Momentum & Impulse

A baseball leaves a pitcher's hand at 40 m/s and a catcher stops it dead. For the few milliseconds the ball is squashing into the leather, the force between them is enormous, wildly non-constant, and completely unknown. You cannot write $F = ma$ for that, because you do not know $F$ at any instant, and $a$ changes from zero to something huge and back again in the time it takes a fly to beat its wings.

And yet a catcher can tell you exactly how hard the catch felt, and you can predict it. The trick is to stop asking what the force is at each instant and ask instead what the force does in total, over the whole contact. That total turns out to be an easy number even when the force is a mess. This lesson is about that number.

The honest form of Newton's second law

In Power you tracked how fast energy moved. Here you track something else that a force delivers: motion itself, weighted by mass.

Define the momentum of an object of mass $m$ moving with velocity $\vec v$ as

$\vec p = m\vec v$

It is a vector, pointing wherever the velocity points, and its units are $\mathrm{kg\cdot m/s}$. A 72 kg runner at 6.5 m/s carries $72 \times 6.5 = 468\ \mathrm{kg\cdot m/s}$.

Now look at what Newton actually wrote. Not $F = ma$, but

$\vec F_{\text{net}} = \frac{d\vec p}{dt}$

The net force on something is the rate at which its momentum changes. If the mass is constant this collapses to the familiar version, since $d(m\vec v)/dt = m\, d\vec v/dt = m\vec a$. But keep the momentum form and you keep problems where the mass is not constant: a rocket throwing fuel out the back, a chain landing on a scale, a jet of water hitting a wall. Those all break $F = ma$ and none of them break $\vec F = d\vec p/dt$.

One more thing worth having in your pocket. Kinetic energy is $K = \tfrac12 mv^2$ and $p = mv$, so $p^2 = m^2v^2 = 2m\left(\tfrac12 mv^2\right)$, giving

$p = \sqrt{2mK}$

A 2.0 kg object with 36 J of kinetic energy therefore has $p = \sqrt{2 \times 2.0 \times 36} = \sqrt{144} = 12\ \mathrm{kg\cdot m/s}$.

Impulse is the area under the force-time curve

Take $\vec F_{\text{net}} = d\vec p/dt$, multiply both sides by $dt$, and add up every instant from the start of the contact to the end:

$\vec J \equiv \int_{t_1}^{t_2} \vec F\, dt = \vec p_2 - \vec p_1 = \Delta\vec p$

That is the impulse-momentum theorem, and $\vec J$ is called the impulse. In words: the total accumulated force-times-time is exactly the change in momentum, and nothing else about the force matters. The spike can be any shape at all. Only the area under it counts.

It has the same relationship to $\vec F = d\vec p/dt$ that the work-energy theorem has to $F = ma$: one integrates force over distance and gives you a scalar change in energy, the other integrates force over time and gives you a vector change in momentum. Which one a problem wants is usually decided by whether it hands you a distance or a duration.

If the force is constant, the integral is a rectangle and $\vec J = \vec F\Delta t$. A 40 N force acting for 0.25 s delivers $40 \times 0.25 = 10\ \mathrm{N\cdot s}$. And $\mathrm{N\cdot s}$ is the same unit as $\mathrm{kg\cdot m/s}$, since $1\ \mathrm{N} = 1\ \mathrm{kg\cdot m/s^2}$. That is not a coincidence; it is the theorem.

If the force is not constant, you still want the area, and you can define an average force $\bar F$ as the constant force that would give the same area:

$\bar F\, \Delta t = \Delta p$

Worked example 1. A collision force is modelled as $F(t) = F_0\sin(\pi t/T)$ with $F_0 = 900\ \mathrm{N}$ and $T = 0.040\ \mathrm{s}$, acting from $t = 0$ to $t = T$. Find the total impulse.

The force is not constant, so integrate:

Now the numbers:

$J = \frac{2 \times 900 \times 0.040}{\pi} = \frac{72.0}{3.14159} = 22.9\ \mathrm{N\cdot s}$

Notice the average force that hides in there: $\bar F = J/T = 22.9/0.040 = 573\ \mathrm{N}$, which is $2/\pi \approx 0.637$ of the peak. A half-sine hump is about two-thirds as effective as a square pulse of the same height.

Stretch the time and the force drops

Rearrange $\bar F \Delta t = \Delta p$ into

$\bar F = \frac{\Delta p}{\Delta t}$

and read it as a bargain you are allowed to strike. The change in momentum is usually fixed by the physics of the situation: a falling egg arrives with the speed it arrives with, and it has to end at rest. You cannot argue with $\Delta p$. What you can control is $\Delta t$. Double the stopping time and you halve the force.

This is every piece of safety equipment ever built. Airbags, crumple zones, crash mats, bending your knees on landing, drawing your hand back as you catch. None of them reduce the momentum you have to get rid of. All of them lengthen the time over which you get rid of it.

Worked example 2. A 0.20 kg ball hits a wall horizontally at 8.0 m/s and rebounds at 6.0 m/s. Contact lasts 0.015 s. Find the impulse on the ball and the average force.

First thing on blank paper: a positive direction. Take towards the wall as positive. Then $v_1 = +8.0\ \mathrm{m/s}$ and $v_2 = -6.0\ \mathrm{m/s}$, because the ball comes back.

$J = \Delta p = p_2 - p_1 = -1.2 - 1.6 = -2.8\ \mathrm{kg\cdot m/s}$

Magnitude 2.8 N·s, directed away from the wall. The signs did the work: if you had subtracted the speeds instead of the velocities you would have got $0.20 \times 2.0 = 0.4\ \mathrm{N\cdot s}$, seven times too small. The average force is

$\bar F = \frac{2.8}{0.015} = 187\ \mathrm{N}$

Every force acting during the interval counts

The theorem is about the net force. If something other than the impact is pushing during the contact time, it contributes its own impulse and you have to include it.

Worked example 3. A 0.055 kg egg falls 1.2 m and is brought to rest by a pillow over 0.15 s. What average force does the pillow exert? Use $g = 9.8\ \mathrm{m/s^2}$.

Two stages. First the fall, to find the arrival speed:

$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.2} = \sqrt{23.52} = 4.850\ \mathrm{m/s}$

Now the stop. Take upward as positive, so the egg arrives with $v_1 = -4.850\ \mathrm{m/s}$ and ends at $v_2 = 0$:

$\Delta p = m(v_2 - v_1) = 0.055 \times (0 - (-4.850)) = +0.2668\ \mathrm{kg\cdot m/s}$

During those 0.15 s, two forces act: the pillow pushing up with $\bar N$, and gravity pulling down with $mg = 0.055 \times 9.8 = 0.539\ \mathrm{N}$. The net impulse is what equals $\Delta p$:

$(\bar N - mg)\,\Delta t = \Delta p$

Dropping the gravity term would have given 1.78 N, about 23 % low. Whether that matters depends on how short the stop is: for a 0.002 s impact against concrete, the weight term is invisible next to a force of hundreds of newtons, and that is exactly why you are allowed to ignore gravity during a genuine collision. Here the stop is slow enough that gravity gets a word in.

When mass arrives instead of speeding up

Now use the form of the law that $F = ma$ cannot reach. Suppose a stream of material arrives steadily at a surface: water from a hose, sand onto a conveyor, air pushed down by a drone. Nothing is accelerating in the ordinary sense. Instead, in every second, a mass $dm$ has its velocity changed by $\Delta v$. The momentum delivered per second is

$F = \frac{dp}{dt} = \frac{dm}{dt}\,\Delta v$

Read $dm/dt$ as "kilograms arriving per second" and $\Delta v$ as "how much each kilogram's velocity changes". Multiply. That is the force.

Water at 4.0 kg/s hitting a wall at 15 m/s and dribbling down gives $4.0 \times 15 = 60\ \mathrm{N}$. Water at 6.0 kg/s at 10 m/s turned completely around by a curved vane changes each kilogram's velocity by $10 - (-10) = 20\ \mathrm{m/s}$, so the force is $6.0 \times 20 = 120\ \mathrm{N}$.

A drone hovering needs thrust equal to its weight, so $\dot m \times 18 = 32\ \mathrm{N}$ gives $\dot m = 1.78\ \mathrm{kg/s}$ of air pushed down. A rocket engine is the same sum run backwards: ejecting exhaust at rate $\dot m$ with speed $u$ relative to the rocket produces thrust $F = \dot m u$, so 3.0 kg/s at 2500 m/s is 7500 N.

Worked example 4. A uniform chain of linear density $\lambda = 1.5\ \mathrm{kg/m}$ is released from rest with its lower end just touching a scale. What does the scale read when 0.80 m has landed?

This one takes a moment. The scale supports two separate things.

The first is the weight of the chain already lying on it: a length $x = 0.80\ \mathrm{m}$ of chain, mass $\lambda x = 1.5 \times 0.80 = 1.2\ \mathrm{kg}$, weighing $1.2 \times 9.8 = 11.76\ \mathrm{N}$.

The second is the force needed to stop the links that are landing right now. The chain fell freely through $x = 0.80$ m, so the arriving links have speed $v = \sqrt{2gx} = \sqrt{2 \times 9.8 \times 0.80} = 3.960\ \mathrm{m/s}$. Chain arrives at a rate $dm/dt = \lambda v$ kilograms per second, and each kilogram has its speed cut from $v$ to zero. So the stopping force is

Since $v^2 = 2gx$, that term is exactly $2\lambda g x$, twice the weight of the landed part. The scale reads

$N = \lambda g x + 2\lambda g x = 3\lambda g x = 3 \times 11.76 = 35.3\ \mathrm{N}$

Three times the weight of what is lying there. The same structure solves the rope pulled off a heap: if you drag rope of density 0.90 kg/m horizontally at a constant 2.0 m/s, you set $\lambda v = 1.8$ kg of rope into motion each second and each kilogram goes from 0 to 2.0 m/s, so $F = \lambda v^2 = 0.90 \times 4.0 = 3.6\ \mathrm{N}$.

Rockets: when the mass leaves

A rocket in deep space throws mass backwards to push itself forwards. In a short time $dt$ it ejects mass $-dm$ (negative because the rocket is losing it) at speed $u$ relative to itself. Total momentum is conserved, so the momentum the exhaust carries backwards equals the momentum the rocket gains forwards:

$m\,dv = -u\,dm \quad\Longrightarrow\quad dv = -u\,\frac{dm}{m}$

Integrate from the initial mass $m_0$ to the final mass $m_f$:

$\Delta v = -u\int_{m_0}^{m_f}\frac{dm}{m} = u\ln\!\frac{m_0}{m_f}$

That is the rocket equation. What it says is brutal: speed grows only with the logarithm of the mass ratio, so doubling your fuel does not double your speed. A 2000 kg rocket burning down to 800 kg with $u = 2800$ m/s, starting from rest, reaches

Launching from the ground, gravity spends the whole burn taking momentum back, delivering an impulse $-mg\,dt$ that costs you $g\,t_{\text{burn}}$ of speed:

$v = u\ln\!\frac{m_0}{m_f} - g\,t_{\text{burn}}$

For 5000 kg down to 2000 kg over 50 s with $u = 2400$ m/s: $2400 \times \ln 2.5 = 2400 \times 0.9163 = 2199.1\ \mathrm{m/s}$, minus $9.8 \times 50 = 490\ \mathrm{m/s}$, leaving $1709\ \mathrm{m/s}$.

How to tell this is an impulse problem

Reach for impulse and momentum when you see any of these:

• A duration, a contact time, or an average force. "Stopped over 0.08 s", "the bat is in contact for 1.2 ms". Energy methods want a distance; this one wants a time.
• A force given as a function of time, or a force-time graph. The question is asking you for an area.
• You are not told the force and do not want to know it. Collisions, catches, kicks, explosions. You know the before and the after, and the middle is a mess you are allowed to skip.
• Mass per second. "kg/s", "20 balls per second", "burns fuel at a constant rate", a chain landing, a belt being loaded. That is $F = \dot m\,\Delta v$.
• Something reverses. The moment a problem says "rebounds" or "is turned around", it is testing whether you will write the minus sign.

And a last check on any answer: momentum is a vector, so every one of these should have been done axis by axis with a chosen positive direction. If you only ever worked with speeds, go back and look for a sign you dropped.

Everything here has been one object and the outside world. Next you point the same theorem at two objects hitting each other, where the internal forces cancel in pairs and the total momentum of the pair is conserved no matter how violent the contact: Collisions in One and Two Dimensions.