Unit 4: Systems of Particles and Linear Momentum

System of particles

A model of an object (or multiple objects) as many masses moving together, allowing analysis of both motion of the system as a whole and internal motion between parts.

Center of mass (COM)

The point that moves as if all external forces act on a single particle of mass equal to the system’s total mass; often the “dot” representing an extended object in a free-body diagram.

Homogeneous (uniform-density) object

An object with uniform density; its center of mass is at the geometric center.

Nonuniform density

A mass distribution where density varies with position, causing the COM to shift toward the denser/heavier regions.

Total mass (M)

The sum of all masses in a system: M = Σ mᵢ (or M = ∫ dm for continuous objects).

COM position vector (discrete)

For particles mᵢ at positions r⃗ᵢ: r⃗_cm = (1/M) Σ (mᵢ r⃗ᵢ).

Weighted average (COM idea)

COM is a weighted average of positions; larger masses contribute more and pull the COM closer to themselves.

Center of mass components

Coordinate-by-coordinate COM: xcm = (1/M)Σ(mᵢxᵢ), ycm = (1/M)Σ(mᵢyᵢ), z_cm = (1/M)Σ(mᵢzᵢ).

1D COM computation routine

Choose an origin/axis, list each xᵢ, compute Σ(mᵢxᵢ), then divide by Σ mᵢ.

Continuous mass distribution

A body whose mass is spread continuously; COM is found using integrals rather than sums.

COM position vector (continuous)

For mass element dm at position r⃗: r⃗_cm = (1/M) ∫ r⃗ dm.

Linear mass density (λ)

Mass per unit length for a 1D object (rod): dm = λ dx.

Surface mass density (σ)

Mass per unit area for a 2D lamina: dm = σ dA.

Volume mass density (ρ)

Mass per unit volume for a 3D object: dm = ρ dV.

COM velocity

v⃗cm = d r⃗cm/dt = (1/M) Σ (mᵢ v⃗ᵢ).

COM acceleration

a⃗cm = d v⃗cm/dt = (1/M) Σ (mᵢ a⃗ᵢ).

Particle equation of motion in a system

For particle i: mᵢ a⃗ᵢ = F⃗{i,ext} + Σ{j≠i} F⃗_{ij} (external plus internal forces).

Internal forces

Forces between particles within the chosen system; they cancel in pairs when summing over the system (Newton’s 3rd law), so they do not change total momentum or COM motion.

Net external force (system)

The vector sum of forces from outside the system acting on all particles; it controls the COM acceleration.

System-level Newton’s 2nd law (COM form)

M a⃗cm = F⃗{ext,net}; COM accelerates as if all mass were concentrated at the COM under the net external force.

Zero net external force implication

If F⃗{ext,net} = 0, then a⃗cm = 0 and the COM remains at rest or moves with constant velocity.

Stability (COM and potential energy)

A system tends to be stable when its COM is at a low point of potential energy and unstable when its COM is at a high point of potential energy.

Center of gravity

The point where the gravitational force (weight) can be treated as acting; depends on the gravitational field as well as mass distribution.

COM vs. center of gravity (uniform field)

In a uniform gravitational field (often a good near-Earth approximation), center of gravity and COM coincide; conceptually they differ because COM depends only on mass distribution.

COM of two masses on a line

For m₁ at x=0 and m₂ at x=L: x_cm = (m₂/(m₁+m₂))L (closer to the larger mass).

Explosion and COM trajectory

During an explosion with negligible external impulse, internal forces change piece motions but the COM continues along the same trajectory dictated by external forces (e.g., gravity only).

Linear momentum (p⃗)

For a particle: p⃗ = m v⃗; a vector in the direction of velocity.

Total momentum of a system (P⃗)

The vector sum of particle momenta: P⃗ = Σ p⃗ᵢ = Σ(mᵢ v⃗ᵢ).

COM–momentum “glue equation”

P⃗ = M v⃗_cm; the system’s momentum equals that of a single mass M moving at the COM velocity.

Newton’s 2nd law (momentum form, particle)

For constant mass: F⃗_net = d p⃗/dt.

Newton’s 2nd law (momentum form, system)

Net external force changes total momentum: F⃗_{ext,net} = d P⃗/dt.

Impulse (J⃗)

The time integral of force: J⃗ = ∫ F⃗ dt; measures the effect of a force over a time interval and has the same units as momentum.

Average force (F⃗_avg)

A constant force that would produce the same impulse over Δt: J⃗ = F⃗_avg Δt.

Impulse–momentum theorem

Net impulse equals change in momentum: J⃗net = Δ p⃗ = p⃗f − p⃗i (and for systems, J⃗ext = ΔP⃗).

Impulse from a force–time graph

Impulse equals the signed area under the F vs. t curve over the interval (positive area increases momentum in the positive direction).

Triangular force pulse impulse

If a force rises to Fmax and returns to 0 forming a triangle over base 2T, the impulse is the triangle area: J = T Fmax.

Sign in rebound average force

If an object reverses direction, Δp is opposite the initial momentum direction, so F_avg = Δp/Δt carries a negative sign in the original + direction.

Conservation of linear momentum

If net external impulse is zero/negligible, total momentum is constant: P⃗f = P⃗i.

Closed/isolated system (momentum)

A chosen system with negligible net external impulse (in the direction of interest), so momentum is conserved.

System boundary (choosing the system)

The definition of “external” depends on what objects are included; internal forces are inside the boundary and cancel in the momentum sum.

Momentum conservation in components

Because momentum is a vector, conservation applies separately in each direction where external impulse is negligible: Px,f = Px,i and Py,f = Py,i.

Recoil

Motion of objects in opposite directions after an internal interaction (e.g., explosion) so that total momentum remains unchanged.

Spring push-apart (explosion-like) relation

Two carts initially at rest pushed apart satisfy m₁v₁ + m₂v₂ = 0, so v₂ = −(m₁/m₂)v₁.

Elastic collision

A collision in which (for an isolated system) both momentum and total kinetic energy are conserved.

Inelastic collision

A collision in which momentum is conserved (if isolated) but total kinetic energy decreases (converted to internal energy, deformation, heat, sound, etc.).

Perfectly inelastic collision

An inelastic collision where objects stick together and move with a common final velocity; momentum conserved but kinetic energy not conserved.

1D elastic collision equations

Solve using both momentum conservation and kinetic energy conservation: m₁v₁i+m₂v₂i=m₁v₁f+m₂v₂f and (1/2)m₁v₁i²+(1/2)m₂v₂i²=(1/2)m₁v₁f²+(1/2)m₂v₂f².

Relative speed reversal (1D elastic)

In a 1D elastic collision: v₁i − v₂i = −(v₁f − v₂f); relative speed of approach equals relative speed of separation.

Elastic collision with stationary target (results)

If m₁ moves at v₀ and m₂ is at rest: v₁f = ((m₁−m₂)/(m₁+m₂))v₀ and v₂f = (2m₁/(m₁+m₂))v₀.

Kinetic energy loss in a perfectly inelastic collision

Compute Ki and Kf using vf = (m₁v₁i+m₂v₂i)/(m₁+m₂); the loss Ki − K_f becomes internal energy (not all kinetic energy is necessarily lost).

Ballistic pendulum (two-stage method)

A problem with (1) a perfectly inelastic collision analyzed by momentum conservation, then (2) a rise/swing analyzed by mechanical energy conservation (e.g., (1/2)(M+m)V²=(M+m)gh).