Vedatom Physics Mechanics / Momentum & Collisions
◈ Simulations ▤ Full book
Reading momentum right
Momentum is a vector. Conserve it component by component; a collision that looks symmetric in x may not be in y.
It survives every collision. During brief contact the internal forces are equal and opposite and external impulse is negligible, so total p is unchanged — even when kinetic energy is not.
Only elastic collisions keep KE. Measure the loss with the restitution e = (relative speed after)/(relative speed before): e = 1 elastic, e = 0 perfectly inelastic.
The COM is unmoved by internal forces. An explosion or collision cannot shift the centre of mass; only external forces can.
try the collision lab · §05
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Mechanics · all fourteen modules inked · all lit
§01

Centre of mass — the balance point

The centre of mass of two bodies sits at xcm = (m₁x₁ + m₂x₂)/(m₁+m₂) — always nearer the heavier one, dividing the gap in the inverse ratio of the masses. Slide the masses or re-weight them and watch the balance point shift.

Bench · two masses on a rod
0510 m m₁ m₂ COM
m₁{{ cmM1 }} kg
m₂{{ cmM2 }} kg
position x₁{{ cmX1 }} m
position x₂{{ cmX2 }} m
centre of mass x_cm
{{ cmXcom }} m
{{ cmNote }}
§03

Bodies with a hole — the negative-mass trick

Cut a piece out of a uniform body and you needn't integrate again. Treat the missing piece as a body of negative mass sitting at its own centre: rcm = (Afull rfull − Ahole rhole)/(Afull − Ahole). The balance point always slides away from the hole. Move the cut and watch.

Bench · a 6 m × 6 m plate with a circular cut
full plate × hole COM
hole radius r{{ hoR }} m
hole centre x{{ hoX }} m
hole centre y{{ hoY }} m
remaining plate · x_cm , y_cm
( {{ hoComX }} , {{ hoComY }} ) m
{{ hoNote }}
§02

The centre of mass by integration

calculus

For a continuous body the discrete sum becomes an integral, rcm = (1/M)∫ r dm. Slice the body into elements whose position and mass can both be written in one variable, then integrate. Choose a body and watch the balance point emerge.

{{ ciBenchLabel }}
x=0 x=L λ = λ₀(1 + x/L) x_cm base b y_cm = h/3 centre 2R/π centre 4R/3π
the integral
dm = λ₀(1 + x/L) dx
xcm = ∫₀ᴸ x(1+x/L)dx∫₀ᴸ (1+x/L)dx = 5L²/63L/2 = 5L9
strip at height y: dm = σ·b(1 − y/h) dy
ycm = ∫₀ʰ y(1−y/h)dy∫₀ʰ (1−y/h)dy = h²/6h/2 = h3
dm = λR dθ,   y = R sinθ
ycm = ∫₀ᵖ R sinθ·λR dθλRπ = 2Rπ
strip at height y: dm = σ·2√(R²−y²) dy
ycm = ∫₀ᴿ y·2√(R²−y²) dyπR²/2 = 4R3π
size {{ ciParamLabel }}{{ ciParam }} m
centre of mass = {{ ciComSym }}
{{ ciComReadout }}