = \vec{P}_{total} Why this step? This reveals a gem: total momentum equals total mass times CM velocity, P⃗=MV⃗CM\vec{P} = M\vec{V}_{CM}.

Step 3 — Differentiate again (acceleration): Ma⃗CM=∑imia⃗i=∑iF⃗iM\vec{a}_{CM} = \sum_i m_i \vec{a}_i = \sum_i \vec{F}_i where F⃗i\vec{F}_i is the total force on particle ii (Newton's 2nd law per particle).

Step 4 — Split each force into internal + external: ∑iF⃗i=∑iF⃗iext⏟F⃗ext+∑i∑j≠if⃗ij⏟internal pairs\sum_i \vec{F}_i = \underbrace{\sum_i \vec{F}_i^{ext}}_{\vec{F}_{ext}} + \underbrace{\sum_{i}\sum_{j\ne i}\vec{f}_{ij}}_{\text{internal pairs}}

Step 5 — Kill the internal sum using Newton's 3rd law: For every pair, f⃗ij=−f⃗ji\vec{f}_{ij} = -\vec{f}_{ji}. So they cancel in pairs: ∑i∑j≠if⃗ij=0\sum_{i}\sum_{j\ne i}\vec{f}_{ij} = 0 Why this step? — the heart of the proof. Internal forces always come in equal-and-opposite couples, so they sum to zero and cannot accelerate the CM.

Corollary: If F⃗ext=0⇒a⃗CM=0⇒V⃗CM=constant\vec{F}_{ext} = 0 \Rightarrow \vec{a}_{CM}=0 \Rightarrow \vec{V}_{CM} = \text{constant} → conservation of momentum.

Figure — Motion of centre of mass — external force determines a_CM

WORKED EXAMPLES

COMMON MISTAKES (Steel-man + fix)

Active Recall

Recall Feynman: explain to a 12-year-old

Imagine a bunch of friends tied together by springs, floating in space. They can yank and push each other all they like — but the average spot of the whole gang won't drift. To make the whole gang's "centre" move, someone OUTSIDE (gravity, a wall, a rocket) has to push. Their own pushing cancels out, because every push has an equal push-back. So the "balance point" only listens to outside forces.

Flashcards

What does F⃗ext=Ma⃗CM\vec F_{ext}=M\vec a_{CM} say?
The CM accelerates as a single particle of total mass MM driven only by the net external force.
Why do internal forces drop out of the CM equation?
By Newton's 3rd law they occur in equal-and-opposite pairs that sum to zero.
Define centre of mass.
R⃗CM=1M∑imir⃗i\vec R_{CM}=\frac{1}{M}\sum_i m_i\vec r_i, the mass-weighted average position.
Relation between total momentum and CM velocity?
P⃗total=MV⃗CM\vec P_{total}=M\vec V_{CM}.
If F⃗ext=0\vec F_{ext}=0, what happens to the CM?
a⃗CM=0\vec a_{CM}=0, so V⃗CM\vec V_{CM} is constant (momentum conserved).
A shell explodes mid-flight; what does the CM do?
It continues on the original parabola (until a piece lands), since explosion is internal.
Why is CM mass-weighted, not the midpoint?
Heavier masses anchor the balance point more; equal averaging would mislocate it.

Connections

Concept Map

justified by

differentiate once

differentiate twice

forces split

internal pairs cancel via

so

leaves only external

derived from

if F_ext = 0

explains

applies to

Centre of mass
mass-weighted position

Mass weighting

P total = M V_CM

Newton 2nd law per particle

Split forces:
internal + external

Newton 3rd law
f_ij = -f_ji

Internal sum = 0

F_ext = M a_CM

Conservation of momentum

Exploding projectile
CM stays on parabola

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Centre of mass (CM) ka matlab hai system ka "average mass-weighted point" — yaani jahan saara mass concentrated maan lo. Ek system mein particles aapas mein kitna bhi push-pull kare (ye internal forces hain), CM ko move nahi kara sakte. Reason simple hai: Newton ke third law se har internal force ka ek equal-and-opposite partner hota hai, to sab cancel ho jaate hain. Sirf BAHAR se aane wala (external) force hi CM ko accelerate karta hai, aur wo bhi is formula se: F⃗ext=Ma⃗CM\vec F_{ext} = M\vec a_{CM}.

Iska sabse mast example hai exploding shell. Ek bomb upar parabola mein jaa raha hai, top par phat jaata hai. Tukde alag-alag udte hain, lekin unka CM wahi original parabola follow karta rehta hai — kyunki explosion ek internal force hai, gravity hi ek external force hai. Isliye aCM=ga_{CM}=g rehta hai jab tak koi tukda zameen na chhu le.

Ek aur key relation yaad rakho: P⃗=MV⃗CM\vec P = M\vec V_{CM} — total momentum equals total mass × CM velocity. Agar external force zero hai (jaise frictionless ice par do skater push karte hain), to VCMV_{CM} constant rehta hai — yahi momentum conservation hai. Skater example mein dono rest se start karte hain, to CM hilta hi nahi, bhale dono opposite directions mein chale jaayein.

Exam tip / 80-20: bas do cheezein pakdo — (1) internal forces always cancel (third law), (2) F⃗ext=Ma⃗CM\vec F_{ext}=M\vec a_{CM} aur P⃗=MV⃗CM\vec P=M\vec V_{CM}. Inhi se boat-man, explosion, aur collision ke saare numericals nikal jaayenge. CM ko geometric midpoint mat samajhna — wo mass ki taraf jhukta hai jo heavier hoti hai.

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