Appearance
1.7 — Momentum, Collisions and the Rocket Equation
Two cars collide head-on and crumple into a tangled mass. Enormous amounts of energy vanish into bending steel, breaking glass and heat. Nothing about the wreck is predictable from energy, because you cannot know in advance how much energy the crumpling will absorb.
And yet there is a quantity that comes through the collision completely untouched, whose value one microsecond before the impact is exactly its value one microsecond after, no matter how violent the crash or how much metal is destroyed. Add it up for both cars before, add it up after, and the two totals are identical.
That quantity is momentum, and its conservation is the single most reliable tool in mechanics.
Momentum is mass times velocity
\vec{p} = m\vec{v}
It is a vector, pointing the way the object is moving, with units \text{kg}\cdot\text{m/s} and dimension \mathrm{M\,L\,T^{-1}}. Newton called it the quantity of motion, and he built his second law on it directly:
\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}
Read aloud: net force equals the rate of change of momentum. Chapter 1.4 showed that this reduces to \vec{F} = m\vec{a} when the mass is constant. Here we use the fuller form, because half of this chapter is about situations where the mass is not constant.
Momentum and kinetic energy are different quantities and it matters. A 1000 kg car at 10 m/s and a 10 kg cannonball at 1000 m/s:
| Car | Cannonball | |
|---|---|---|
| Momentum | 10 000 kg·m/s | 10 000 kg·m/s |
| Kinetic energy | 50 000 J | 5 000 000 J |
Identical momentum, and a hundredfold difference in energy. Momentum is linear in speed, energy is quadratic, so speed matters far more for energy than for momentum. The relationship between them, worth having, is:
K = \frac{p^2}{2m}
which follows from substituting v = p/m into K = \tfrac12mv^2. For a fixed momentum, a lighter object carries more energy. This turns up again in Chapter 7.2, where it decides how a particle's wavelength relates to its energy.
Impulse: momentum delivered over time
Rearrange the second law and integrate over the duration of a force:
\vec{F} = \frac{d\vec{p}}{dt} \quad\Longrightarrow\quad d\vec{p} = \vec{F}\,dt \quad\Longrightarrow\quad \Delta\vec{p} = \int_{t_1}^{t_2}\vec{F}\,dt
That integral is the impulse, \vec{J}, and the result is the impulse–momentum theorem:
\boxed{\vec{J} = \Delta\vec{p}}
Impulse is force integrated over time, exactly as work was force integrated over distance. The two integrals of the same second law give you the two great conservation laws, and which one you want depends on whether the problem hands you a time or a distance.
For a constant force, J = F\Delta t. For a varying one, it is the area under the force–time graph.
Why this is the most practically important idea in safety engineering
Rearrange it: F_{\text{avg}} = \Delta p/\Delta t.
The change in momentum \Delta p is usually fixed by the situation. A car crashing from 60 km/h to zero must shed all of its momentum, and nothing can change that total. What can be changed is the time over which it is shed — and the force is inversely proportional to that time.
Stretch the collision from 0.01 s to 0.10 s and the average force drops by a factor of ten. That is what every piece of crash safety equipment does:
- Crumple zones deliberately fail slowly, stretching the stopping time from milliseconds to tenths of a second. A car that bounced off a wall undamaged would kill its occupants.
- Airbags add a few more centimetres and a few more milliseconds between your head and the steering column.
- Helmets crush their foam liner progressively, converting a sudden stop into a slower one.
- A boxer rolling with a punch moves their head backwards, adding time to the impact.
- Bending your knees when you land stretches the stop from 2 cm to 40 cm, cutting the force by twenty.
Worked example: catching a cricket ball
A 160 g cricket ball arrives at 30 m/s. Catch it with rigid hands, stopping it in 0.02 s; or draw your hands back, stopping it in 0.20 s.
The momentum change is the same either way:
\Delta p = 0.160 \times 30 = 4.8\ \text{kg·m/s}
Rigid hands:
F = \frac{4.8}{0.02} = 240\ \text{N}
That is like catching a 24 kg weight — it stings and can break a finger.
Hands drawn back:
F = \frac{4.8}{0.20} = 24\ \text{N}
Two and a half kilograms. Every fielder learns this by pain long before they learn it by equation.
Conservation of momentum, derived from the third law
Take two bodies interacting with each other and nothing else — two billiard balls, two cars, a gun and a bullet. Call the force body 1 exerts on body 2 \vec{F}_{12}, and the reverse \vec{F}_{21}.
Newton's third law says these are equal and opposite:
\vec{F}_{12} = -\vec{F}_{21}
Apply the second law to each body separately:
\frac{d\vec{p}_2}{dt} = \vec{F}_{12}, \qquad \frac{d\vec{p}_1}{dt} = \vec{F}_{21}
Add the two equations:
\frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = \vec{F}_{21} + \vec{F}_{12} = \vec{F}_{21} - \vec{F}_{21} = 0
The left side is the rate of change of the total momentum, so:
\frac{d}{dt}(\vec{p}_1 + \vec{p}_2) = 0 \quad\Longrightarrow\quad \boxed{\vec{p}_1 + \vec{p}_2 = \text{constant}}
The total momentum of an isolated system never changes. "Isolated" means no external forces — internal forces, however violent, cancel in pairs and cannot shift the total.
Three things to notice about this result.
It is a vector equation, so it holds separately in each direction. In a two-dimensional collision you get one equation for x and one for y, and that is usually exactly the two equations you need.
It does not care what happens during the collision. The bodies can crumple, explode, stick, heat up, shatter. None of it appears in the derivation, which used only the third law. This is why momentum works where energy fails.
"Isolated" needs care in practice. During a car crash, gravity and friction are acting. But the collision lasts perhaps 100 ms and the collision forces are of order 10^5 N, while friction is of order 10^4 N. The impulse from friction over 0.1 s is negligible against the impulse from the impact. So momentum is conserved to an excellent approximation over the short window of the collision, and that is the standard justification for every collision problem you will solve.
Recoil: the gun and the bullet
A 4.0 kg rifle fires a 20 g bullet at 600 m/s. What is the recoil speed?
Before firing, everything is at rest, so the total momentum is zero. After firing it must still be zero:
0 = m_b v_b + m_r v_r
0 = (0.020)(600) + (4.0)v_r
v_r = -\frac{12}{4.0} = -3.0\ \text{m/s}
The minus sign means backwards, as expected. Note the energies: the bullet carries \tfrac12(0.020)(600)^2 = 3600\ \text{J} and the rifle only \tfrac12(4.0)(3.0)^2 = 18\ \text{J}. Equal and opposite momentum, two hundredfold difference in energy — because K = p^2/2m and the rifle is two hundred times heavier. This is why the shooter is not thrown across the room, and it is the whole design principle of recoil management.
Collisions
A collision is any brief, strong interaction. Momentum is always conserved. Kinetic energy may or may not be, and that is how collisions are classified.
Perfectly inelastic: they stick together
Maximum kinetic energy is lost. The two bodies leave as one.
Momentum conservation in one dimension:
m_1u_1 + m_2u_2 = (m_1+m_2)v
\boxed{v = \frac{m_1u_1+m_2u_2}{m_1+m_2}}
Worked example. A 1500 kg car at 20 m/s runs into a stationary 1000 kg car and they lock together.
v = \frac{1500 \times 20 + 0}{2500} = \frac{30\,000}{2500} = 12\ \text{m/s}
Now check the energy books:
K_i = \tfrac12(1500)(20)^2 = 300\,000\ \text{J}
K_f = \tfrac12(2500)(12)^2 = 180\,000\ \text{J}
120 000 J — 40% of the total — has gone. Into bent metal, heat, sound and broken glass. Momentum did not notice.
Perfectly elastic: kinetic energy is conserved too
This is the idealisation for billiard balls, gas molecules and subatomic particles. Both quantities are conserved:
m_1u_1 + m_2u_2 = m_1v_1+m_2v_2 \tag{momentum}
\tfrac12m_1u_1^2 + \tfrac12m_2u_2^2 = \tfrac12m_1v_1^2+\tfrac12m_2v_2^2 \tag{energy}
Solving these two simultaneously is messy if done head-on, so here is the tidy route. Rearrange each to put body 1 on the left and body 2 on the right:
m_1(u_1-v_1) = m_2(v_2-u_2) \tag{1}
m_1(u_1^2-v_1^2) = m_2(v_2^2-u_2^2) \tag{2}
Factor the differences of squares in (2), using a^2-b^2 = (a-b)(a+b):
m_1(u_1-v_1)(u_1+v_1) = m_2(v_2-u_2)(v_2+u_2)
Now divide this by equation (1). The bracketed factors m_1(u_1-v_1) and m_2(v_2-u_2) cancel completely, leaving:
u_1+v_1 = v_2+u_2
Rearranged:
\boxed{u_1 - u_2 = -(v_1-v_2)}
In an elastic collision, the relative speed of approach equals the relative speed of separation. That is a remarkably clean result, and it is much easier to use than the energy equation it replaced. Solving it together with momentum conservation gives:
v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2, \qquad v_2 = \frac{2m_1}{m_1+m_2}u_1 + \frac{m_2-m_1}{m_1+m_2}u_2
Rather than memorise these, look at the three special cases, which are the ones that carry physical meaning. Take u_2 = 0 throughout — body 2 initially at rest.
Equal masses (m_1 = m_2). The first fraction is zero, so v_1 = 0 and v_2 = u_1. The moving ball stops dead and the struck ball leaves at the full incoming speed. This is the Newton's cradle result and any billiards player's stop shot.
Heavy hits light (m_1 \gg m_2). Then v_1 \approx u_1 and v_2 \approx 2u_1. The heavy body barely notices; the light one is fired off at twice the incoming speed. A moving truck barely slows when it clips a football, and the football departs at twice the truck's speed.
Light hits heavy (m_1 \ll m_2). Then v_1 \approx -u_1 and v_2 \approx 0. The light body bounces straight back at its original speed and the heavy one hardly moves. A ball bouncing off a wall — the wall is attached to the Earth, so m_2 is effectively the mass of the planet.
That last case has a beautiful application. In a nuclear reactor you need to slow fast neutrons down. Hitting them against heavy nuclei is useless — they bounce back with the same speed and lose nothing. To take energy out of a neutron you want a target of similar mass, and the closest available is a proton, which is a hydrogen nucleus. That is why reactors use water or graphite as a moderator: the neutrons are slowed by billiard-ball collisions with light nuclei. It is the equal-mass case, deliberately engineered.
The coefficient of restitution
Real collisions sit between the two extremes. Define:
e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{v_2-v_1}{u_1-u_2}
e = 1 is perfectly elastic, e = 0 is perfectly inelastic. Steel on steel is about 0.95, a tennis ball about 0.75, a basketball about 0.85, putty about 0.
A dropped ball gives you a way to measure e with nothing but a tape measure. Drop it from height h. It arrives at u = \sqrt{2gh} by the third kinematic equation, leaves the floor at v = eu by the definition above, and rises to a height h' given by v^2 = 2gh':
h' = \frac{v^2}{2g} = \frac{e^2u^2}{2g} = \frac{e^2(2gh)}{2g} = e^2h
So the rebound height is e^2h, not eh — the coefficient is squared because it applies to the speed, and height depends on the square of speed. A basketball with e = 0.85 dropped from 2 m comes back to 0.7225 \times 2 = 1.44 m. Measure the two heights, take the square root of their ratio, and you have e.
Two-dimensional collisions
Momentum conservation holds in each direction independently:
\sum p_x \text{ before} = \sum p_x \text{ after}, \qquad \sum p_y \text{ before} = \sum p_y \text{ after}
Worked example. A 2.0 kg ball moving at 5.0 m/s along x strikes a stationary 3.0 kg ball. Afterwards the 2.0 kg ball moves at 3.0 m/s at 30° above the x-axis. Find the other ball's velocity.
Before: total p_x = 2.0\times5.0 = 10.0, total p_y = 0.
After, ball 1: p_x = 2.0\times3.0\cos30° = 2.0\times2.598 = 5.196, and p_y = 2.0\times3.0\sin30° = 3.0.
So ball 2 must carry:
p_{2x} = 10.0 - 5.196 = 4.804, \qquad p_{2y} = 0 - 3.0 = -3.0
The negative p_y is required: ball 1 went up, so ball 2 must go down by exactly as much, because there was no y-momentum to begin with and there cannot be any afterwards.
v_{2x} = \frac{4.804}{3.0} = 1.601, \qquad v_{2y} = \frac{-3.0}{3.0} = -1.0
v_2 = \sqrt{1.601^2+1.0^2} = \sqrt{2.563+1.0} = 1.888\ \text{m/s} \quad\text{at}\quad \tan^{-1}\!\left(\frac{-1.0}{1.601}\right) = -32.0°
Is it elastic? Check the energy:
K_i = \tfrac12(2.0)(5.0)^2 = 25.0\ \text{J}
K_f = \tfrac12(2.0)(3.0)^2 + \tfrac12(3.0)(1.888)^2 = 9.0 + 5.35 = 14.35\ \text{J}
Over 10 J lost, so this collision was inelastic — and notice that we solved it completely without needing to know that. Momentum alone was enough.
The ballistic pendulum
This device is worth studying because it is the classic trap: it uses both conservation laws, in different phases, and using the wrong one in the wrong phase gives a badly wrong answer.
A bullet of mass m at speed u embeds in a hanging block of mass M. The block swings up to height h. Find u.
Phase 1 — the embedding. This is a perfectly inelastic collision, over in microseconds. Momentum is conserved; kinetic energy is emphatically not, because the bullet is being stopped by friction inside wood and most of its energy becomes heat.
mu = (m+M)V \quad\Longrightarrow\quad V = \frac{mu}{m+M}
Phase 2 — the swing. Now the block-plus-bullet rises on a string. This takes about a second, gravity acts throughout, so momentum is not conserved (there is a large external force). But no energy is being dissipated, so mechanical energy is conserved:
\tfrac12(m+M)V^2 = (m+M)gh \quad\Longrightarrow\quad V = \sqrt{2gh}
Combine:
\boxed{u = \frac{m+M}{m}\sqrt{2gh}}
With m = 10 g, M = 2.0 kg and h = 8.0 cm:
u = \frac{2.010}{0.010}\sqrt{2\times9.81\times0.08} = 201 \times \sqrt{1.570} = 201 \times 1.253 = 252\ \text{m/s}
If you had wrongly used energy conservation in phase 1 you would get u = \sqrt{2gh(m+M)/m} \approx 17.8\ \text{m/s} — off by a factor of fourteen. The whole art is knowing which law applies to which phase, and the rule is simple: momentum during the impact, energy during the smooth motion afterwards.
Centre of mass
For a system of particles, the centre of mass is the mass-weighted average position:
\vec{R}_{\text{cm}} = \frac{\sum m_i\vec{r}_i}{\sum m_i}
Differentiate with respect to time:
\vec{V}_{\text{cm}} = \frac{\sum m_i\vec{v}_i}{M_{\text{total}}} = \frac{\vec{p}_{\text{total}}}{M_{\text{total}}}
So the total momentum of any system is simply its total mass times the velocity of its centre of mass. Differentiate once more:
M\vec{A}_{\text{cm}} = \frac{d\vec{p}_{\text{total}}}{dt} = \vec{F}_{\text{external}}
The centre of mass moves exactly as a single particle of the total mass would, driven only by the external forces. Internal forces are completely invisible to it.
This is why a spinning hammer thrown across a room has one point that traces a clean parabola while the rest of it tumbles chaotically — that point is the centre of mass, and the parabola is Chapter 1.3's projectile result applying to it alone. It is also why a firework shell, once it explodes, sends fragments whose centre of mass carries on along the original parabola as though nothing had happened. The explosion was internal.
A high jumper exploits the same fact in reverse. In the Fosbury flop the body arches so extremely that the centre of mass passes underneath the bar while every part of the body passes over it. The athlete only has to raise their centre of mass to the height the physics demands, and the arch buys several centimetres for free.
The rocket equation
Now the case where mass is not constant, which is the reason Newton's second law is properly written in terms of momentum.
A rocket of mass m moving at speed v ejects a small mass dm_e of exhaust backwards at speed v_e relative to the rocket. There is nothing outside pushing, so total momentum is conserved.
In a short time dt the rocket's mass falls by |dm| and its speed rises by dv. Take the exhaust mass to be -dm (positive, since dm is negative).
Momentum before, in the ground frame:
p_i = mv
Momentum after — the rocket, now lighter and faster, plus the exhaust, which is moving at the rocket's speed minus its exhaust speed:
p_f = (m+dm)(v+dv) + (-dm)(v-v_e)
Expand the first bracket:
(m+dm)(v+dv) = mv + m\,dv + v\,dm + dm\,dv
The term dm\,dv is a product of two infinitesimals and vanishes in the limit — it is second order, meaning if both dm and dv are one-thousandth of their quantities, their product is one-millionth, negligible against everything else.
Expand the second piece:
(-dm)(v-v_e) = -v\,dm + v_e\,dm
Add them:
p_f = mv + m\,dv + v\,dm - v\,dm + v_e\,dm = mv + m\,dv + v_e\,dm
Set p_f = p_i:
mv + m\,dv + v_e\,dm = mv
m\,dv = -v_e\,dm
dv = -v_e\frac{dm}{m}
Integrate from the initial mass m_0 (fuelled) to the final mass m_f (empty), and from initial speed v_0 to final v_f:
\int_{v_0}^{v_f}dv = -v_e\int_{m_0}^{m_f}\frac{dm}{m}
v_f - v_0 = -v_e\left[\ln m\right]_{m_0}^{m_f} = -v_e(\ln m_f - \ln m_0) = v_e\ln\frac{m_0}{m_f}
\boxed{\Delta v = v_e \ln\frac{m_0}{m_f}}
This is the Tsiolkovsky rocket equation, published in 1903 by a deaf Russian schoolteacher working alone, and it is the equation that governs every space mission ever flown. Part 11 uses it constantly.
Read what it says. The speed you can gain depends on only two things: how fast you throw your exhaust (v_e), and the ratio of your fuelled mass to your empty mass. And it depends on that ratio logarithmically, which is brutal. To go twice as fast you do not need twice the fuel — you need to square the mass ratio.
Put numbers on it. A good chemical rocket has v_e \approx 4500\ \text{m/s} (this is usually quoted as specific impulse, I_{sp} = v_e/g \approx 450 s). Reaching low Earth orbit needs \Delta v \approx 9400 m/s once you include gravity and air drag losses. So:
\frac{m_0}{m_f} = e^{9400/4500} = e^{2.09} = 8.1
The rocket must be eight parts fuel to one part everything else — structure, engines, and payload combined. That is why rockets are enormous thin-walled tanks with a tiny payload on top, why staging exists (throw away the empty tank so you stop accelerating dead weight), and why getting to orbit is genuinely hard rather than merely expensive.
It is also why chemical rockets will never reach another star. To get \Delta v = 0.1c = 3\times10^7 m/s you would need a mass ratio of e^{6667}, a number with about 2900 digits. The only escape is a larger v_e, which is why ion drives (with v_e around 30 000 m/s) are used for deep space despite their feeble thrust.
Where this shows up in your life
Every time you push off from a wall in a swimming pool you are doing the recoil calculation: you push the Earth one way and go the other, and because the Earth is 10^{23} times heavier, all the speed ends up on your side.
A fire hose needs two people not because it is heavy but because of momentum. Water leaving at 25 m/s from a nozzle passing 20 kg/s carries away momentum at 20 \times 25 = 500 N, and by the third law the hose is pushed backwards with exactly that force — about the weight of a 50 kg person, applied continuously and unpredictably.
And the reason a karate strike breaks a board is entirely impulse. The hand carries a modest momentum, but it delivers it in about 5 milliseconds, so the force is \Delta p/\Delta t with a very small denominator. Strikers are taught to aim through the board rather than at it, which keeps the hand accelerating and keeps \Delta t short.
What the next chapter fixes
Everything so far has treated objects as points. Real objects spin, and spinning turns out to need its own complete set of quantities: an angular version of force (torque), of mass (moment of inertia), and of momentum (angular momentum), each with its own conservation law. Chapter 1.8 builds all of them, explains why a figure skater speeds up when they pull their arms in, and why a spinning bicycle wheel refuses to fall over.