← All printables Chapter 07 · Mechanics · 2 sheets · read the chapter →
Vedatom Physics
FORMULA CARD · PHYSICS.VEDATOM.COM
Chapter 07 · Part I · Mechanics
Revised on ____________________
Chapter 07 · Centre of Mass, Momentum & Collisions

One point moves as if the whole world were there

2 sheets
2 diagrams
28 results

Formulas and conditions only — no derivations, no solved numbers. The diagrams are the reader's own.

I

Centre of mass

IT MOVES ONLY IF AN EXTERNAL FORCE ACTS
I.Discrete bodies  rcm = Σ mi riΣ mi
II.Continuous body  rcm = 1M ∫ r dm
III.Its velocity  vcm = Σ mi viM = PM
IV.Its acceleration  M acm = ΣFext internal forces never move it
V.Two bodies  the cm divides the join as m2 : m1
VI.Removed mass  rcm = M rM − m rmM − m the cavity trick
VII.Semicircular wire  2Rπ · disc 4R3π
VIII.Solid cone / hemisphere  h4 · 3R8 from the base
OCOM4R/3πR
II

Momentum and impulse

THE CONSERVATION LAW THAT SURVIVES A COLLISION
I.Momentum  p = m v
II.Second law  F = dpdt
III.Impulse  J = ∫ F dt = Δp area under an F–t graph
IV.Conservation  ΣFext = 0 ⇒ p constant component-wise, axis by axis
V.During a collision  external forces are negligible the impact force dominates
VI.Recoil  m1 v1 = m2 v2 from rest, opposite directions
Momentum is conserved in every collision. Kinetic energy is conserved only in the elastic ones — and a question that says “collision” has told you nothing about energy.
III

Collisions in one dimension

e DOES ALL THE WORK
I.Restitution  e = speed of separationspeed of approach 0 ≤ e ≤ 1
II.Final velocities  v1 = (m1 − e m2) u1 + (1 + e) m2 u2m1 + m2
III.Perfectly inelastic  v = m1 u1 + m2 u2m1 + m2 e = 0, they move together
IV.Energy lost then  ΔK = 12 m1 m2m1 + m2 (u1 − u2)² μ is the reduced mass
V.Elastic, equal masses  the velocities are exchanged
VI.Elastic, heavy on light  v2 ≈ 2 u1 the light one leaves at twice the speed
VII.Elastic, light on heavy  v1 ≈ − u1 bounces straight back
VIII.Ball dropped from h  hn = e2n h total path = h 1 + e²1 − e²
IV

Oblique impact and explosions

RESOLVE ALONG THE LINE OF IMPACT
I.Along the line of impact  momentum conserved, e applies
II.Equal masses, elastic  the two paths separate at 90°
III.Explosion  Σ p = 0 if it began at rest
IV.Projectile that bursts  the cm keeps the original parabola
uat restv1v290°
V

Variable mass

ROCKETS AND FALLING CHAINS
I.Thrust  Fth = vrel dmdt
II.Rocket equation  v = u ln( M0M ) − g t
Where marks are lost in this chapter
1Conserving momentum through the sliding after a collision. Friction is negligible only during the impact.
2Writing both Pi = Pf and Ki = Kf for an inelastic collision. They contradict.
3Using scalar momentum in two dimensions. Conserve x and y separately.
4Forgetting that e is defined along the line of impact, not along the original velocity.