Part 1 · Chapter 09

Center of Mass and Linear Momentum

Why a complicated system moves like a single point — and the conserved quantity that rules every collision

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 45 min
i What you'll learn
  • How to locate the center of mass of a group of particles and of an extended solid body.
  • Newton's second law for a system of particles — \(F_{\text{net}} = Ma_{\text{com}}\) — and why the com moves so simply.
  • The definition of linear momentum \(p = mv\), and momentum's role as the quantity Newton's law really governs.
  • Impulse and the impulse–momentum theorem, and how forces during brief collisions are tamed.
  • The conservation of linear momentum, and how to analyze elastic, inelastic, and two-dimensional collisions.
  • How to handle systems of varying mass, leading to the rocket equation.
Section 9-1

What Is Physics?

Every mechanical system we have studied so far — the projectile, the sliding block — we quietly treated as a single particle. But a real baseball bat tumbles end over end, a diver pikes and twists, and an exploding firework scatters in all directions. How can Newton's laws possibly apply to such messy motion? The answer is one of the most elegant ideas in mechanics: hidden inside every object and every system of objects is a special point, the center of mass, that moves as though all the mass were concentrated there and all the external forces acted there. Master that point, and the chaos becomes simple.

Section 9-2

The Center of Mass

For a collection of particles, the center of mass (com) is the mass-weighted average position. In one dimension, and then as a single vector equation for any number of particles in three dimensions:

Center of mass of a system of particles
\[ x_{\text{com}} = (1/M) \sum m_{i} x_{i} \qquad \vec{r}_{\text{com}} = (1/M) \sum m_{i} \vec{r}_{i} \]
M = Σmi is the total mass. A heavier particle pulls the com toward itself; the com need not lie on any particle.

For a solid body, the sum becomes an integral over the mass elements. If the body has uniform density, mass ratios become volume ratios:

Center of mass of an extended body
\[ \vec{r}_{\text{com}} = (1/M) \int \vec{r} dm \qquad (uniform: dm = (M/V) dV) \]
By symmetry, the com of a uniform sphere, ring, or rod lies at its geometric center — even where there is no material at all.
Symmetry is your shortcut. If a body has a point, line, or plane of symmetry, the com lies on it. You can also treat a complicated shape as several simple pieces, find each piece's com, then combine them as if each were a single particle.
Section 9-3

Newton's Second Law for a System of Particles

🎯
The motion of the center of mass
The center of mass of a system moves as if all the mass were concentrated there and all external forces were applied there.

\(\vec{F}_{\text{net}} = M \vec{a}_{\text{com}}\), where \(\vec{F}_{\text{net}}\) is the vector sum of all external forces. Internal forces between the parts cancel in pairs (Newton's third law) and never affect the com. This is why an exploding shell's fragments still, collectively, follow the original parabola — until a piece hits the ground.

Section 9-4

Linear Momentum

Newton originally wrote his second law not as \(F = ma\) but in terms of linear momentum \(\vec{p} = m \vec{v}\) — a vector pointing along the velocity. For a single particle, and then for an entire system whose total momentum is just \(\vec{P} = M \vec{v}_{\text{com}}\):

Momentum form of Newton's second law
\[ \vec{F}_{\text{net}} = d\vec{p}/dt \qquad \text{(particle)} \qquad \vec{F}_{\text{net}} = d\vec{P}/dt \qquad \text{(system)} \]
The net external force equals the rate of change of total momentum. This form survives even when mass changes — which \(F = ma\) cannot handle.
Section 9-5

Collision and Impulse

In a collision a large force acts for a very short time, and we rarely know its detailed shape. Integrating Newton's law over the contact time gives the impulse \(\vec{J}\), which equals the change in momentum — the impulse–momentum theorem:

Impulse–momentum theorem
\[ \vec{J} = \int \vec{F}(t) dt = \Delta \vec{p} \qquad \vec{J} = \vec{F}_{\text{avg}} \Delta t \]
We trade an unknown time-varying force for an average force times the contact time. Same impulse, gentler force, if Δt is stretched — the principle behind airbags, crumple zones, and bending your knees on landing.
⏱ Long contact, soft force

For a fixed Δp, lengthening the collision time lowers the average force. A boxer rolling with a punch and a catcher giving with the ball both extend Δt to spare themselves.

↩ Bouncing beats sticking

A ball that rebounds reverses its momentum, so \(|\Delta p|\) — and the impulse it delivers — is larger than if it merely stopped. Rebounding hits harder.

Section 9-6

Conservation of Linear Momentum

⚖️
Conservation of linear momentum
If the net external force on a system is zero, the total linear momentum cannot change.

\(\vec{P} = \text{constant}\), or \(\vec{P}_{i} = \vec{P}_{f}\). The system need not be isolated in every direction — if the net external force has zero component along some axis, momentum is conserved along that axis. In any collision or explosion the internal forces are huge but always cancel, so total momentum carries straight through.

The recoil rule. A rifle kicks back, a swimmer pushes water backward to glide forward, a rocket throws gas down to climb up. Each is conservation of momentum: a system starting at rest must keep \(\vec{P} = 0\), so forward momentum of one part is paid for by backward momentum of the other.
Section 9-7

Momentum and Kinetic Energy in Collisions

Momentum is conserved in every collision. Kinetic energy is the discriminator: it sorts collisions into types.

⚡ Elastic

Kinetic energy is conserved. \(K_{i} = K_{f}\). An idealization — billiard balls and atoms come close.

🔥 Inelastic

Some kinetic energy is lost to heat, sound, and deformation. Momentum still holds; \(K_{f} \lt K_{i}\).

🧲 Completely inelastic

The bodies stick and move off together, losing the maximum K allowed by momentum conservation.

Completely inelastic collision (target at rest)
\[ m_{1} v_{1i} = (m_{1} + m_{2}) V \qquad \;\Longrightarrow\; \qquad V = m_{1} v_{1i} / (m_{1} + m_{2}) \]
The combined body always moves slower than the incoming one, since the same momentum is now shared by more mass.
Section 9-8

Elastic Collisions in One Dimension

When both momentum and kinetic energy are conserved in 1D, the two conservation laws solve completely for the final velocities. With the target (body 2) initially at rest:

Final velocities — target at rest
\[\begin{gathered} v_{1f} = [(m_{1} - m_{2}) / (m_{1} + m_{2})] v_{1i} \\ v_{2f} = [2m_{1} / (m_{1} + m_{2})] v_{1i} \end{gathered}\]
Equal masses: the projectile stops dead and the target leaves with the full speed (Newton's cradle). A light body hitting a massive wall rebounds with nearly its original speed.
Section 9-9

Collisions in Two Dimensions

Momentum is a vector, so in a glancing collision it is conserved component by component. Resolve along two perpendicular axes and write one equation for each:

Two-dimensional momentum conservation
\[\begin{gathered} x: m_{1}v_{1i} = m_{1}v_{1f}\cos \theta _{1} + m_{2}v_{2f}\cos \theta _{2} \\ y: 0 = -m_{1}v_{1f}\sin \theta _{1} + m_{2}v_{2f}\sin \theta _{2} \end{gathered}\]
If the collision is also elastic, add \(K_{i} = K_{f}\) as a third equation. Two equal masses in an elastic 2D collision fly off at 90° to each other.
Section 9-10

Systems with Varying Mass: A Rocket

A rocket is a system that loses mass continuously, ejecting burned fuel at relative speed \(v_{\text{rel}}\). Applying momentum conservation to the rocket-plus-exhaust system gives the two rocket equations:

First and second rocket equations
\[\begin{gathered} R v_{\text{rel}} = M a \qquad (thrust T = R v_{\text{rel}}) \\ v_{f} - v_{i} = v_{\text{rel}} \ln (M_{i} / M_{f}) \end{gathered}\]
R = −dM/dt is the fuel-burn rate. The gain in speed depends on the exhaust speed and the natural log of the mass ratio — which is why staging, shedding empty tanks, pays off so handsomely.
Why the logarithm matters. Because Δv grows only as the log of the mass ratio, doubling the speed gain demands squaring the fuel ratio. Big velocities are expensive — the tyranny of the rocket equation, and the reason spacecraft are mostly propellant.
Worked Examples

Putting It to Work

1 Three particles — locating the center of mass

Problem. Particles of mass 1.2 kg, 2.5 kg, and 3.4 kg sit at (0, 0), (0.14 m, 0), and (0.07 m, 0.12 m). Find the center of mass.

Solution. Total mass \(M = 1.2 + 2.5 + 3.4 = 7.1 \mathrm{kg}\). Take the mass-weighted average of each coordinate.

Weighted averages
\[\begin{gathered} x_{\text{com}} = [1.2(0) + 2.5(0.14) + 3.4(0.07)] / 7.1 \approx 0.083 m \\ y_{\text{com}} = [1.2(0) + 2.5(0) + 3.4(0.12)] / 7.1 \approx 0.057 m \end{gathered}\]

The com lies at about (0.083, 0.057) m — pulled toward the heavy 3.4 kg particle, yet sitting in empty space where no particle is.

2 A bouncing ball — impulse and average force

Problem. A 0.15 kg ball strikes a wall horizontally at 25 m/s and rebounds at 22 m/s in the opposite direction. The contact lasts 4.0 ms. Find the impulse on the ball and the average force.

Solution. Take the rebound direction as positive. The momentum reverses, so the change is large.

Impulse, then average force
\[\begin{gathered} J = \Delta p = m(v_{f} - v_{i}) = 0.15[22 - (-25)] = 7.05 \mathrm{kg}\cdot m/s \\ F_{\text{avg}} = J / \Delta t = 7.05 / 0.0040 \approx 1.8 \times 10^{3} N \end{gathered}\]

Nearly 1800 N from a small ball — and note that bouncing (not just stopping) is what makes Δp, and the force, so large.

3 Ballistic pendulum — momentum then energy

Problem. A 9.5 g bullet embeds in a 5.4 kg block hanging on a cord; the block then swings up by 6.3 cm. Find the bullet's speed.

Solution. The collision is completely inelastic — use momentum there. The swing afterward conserves energy. Do not mix the two: kinetic energy is lost in the embedding.

Two stages, two laws
\[\begin{gathered} embed: m v = (m + M) V \\ swing: \tfrac{1}{2}(m + M)V^{2} = (m + M)g h \;\Longrightarrow\; V = \sqrt{2gh} \\ v = [(m + M)/m] \sqrt{2gh} = (5.4095/0.0095)\sqrt{2\cdot 9.8\cdot 0.063} \approx 633 m/s \end{gathered}\]

The block's gentle 6.3 cm rise reveals a bullet speed of about 630 m/s — the classic way to measure muzzle velocity with a ruler.

4 Elastic collision — heavy ball strikes a light one

Problem. A 0.50 kg ball moving at 4.0 m/s elastically strikes a stationary 0.10 kg ball head-on. Find both final velocities.

Solution. Use the 1D elastic-collision formulas with the target at rest.

Final velocities
\[\begin{gathered} v_{1f} = [(0.50 - 0.10)/(0.60)](4.0) = +2.7 m/s \\ v_{2f} = [2(0.50)/(0.60)](4.0) = +6.7 m/s \end{gathered}\]

Both move forward; the light target shoots off faster than the incoming ball. A quick check confirms both momentum (2.0 kg·m/s) and kinetic energy (4.0 J) are conserved.

5 Rocket in deep space — the second rocket equation

Problem. A rocket of initial mass 850 kg (including 720 kg of fuel) ejects exhaust at 2.5 km/s relative to itself. Starting from rest in deep space, what speed does it reach when the fuel is spent?

Solution. Final mass \(M_{f} = 850 - 720 = 130 \mathrm{kg}\). Apply \(\Delta v = v_{\text{rel}} \ln (M_{i}/M_{f})\).

Velocity gain
\[ \Delta v = (2500) \ln (850 / 130) = 2500 \cdot \ln (6.54) \approx 2500 \cdot 1.878 \approx 4.7 \times 10^{3} m/s \]

Burning fuel that is more than 6× the dry mass yields a final speed of about 4.7 km/s — still under twice the exhaust speed, a direct cost of the logarithm.

Review

Chapter Summary

✓ Center of mass

Mass-weighted average position: \(\vec{r}_{\text{com}} = (1/M)\sum m_{i}\vec{r}_{i}\). Use symmetry to find it fast.

✓ Newton for a system

\(\vec{F}_{\text{net}} = M\vec{a}_{\text{com}}\). Only external forces matter; internal forces cancel in pairs.

✓ Momentum & impulse

\(\vec{p} = m\vec{v}\), \(\vec{F}_{\text{net}} = d\vec{p}/dt\), and \(\vec{J} = \Delta \vec{p} = \vec{F}_{\text{avg}}\Delta t\).

✓ Conservation

Zero net external force ⟹ \(\vec{P}\) constant. Conserved per-axis if that component of force is zero.

✓ Collisions

Momentum always conserved. Elastic also conserves K; completely inelastic bodies stick and lose the most K.

✓ Varying mass

Thrust \(T = Rv_{\text{rel}}\); speed gain \(\Delta v = v_{\text{rel}} \ln (M_{i}/M_{f})\).

Practice

Problems

Momentum bookkeeping is the theme: pick a system, choose a positive direction, and decide which conservation law applies — momentum holds whenever external forces cancel, but kinetic energy holds only in elastic collisions. Take \(g = 9.8 m/s^{2}\).

  1. Two particles of mass 2.0 kg and 6.0 kg sit 0.80 m apart on the x-axis. How far from the lighter particle is their center of mass?
  2. A uniform square plate of side 0.40 m has a square hole of side 0.20 m cut from one corner. Locate the center of mass of the remaining plate. (Treat the hole as negative mass.)
  3. A 0.42 kg soccer ball moving at 8.0 m/s is kicked back the way it came at 14 m/s. If the foot is in contact for 8.0 ms, find the impulse on the ball and the average force.
  4. A 75 kg skater throws a 3.0 kg ball horizontally at 12 m/s. Starting from rest on frictionless ice, how fast does the skater recoil?
  5. A 1200 kg car at 18 m/s collides head-on and sticks to a 1800 kg car at rest. Find their common velocity and the fraction of kinetic energy lost.
  6. A 0.20 kg cart at 3.0 m/s elastically strikes a stationary 0.30 kg cart head-on. Find both final velocities and verify that kinetic energy is conserved.
  7. A 12 g bullet at 380 m/s embeds in a 2.0 kg block resting on a frictionless surface attached to a spring (k = 250 N/m). Find the maximum compression of the spring.
  8. An alpha particle (mass 4u) elastically strikes a stationary helium nucleus (mass 4u) head-on. Describe the final motion, and explain why this differs from striking a much heavier gold nucleus.
  9. In a 2D collision, a 0.16 kg puck moving at 2.0 m/s east strikes an identical stationary puck; afterward the first moves at 30° north of east. If the collision is elastic, find the direction of the second puck and confirm the 90° rule.
  10. A rocket ejects gas at 2.0 km/s relative to itself, burning fuel at 480 kg/s. Find the thrust, and the acceleration when the rocket's instantaneous mass is 3.0 × 10⁴ kg (ignore gravity).
Tip: the decisive first move is to draw the system before and after and label every momentum vector. Once the picture is clear, "momentum in = momentum out" along each axis writes the equations for you — and only then ask whether kinetic energy is also conserved.