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Topic 9 of 19

Linear Momentum & Impulse

Linear momentum is the "quantity of motion" of an object. Newton's Second Law is most generally stated as: net force equals the rate of change of momentum. This leads to the impulse–momentum theorem and, crucially, to conservation of momentum — one of the most broadly useful principles in physics.

9.1 Center of Mass

When we study a complex system — a spinning gymnast, an exploding firework, two colliding billiard balls — we need a single representative point whose motion we can track. That point is the center of mass (com): the mass-weighted average position of all the particles in the system.

Two-Particle System

Place two particles on the x-axis: mass m₁ at position x₁ and mass m₂ at position x₂. The center of mass lies at:

x_{\text{com}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

The com is always between the two particles, and it lies closer to the more massive one. If m₁ = m₂, the com sits exactly at the midpoint.

Center of mass for two unequal masses. The com is closer to the heavier mass (m₂ = 3 kg here).

Check: x_com = (1×1 + 3×5)/4 = 16/4 = 4 m. Three-quarters of the way from m₁ to m₂, as expected since m₂ is three times heavier.

Many Particles — 3D

For a system of n particles with total mass M = Σmᵢ, the com position in three dimensions is:

\vec{r}_{\text{com}} = \frac{1}{M}\sum_{i=1}^{n} m_i \vec{r}_i

Written component by component:

x_{\text{com}} = \frac{\sum m_i x_i}{M}, \quad y_{\text{com}} = \frac{\sum m_i y_i}{M}, \quad z_{\text{com}} = \frac{\sum m_i z_i}{M}

Solid Bodies — Continuous Mass

For a solid object with continuous mass distribution ρ(r), replace the sum with an integral:

x_{\text{com}} = \frac{1}{M}\int x\,dm

In practice this is usually evaluated by choosing dm = ρ dV for a volume element, or dm = λ dx for a thin rod (λ = linear mass density), or dm = σ dA for a flat plate.

Symmetry shortcut: If an object has a point, line, or plane of symmetry, the com lies on that symmetry element. A uniform sphere's com is at its geometric center. A uniform rod's com is at its midpoint. You never need to integrate for symmetric objects — use the shortcut.

The com Can Lie Outside the Object

For a donut (torus), the com is at the center of the hole — a point with no mass at all. For a boomerang or a bent rod, the com lies in empty space. The com is a mathematical point, not necessarily a material point.

Center of mass for common uniform shapes

Object	Location of com
Uniform rod (length L)	Midpoint — L/2 from either end
Uniform rectangular plate	Geometric center (intersection of diagonals)
Uniform disk or solid sphere	Geometric center
Thin hemispherical shell (radius R)	3R/8 from the base along symmetry axis
Solid hemisphere (radius R)	3R/8 from the flat face
Right triangle (legs a, b)	At (a/3, b/3) from the right-angle vertex

9.2 Newton's Second Law for a System of Particles

The real power of the center-of-mass concept emerges when we differentiate the com position twice with respect to time. Start from:

M\vec{r}_{\text{com}} = m_1\vec{r}_1 + m_2\vec{r}_2 + \cdots + m_n\vec{r}_n

Differentiating once gives the com velocity:

M\vec{v}_{\text{com}} = m_1\vec{v}_1 + m_2\vec{v}_2 + \cdots + m_n\vec{v}_n = \vec{P}_{\text{total}}

Differentiating again:

M\vec{a}_{\text{com}} = m_1\vec{a}_1 + m_2\vec{a}_2 + \cdots + m_n\vec{a}_n

By Newton's Second Law, mᵢaᵢ = Fᵢ (net force on particle i). The forces on each particle include internal forces (from other particles in the system) and external forces (from outside the system). By Newton's Third Law, every internal force has an equal and opposite reaction within the system — they sum to zero. So:

\vec{F}_{\text{net,ext}} = M\vec{a}_{\text{com}}

Newton's Second Law for a system: The net external force on a system of particles equals the total mass M times the acceleration of the center of mass. Internal forces — no matter how large — do not contribute.

Why This Is Profound

This single equation lets us treat arbitrarily complex systems as if they were point particles. Three examples:

Exploding firework: Once launched, the only external force is gravity (F_ext = −Mg ĵ). The com follows a perfect parabola — even as the shell shatters into hundreds of fragments spiraling in every direction.
Gymnast: While airborne, the com traces a parabola. The athlete can twist and tuck (redistributing mass internally) but cannot alter the com's trajectory.
Two skaters pushing off: If the floor exerts no net horizontal friction, the com of the two-skater system remains stationary — even as both skaters slide apart.

An exploding firework: fragments fly chaotically, but the center of mass (green cross) continues along the original parabolic trajectory determined by gravity alone.

9.3 Linear Momentum

The linear momentum of a single particle is defined as the product of its mass and velocity:

\vec{p} = m\vec{v}

Momentum is a vector with SI units of kg·m/s. Its direction is always the same as the velocity. Newton's original formulation of his second law was in terms of momentum — force is the rate of change of momentum:

\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}

When mass is constant, dp/dt = m(dv/dt) = ma, recovering F = ma. But Newton's momentum form is more general — it applies even when mass changes (like a rocket expelling fuel).

System of Particles — Total Momentum

The total linear momentum of a system of n particles is the vector sum of all individual momenta:

\vec{P} = \sum_{i=1}^{n} m_i\vec{v}_i = M\vec{v}_{\text{com}}

The total momentum equals the total mass times the velocity of the center of mass. Newton's second law for the system then reads:

\vec{F}_{\text{net,ext}} = \frac{d\vec{P}}{dt}

Two equivalent statements:
① F_net,ext = Ma_com (com acceleration form)
② F_net,ext = dP/dt (total momentum form)

They are the same equation — just written differently. Use whichever is more convenient for the problem.

Momentum vs. Kinetic Energy

Students sometimes confuse momentum and kinetic energy because both involve mass and speed. Key differences:

Momentum versus kinetic energy

Property	Momentum p	Kinetic Energy K
Formula	p = mv	K = ½mv²
Type	Vector	Scalar
Units	kg·m/s	J = kg·m²/s²
Conserved when…	No net external force	No nonconservative work
Connection	K = p²/(2m)	p = √(2mK)

The relation K = p²/(2m) is often useful: if you know the momentum, you can find the kinetic energy without needing the velocity explicitly.

9.4 Collision and Impulse

When a bat strikes a baseball, or a car airbag cushions a crash, an enormous force acts for a very brief time. We define the impulse J of such a force as the integral of the force over the duration of the interaction:

\vec{J} = \int_{t_i}^{t_f} \vec{F}(t)\,dt

Since F_net = dp/dt, integrating both sides from tᵢ to t_f gives:

\vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i

Impulse–Momentum Theorem: The impulse delivered to a particle equals the change in its momentum. This holds regardless of how the force varies in time.

Average Force

The time-averaged force F_avg is the constant force that would produce the same impulse over the same interval Δt:

\vec{J} = \vec{F}_{\text{avg}}\,\Delta t

This is enormously useful: if you know the impulse (from momentum change) and the contact time Δt, you can find the average impact force. Conversely, extending Δt reduces F_avg — the engineering principle behind airbags, padding, crumple zones, and catching a ball with a relaxed arm.

Force vs. time during a collision. The area under the curve (shaded) equals the impulse J = Δp. The dashed rectangle has the same area — its height is F_avg.

Series of Collisions

When a stream of n projectiles (each mass m, speed v) strikes a target in time Δt, reversing direction, the average force on the target is:

F_{\text{avg}} = \frac{n}{\Delta t}\,|\Delta p| = \frac{n}{\Delta t}\,(2mv)\quad\text{(perfectly elastic reversal)}

This is the physics of a water jet hitting a wall, a machine gun's recoil, or the pressure of gas molecules bouncing off a container wall (which leads directly to the kinetic theory of gases).

9.5 Conservation of Linear Momentum

We have shown that F_net,ext = dP/dt. If the net external force on a system is zero, then:

\frac{d\vec{P}}{dt} = 0 \implies \vec{P} = \text{constant}

Law of Conservation of Linear Momentum: If no net external force acts on a system, the total linear momentum of the system remains constant in both magnitude and direction.

In equation form for a "before and after" scenario:

\vec{P}_i = \vec{P}_f \qquad \text{(closed, isolated system)}

Component Form — Apply Axis by Axis

Because momentum is a vector, conservation applies independently in each coordinate direction. Even if there is an external force in the y-direction (like gravity), momentum may still be conserved in the x-direction if no external force has an x-component:

P_{x,i} = P_{x,f} \quad \text{(if }F_{\text{ext},x}=0\text{)}

P_{y,i} = P_{y,f} \quad \text{(if }F_{\text{ext},y}=0\text{)}

Why Internal Forces Don't Matter

Internal forces always come in Newton's Third Law pairs: if particle 1 pushes particle 2 with force F, particle 2 pushes particle 1 with force −F. These cancel exactly in the sum for F_net,ext. No matter how violent the internal explosion or collision, it cannot change the total momentum of the system.

Space hauler example: initially at rest. Internal explosive pushes two halves apart. Total momentum remains zero — the halves have equal and opposite momenta.

Scope of the Law

Conservation of momentum is believed to hold exactly in all known physical interactions — classical mechanics, electromagnetism, quantum mechanics, special relativity, and the Standard Model of particle physics. It is a consequence of the homogeneity of space (Noether's theorem): the laws of physics are the same at all locations.

When is momentum conserved?

Situation	x-momentum	y-momentum
Ball in free fall (gravity only)	Conserved	NOT conserved
Horizontal explosion on frictionless floor	Conserved	NOT conserved (normal force)
Collision in deep space (isolated)	Conserved	Conserved
Car crash on flat road (brief impact)	Approx. conserved (impulse approx.)	Approx. conserved

The impulse approximation: During a short, violent collision, internal collision forces (impulse ≫ external force × Δt) dominate. We treat momentum as conserved even when weak external forces (gravity, friction) are technically present, because their impulse during the brief collision is negligible.

9.6–9.7 Collisions — Elastic, Inelastic, and Elastic Formulas

A collision is a brief, strong interaction between objects in which we can apply conservation of momentum. We classify collisions by whether kinetic energy is also conserved:

Three types of collision

Type	Momentum conserved?	KE conserved?	Example
Elastic	Yes	Yes	Billiard balls, atomic collisions, ideal gas
Inelastic	Yes	No (KE → heat, sound, deformation)	Most everyday collisions
Perfectly inelastic	Yes	Maximum KE lost	Objects stick together after impact

Key rule: Momentum is ALWAYS conserved in any collision (by Newton's 3rd law). Kinetic energy is only conserved in elastic collisions. Energy that "disappears" in an inelastic collision has been converted to internal energy (thermal energy, deformation, sound), not lost from the universe.

Perfectly Inelastic Collisions

When objects stick together, they move with a common final velocity v_f. Applying P conservation:

m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f

v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}

The fraction of kinetic energy lost can be substantial — for a bullet (mass m) embedding in a block (mass M initially at rest):

\frac{\Delta K}{K_i} = \frac{M}{m+M}

A 5 g bullet hitting a 1 kg block loses 99.5% of its kinetic energy to heat and deformation.

Elastic Collisions — 1D Formulas

In one dimension, if particle 1 (mass m₁, initial velocity v₁ᵢ) hits stationary particle 2 (mass m₂, v₂ᵢ = 0), applying both momentum conservation and kinetic energy conservation yields:

v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1i}

v_{2f} = \frac{2m_1}{m_1 + m_2}\,v_{1i}

These are exact results for elastic collisions with a stationary target. Three instructive special cases:

Special cases of elastic collision (m₂ initially at rest)

Case	Result
Equal masses (m₁ = m₂)	v₁f = 0, v₂f = v₁ᵢ — particle 1 stops dead; particle 2 takes all the velocity. Classic billiard ball result.
Massive projectile (m₁ ≫ m₂)	v₁f ≈ v₁ᵢ (barely slows), v₂f ≈ 2v₁ᵢ — like a bowling ball hitting a ping-pong ball.
Massive target (m₁ ≪ m₂)	v₁f ≈ −v₁ᵢ (bounces back), v₂f ≈ 0 — like a tennis ball hitting a wall.

Equal-mass elastic collision: particle 1 stops, particle 2 moves at the original speed. Momentum and kinetic energy are both conserved.

The Ballistic Pendulum

A classic demonstration combining two conservation laws. A bullet (mass m, speed v₀) embeds in a suspended block (mass M). The collision is perfectly inelastic — momentum conservation gives the common speed V just after impact. Then the block swings up — energy conservation gives the height h reached:

mv_0 = (m+M)V \quad \Rightarrow \quad V = \frac{m}{m+M}\,v_0

\tfrac{1}{2}(m+M)V^2 = (m+M)gh \quad \Rightarrow \quad h = \frac{V^2}{2g}

Combining: $v_0 = \frac{m+M}{m}\sqrt{2gh}$ . Measuring h with a ruler gives the bullet speed — this was the standard method before high-speed electronics.

Why Elastic Collisions Require Two Equations

A perfectly inelastic collision has one unknown (v_f) and one equation (momentum). An elastic collision has two unknowns (v₁f and v₂f) and two equations (momentum conservation + kinetic energy conservation). This is why elastic collisions have unique solutions.

Trick for elastic collisions: Instead of solving the quadratic KE equation, use the equivalent linear relation that holds for elastic collisions in 1D:

v_{1i} - v_{2i} = -(v_{1f} - v_{2f})

The relative velocity of approach equals the relative velocity of separation (sign reversed). This replaces the KE equation with a simple linear one.

9.8 Coefficient of Restitution, 2D Collisions, and the CM Frame

Coefficient of Restitution

Real collisions fall between the two ideals (perfectly inelastic e = 0 and elastic e = 1). The coefficient of restitution e quantifies how "bouncy" a collision is:

e = \frac{|v_{1f} - v_{2f}|}{|v_{1i} - v_{2i}|} = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}

Coefficient of restitution for various collisions

Type	Value of e	Examples
Perfectly elastic	e = 1	Ideal billiard balls, atomic collisions
Partially inelastic	0 < e < 1	Steel balls (~0.95), rubber balls (~0.85), baseballs (~0.55)
Perfectly inelastic	e = 0	Clay, objects that stick together

The coefficient of restitution is a measured property of the materials involved. A superball has e ≈ 0.90; a beanbag has e ≈ 0.05. Used together with momentum conservation, e determines both final velocities uniquely for any 1D collision:

v_{1f} = \frac{m_1 v_{1i} + m_2 v_{2i} + m_2 e(v_{2i}-v_{1i})}{m_1+m_2}

v_{2f} = \frac{m_1 v_{1i} + m_2 v_{2i} + m_1 e(v_{1i}-v_{2i})}{m_1+m_2}

(Setting e = 1 recovers the elastic formulas; setting e = 0 recovers the perfectly inelastic formula.)

2D Collisions

When a collision does not occur head-on, momentum must be conserved independently in each direction. Let particle 1 have initial velocity along the x-axis and particle 2 be at rest:

x:\quad m_1 v_{1i} = m_1 v_{1f}\cos\theta_1 + m_2 v_{2f}\cos\theta_2

y:\quad 0 = m_1 v_{1f}\sin\theta_1 - m_2 v_{2f}\sin\theta_2

For elastic collisions, add the kinetic energy equation as a third constraint. This gives three equations for four unknowns (v₁f, v₂f, θ₁, θ₂) — so one quantity must be measured (e.g., one angle) and the rest are determined.

2D elastic shortcut: For an elastic collision between equal masses where particle 2 is initially at rest, the two particles always move off at exactly 90° to each other after the collision. This is a consequence of both momentum and KE conservation simultaneously.

2D elastic collision: equal masses. Particle 1 deflects at angle θ₁, particle 2 recoils at θ₂. For equal masses, θ₁ + θ₂ = 90° always.

The Center-of-Mass (CM) Frame

For any two-body problem, there exists a reference frame in which the total momentum is zero — the center-of-mass frame. This frame moves with velocity v_cm = P_total/M relative to the lab frame.

In the CM frame, the two particles always approach each other with equal and opposite momenta. After an elastic collision, they simply reverse their directions — the analysis reduces to a trivially symmetric problem. The results are then transformed back to the lab frame.

This is why the CM frame is so powerful in particle physics: a symmetric collider (particles of equal momentum approaching head-on) is far more efficient than a fixed-target experiment, because all the collision energy is available for creating new particles rather than being "wasted" in the kinetic energy of the center of mass.

Threshold energy: In a fixed-target experiment with a moving projectile (mass m, KE = K) hitting a stationary target (mass M), the kinetic energy available for creating new particles is only K_cm = K · M/(m+M). A collider with both beams of energy K has K_cm ≈ 2K. This is why the LHC uses colliding beams — otherwise it would need to be hundreds of times more powerful to achieve the same physics.

Key Concepts

Linear Momentum

The product of mass and velocity:

\vec{p} = m\vec{v}

. A vector quantity with units of kg·m/s. More massive objects and faster objects have more momentum.

Newton's Second Law (momentum form)

The most general statement:

\vec{F}_\text{net} = d\vec{p}/dt

. When mass is constant this reduces to

\vec{F}_\text{net} = m\vec{a}

Impulse

Impulse is the integral of force over time:

\vec{J} = \int\vec{F}\,dt

. For a constant force:

\vec{J} = \vec{F}\Delta t

. Impulse equals the change in momentum.

Impulse–Momentum Theorem

\vec{J} = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i

. A large force over a short time produces the same impulse as a smaller force over a longer time.

Conservation of Momentum

If the net external force on a system is zero, total momentum is conserved:

\vec{p}_\text{total} = \text{const}

. Apply separately in each direction. Internal forces do not change total momentum.

Center of Mass

The mass-weighted average position of a system:

\vec{r}_\text{cm} = \sum m_i\vec{r}_i / M

. The total external force equals

M\vec{a}_\text{cm}

— the CM moves as if all mass were concentrated there.

Key Equations

Linear momentum

\vec{p} = m\vec{v}

Momentum is a vector; direction matches velocity.

Impulse–Momentum Theorem

\vec{J} = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i

Impulse (area under F–t graph) equals change in momentum.

Conservation of momentum

\vec{p}_{1i} + \vec{p}_{2i} = \vec{p}_{1f} + \vec{p}_{2f} \quad (\vec{F}_\text{ext}=0)

Total momentum is constant when there is no net external force.

Center of mass (two bodies)

x_\text{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

Position of the center of mass; generalizes to any number of bodies.

CM velocity

\vec{v}_\text{cm} = \frac{\sum m_i \vec{v}_i}{M}

Velocity of the center of mass; constant when net external force is zero.

Worked Example

Recoil of a Rifle

Problem

A 4 kg rifle fires a 10 g bullet at 600 m/s. Find the recoil speed of the rifle.

Solution

System: rifle + bullet. Initially both at rest, so $p_\text{total,i} = 0$ .

Conservation of momentum (no external horizontal force):

0 = m_\text{bullet}\,v_\text{bullet} + m_\text{rifle}\,v_\text{rifle}

v_\text{rifle} = -\frac{m_\text{bullet}}{m_\text{rifle}}\,v_\text{bullet} = -\frac{0.010}{4}\times 600

v_\text{rifle} = -1.5 \text{ m/s}

Answer The rifle recoils at 1.5 m/s in the direction opposite the bullet.

Practice

Exercises

14 problems

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Exercise 1 / 14 Free

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The graph shows a triangular force-time pulse. Impulse = area under the F-t graph. F peaks at 10 N and the pulse lasts 2 s. Find the impulse J.

J = N·s

Exercise 2 / 14 Free

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A cart of mass 3 kg moves at 5 m/s. Press ▶ to watch it roll. What is its momentum p = mv?

p = kg·m/s

3 of 14

A $1500 \text{ kg}$ car decelerates from $20 \text{ m/s}$ to rest in $5 \text{ s}$ . What is the magnitude of the average braking force?

Unlock Exercise 3