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Physics notesMechanicsCOM & Collisions

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Centre of Mass & Collisions

Track COM under external force, then switch to impulse–momentum and restitution along the line of impact; most collision stems close without full trajectory integration.

Portion MechanicsFocus Reliable Advanced marks from COM motion, impulse and elastic/inelastic collision algebraTarget 60–90sPortion share 32.3%Trend Falling share
Systems of particles: COM moves as if all mass sat there with ΣF_ext. Collisions are instantaneous impulses—external impulses from the floor may matter. Elastic means e=1 and kinetic energy conserved along the interaction; inelastic has e<1; perfectly inelastic sticks (common velocity). Collision template: free vs constrained → impact normal → momentum (± wall) → e along normal → energy only if e=1 or asked separately.

How Advanced asks this

  • Locate COM of discrete masses, rods, plates or explosion fragments
  • COM motion under gravity / on smooth surface while internal motions occur
  • 1D/2D collisions with e given; find speeds or % KE loss
  • Bullet–block, ballistic, or perfectly inelastic catch
  • Impulse and average force from Δp and Δt
  • Collision with wall/floor with restitution; bounce heights

Traps that leak marks

  • Conserving mechanical energy through a perfectly inelastic collision
  • Using 1D elastic formulas for oblique impacts without components
  • Conserving momentum against a fixed wall (wall impulse is external)
  • Forgetting that friction impulse can change tangential momentum
  • Mixing lab-frame and COM-frame velocities
  • Taking e as a ratio of kinetic energies
  • Using vector e without projecting along the contact normal
  • Conserving KE for a clearly inelastic stem
  • Applying e to the wrong component (tangential vs normal)
  • Forgetting external impulse from a wall or hinge during contact

01. R_COM = Σm_i r_i / M; V_COM = Σm_i v_i / M; A_COM = ΣF_ext / M. Internal forces cancel in pairs for the COM theorem (ideal Newton III).

02. Impulse J=∫F dt = Δp. During a collision, finite forces are negligible compared with impulsive contact forces if the duration is tiny.

03. Coefficient of restitution e = −(relative speed after)/(relative speed before) along the common normal at contact (with sign convention).

04. 1D elastic two-mass formulas follow from momentum + e=1; equal masses exchange velocities. In COM frame, elastic collision reverses each mass’s velocity along the line of impact.

05. Centre-of-mass motion follows external force only: M a_cm = F_ext. Internal impulses rearrange relative motion but leave the CM velocity unchanged if F_ext≈0 during the hit.

06. Collision bookkeeping: impulse along the line of impact, momentum along that line if external impulse is negligible, and e relating separation to approach speeds along the line. Energy is conserved only for e=1.

07. 2D collisions need the impact normal; tangential momentum of each body may stay continuous if frictionless contact. CM frame often turns an oblique hit into a cleaner algebra.

Official · 2026

Syllabus for this topic

Official syllabus points for Centre of Mass & Collisions only — not the full Mechanics portion map. Portion hub: Mechanics · Mechanics analysis.

Syllabus on this page is topic-scoped: Centre of Mass & Collisions. The parent portion (Mechanics) also covers other topics — use the portion hub for the full map.

Must cover (this topic)

  • COM collisions; impulse

Gaps / watch-outs

No topic-specific exclusions flagged — finish the unit points below for COM & Collisions.

Source: https://jeeadv.ac.in/documents/jee-advanced-2026-syllabus.pdf

Unit 01

Centre of Mass & Collisions

high

Exam focusCOM motion, impulse, and elastic/inelastic collisions.

Syllabus points

  • Systems of particles; COM; impulse; elastic/inelastic collisions

Full Physics syllabus →

Cross-topic stems

Mixed joins for this topic

Advanced rarely asks a chapter in isolation. These are tagged multi-idea Physics stems that touch Centre of Mass & Collisions — practise the knowledge bridge, not only the single-chapter drill.

Physics2026 · Paper 2 · Q17–Q18linked-numerical

Disk collision + rotation about pivot + energy loss

Mechanics – Collision+Rotation about a fixed point+Energy & angular momentum

Knowledge bridge

Step 1: apply angular impulse about the pivot C for the particle–disk collision. Step 2: convert post-collision ω into maximum rise of centre O (energy after collision). Step 3: compare mechanical energy before vs after to get the collision energy loss.

Why it feels hard

Choosing the wrong axis or mixing linear KE with rotational KE about C breaks both answers.

What to practise

Linked collision–rotation: conserve angular momentum about the eventual pivot, then energy separately.

2026 year review →Physics notes →Quick card →Syllabus →

All mixed joins →

Why methods matter

Common methods that change your score

COM tracking plus momentum and restitution replace case-by-case collision folklore.

01

COM as free particle

Decouples messy relative motion from the easy COM trajectory.

Do thisIf ΣF_ext=0 horizontally, V_COM constant; solve internal motions relative to a steadily moving COM.

Why it works

Decouples messy relative motion from the easy COM trajectory.

See it once

Two skaters push apart on ice: COM stays put; speeds inverse to masses.

02

Impulse axis only

Walls and rough floors break momentum conservation in some components.

Do thisConserve momentum only along directions with zero external impulse; apply e only along the contact normal.

Why it works

Walls and rough floors break momentum conservation in some components.

See it once

Ball hitting rough ground: horizontal friction impulse can change p_x; vertical uses N impulse + e.

03

COM-frame elastic flip

Replaces memorising both v₁,v₂ formulas under time pressure.

Do thisTransform to COM frame; for 1D elastic, reverse velocities; transform back.

Why it works

Replaces memorising both v₁,v₂ formulas under time pressure.

See it once

Equal mass: COM frame velocities ±u_rel/2; reverse → exchange in lab frame.

04

e and momentum pair

Stops under-determined guessing from energy alone when e<1.

Do thisAlways write two equations for two unknowns after 1D collision: Σp and e-definition.

Why it works

Stops under-determined guessing from energy alone when e<1.

See it once

Inelastic: momentum + e=0.5 gives both final speeds; energy is a check, not a third independent unknown solver unless e unknown.

Processed examples

Sample questions · full working

Read the impact line first. Then notice which family/model was chosen. The solution is written the way your rough sheet should look in the hall.

01

bankFamily · 1D elastic / inelastic collision

Sample question

Masses 1 kg and 2 kg move at 6 m/s and 0 toward each other? They move along a line: 1 kg at 6 m/s hits 2 kg at rest, e=1. Find final speeds.

This stem rewards early recognition of the 1d elastic / inelastic collision model.

Model chosenEqual-mass formulas generalize via momentum + e.

Processed solution

  1. Momentum: v₁+2v₂=6.
  2. e: v₂−v₁=6.
  3. v₂=4 m/s, v₁=−2 m/s.

Answerv₁=−2 m/s; v₂=4 m/s

Takeaway: Elastic 1D collisions need both p and e.

02

standardFamily · Centre-of-mass kinematics

Sample question

Particles of 1 kg and 3 kg are at x=0 and x=4 m. Find the COM position.

This stem rewards early recognition of the centre-of-mass kinematics model.

Model chosenX_cm=Σmᵢxᵢ/M.

Processed solution

  1. X_cm=(0+12)/4=3 m.

Answer3 m

Takeaway: COM is the mass-weighted average position.

03

twistFamily · Oblique collision and impulse

Sample question

A smooth sphere of mass m and speed u strikes a stationary identical smooth sphere head-on offset so that the line of centres makes 30° with u. Take e=1. Find the speeds after impact.

This Advanced-level variation couples hidden constraints with the oblique collision and impulse model.

Model chosenImpulse along the line of centres; tangential components unchanged.

Processed solution

  1. Resolve u into normal (along centres) and tangential.
  2. u_n=u cos30°=u√3/2 and u_t=u/2.
  3. For equal masses, e=1, normal speeds exchange: first sphere’s u_n′=0, second gets u√3/2.
  4. First keeps tangential u/2; second has only normal speed u√3/2.

Answerv₁=u/2 (tangential); v₂=(u√3)/2 along line of centres

Takeaway: Smooth oblique collisions separate into independent n and t problems.

Reusable models

Common question families

Advanced stems cluster. Name the family in 10–15s, write the fast model, then run the same path. Do not invent a new plan each time.

01

Centre-of-mass kinematics

Looks likeSystem of particles, explosion, or variable mass pieces

Fast modelTrack R_cm with M R_cm = Σ m_i r_i; F_ext = M a_cm.

Same solve path

  1. Locate or update COM.
  2. Apply momentum of system = M V_cm.
  3. Use only external impulse for ΔP_cm.
  4. Relate individual motions to COM frame if useful.

Rough skeleton

  • R_cm
  • P_cm
  • frame

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

02

1D elastic / inelastic collision

Looks likeTwo masses collide; e given or perfectly elastic

Fast modelUse momentum + restitution (or energy if e=1).

Same solve path

  1. Write momentum conservation along the line.
  2. Write e = relative-separation / relative-approach.
  3. Solve u₁′, u₂′.
  4. Check energy only if e=1 or asked.

Rough skeleton

  • p conserve
  • e equation
  • speeds

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

03

Oblique collision and impulse

Looks likeGlancing blow, wall, or impulse J along normal

Fast modelResolve into normal/tangential; impulse only along normal (smooth).

Same solve path

  1. Choose n–t axes at contact.
  2. Apply J along n; tangential momentum unchanged if smooth.
  3. Use e along the normal.
  4. Recombine velocity components.

Rough skeleton

  • n–t
  • J/e
  • components

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

04

Variable-mass / rocket thrust

Looks likeSand on conveyor, chain, or rocket equation idea

Fast modelUse thrust = u_rel (dm/dt) with careful mass-flow sign.

Same solve path

  1. Identify the system and dm/dt.
  2. Write momentum balance with relative velocity.
  3. Include external forces.
  4. Integrate if mass changes with time.

Rough skeleton

  • dm/dt
  • thrust
  • equation

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

Hall timing

60–90s process

  1. 0–15s
    Classify: COM location/motion, impulse Δp, or collision (elastic/inelastic/2D).

    Underline given data, unknowns and constraints. Name the mechanical model before recalling any formula; discard relations whose assumptions are already broken.

  2. 15–35s
    Write momentum along impact line; write restitution for relative velocities; note external impulse.

    Draw axes, free-body or energy/impulse diagram. Write governing equations symbolically with consistent sign and reference.

  3. 35–75s
    Solve for unknowns; compute KE loss if asked; track COM velocity separately.

    Solve algebraically as far as possible; substitute numbers only once. Keep vectors, components and units on the rough sheet.

  4. 75–90s
    Check mass units, e∈[0,1], and that tangential momentum is untouched if frictionless impact.

    Check dimensions, limiting cases and physical sense. For multi-correct, test every option independently rather than forcing one narrative.

After the model is locked

Quick formula bank

COM position

R_COM = Σm_i r_i / Σm_i

Discrete system

COM dynamics

M A_COM = ΣF_ext

System translation

Impulse

J = Δp = F_avg Δt

Collision duration or kick

Restitution

e = (v₂−v₁)_after / (u₁−u₂)_before along normal (speed form with care)

Impact line separation/approach

1D elastic

v₁=(u₁(m₁−m₂)+2m₂u₂)/(m₁+m₂); v₂=(u₂(m₂−m₁)+2m₁u₁)/(m₁+m₂)

Head-on elastic

Equal mass elastic 1D

velocities exchanged

m₁=m₂, head-on, e=1

Perfectly inelastic

v = (m₁u₁+m₂u₂)/(m₁+m₂)

Bodies stick; e=0

KE in COM frame

K = K_COM + K_rel; inelastic loses K_rel

Energy bookkeeping in collisions

Mental bank

  • COM: M A = ΣF_ext
  • Internal forces do not accelerate COM
  • J=Δp
  • e=1 elastic; e=0 perfectly inelastic
  • Equal mass head-on elastic: velocity exchange
  • KE loss occurs in inelastic (and rough) collisions
  • Conserve p only where external impulse vanishes
  • Explosion: internal impulses; COM velocity continuous if no ext impulse
  • CM ignores internal forces for a_cm
  • Impulse || impact line; stamp e on that line only
  • e=1 ⇒ KE conserved (ideal); e=0 ⇒ common normal speed
  • Explosion/recoil: CM velocity fixed if F_ext≈0

Make it automatic

Practice · checklist

Daily

  • Two 1D collision (e given) and one COM location/motion drill.
  • One ballistic/inelastic and one wall-bounce with restitution.

Weekly

  • One mixed 2D collision + COM explosion set.
  • Re-derive equal-mass exchange from momentum+e=1 once without notes.

100% checklist

  • COM vs collision mode identified
  • External impulse directions named
  • Momentum written only on allowed axes
  • Restitution along contact normal only
  • Elastic/inelastic energy treatment correct
  • KE loss computed from before/after when asked
  • Final relative velocity sign matches separation
  • Impact line drawn; components named
  • Momentum and e used only where justified
  • CM check: internal impulses did not fake F_ext
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