Class 11 · Chapter 6

Centre of Mass & Collisions

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Centre of Mass & Collisions Formula Sheet

10 formulas across 3 topics in Centre of Mass & Collisions.

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Updated 2026-07-03 · v1.0.0

Centre of Mass

The centre of mass is the mass-weighted average position — the single point where the whole system's translational story can be told.

x(cm) = (m₁x₁ + m₂x₂ + ...) / (m₁ + m₂ + ...)

Position of centre of mass (particles)

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Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
x(cm)	centre of mass coordinate	m	[L]
mᵢ	mass of the ith particle	kg	[M]
xᵢ	position of the ith particle	m	[L]

Valid when

Apply separately for each axis (x, y, z)

Common mistakes

The centre of mass needs NO mass at its own location — for a ring it is at the empty centre
For two particles, the COM divides the line joining them in the INVERSE ratio of masses (closer to the heavier one)

Read the concept →

x(remaining) = (M x(M) − m x(m)) / (M − m)

COM after removing a piece (cavity problems)

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Treat the removed piece as negative mass at its own COM

Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
M	mass of the original full body	kg	[M]
m	mass of the removed piece	kg	[M]

Valid when

x(M), x(m) are the COMs of the full body and the removed piece

Common mistakes

Using areas or volumes directly as masses is fine ONLY for uniform density

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v(cm) = (m₁v₁ + m₂v₂) / (m₁ + m₂) = p(total)/M

Velocity of the centre of mass

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Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
v(cm)	velocity of centre of mass	m/s	[LT⁻¹]
p(total)	total momentum of the system	kg·m/s	[MLT⁻¹]
M	total mass	kg	[M]

Valid when

With zero external force, v(cm) stays constant no matter what happens internally

Common mistakes

In explosion problems the COM keeps moving on the ORIGINAL trajectory — fragments rearrange around it

Read the concept →

Worth remembering

Standard COM results: uniform semicircular ring → 2R/π from centre; semicircular disc → 4R/3π; solid hemisphere → 3R/8; hollow hemisphere → R/2 (all along the symmetry axis)
Internal forces can never move the centre of mass of an isolated system

JEE loves the man-walking-on-boat setup — the COM of (man + boat) stays fixed when water friction is neglected.

Conservation of Linear Momentum

When the net external force is zero, total momentum cannot change. This survives even when kinetic energy is destroyed — which is why it rules every collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Conservation of linear momentum

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Newton's third law: internal impulses cancel in pairs

Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
u₁, u₂	velocities before the event	m/s	[LT⁻¹]
v₁, v₂	velocities after the event	m/s	[LT⁻¹]

Valid when

Net EXTERNAL force zero (or the event is so brief that external impulse is negligible)
It is a VECTOR equation — conserve each component separately in 2-D

Common mistakes

Momentum is conserved in every collision — elastic or not; it is kinetic energy that may be lost

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v(gun) = − (m(bullet) / M(gun)) v(bullet)

Recoil velocity of a gun

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Total momentum zero before and after firing

Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
v(gun)	recoil velocity	m/s	[LT⁻¹]
m(bullet), M(gun)	bullet and gun masses	kg	[M]

Common mistakes

KE is NOT shared equally — the lighter bullet carries almost all the kinetic energy (K = p²/2m, same p, smaller m)

Read the concept →

Worth remembering

In 2-D explosion/collision problems, draw the momentum vector triangle — if total momentum is zero, the three momenta close a triangle

Collisions

Every collision conserves momentum. What distinguishes them is the coefficient of restitution — how much of the approach speed survives as separation speed.

e = (v₂ − v₁) / (u₁ − u₂)

Coefficient of restitution

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Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
e	coefficient of restitutione = 1 elastic, e = 0 perfectly inelastic, 0 < e < 1 real collisions	dimensionless	[M⁰L⁰T⁰]
u₁ − u₂	velocity of approach	m/s	[LT⁻¹]
v₂ − v₁	velocity of separation	m/s	[LT⁻¹]

Valid when

Velocities along the line of impact

Common mistakes

Getting the order wrong: e = separation speed / approach speed — both taken along the impact line

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

Final velocities — 1-D elastic collision

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Solving momentum conservation together with e = 1

Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
u₁, u₂	initial velocities	m/s	[LT⁻¹]
v₁, v₂	final velocities	m/s	[LT⁻¹]

Valid when

Equal masses → velocities EXCHANGE
Heavy hits light at rest → light leaves at ≈ 2u₁, heavy barely slows
Light hits heavy at rest → light bounces back with ≈ same speed

Common mistakes

Memorise the three limiting cases above — most exam questions are one of them in disguise

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v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Common velocity — perfectly inelastic collision

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Momentum conservation with both bodies moving together after impact

Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
v	common final velocity	m/s	[LT⁻¹]

Valid when

Bodies stick together (e = 0)

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ΔK = ½ (m₁m₂/(m₁+m₂)) (u₁ − u₂)²

KE lost in a perfectly inelastic collision

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K(initial) − K(final) using the common-velocity result; μ = m₁m₂/(m₁+m₂) is the reduced mass

Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
ΔK	kinetic energy converted to heat/deformation	J	[ML²T⁻²]
u₁ − u₂	relative velocity of approach	m/s	[LT⁻¹]

Valid when

Perfectly inelastic (e = 0); for general e, multiply by (1 − e²)

Common mistakes

The ballistic pendulum needs BOTH tools: momentum conservation for the impact, energy conservation for the swing after — never energy across the impact itself

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h(n) = e²ⁿ h₀, v(after) = e v(before)

Ball bouncing on the floor (nth bounce)

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Each impact scales speed by e; height goes as speed squared

Variables used in this formula, with units and dimensions
Symbol	Meaning	Unit	Dimension
h₀	initial drop height	m	[L]
h(n)	height after the nth bounce	m	[L]
n	bounce number	dimensionless	[M⁰L⁰T⁰]

Common mistakes

Height scales as e²ⁿ (not eⁿ) because h ∝ v²

Read the concept →

Worth remembering

In a 2-D elastic collision between EQUAL masses (one initially at rest), the two final velocities are perpendicular — a JEE classic
Oblique collisions: the velocity component PERPENDICULAR to the line of impact never changes for smooth spheres

NEET stays with 1-D collisions and the limiting cases; JEE Advanced moves to oblique impacts and collision-with-spring problems where maximum compression occurs at common velocity.

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