Classical Mechanics

Before anything can be said about why things move, there has to be a language precise enough to say how they move, and that language turns out to be one already built: a single function of time, differentiated twice and integrated back.

This course assumes the calculus in Calculus: derivatives, the chain rule, definite integrals, separation of variables in a simple first order equation, and Taylor expansion about a point. No vector calculus is needed anywhere, and vectors themselves are built from components in the next lesson.

Position, and the trouble with average speed

Fix an origin and a positive direction along a line. The position $x(t)$ of a particle is its signed distance from that origin at time $t$, in metres, and everything in this lesson is a statement about that one function. Choosing the origin is free: physics never depends on where the zero sits, only on differences, which is why $x$ appears in real equations almost exclusively as $\Delta x=x_2-x_1$.

The obvious summary of a motion is its average velocity, $\Delta x/\Delta t$. Usain Bolt covered 100 m in 9.58 s in Berlin in 2009, so his average velocity was $100/9.58=10.44$ m s⁻¹. The published 10 metre splits for that run, which sum to exactly 9.58 s, are 1.89, 0.99, 0.90, 0.86, 0.83, 0.82, 0.81, 0.82, 0.83, 0.83. The first ten metres took 1.89 s, an average of only 5.29 m s⁻¹, and the stretch from 60 to 70 m took 0.81 s, an average of 12.35 m s⁻¹. The single number 10.44 describes no instant of the race. It is a fact about the two endpoints and nothing else.

Shortening the interval helps. The 10 m splits are better than the one number, 1 m splits would be better still, and the question that produces calculus is what happens as the interval shrinks towards nothing. On a graph of $x$ against $t$, the average velocity over an interval is the slope of the chord joining its ends, and the chords approach the tangent.

Velocity as a derivative

Define the instantaneous velocity as the limit of the average:

Velocity is signed: $v<0$ means moving towards decreasing $x$. Speed is $|v|$, and losing the distinction is a reliable way to get a sign wrong later. A ball thrown upward has positive velocity going up, negative coming down, and a speed that is zero only at the single instant at the top.

The derivative is what makes a speedometer meaningful. There is no interval short enough to be "the instant", so the reading has to be defined as a limit rather than measured as a ratio, and the limit exists as long as the position function is smooth. It is worth noticing how much physics is packed into the assumption that it is smooth: a particle that jumped from one place to another would have no derivative at the jump, and Newtonian mechanics simply asserts that this does not happen.

Example. A trolley moves along a track with $x(t)=12t-1.5t^2$, in metres, with $t$ in seconds. Where is it at $t=2$ s, and how fast is it going?

Differentiate: $v(t)=12-3t$. At $t=2$, the position is $x=24-6=18$ m and the velocity is $v=12-6=6$ m s⁻¹. The trolley is still moving forward but has lost half the speed it started with, and $v=0$ at $t=4$ s, which is where it turns around.

Now you. A cart has $x(t)=5+8t-0.4t^3$ metres. Find its position, velocity and acceleration at $t=2$ s.

Acceleration, and why it is the physical one

Differentiate again and you have the acceleration:

in metres per second per second, m s⁻². A car going at a steady 30 m s⁻¹ has zero acceleration; one speeding up from rest to 30 m s⁻¹ in 8 s has an average acceleration of 3.75 m s⁻². Acceleration is also signed, and its sign is not "speeding up" or "slowing down" on its own: a car reversing and speeding up has negative velocity and negative acceleration. The rule is that speed increases when $v$ and $a$ share a sign and decreases when they do not.

Nothing so far singles out acceleration as more important than the third derivative or the fourth. What singles it out is a physical fact that this lesson cannot yet prove: forces determine acceleration, and nothing else. That is why the chain stops at two derivatives and why the state of a particle is its position and velocity together. Give both now, plus the forces from now on, and the entire future follows. That claim is the content of the next few lessons.

Two derivatives is also exactly the number a person can feel. Sitting in a cruising aircraft at 250 m s⁻¹ feels like sitting in a chair, because velocity is not detectable from inside; the takeoff run is unmistakable, because acceleration is.

Running the calculus backwards

If acceleration is what physics hands you, the useful direction is the other one. Integrating once recovers velocity and integrating again recovers position:

Each integration introduces one constant, fixed by an initial condition, and this is where $v_0$ and $x_0$ come from. Two integrations, two constants: the future needs the initial position and the initial velocity, and nothing more. A law of motion that gave the third derivative would need three, and a body's history would not be settled by where it is and how fast it is going, which is not the world we live in.

Geometrically, integration says that displacement is the area under the velocity graph, counting area below the axis as negative. A car that drives forward at 20 m s⁻¹ for 30 s and back at 20 m s⁻¹ for 30 s covers 1200 m of road with a displacement of zero, and the two areas cancel exactly. Distance travelled is $\int |v|dt$, which is a different integral, and confusing the two is the second most common error in kinematics after sign confusion.

The constant acceleration formulas, derived

The special case worth memorising is constant $a$, because it covers free fall near the ground, braking with a locked wheel, and any short interval over which the force barely changes. With $a$ constant, the first integral is immediate:

and integrating that gives

These two are the whole set. Everything else is algebra: eliminating $t$ between them gives $v^2=v_0^2+2a(x-x_0)$, and averaging the velocity gives $x-x_0=\frac{1}{2}(v_0+v)t$, which is valid only because a linear velocity has its mean at the midpoint of the interval. There are four formulas in most textbooks and two independent facts.

Free fall is the standard instance. Near the Earth's surface every body released from rest falls with the same downward acceleration $g=9.81$ m s⁻², independent of its mass, provided air resistance can be ignored. Drop a stone from rest and after 3.0 s it has fallen $\frac{1}{2}(9.81)(9)=44.1$ m and is moving at $9.81\times 3=29.4$ m s⁻¹. Galileo could not time a fall accurately with the clocks of 1600, so he diluted gravity by rolling balls down inclined planes and found the distance growing as the square of the time, which is the same claim in slow motion. The cleanest demonstration is David Scott's on Apollo 15 in 1971: a hammer and a falcon feather released together on the Moon, where there is no air, hit the surface together, in a fall from 1.6 m taking about 1.41 s at the lunar $g$ of 1.62 m s⁻².

Example. A car brakes from 28 m s⁻¹, roughly 100 km/h, at a steady 7.5 m s⁻². How far does it travel before stopping, and how long does it take?

Take the initial direction as positive, so $a=-7.5$ m s⁻² and the final $v$ is zero. From $v^2=v_0^2+2a\Delta x$ with $v=0$, $\Delta x=-v_0^2/2a=784/15=52.3$ m. From $v=v_0+at$, $t=28/7.5=3.73$ s. Note that halving the speed quarters the distance, since $\Delta x$ goes as $v_0^2$: that quadratic is the whole argument for speed limits.

Now you. The same car brakes from 20 m s⁻¹ at 6.0 m s⁻². Find the stopping distance and the stopping time.

When time is not wanted

Often the question is about position and speed with no interest in when. The formula $v^2=v_0^2+2a\Delta x$ answers it for constant $a$, but the trick behind it survives when $a$ is not constant, and it is worth seeing once. Write acceleration using the chain rule with position as the intermediate variable:

so that $adx=vdv$. Integrating both sides between two positions gives

For constant $a$ the left side is $a\Delta x$ and the familiar formula drops out. For any $a$ that is known as a function of position, the same integral still works, and time never appears. Multiply through by mass and this identity becomes the work energy theorem, which is where this course goes in a few lessons: the shortcut is not a trick at all but energy conservation in disguise.

Example. A ball is thrown straight up at 22 m s⁻¹. How high does it go, and when does it reach the top?

At the top $v=0$. Using $v^2=v_0^2-2gh$ with $v=0$ gives $h=v_0^2/2g=484/19.62=24.7$ m. The time is $t=v_0/g=22/9.81=2.24$ s. The ball then falls back, reaching the thrower's hand at the same speed it left, which the symmetry of the equations guarantees and air resistance spoils by a few per cent.

Now you. A ball is thrown straight up at 16 m s⁻¹. Find its maximum height and the time to reach it.

What kinematics cannot do

Everything above is mathematics. Given $a(t)$, the motion follows; given the motion, the acceleration follows. There is no physics in it, and the proof is that the equations are just as happy with a body whose acceleration is $-9.81$ m s⁻² as with one whose acceleration is $+400$ m s⁻² upward, or one that reverses every second for no reason.

That is the gap. Kinematics is a bookkeeping system with an empty input slot, and the whole of the next lesson but one is about what fills it. The answer, that acceleration is caused by forces and is proportional to their sum divided by the mass, is not derivable from any amount of graph reading. It had to be discovered, and it was discovered late, because the obvious guess, that force determines velocity rather than acceleration, is what everyday experience with friction actually suggests.

First, though, the description has to escape the straight line. Real motion happens in a plane or in space, and the next lesson shows that the escape costs almost nothing: a vector is a bundle of independent copies of what has just been built.

Real bodies do not move along a line, and the cost of escaping the line turns out to be almost nothing, because motion in a plane is two straight line problems that happen to share a clock.

The previous lesson built the whole of one dimensional kinematics out of one function $x(t)$ and its two derivatives. Everything here is that machinery run twice, once per axis, so a reader arriving cold needs only that velocity is $dx/dt$, that acceleration is $dv/dt$, and that constant acceleration gives $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$.

Two numbers instead of one

Fix an origin and two perpendicular axes. The position of a particle is now the pair $(x,y)$, and it is convenient to write the pair as a single object, the position vector $\overrightarrow{r}=(x,y)$. A vector is a quantity with a magnitude and a direction, and in practice it is a list of components, one per axis. Its magnitude is $|\overrightarrow{r}|=\sqrt{x^2+y^2}$ by Pythagoras, and its direction can be reported as the angle $\theta =\arctan (y/x)$ measured from the positive $x$ axis, taking care with the quadrant, since $\arctan$ cannot tell $(3,4)$ from $(-3,-4)$.

Two vectors add by adding components: $(a_x,a_y)+(b_x,b_y)=(a_x+b_x,a_y+b_y)$. Geometrically this is the tip to tail rule, and the two descriptions are the same statement. Multiplying by a number scales every component, so $2\overrightarrow{a}$ points the same way and is twice as long, and $-\overrightarrow{a}$ points the opposite way. There is no operation that divides one vector by another, and looking for one is a sign that the wrong quantity is being sought.

The choice of axes is free, and choosing them well is most of the skill. Motion on a slope is usually easier with one axis along the slope; a projectile is easiest with one axis vertical. What must not change under that choice is any physical answer, and that invariance is worth testing on a problem occasionally, because it catches errors nothing else does.

Differentiating a vector

Velocity in a plane is defined exactly as on a line, as the limit of displacement over elapsed time:

Because the axes are fixed, differentiating a vector means differentiating each component separately, and the same goes for acceleration, $\overrightarrow{a}=d\overrightarrow{v}/dt$. That single sentence is the whole content of two dimensional kinematics: the components do not talk to each other. Whatever is happening along $x$ has no influence on what is happening along $y$, because the derivative of $x$ never contains $y$.

Two consequences are worth stating separately because they are where intuition fails. First, velocity is a vector, so it can change without the speed changing, simply by turning. Second, acceleration need not point along the velocity: it is the rate of change of the velocity vector, not of its length. Split $\overrightarrow{a}$ into a part along $\overrightarrow{v}$ and a part perpendicular to it, and the parallel part changes speed while the perpendicular part changes direction only. A car that accelerates in a straight line uses only the first; a car cornering at constant speed uses only the second.

Speed is $|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$, and it is worth noticing that this is not the derivative of $|\overrightarrow{r}|$. A satellite in a perfectly circular orbit has a constant $|\overrightarrow{r}|$ and a speed of nearly eight kilometres per second.

Projectiles: the same lesson twice

Take a body moving near the ground with air resistance neglected. Every experiment since Galileo says the acceleration is $g=9.81$ m s⁻² straight down and nothing else, so in components $a_x=0$ and $a_y=-g$. The two axes are now separate one dimensional problems with a shared time:

The horizontal motion has no acceleration and so is uniform. The vertical motion is free fall, identical to a dropped stone. This is the substance of Galileo's claim in the Two New Sciences of 1638, and the memorable form of it is that a bullet fired horizontally and a bullet dropped from the same height at the same instant hit the ground together, because their $y$ equations are identical and their $x$ equations are irrelevant to the landing.

For a launch from the ground at speed $v_0$ and angle $\theta$, the components are $v_{0x}=v_0\cos \theta$ and $v_{0y}=v_0\sin \theta$. Setting $y=0$ gives the time of flight $T=2v_0\sin \theta /g$, and multiplying by the horizontal speed gives the range

using $2\sin \theta \cos \theta =\sin 2\theta$. Three things fall out at once. The range is maximised at $\theta ={45}^{\circ }$, since that is where $\sin 2\theta =1$. Angles either side of ${45}^{\circ }$ that are equally far from it give the same range, so ${30}^{\circ }$ and ${60}^{\circ }$ land in the same place, one on a flat trajectory and one on a lofted one. And $R$ goes as $v_0^2$, so a ten per cent faster throw goes twenty one per cent further.

Example. A ball is launched from ground level at 25 m s⁻¹ at 40° above the horizontal. Find its time of flight, range and greatest height.

The components are $v_{0x}=25\cos {40}^{\circ }=19.15$ m s⁻¹ and $v_{0y}=25\sin {40}^{\circ }=16.07$ m s⁻¹. The vertical problem is a stone thrown up at 16.07 m s⁻¹, so it rises for $16.07/9.81=1.638$ s and the flight lasts $T=3.28$ s. The range is $19.15\times 3.28=62.7$ m, which the formula confirms: $625\sin {80}^{\circ }/9.81=62.7$ m. The greatest height is $v_{0y}^2/2g=258.2/19.62=13.2$ m.

Now you. A ball is launched from ground level at 18 m s⁻¹ at 30°. Find its time of flight, range and greatest height.

Where the parabola stops being true

Two assumptions are buried in that derivation, and both fail in ordinary cases.

The first is that the launch and landing heights are equal, which is what let $y=0$ close the problem. A shot putter releases the shot from about 2.1 m above the ground, and the extra fall changes the arithmetic. Launching at 14 m s⁻¹ and 40° from that height, the vertical equation $2.1+9.00t-4.905t^2=0$ gives $t=2.044$ s rather than the 1.834 s of a ground level launch, and the range becomes $10.72\times 2.044=21.9$ m. More interestingly, the best angle is no longer 45°: scanning the angles for this speed and height puts the maximum at 42.3°, giving 21.98 m. Whenever a projectile lands below its launch point, the optimum tilts flatter, because time in the air is partly free.

The second assumption is the fatal one. Air resistance is not small for most things that fly. A baseball leaving the bat at 45 m s⁻¹ at 35° would, in a vacuum, travel $2025\sin {70}^{\circ }/9.81=194$ m. Real home runs of that launch condition go about 120 to 135 m, so the vacuum formula overestimates by roughly half. The trajectory is not a parabola either: it is steeper on the way down than on the way up, because the ball loses horizontal speed throughout the flight and never gets it back. Drag depends on speed, so the horizontal and vertical equations stop being independent, and the whole method of this section collapses. Solving that case needs the differential equation of a later lesson, and even then usually a computer.

The parabola is therefore a good model for a thrown stone, a poor one for a badminton shuttlecock, and a catastrophic one for artillery, which is why range tables were compiled by firing rather than by algebra.

Relative velocity

If a boat moves at ${\overrightarrow{v}}_{BW}$ with respect to the water and the water moves at ${\overrightarrow{v}}_{WG}$ with respect to the ground, the boat's velocity over the ground is the sum:

The subscripts chain, and reversing a pair negates it, ${\overrightarrow{v}}_{WB}=-{\overrightarrow{v}}_{BW}$. That is the entire theory of relative motion at everyday speeds, and it is exactly what fails at speeds near light, where velocities do not simply add. Nothing in this course goes near that regime.

Example. A river 80 m wide flows at 1.2 m s⁻¹. A boat that can do 2.5 m s⁻¹ through the water points straight across. How long does the crossing take, where does it land, and how fast is it moving over the ground?

Point the $x$ axis across the river and $y$ downstream. The boat's velocity through the water is $(2.5,0)$ and the water's over the ground is $(0,1.2)$, so over the ground the boat does $(2.5,1.2)$. The crossing is governed by the $x$ component alone: $t=80/2.5=32$ s, unaffected by the current, which is the same independence that governed the projectile. In that time it drifts $1.2\times 32=38.4$ m downstream, and its ground speed is $\sqrt{{2.5}^2+{1.2}^2}=2.77$ m s⁻¹ at $\arctan (1.2/2.5)={25.6}^{\circ }$ from straight across.

Now you. The same boat on the same river wants to land directly opposite its start. At what angle upstream must it point, and how long does the crossing now take?

Going round in a circle

Now the case that breaks intuition. A particle moves round a circle of radius $r$ at constant speed $v$. Its speed never changes, yet it is accelerating, and the size of that acceleration can be got without any calculus beyond similar triangles.

Over a short interval $\Delta t$ the particle turns through an angle $\Delta \theta =v\Delta t/r$, because it covers an arc of length $v\Delta t$. Its velocity vector has the same length before and after but has turned through that same angle $\Delta \theta$, since velocity is always tangent to the circle. Draw the two velocity vectors from a common point: they form an isosceles triangle with two sides of length $v$ and an apex angle $\Delta \theta$, so the third side, which is $|\Delta \overrightarrow{v}|$, is $2v\sin (\Delta \theta /2)$, and for small angles that is $v\Delta \theta$. Therefore

The direction takes one more step. As $\Delta \theta$ shrinks, $\Delta \overrightarrow{v}$ becomes perpendicular to $\overrightarrow{v}$, and of the two perpendicular directions it points to the inside of the turn. So the acceleration has magnitude $v^2/r$ and points at the centre. It is called the centripetal acceleration, from centrum petere, to seek the centre. Since the time for one lap is $T=2\pi r/v$, the acceleration can also be written $4{\pi }^{2}r/T^2$, which is the form to use when a period is known and a speed is not.

Nothing here says what causes the acceleration. Centripetal acceleration is not a force and not a cause; it is a description of what the velocity vector is doing. The next lessons ask what has to push.

Example. The Moon orbits the Earth at a mean radius of $3.844\times {10}^8$ m with a period of 27.32 days. What is its centripetal acceleration?

The period is $27.32\times 86400=2.361\times {10}^6$ s, so the orbital speed is $v=2\pi r/T=1023$ m s⁻¹. Then $a=v^2/r={1023}^2/(3.844\times {10}^8)=2.72\times {10}^{-3}$ m s⁻². That is smaller than $g$ at the Earth's surface by a factor of 3600, and the Moon's orbit is larger than the Earth's radius by a factor of 60.3, whose square is 3640. The agreement of those two numbers is the single most important coincidence in the history of physics, and a later lesson spends itself on what it means.

Now you. A centrifuge rotor spins at 12000 revolutions per minute. What is the centripetal acceleration at a radius of 0.10 m, and how many times $g$ is it?

What has been gained, and what has not

Kinematics is now complete. Position, velocity and acceleration are defined in any number of dimensions, the constant acceleration case is solved, projectiles and circular motion are worked out, and relative motion is a sum. Given the acceleration of a body at every instant, its whole future can be written down.

That last sentence still contains the word "given". Nothing so far predicts an acceleration; every result above took one as input. The equations are equally content with a stone that falls at 9.81 m s⁻² and one that drifts sideways at 400 m s⁻² for no reason, and no amount of graph reading will separate them.

Filling that gap is the business of the next lesson, and it took the better part of two thousand years, because the natural guess is wrong. Everyday experience with sledges and carts suggests that force sets velocity, since a cart stops when you stop pushing. The right answer, that force sets acceleration and that the stopping is itself a force, requires believing in something nobody has ever seen: a body moving forever with nothing pushing it.

Kinematics can describe any motion whatsoever and predict none of them, because the acceleration has to be supplied from outside, and the three laws published by Newton in 1687 are the supply.

This lesson assumes only that acceleration is the second derivative of position, which the first two lessons built. It is the least computational lesson in the course and the most important, because every calculation after it is an application of one equation stated here.

The wrong answer that lasted two thousand years

Aristotle's physics held that a body's natural state is rest, and that continued motion requires a continued cause. It is a reasonable summary of the evidence available to anyone pushing a cart on a road: stop pushing and the cart stops. It also has an obvious embarrassment, the thrown stone, which keeps moving after the hand has let go. The standard repair was that the air closes behind the stone and drives it forward, which is worse than the problem it solves.

The medieval alternative, developed by Jean Buridan in Paris around 1350, was impetus: the thrower imparts to the stone a quantity that keeps it going and gradually runs out. This is closer, and it survived because it makes roughly the right predictions for real projectiles in real air. What it gets wrong is the running out. Impetus is spent by the motion itself; momentum is not spent at all, and is only changed by something else pushing.

Galileo got the decisive result by a thought experiment about inclined planes. A ball rolling down one incline and up another rises to nearly the height it started from, whatever the second incline's slope; make the second incline gentler and the ball travels further to reach the same height; make it horizontal and there is no height to reach, so the ball should travel forever. He could not test this, because there is always friction, and the argument works by extrapolating friction away. That extrapolation is the birth of theoretical physics: the claim is not about any ball anyone has rolled, but about the limit that no experiment can reach.

The first law: inertial frames exist

Newton's first law, in the Principia of 1687, states that a body continues in its state of rest, or of uniform motion in a straight line, unless compelled to change that state by forces impressed upon it.

Read carelessly, this is a special case of the second law with $\overrightarrow{F}=0$, and therefore redundant. Read properly it is a separate and stronger claim, because both laws are silently about some frame of reference, and the first law is what says a suitable frame exists.

Consider a ball resting on the floor of a train. If the train brakes, the ball rolls forward with no one touching it. In the frame of the train, a body with no force on it has accelerated, so the first law is false in that frame. The law is therefore not a universal truth about bodies; it is a test that selects frames. A frame in which an isolated body stays at uniform velocity is called an inertial frame, and the first law asserts that such frames exist. Once one is found, any frame moving at constant velocity with respect to it is another, since a constant added to a velocity vanishes on differentiation.

This matters because the second law is only true in an inertial frame. Everything else in this course is a calculation done in one, and the honest question is whether the ground under our feet qualifies. It does not exactly. The Earth spins, so a point on the equator is accelerating towards the axis at ${\omega }^{2}R$, with $\omega =2\pi /86164=7.292\times {10}^{-5}$ radians per second and $R=6.378\times {10}^6$ m, giving $0.0339$ m s⁻², about 0.35 per cent of $g$. The Earth also orbits the Sun, adding $5.9\times {10}^{-3}$ m s⁻². Both are far too small to notice while dropping a stone, and far too large to ignore in a Foucault pendulum, in ballistics at long range, or in the circulation of the atmosphere.

The second law: force and mass, defined together

The second law is the working equation of the whole subject:

the vector sum of all forces on a body equals its mass times its acceleration. Being a vector equation, it is really one equation per axis, $\sum F_x=m{a}_x$ and $\sum F_y=m{a}_y$, which is what makes the component method of the previous lesson pay off. The unit of force follows: one newton is the force that gives one kilogram an acceleration of one metre per second squared, so 1 N = 1 kg m s⁻².

There is a circularity here that textbooks often slide past, and it is worth facing. What is a force? Something that causes acceleration. What is mass? The resistance to acceleration by a force. Each term is defined by the other, so as it stands the law says nothing falsifiable.

The way out is experimental, and it comes in two steps. First, mass can be compared without knowing anything about forces at all: let two bodies interact with each other alone, on an air track, and measure the two accelerations. Experiment says the ratio $a_1/a_2$ is always the same for the same pair of bodies, whatever the interaction is, whether they collide, or repel by magnets, or are joined by a spring. Define the mass ratio as the inverse of the acceleration ratio, $m_2/m_1=a_1/a_2$, pick one body as the standard, and every mass in the world follows by comparison. Since May 2019 the standard is not a metal cylinder in Sèvres but the fixed value of the Planck constant, $6.62607015\times {10}^{-34}$ J s, which fixes the kilogram through measurement rather than through an object that could be scratched.

Second, with masses known, the law becomes a real claim about forces: measure the acceleration a spring stretched by 2 cm gives to a known mass, and the law predicts what it gives to any other. That prediction can fail, and does not. The content of the second law is not the algebra but the discovery that the same number $m$ works for every kind of force applied to a given body, and that the same force gives a given body the same acceleration regardless of what else is happening to it. Forces add as vectors: that is a physical finding, not a definition.

Example. A 3.0 kg block on a frictionless horizontal surface is pulled by a 12 N force along the positive $x$ axis and simultaneously by a 9.0 N force at 60° to it. Find the acceleration.

Resolve into components. The 12 N force is $(12,0)$. The 9.0 N force is $(9\cos {60}^{\circ },9\sin {60}^{\circ })=(4.50,7.79)$ N. The sum is $(16.50,7.79)$ N, whose magnitude is $\sqrt{{16.50}^2+{7.79}^2}=18.25$ N at $\arctan (7.79/16.50)={25.3}^{\circ }$ from the $x$ axis. Dividing by the mass, $a=18.25/3.0=6.08$ m s⁻² in that same direction. The direction of the acceleration is the direction of the net force, always, and never the direction of the largest single force.

Now you. A 2.5 kg block is pulled by 15 N along $x$ and 8.0 N along $y$, with no friction. Find the magnitude and direction of its acceleration.

Mass, weight, and what a scale measures

Mass is a property of a body and is the same everywhere. Weight is the gravitational force on it, $W=mg$, and depends on where it is. A 70 kg astronaut weighs 687 N on Earth, 114 N on the Moon where $g=1.62$ m s⁻², and has a mass of 70 kg in both places. Confusing the two is harmless in ordinary speech and fatal in a calculation, because it is $m$ and not $W$ that appears on the right of the second law.

A bathroom scale does not measure either. It measures the normal contact force it exerts on your feet, and reports that force divided by 9.81 as though you were not accelerating. When you are, the reading is wrong in an informative way, and the whole of the sensation of a lift is contained in this.

Example. A 70 kg person stands on a scale in a lift accelerating upward at 2.0 m s⁻². What does the scale read, in newtons?

Two forces act on the person: gravity $mg=687$ N down, and the normal force $N$ up. Take up as positive, so the second law gives $N-mg=ma$, and $N=m(g+a)=70(9.81+2.0)=827$ N. The scale reads 827 N, twenty per cent above the standing value, which is the pressed-into-the-floor feeling on starting to rise. If instead the cable snapped and the lift fell freely at $a=-g$, then $N=0$: the person floats, weightless, while the gravitational force on them has not changed at all. Weightlessness in orbit is exactly this, and not the absence of gravity.

Now you. The same person is in a lift accelerating downward at 1.5 m s⁻². What does the scale read?

The third law and the pair it refers to

If body A exerts a force on body B, then B exerts on A a force equal in magnitude and opposite in direction, along the same line. The two forces are of the same kind, they act at the same instant, and, decisively, they act on different bodies.

That last clause is what makes the law useful and what makes it constantly misapplied. The stock objection is that a horse cannot pull a cart, since the cart pulls back equally hard and the two cancel. They do not cancel, because they never appear in the same equation: the horse's pull acts on the cart, and the cart's pull acts on the horse. Ask what accelerates the cart and the answer involves only forces on the cart. Ask what accelerates the horse and the answer involves the cart's backward pull and the ground's forward push on its hooves, which is a different pair.

The reliable test is to name both bodies for every force: not "the weight of the book" but "the Earth pulls the book down". Its partner is then automatic: the book pulls the Earth up with an equal force. The normal force of the table on the book is not that partner, despite being equal and opposite in the common case of a book at rest, and the giveaway is that it stops being equal the moment you press down on the book, while the true third law partner remains equal always.

The Earth really does accelerate upward towards a dropped stone. For a 1 kg stone the force on the Earth is 9.81 N, and dividing by $5.97\times {10}^{24}$ kg gives $1.6\times {10}^{-24}$ m s⁻², which is why nobody notices.

Momentum, and the form Newton actually used

Newton did not write $F=ma$. He defined the quantity of motion as mass times velocity, what we call the momentum $\overrightarrow{p}=m\overrightarrow{v}$, and stated the second law as the claim that the change of motion is proportional to the impressed force:

For constant mass this gives $md\overrightarrow{v}/dt=m\overrightarrow{a}$ and the two forms agree. When mass is not constant, as for a rocket burning fuel or a rope being lifted onto a table link by link, the momentum form is the one to reason with, though even then it must be applied to a fixed collection of matter rather than to a shrinking body, a subtlety a later lesson handles properly.

The momentum form also makes the third law say something startling. If A and B interact and nothing else acts, then $d{\overrightarrow{p}}_A/dt=-d{\overrightarrow{p}}_B/dt$, so the total momentum ${\overrightarrow{p}}_A+{\overrightarrow{p}}_B$ has zero derivative and never changes. Conservation of momentum is not an extra law; it is the third law rewritten. Two lessons from now that observation carries the whole of collision theory.

Example. A rifle of mass 4.0 kg fires a 12 g bullet at 850 m s⁻¹. What is the recoil speed of the rifle, and what average force acts on the shoulder if the recoil is stopped in 0.15 s?

Total momentum starts at zero and the third law keeps it there, so $m_b{v}_b=m_r{v}_r$, giving $v_r=(0.012)(850)/4.0=2.55$ m s⁻¹ backward. Stopping that in 0.15 s requires a force of $\Delta p/\Delta t=(4.0)(2.55)/0.15=68$ N, which is comfortable. The bullet leaves with the same 10.2 kg m s⁻¹ of momentum, but with vastly more energy, and the reason for that asymmetry is the subject of a later lesson.

Now you. A 60 kg skater standing at rest on frictionless ice throws a 3.0 kg ball forward at 8.0 m s⁻¹. How fast does the skater move backward?

An equation with an empty right hand side

Write the second law out as what it is, a differential equation:

Given the forces as functions of position, velocity and time, plus the initial position and velocity, the solution is unique and the entire future of the body follows. That is the claim of determinism that made mechanics the model for every science that came after, and it survived intact until quantum mechanics and, in a different way, until the discovery that some solutions depend so sensitively on the initial conditions that predicting them requires knowing the start to impossible precision.

None of that power is available yet, because the right hand side is empty. The laws say what force does, not what forces there are. Gravity near the ground, the push of a surface, the pull of a rope, the resistance of friction and the drag of air are separate empirical discoveries, each with its own formula and its own range of validity, and assembling that catalogue is the next lesson's work. It is also where the method that makes mechanics tractable appears: draw one body, name every force on it, and turn the picture into two equations.

Newton's second law is a machine with an empty input slot: it says what a net force does but never what forces are present, and filling that slot is a separate, empirical business.

The previous lesson established $\sum \overrightarrow{F}=m\overrightarrow{a}$ and that it holds one component at a time. This lesson names the forces that actually appear in mechanical problems, and sets out the procedure that turns a physical situation into algebra.

The free body method

Almost every mistake in elementary mechanics is a bookkeeping mistake, and the free body diagram exists to prevent it. The procedure has four steps and no shortcuts.

Choose one body and draw it alone, detached from everything it touches. Draw every force acting on it, each as an arrow from the body, and name the agent of each: the Earth pulls it down, the table pushes it up, the rope pulls it along the rope. If no agent can be named, the force is imaginary, and this test alone kills most of the spurious forces beginners draw, including the mysterious forward force on a coasting ball. Choose axes, preferring one along the acceleration if the direction is known. Then write $\sum F_x=m{a}_x$ and $\sum F_y=m{a}_y$ and solve.

Forces on other bodies never appear in this diagram. If two bodies are connected, draw two diagrams and let the connection appear once in each, with opposite signs, which is the third law doing its work.

Weight, normal contact and tension

Near the ground, gravity pulls every body straight down with a force $mg$, with $g=9.81$ m s⁻² in Britain and varying from about 9.78 at the equator to 9.83 at the poles. Nothing in this lesson depends on why: the inverse square law that explains it comes much later.

The normal force is the push a surface exerts perpendicular to itself. It is not $mg$ except by accident, and treating it as though it were is the most common single error in the subject. A surface pushes exactly as hard as it must to stop the body sinking into it, and that requirement changes when the surface is tilted, when the body accelerates, or when something else presses on it. On a slope of angle $\theta$ with no other vertical force, resolving perpendicular to the surface gives $N=mg\cos \theta$; in a lift accelerating up at $a$ it is $m(g+a)$; and the moment a surface would have to pull rather than push, contact is lost and $N=0$, which is how a problem tells you a body has left the ground.

Tension is the pull transmitted along a rope or rod. An idealised string is massless and inextensible, and both idealisations do real work. Massless means the tension is the same at both ends, since a massless segment with unequal pulls would have infinite acceleration. Inextensible means the two bodies it connects have accelerations of equal magnitude, which is the extra equation that closes most connected body problems. A real rope has mass, so the tension in a hanging rope is larger at the top than the bottom, by exactly the weight of the rope below the point in question.

Friction, which is not a law

Slide one dry surface over another and it resists. The standard description, due to Amontons in 1699 and refined by Coulomb in 1785, is two rules.

While the surfaces are not sliding, static friction takes whatever value it must to prevent sliding, up to a limit:

Once sliding, kinetic friction acts backward along the motion with a roughly constant magnitude:

The first is an inequality, not an equation, and writing $f_s={\mu }_{s}N$ for a body that is not on the verge of slipping is wrong in a way that produces plausible nonsense. A 10 kg crate that nobody is pushing has zero friction on it, not ${\mu }_{s}mg$.

Both rules are fits to data, not laws, and their strangest feature is what is missing: the contact area. A brick slides no more easily on its side than on its end. The accepted explanation is that surfaces touch only at microscopic asperities whose true contact area is a tiny fraction of the apparent area and grows in proportion to the load, so the two effects cancel. That explanation also predicts where the rules fail, and they do fail: for very light loads, for very clean surfaces in vacuum, which can weld, and for polymers, where contact area does matter and $f$ is not proportional to $N$. Rubber on dry road has ${\mu }_s$ near 1.0, steel on steel about 0.6, ice on ice about 0.1, and PTFE on PTFE about 0.04. Kinetic values run slightly below static ones, which is why a stuck drawer jerks free.

The braking distance of the previous lessons can now be predicted rather than assumed. With a locked wheel, the deceleration is ${\mu }_{k}g$, so stopping from 28 m s⁻¹ on dry road with ${\mu }_k=0.80$ needs $v^2/(2{\mu }_{k}g)=784/15.70=49.9$ m. On ice with ${\mu }_k=0.15$ the same stop needs 266 m, more than five times as far, and no amount of care by the driver alters that number.

The inclined plane

Put a block of mass $m$ on a slope at angle $\theta$ and choose axes along and perpendicular to the surface, which is the choice that makes the acceleration lie on one axis. Gravity, of magnitude $mg$ straight down, resolves into $mg\sin \theta$ down the slope and $mg\cos \theta$ into it. Perpendicular to the slope there is no acceleration, so $N=mg\cos \theta$. Along the slope, with friction opposing the sliding,

and the mass cancels: $a=g(\sin \theta -{\mu }_k\cos \theta )$. Heavy and light blocks of the same material slide alike, which is the same insensitivity to mass that makes all bodies fall at $g$.

Setting $a=0$ at the point of slipping gives the angle of repose, $\tan \theta ={\mu }_s$, and this is how coefficients are measured in the simplest laboratory: tilt the plane until the block moves and take the tangent. For ${\mu }_s=0.30$ that angle is 16.7°, and for 0.60 it is 31.0°, which is roughly the steepest slope a pile of dry sand will hold.

Example. A block slides down a 25° slope with ${\mu }_k=0.30$. What is its acceleration?

$a=9.81(\sin {25}^{\circ }-0.30\cos {25}^{\circ })=9.81(0.4226-0.2719)=9.81\times 0.1507=1.48$ m s⁻². The friction has removed nearly two thirds of the driving component. Had ${\mu }_k$ been 0.47 the bracket would have vanished and the block would have slid at constant speed.

Now you. A block slides down a 35° slope with ${\mu }_k=0.25$. What is its acceleration?

Connected bodies

When two bodies are joined, each gets its own free body diagram and its own equation, and the string supplies the link. The Atwood machine, two masses over a frictionless massless pulley, is the classic case, and George Atwood built it in 1784 precisely to slow gravity down to a speed his clocks could measure.

Take $m_2>m_1$, with $m_2$ descending. For $m_1$, taking up as positive, $T-m_{1}g=m_{1}a$. For $m_2$, taking down as positive, $m_{2}g-T=m_{2}a$, using the same $a$ because the string is inextensible. Add the two equations and $T$ vanishes:

Two checks say the algebra is right. With $m_1=m_2$ the acceleration is zero and $T=mg$, as it must be. With $m_1=0$ the acceleration is $g$ and the tension is zero, which is free fall. The device dilutes gravity by the ratio of the mass difference to the total, so nearly equal masses fall arbitrarily slowly while accelerating uniformly, and that is what made $g$ measurable in 1784.

Example. An Atwood machine carries 3.0 kg and 5.0 kg. Find the acceleration and the tension.

$a=(5.0-3.0)(9.81)/8.0=2.45$ m s⁻², and $T=2(3.0)(5.0)(9.81)/8.0=36.8$ N. Notice that the tension lies between the two weights, 29.4 N and 49.1 N, as it must: the string pulls the light mass up faster than gravity alone and holds the heavy one back.

Now you. An Atwood machine carries 2.0 kg and 6.0 kg. Find the acceleration and the tension.

Drag, and the first real differential equation

Everything above has constant forces, so the second law never had to be integrated. Air resistance breaks that, because the force depends on the speed, which depends on the force.

At low speeds, for small slow objects in viscous flow, drag is proportional to speed, $f=bv$, which is Stokes' law. At everyday speeds for everyday objects the flow is turbulent and drag goes as the square of the speed:

with $\rho$ the air density, 1.2 kg m⁻³ at sea level, $A$ the frontal area and $C_d$ a dimensionless drag coefficient, about 1.0 for a person, 0.3 for a car, 0.47 for a sphere. The crossover between the two regimes is governed by the Reynolds number, which is not needed here beyond knowing that the linear law is for dust and mist and the quadratic one for cars, balls and people.

Take the linear case, since it can be solved in closed form. A body falling from rest obeys

Separate the variables and integrate:

which gives $-(m/b)\ln [1-(b/mg)v]=t$, and rearranging,

Two features matter. The speed approaches a terminal speed $v_t=mg/b$ but never reaches it, because the exponential never vanishes; at $t=m/b$ the body has 63 per cent of it and at three times that, 95 per cent. And the terminal speed is where the drag exactly balances the weight, which can be read off the original equation by setting $dv/dt=0$ without solving anything. That shortcut works for any drag law, including the quadratic one that has no such tidy solution.

For quadratic drag the balance gives

Example. A skydiver of mass 80 kg falls head down with a frontal area of 0.70 m² and $C_d=1.0$. What is the terminal speed?

$v_t=\sqrt{2(80)(9.81)/(1.2\times 1.0\times 0.70)}=\sqrt{1569.6/0.84}=\sqrt{1868.6}=43.2$ m s⁻¹, about 156 km/h. The dependence on the square root of everything is why the answer is so insensitive: doubling the mass raises the terminal speed by only 41 per cent.

Now you. The same skydiver spreads out, raising the area to 1.0 m² and $C_d$ to 1.2. What is the new terminal speed?

The formula is honest about its own limits. It assumes constant air density, which is why Felix Baumgartner, jumping from 39 km in 2012 where the air is about one per cent as dense, passed 377 m s⁻¹ and went supersonic before the thickening air slowed him to an ordinary terminal speed lower down.

Where this method runs out

The free body method solves any problem in which the forces are known at every instant. That covers a great deal, but it has a structural weakness that the next lessons attack from two directions.

The first is that a body moving on a curved path has an acceleration that is not simply along one axis, and the natural question becomes what force is required to produce a given turn rather than what acceleration follows from a given force. Circular motion is where that inversion is worked out, and it is next.

The second weakness is deeper. Solving $m{d}^{2}x/d{t}^2=F$ requires knowing $F$ as a function of time, and the interesting forces are known as functions of position instead: a spring depends on its extension, gravity on the separation. The whole apparatus of energy exists because there is a way to integrate over position rather than time, and it answers questions about speed and place without ever finding out when.

A body going round a circle at constant speed is accelerating towards the centre at $v^2/r$, and since acceleration requires force, something real has to supply it, which turns a geometric fact into an engineering constraint.

Two results are carried in from earlier lessons: the centripetal acceleration of a body moving at speed $v$ on a circle of radius $r$ has magnitude $v^2/r$ and points at the centre, and the net force on a body of mass $m$ equals $m$ times its acceleration.

Inverting the question

Every problem so far has run forwards: given the forces, find the motion. Circular motion is naturally run backwards, because the path is usually known in advance. A car must follow the road, a satellite must stay in its orbit, a conker on a string must go round in a circle because the string is that long. The path fixes the acceleration, and the second law then says what the forces must add up to:

That is the entire content of this lesson. The equation is not a new law and $m{v}^2/r$ is not a new force: it is the right hand side of $\sum \overrightarrow{F}=m\overrightarrow{a}$ with the known acceleration substituted in. The skill is entirely in the free body diagram, and the discipline is to write only forces with named agents on the left and put $m{v}^2/r$ on the right, never both on the same side.

The pattern is worth noticing before the examples. The required force grows with the square of the speed and falls with the radius, so the dangerous combination is fast and tight. Real forces have limits, and where the demand exceeds the supply, the body does not go round. It leaves.

The flat curve, and the limit of friction

A car of mass $m$ takes a level bend of radius $r$ at speed $v$. Draw its free body diagram: weight $mg$ down, normal force $N$ up, and friction from the road, horizontally. Nothing else touches it.

Vertically there is no acceleration, so $N=mg$. Horizontally the only force available is friction, and it must supply the whole centripetal requirement: $f=m{v}^2/r$. Notice that this is static friction, not kinetic, because a rolling tyre is not sliding across the road: the contact patch is momentarily at rest. Static friction is capped at ${\mu }_{s}N={\mu }_{s}mg$, so

The mass cancels, so a loaded lorry and an empty one have the same cornering limit, at least until tyre behaviour stops being ideal. Above $v_{\max }$ the friction available is less than the friction required, and the car travels on a path of larger radius than the bend, which is to say it runs wide off the outside of the corner. It is not "thrown outward": nothing pushes it out, and what actually happens is that the inward force ran out and the car went comparatively straight.

Example. A bend of radius 60 m has ${\mu }_s=0.80$ in the dry. What is the fastest speed a car can take it?

$v_{\max }=\sqrt{0.80\times 9.81\times 60}=\sqrt{470.9}=21.7$ m s⁻¹, which is 78 km/h. The square root is unforgiving in both directions: halving the grip costs only 29 per cent of the speed, but a corner posted for 78 km/h taken at 95 km/h needs a coefficient of 1.2, which no ordinary tyre and road can deliver.

Now you. The same road is wet, with ${\mu }_s=0.35$, on a bend of radius 120 m. What is the fastest safe speed?

Banking, which removes the need for friction

Tilt the road inward by an angle $\theta$ and the normal force, which is perpendicular to the surface, acquires a horizontal component pointing at the centre of the bend. That component can do the whole job.

Assume no friction at all, which is the design case. The forces are $N$ perpendicular to the road and $mg$ down. Resolve into horizontal and vertical, not along the slope, because the acceleration is horizontal. Vertically there is no acceleration, so $N\cos \theta =mg$. Horizontally, $N\sin \theta =m{v}^2/r$. Divide the second by the first and $N$ and $m$ both disappear:

A banked curve has exactly one design speed, at which no friction is needed. Below it the car tends to slide down the bank and friction must hold it up; above it friction must hold it in. Real roads are banked for a chosen speed and rely on friction for the spread of speeds around it.

The numbers are checkable. Daytona's turns, banked at 31° with a radius of about 300 m, have a design speed of $\sqrt{300\times 9.81\times \tan {31}^{\circ }}=\sqrt{1768}=42.1$ m s⁻¹, or 151 km/h, and cars lap far faster than that, which is why they need enormous downforce and enormous tyres: everything above the design speed is paid for by friction. An ordinary motorway curve of radius 200 m intended for 30 m s⁻¹ would need $\tan \theta =900/(200\times 9.81)=0.459$, an angle of 24.6°, which is far steeper than any road is built, and so real motorway curves are much larger in radius instead.

The conical pendulum

Hang a mass on a string of length $L$ and swing it so that it travels in a horizontal circle with the string making a constant angle $\theta$ to the vertical. The string sweeps out a cone, and the analysis is the banked curve with tension in place of the normal force.

The radius of the circle is $r=L\sin \theta$. Vertically, $T\cos \theta =mg$; horizontally, $T\sin \theta =m{v}^2/r$. Dividing gives $\tan \theta =v^2/(rg)$ again, and substituting $r=L\sin \theta$ and $v=2\pi r/\tau$ for a period $\tau$ gives the tidy result

The tension is $T=mg/\cos \theta$, which exceeds the weight always and diverges as $\theta$ approaches 90°: a string can never be pulled horizontal by a mass on its end, however fast it is swung, because a horizontal string has no vertical component to hold the weight up.

That period formula is also the first appearance of a result the course returns to. As $\theta$ becomes small, $\cos \theta \to 1$ and $\tau \to 2\pi \sqrt{L/g}$, which is the period of an ordinary pendulum swinging back and forth. The two motions are the same motion seen from different sides, and the connection is made properly in the lesson on oscillations.

Example. A 0.25 kg bob on a 1.2 m string swings in a horizontal circle at 30° to the vertical. Find the radius, the speed, the period and the tension.

The radius is $1.2\sin {30}^{\circ }=0.600$ m. From $\tan \theta =v^2/(rg)$, $v=\sqrt{9.81\times 0.600\times \tan {30}^{\circ }}=\sqrt{3.399}=1.84$ m s⁻¹. The period is $2\pi \sqrt{1.2\cos {30}^{\circ }/9.81}=2\pi \sqrt{0.1059}=2.05$ s, which the check $\tau =2\pi r/v=3.770/1.843$ confirms. The tension is $mg/\cos {30}^{\circ }=(0.25)(9.81)/0.866=2.83$ N, some 15 per cent more than the bob's weight of 2.45 N.

Now you. A bob on a 0.80 m string swings in a horizontal circle at 40° to the vertical. Find the radius, the speed and the period.

The vertical loop

Now let the circle be vertical, so gravity is sometimes towards the centre and sometimes away from it. The speed is no longer constant, but at any instant the components of force towards the centre must still sum to $m{v}^2/r$.

At the top of the loop, both the weight and the track's push point downward, which is to say towards the centre:

so $N=m{v}^2/r-mg$. As the speed falls, $N$ falls, and it reaches zero when $v^2=gr$. Below that speed the equation demands a negative $N$, meaning the track would have to pull the car inward, which a track that only pushes cannot do. So the minimum speed at the top of a loop is

and it is independent of mass. For a loop of radius 8.0 m that is $\sqrt{9.81\times 8.0}=8.86$ m s⁻¹. At the bottom of the same loop, energy conservation, which the next lesson derives properly, gives $v_{\text{bot}}^2=v_{\text{top}}^2+4gr=5gr$, so $v_{\text{bot}}=19.8$ m s⁻¹, and there the normal force is $N=m{v}^2/r+mg=6mg$: the rider is pressed into the seat at six times their weight. That is why real roller coaster loops are not circles but clothoids, tightening as they rise, so that the radius is small where the speed is low and large where the speed is high, and the load on the rider stays near 4$g$ throughout instead of spiking at the bottom.

The same equation describes a bucket of water swung in a vertical circle. The water stays in because at the top the bucket's base is pushing it downward, adding to gravity to supply $m{v}^2/r$. For an arm and bucket of radius 1.0 m, the minimum speed is $\sqrt{9.81}=3.13$ m s⁻¹, a period of 2.0 s. Slower than that and the water leaves the bucket in a parabola, exactly as the lesson on projectiles said it would.

Example. A stone of mass 0.40 kg is whirled on a string in a vertical circle of radius 0.90 m. What is the minimum speed at the top, and what is the tension at the top if it moves at 5.0 m s⁻¹ there?

The minimum is $\sqrt{9.81\times 0.90}=2.97$ m s⁻¹. At 5.0 m s⁻¹, the required centripetal force is $m{v}^2/r=(0.40)(25)/0.90=11.1$ N, of which gravity supplies $mg=3.92$ N, so the string supplies $T=11.1-3.92=7.19$ N.

Now you. The same stone is at the bottom of the circle moving at 6.0 m s⁻¹. What is the tension there?

Centrifugal force, and what it really is

A passenger in a cornering car feels thrown against the door, and calls the sensation centrifugal force. In the road's frame there is no such force. What acts on the passenger is the door pushing them inward, supplying their share of $m{v}^2/r$; without the door they would carry on in a straight line, which relative to the turning car means moving outward. The feeling is the door, and the outward tendency is the absence of a force rather than the presence of one.

That said, the term is not simply an error. In a frame that is itself rotating, the first law fails, and the failure can be repaired by inventing forces. Add an outward $m{\omega }^{2}r$ on every body, plus a velocity dependent Coriolis term, and the second law works again inside the rotating frame. These are fictitious or inertial forces: they have no agent, no third law partner, and they vanish on returning to an inertial frame. They are also indispensable in practice, because meteorology, oceanography and long range gunnery are all done in the Earth's rotating frame and would be unmanageable otherwise.

The Earth supplies a measurable instance. The centrifugal effect of its spin at the equator is ${\omega }^{2}R=0.0339$ m s⁻², so a body there weighs about 0.35 per cent less than the same body at the pole, part of the reason $g$ measures 9.78 m s⁻² at the equator and 9.83 at the poles, the rest coming from the equatorial bulge that the same spin produced.

When the speed is not constant

Uniform circular motion is the special case where only the direction changes. In general a body on a curved path has both: an acceleration $v^2/r$ towards the centre, changing direction, and an acceleration $dv/dt$ along the path, changing speed. The two are perpendicular, so the total has magnitude $\sqrt{(v^2/r{)}^2+(dv/dt{)}^2}$, and a car accelerating out of a corner is using both at once. Since the tyre's total grip is limited by ${\mu }_{s}g$ however it is spent, spending some on speeding up leaves less for turning, which is the entire content of the friction circle that racing drivers work with.

Every problem in this lesson has been solved by knowing the force at each instant of time. That has been the method since the second law appeared, and it is about to become inadequate. Ask how fast a roller coaster is going at the top of a loop given its speed at the bottom, and the time it took is neither known nor wanted. Answering it requires integrating force over distance instead of over time, which produces a quantity that is conserved, and that is the next lesson.

Solving the second law requires knowing the force at every instant of time, and most interesting forces are known instead as functions of position, which is the gap this lesson closes.

What is carried in is $\sum \overrightarrow{F}=m\overrightarrow{a}$ and the identity from the first lesson that $a=vdv/dx$. Nothing else is assumed; energy is built here from scratch.

Integrating over distance

The first lesson noticed a trick and promised to come back to it. Write acceleration using the chain rule with position as the intermediate variable, $a=dv/dt=(dv/dx)(dx/dt)=vdv/dx$, so that $adx=vdv$. Multiply the second law by $dx$ and integrate from one position to another:

Time has vanished. What is left on the left is a new quantity, the integral of force over distance, called the work. What is left on the right is another new quantity, $\frac{1}{2}m{v}^2$, evaluated at the two ends, called the kinetic energy. The equation says that the work done equals the change in kinetic energy, and it is not an extra physical law: it is the second law integrated once, so anything it predicts could in principle be got the long way round.

The reason to bother is that the long way round is often impossible. Ask how fast a roller coaster reaches the bottom of a drop and the answer follows in one line; ask for the same thing by integrating the second law and you need the shape of the track, the time of every instant, and a numerical solver.

Work, and what the angle does

For a constant force $\overrightarrow{F}$ moving its point of application through a displacement $\overrightarrow{s}$, the work is the product of the displacement with the component of the force along it:

where $\phi$ is the angle between the force and the displacement. This combination is common enough to have a name, the scalar product of the two vectors, written $\overrightarrow{F}\cdot \overrightarrow{s}$, and in components it is $F_x{s}_x+F_y{s}_y$. It turns two vectors into a single number, and that number has no direction: work is a scalar, and a body has no notion of which way its kinetic energy points. The unit is the joule, one newton metre, and 1 J = 1 kg m² s⁻².

The cosine carries three cases worth separating. When the force has a component along the motion, $\phi <{90}^{\circ }$ and the work is positive: energy goes in. When it opposes the motion, as friction always does, the work is negative: energy comes out. And when the force is perpendicular to the motion, the work is exactly zero.

That last case is not a curiosity, it is most of mechanics. The normal force on a block sliding along a floor does no work, because the floor pushes up while the block moves sideways. The tension in a string does no work on a conical pendulum bob. The gravitational force on a satellite in a circular orbit does no work, which is why its speed is constant. Static friction on the driving wheel of a car does no work either, since the contact patch is instantaneously at rest, and the energy that accelerates the car comes from the fuel rather than from the road.

Example. A 20 kg crate is dragged 12 m along a floor by a rope pulling at 50 N, 30° above the horizontal, against ${\mu }_k=0.15$. Find the work done by each force and the final speed, starting from rest.

The rope does $W=(50)(12)\cos {30}^{\circ }=519.6$ J. The normal force must be found before friction can be: vertically, $N+50\sin {30}^{\circ }=mg$, so $N=196.2-25.0=171.2$ N, less than the weight because the rope lifts as well as pulls. Friction is then $0.15\times 171.2=25.68$ N opposing the motion, doing $-25.68\times 12=-308.2$ J. Gravity and the normal force do no work, being perpendicular to the displacement. The net work is $519.6-308.2=211.5$ J, so $\frac{1}{2}(20)v^2=211.5$ and $v=4.60$ m s⁻¹.

Now you. The same crate is dragged 6.0 m by an 80 N rope at 25° above the horizontal, with the same ${\mu }_k=0.15$. How much work does the rope do, and what is the work done against friction?

The work energy theorem

Collecting the result and stating it as a theorem: for any body, the total work done by all forces acting on it equals the change in its kinetic energy.

Kinetic energy is never negative and does not depend on the direction of travel, which is both the strength and the weakness of the method. It is a strength because a curved, complicated path can be handled without tracking the direction at every point. It is a weakness because one scalar equation cannot determine a two dimensional velocity: the theorem gives the speed at the bottom of a slide and says nothing about which way the rider is facing.

The dependence on $v^2$ is where the practical bite is. A 1500 kg car at 28 m s⁻¹ carries $\frac{1}{2}(1500)(784)=588$ kJ. Stopping it in 50 m requires the brakes to do $-588$ kJ of work, an average force of 11.8 kN, which is exactly what a coefficient of friction of 0.80 supplies, since $0.80\times 1500\times 9.81=11.8$ kN. Double the speed and the energy quadruples, so the same brakes need four times the distance. Every argument about speed limits and stopping distances is this quadratic, and it is worth noting that the energy has to go somewhere: 588 kJ deposited into brake discs is enough to raise 2 kg of steel by roughly 600 K, which is why brakes fade on long descents.

Variable forces, and the spring

The theorem was derived as an integral, so a force that varies with position costs nothing extra. The standard case is a spring. Robert Hooke published the relation in 1678 as the anagram ceiiinosssttuv, unscrambled as ut tensio sic vis, as the extension so the force:

with $x$ the displacement from the natural length and $k$ the stiffness in newtons per metre. The minus sign says the force opposes the displacement, which is what makes a spring restore rather than run away. To stretch a spring you must pull with $+kx$, so the work you do is

That is the area of a triangle under the straight line $F=kx$, which is a useful check: for a linear force law the work is the average force times the distance, and the average of a force rising from 0 to $kx$ is $\frac{1}{2}kx$.

Hooke's law is a first order approximation and fails for large extensions, where a real spring stiffens and then yields permanently. The reason it works at all for small ones is the same reason it will reappear in the lesson on oscillations: any smooth restoring force looks linear close enough to equilibrium.

Example. A spring of stiffness 800 N m⁻¹ is compressed 0.15 m and released against a 0.50 kg block on a frictionless surface. How fast does the block leave the spring?

The spring does $\frac{1}{2}k{x}^2=\frac{1}{2}(800)(0.0225)=9.0$ J of work on the block as it returns to its natural length. All of it becomes kinetic energy, so $\frac{1}{2}(0.50)v^2=9.0$ and $v=\sqrt{36}=6.0$ m s⁻¹.

Now you. A spring of stiffness 250 N m⁻¹ is stretched 0.20 m. How much work was done stretching it, and what speed would that give a 0.40 kg block?

Power

Work says how much energy moved; power says how fast. It is the rate of doing work,

measured in watts, one joule per second. The second form follows because $dW=\overrightarrow{F}\cdot d\overrightarrow{s}$ and $d\overrightarrow{s}/dt=\overrightarrow{v}$, and it is usually the more useful one, since it needs no integration.

James Watt, needing to sell engines to people who owned horses, defined the horsepower in the 1780s by measuring how much a mill horse could lift, and set it at 550 foot-pounds per second, which is 746 W. The figure is generous: a horse sustains rather less, and a fit human cyclist sustains about 200 to 300 W for an hour, with sprint peaks above 1000 W.

The $\overrightarrow{F}\cdot \overrightarrow{v}$ form explains the shape of a car's performance. A 100 kW engine at 30 m s⁻¹ can deliver at most $100000/30=3.3$ kN of force, and at 60 m s⁻¹ only 1.7 kN, which is why acceleration falls away at speed even with the throttle wide open. Top speed arrives when all the power is being spent against drag: with $C_{d}A=0.66$ m² and air at 1.2 kg m⁻³, the drag at 30 m s⁻¹ is $\frac{1}{2}(1.2)(0.66)(900)=356$ N, costing 10.7 kW. Since drag force goes as $v^2$, drag power goes as $v^3$, and doubling the top speed of a car requires eight times the power.

Example. A cyclist and machine of total mass 85 kg climb a 6 per cent gradient at a steady 5.0 m s⁻¹. What power is needed against gravity alone?

A 6 per cent gradient rises 6 m in 100 m along the road, so $\sin \theta =0.0599$ once the angle is worked out from $\tan \theta =0.06$. At steady speed the rider's force along the road matches the gravitational component $mg\sin \theta =(85)(9.81)(0.0599)=49.9$ N, and the power is $49.9\times 5.0=250$ W. That is a hard but sustainable effort for a trained rider, and it explains why hills are where cycling races are decided: on the flat the same 250 W buys a much higher speed, because it is fighting drag rather than gravity.

Now you. A lift of total mass 1200 kg rises at a constant 2.0 m s⁻¹. What power must the motor supply, ignoring any counterweight and friction?

What the theorem cannot do

The work energy theorem holds for every force, including friction, and that generality hides a distinction which the next lesson makes central.

Slide a block round a closed loop on a rough table and return it to its start. Gravity did no net work, since the height did not change. The normal force did none, being perpendicular throughout. Friction, though, did negative work on every centimetre of the journey, because it always opposes the motion, and over the whole loop it took out an amount proportional to the path length. Take a longer route and it takes out more.

So some forces have the property that the work they do between two points is the same for every route, and others do not. That distinction is not a technicality: it is the difference between a force for which a potential energy can be defined and one for which it cannot, and therefore between energy that can be recovered and energy that is gone. Where the energy goes when friction takes it is a question that mechanics cannot answer at all and thermodynamics was invented to answer.

The next lesson develops the recoverable case and gets a conserved total energy out of it, which turns out to be the single most useful equation in the subject.

Friction takes more energy out of a body the longer the route it travels, while gravity takes the same amount whatever route is taken between two heights, and that difference is what makes energy conservation possible.

Carried in from the previous lesson: work is the integral of force over distance, the net work on a body equals its change in kinetic energy $\frac{1}{2}m{v}^2$, and stretching a spring by $x$ takes $\frac{1}{2}k{x}^2$ of work.

The test that divides forces in two

Take a body from point A to point B by two different routes and compute the work done by one particular force on each route. For some forces the two answers agree, always, for every pair of routes. Such a force is called conservative, and an equivalent statement is that the work it does round any closed loop is zero.

Gravity near the ground passes the test immediately. The force is $mg$ downward, so the work it does is $-mg$ times the height gained, whatever the horizontal wandering, because horizontal displacement is perpendicular to the force and contributes nothing. Walk to the top of a hill by the steep path or the long zigzag and gravity has taken the same $mgh$ from you. An ideal spring passes too, since its work depends only on the extensions at the two ends.

Friction fails, and fails structurally rather than by accident. Kinetic friction always opposes the motion, so its work is negative on every segment of every path and can never cancel on the way back. Round a closed loop it gives $-{\mu }_{k}NL$ where $L$ is the total path length, which is not zero and grows without limit as the path is made longer. Air resistance fails for the same reason.

The distinction is the whole lesson. For a conservative force, the work done between two points is a property of the two points, so it can be tabulated in advance as a function of position and looked up rather than integrated. For a non-conservative force it cannot, because there is no such function to tabulate.

Potential energy

For a conservative force, define the potential energy $U$ so that the work the force does going from 1 to 2 equals the drop in $U$:

Equivalently, $U(x)=-\int Fdx$, and differentiating back,

Force is minus the slope of the potential energy curve. That single relation replaces a great deal of reasoning about directions: a body is pushed towards lower potential energy, downhill on the graph, and the steeper the graph the harder the push.

Two instances cover almost everything in this course. Near the ground, $F=-mg$ taking up as positive, so $U=mgy$, with the zero of height chosen wherever it is convenient. For a spring, $F=-kx$ gives $U=\frac{1}{2}k{x}^2$, measured from the natural length.

The arbitrary constant is not a defect. Only differences in $U$ appear in any physical result, so choosing the floor, the table top or sea level as the zero changes every value of $U$ and no answer. It is also worth being careful about where the energy lives. Potential energy is not stored in the body; it belongs to the configuration of the interacting system, the book and the Earth together. Saying that a raised book has potential energy is shorthand, and the shorthand becomes misleading in the lesson on gravitation, where the two bodies are of comparable importance.

Conservation of mechanical energy

Now put the two halves together. The work energy theorem says $W_{\text{net}}=\Delta K$. If every force doing work is conservative, then $W_{\text{net}}=-\Delta U$, so $\Delta K+\Delta U=0$, which means the sum

does not change as the body moves. This is conservation of mechanical energy, and it is a consequence of the second law rather than an addition to it.

Its practical value is that it relates speed to position directly, skipping the trajectory. A ball dropped 40 m arrives at $\sqrt{2gh}=\sqrt{2(9.81)(40)}=28.0$ m s⁻¹, and so does a ball that slid down a frictionless curved chute of the same height, or one that swung down on a string, because the equation contains only the height. That indifference to the path is what makes it powerful, and it is the same indifference that makes it silent about direction and about time.

Example. A roller coaster car starts from rest at the top of a 40 m drop and runs on a frictionless track into a vertical loop of radius 12 m. How fast is it at the bottom, how fast at the top of the loop, and does it make it round?

Take the bottom of the track as the zero of height. From the start to the bottom, $mgh=\frac{1}{2}m{v}^2$ gives $v=28.0$ m s⁻¹, independent of mass, which is why coasters do not need to be weighed. At the top of the loop the car is $2\times 12=24$ m up, so $\frac{1}{2}m{v}^2=mg(40-24)$ and $v=\sqrt{2(9.81)(16)}=17.7$ m s⁻¹. The previous lesson's condition for staying on the track is $v\ge \sqrt{gr}=\sqrt{9.81\times 12}=10.85$ m s⁻¹, so 17.7 is comfortable. Note that the two lessons were needed together: energy gave the speed, and circular dynamics said what speed was enough.

Now you. The same car starts from rest at the top of a 25 m drop, on a frictionless track. What is its speed at the bottom, and could it get round a loop of radius 12 m?

Energy diagrams

Draw $U(x)$ against $x$ and add a horizontal line at the total energy $E$. Since $K=E-U$ and kinetic energy cannot be negative, the body is confined to the regions where the curve lies below the line, and the points where the curve meets the line are turning points, where the speed is zero and the motion reverses.

The picture answers questions that would otherwise take a calculation. A dip in $U$ with the line cutting both sides gives motion trapped between two turning points, which is bound: a pendulum, a mass on a spring, a planet in orbit. Raise the line above the lip on one side and the body escapes that way, which is how escape velocity is defined in a later lesson. Where the curve is flat, $dU/dx=0$ and the force is zero, which is equilibrium: stable at a minimum, since a displacement produces a force back towards it, and unstable at a maximum, since a displacement produces a force away. A ball in a valley and a ball balanced on a hilltop both have zero force on them, and the second derivative is what tells them apart.

That last observation is the seed of a later lesson. Near a minimum, every smooth potential looks like a parabola, and a parabolic potential is exactly a spring, so every stable system oscillates the same way when disturbed gently. Reading it off the graph costs nothing; deriving it takes a Taylor expansion, which is done when the oscillation lesson arrives.

Example. A pendulum of length 2.0 m is released from rest at 40° to the vertical. How fast is the bob moving at the lowest point?

The tension does no work, being perpendicular to the motion throughout, so mechanical energy is conserved with gravity alone. The bob rises above its lowest point by $h=L(1-\cos \theta )=2.0(1-\cos {40}^{\circ })=2.0(1-0.766)=0.468$ m. Then $v=\sqrt{2gh}=\sqrt{2(9.81)(0.468)}=3.03$ m s⁻¹. Nothing about the swing's shape or duration entered.

Now you. A pendulum of length 1.5 m is released from rest at 60° to the vertical. Find the speed at the lowest point.

When friction is there anyway

Real problems have friction, and the method survives with one extra term. Split the forces into conservative and the rest, and the work energy theorem becomes

where $W_{\text{other}}$ is the work done by every non-conservative force, negative for friction and drag, positive for a motor or a person pushing. Mechanical energy is not conserved; it changes by exactly the work of the other forces, and no book-keeping is lost.

Example. A 2.0 kg block slides 5.0 m down a 30° incline with ${\mu }_k=0.25$, starting from rest. How fast is it going at the bottom?

The height dropped is $5.0\sin {30}^{\circ }=2.50$ m, so gravity releases $mgh=(2.0)(9.81)(2.50)=49.05$ J. The normal force is $mg\cos {30}^{\circ }=16.99$ N, so friction is 4.248 N and removes $4.248\times 5.0=21.24$ J over the slide. What is left is $49.05-21.24=27.81$ J of kinetic energy, giving $v=\sqrt{2(27.81)/2.0}=5.27$ m s⁻¹, against the 7.0 m s⁻¹ a frictionless slide would have given.

Now you. A 3.0 kg block slides 4.0 m down a 25° incline with ${\mu }_k=0.35$, from rest. Find its speed at the bottom.

Where the missing energy goes

The 21.24 J that friction removed did not cease to exist. It went into heating the block and the incline, and mechanics has nothing to say about that, because a temperature is not a mechanical variable and the sliding surfaces are not a point particle.

This is the honest boundary of the subject. Mechanical energy is conserved only when the non-conservative forces do no work; total energy is conserved always, but proving that requires counting the energy stored in the disordered motion of enormous numbers of molecules, which is where thermodynamics begins. The joule is the same unit in both subjects, and the reason the two fields use it is Joule's discovery, in the 1840s, that a fixed amount of mechanical work always produces a fixed amount of heating: 4.18 kJ per kilogram of water per kelvin, measured by letting falling weights turn a paddle in an insulated tank.

There is also an asymmetry worth stating plainly. The kinetic energy of an ordered stream of molecules can be turned entirely into disordered motion, as friction does effortlessly; the reverse conversion is limited by a law that mechanics does not contain. Classical mechanics is time reversible, and a film of a block sliding to a halt run backwards shows a block spontaneously cooling and accelerating, which no one has ever seen. The equations of this course permit it.

What one scalar equation cannot do

Energy conservation gives a single equation relating speed and position, and one equation cannot determine a velocity in two dimensions. That is a real limitation, and the case where it bites is the collision.

Two pucks meet on ice and separate. Energy conservation, if it held, would give one relation between the two outgoing speeds, leaving the directions undetermined. Worse, it usually does not hold: a collision that dents, deforms or sticks converts mechanical energy into heat and sound, and there is no way to know in advance how much.

What survives is a different quantity, and it survives precisely because the third law makes the internal forces cancel in pairs. It is a vector, so it carries direction, and it is conserved in every collision whether energy is or not. That quantity is momentum, and it is next.

Energy conservation is one scalar equation and a collision has more unknowns than that, which is why the subject needs a conserved quantity that carries direction.

Two results are carried in. Newton's second law in its original form says $\sum \overrightarrow{F}=d\overrightarrow{p}/dt$ with $\overrightarrow{p}=m\overrightarrow{v}$, and the third law says that when two bodies interact the forces on them are equal and opposite at every instant.

Impulse: integrating over time

The previous two lessons integrated the second law over distance. Integrating it over time instead gives something simpler and just as useful. From $\overrightarrow{F}=d\overrightarrow{p}/dt$,

The integral is the impulse, measured in newton seconds, which are the same as kg m s⁻¹. The statement is that the impulse delivered equals the change in momentum, exactly, whatever shape the force has in between.

That last clause is the point. A collision involves a force that rises from nothing to an enormous peak and falls back within milliseconds, and nobody knows its detailed shape. The impulse does not need it: only the area under the curve matters, and the area is fixed by the momentum change, which is measurable at leisure before and after. Where a shape is wanted, the useful summary is the average force, $\bar{F}=\Delta p/\Delta t$.

The engineering consequence is that a required momentum change can be bought either with a large force over a short time or a small force over a long one, and safety design is entirely the second option. A 70 kg driver at 15 m s⁻¹ has 1050 kg m s⁻¹ of momentum to lose. Stopped by a rigid steering column in 0.10 s, the average force is 10.5 kN, roughly fifteen times body weight and enough to break a sternum. Stopped by an airbag and a crumple zone in 0.50 s, it is 2.1 kN. Nothing about the momentum changed; the collision was merely made longer.

Example. A 0.145 kg baseball arrives at 40 m s⁻¹ and leaves at 50 m s⁻¹ in the opposite direction. What impulse did the bat deliver, and what average force acted if the contact lasted 0.70 ms?

Take the outgoing direction as positive, so the incoming momentum is $-(0.145)(40)=-5.80$ kg m s⁻¹ and the outgoing is $+(0.145)(50)=+7.25$. The impulse is the difference, $+13.05$ N s, and the sign convention has done real work: the speeds add rather than subtract because the direction reversed. The average force is $13.05/0.00070=18.6$ kN, about two and a half tonnes weight, applied for less than a thousandth of a second.

Now you. A 0.058 kg tennis ball arrives at 30 m s⁻¹ and is returned at 45 m s⁻¹ the other way, with 5.0 ms of contact. Find the impulse and the average force.

Conservation, and why it is the third law in disguise

Take two bodies that interact only with each other. By the third law, ${\overrightarrow{F}}_{12}=-{\overrightarrow{F}}_{21}$ at every instant, so

and the total momentum ${\overrightarrow{p}}_1+{\overrightarrow{p}}_2$ has zero derivative. The total momentum of a system is constant whenever the net external force on it is zero. Internal forces, however violent, cancel in pairs and cannot shift the total.

Three features make this more useful than energy conservation in a collision. It is a vector statement, so it is really one conservation law per axis, and each can be applied separately. It holds regardless of what the internal forces are, so nothing need be known about how the bodies deform. And it holds even when mechanical energy does not, because the third law says nothing about energy.

Two qualifications keep it honest. External forces do have to be absent, or at least negligible: gravity acts throughout a collision, but over a millisecond it changes the momentum by an amount too small to matter against the collision forces, which is why momentum is treated as conserved during an impact and not afterwards. And momentum is conserved along an axis only if the external force along that axis vanishes: two cars colliding on a road conserve horizontal momentum while the ground supplies whatever vertical impulse it likes.

Collisions, and the two extreme kinds

A collision is any brief interaction in which the internal forces are much larger than the external ones. Momentum is conserved in all of them; kinetic energy is not.

A perfectly inelastic collision is one in which the bodies move off together, and it is the case that loses the most energy consistent with conserving momentum. With $m_1$ at $u_1$ striking a stationary $m_2$:

Example. A 1200 kg car at 20 m s⁻¹ runs into a stationary 900 kg car and the two lock together. Find the common speed and the fraction of kinetic energy lost.

$v=(1200)(20)/2100=11.43$ m s⁻¹. The kinetic energy before is $\frac{1}{2}(1200)(400)=240$ kJ and after is $\frac{1}{2}(2100)({11.43}^2)=137$ kJ, so 103 kJ, or 42.9 per cent, has gone into crushing metal, heat and noise. That lost energy is not a defect of the calculation: it is the design intent, since a car that bounced off elastically would deliver a far larger impulse to its occupants.

Now you. A 0.40 kg lump of putty at 6.0 m s⁻¹ hits a stationary 1.6 kg block and sticks. Find the common speed and the energy lost.

An elastic collision is one in which kinetic energy is also conserved. No macroscopic collision is exactly elastic, but hard steel balls, billiard balls and gas molecules come close, and collisions between subatomic particles can be exactly so.

Solving the elastic collision

Take $m_1$ moving at $u_1$ into $m_2$ at rest, in one dimension. Two conservation laws give two equations:

Solving these directly involves a quadratic and a wrong root. The elegant route is to rearrange each as a difference: $m_1(u_1-v_1)=m_2{v}_2$ from the first, and $m_1(u_1^2-v_1^2)=m_2{v}_2^2$ from the second. Dividing the second by the first and using the difference of two squares gives $u_1+v_1=v_2$, which rearranges to

once a general $u_2$ is carried through. In an elastic collision the relative speed of separation equals the relative speed of approach. That is one linear equation replacing the quadratic, and with the momentum equation it solves in two lines:

The limits are worth reading off. Equal masses give $v_1=0$ and $v_2=u_1$: the incoming body stops dead and the target leaves with the whole velocity, which is the shot every snooker player learns first. A light body hitting a much heavier one gives $v_1\to -u_1$ and $v_2\to 0$: it bounces back at nearly its original speed, which is why a ball rebounds from a wall. A heavy body hitting a light one gives $v_1\to u_1$ and $v_2\to 2u_1$: the projectile is barely slowed and the target leaves at twice the incoming speed, which is how a golf club moving at 50 m s⁻¹ sends a ball off at nearly 70.

That middle case is also why nuclear reactors are moderated with light nuclei. A neutron loses the largest fraction of its energy to a target of its own mass, which is hydrogen, so water and graphite slow neutrons in a few dozen collisions while lead would take thousands.

Example. A 0.50 kg ball moving at 4.0 m s⁻¹ collides elastically with a stationary 0.30 kg ball. Find both final velocities and check the energy.

$v_1=(0.50-0.30)/(0.80)\times 4.0=1.00$ m s⁻¹ and $v_2=2(0.50)/(0.80)\times 4.0=5.00$ m s⁻¹, both forward. Momentum: before $2.00$, after $(0.50)(1.00)+(0.30)(5.00)=2.00$ kg m s⁻¹. Energy: before $\frac{1}{2}(0.50)(16)=4.00$ J, after $0.25+3.75=4.00$ J. And the separation speed, $5.00-1.00=4.00$ m s⁻¹, equals the approach speed, as it must.

Now you. A 2.0 kg body at 3.0 m s⁻¹ collides elastically with a stationary 6.0 kg body. Find both final velocities.

Two dimensions, and the missing equation

In a plane, momentum conservation supplies two equations, one per axis, and the unknowns are two outgoing speeds and two outgoing directions: four unknowns, two equations. Even adding energy conservation for an elastic collision leaves one short. The missing information is physical, not mathematical: it is the impact parameter, how squarely the bodies hit, and no conservation law can supply it. In practice one outgoing direction is measured and the rest follows.

One result does survive without that measurement. For an elastic collision between equal masses with one initially at rest, the two bodies always separate at 90° to each other. The proof is short: momentum gives $\overrightarrow{u}={\overrightarrow{v}}_1+{\overrightarrow{v}}_2$, and squaring gives $u^2=v_1^2+v_2^2+2{\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2$, while energy for equal masses gives $u^2=v_1^2+v_2^2$. So ${\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2=0$, which for two non-zero velocities means perpendicular. Snooker players rely on it constantly, and it visibly fails when the balls have spin or the collision is not quite elastic.

The ballistic pendulum

The instrument that made this lesson practical was Benjamin Robins' ballistic pendulum of 1742, the first device that could measure the speed of a musket ball. A bullet is fired into a heavy block hanging on a cord; the block swings up by a measured height; the bullet's speed is inferred.

The critical point, and the reason the device is a teaching classic, is that the problem has two stages governed by different laws, and using the wrong one in either stage gives an answer wrong by a factor of hundreds.

Stage one is the embedding, which takes about a millisecond. It is perfectly inelastic, so momentum is conserved and kinetic energy is emphatically not: $mu=(m+M)v$.

Stage two is the swing, which takes about a second. Now the only forces doing work are gravity and the cord tension, both conservative or workless, so mechanical energy is conserved: $\frac{1}{2}(m+M)v^2=(m+M)gh$, giving $v=\sqrt{2gh}$.

Combining, $u=\frac{m+M}{m}\sqrt{2gh}$.

Example. A 10 g bullet is fired into a 2.00 kg block and the block rises 0.202 m. What was the bullet's speed, and what fraction of the kinetic energy survived the impact?

From the swing, $v=\sqrt{2(9.81)(0.202)}=1.99$ m s⁻¹. From the embedding, $u=(2.010/0.010)(1.99)=400$ m s⁻¹, a plausible musket velocity. The energy before is $\frac{1}{2}(0.010)({400}^2)=800$ J and after is $\frac{1}{2}(2.010)({1.99}^2)=3.98$ J, so 99.5 per cent was lost in the first millisecond. Anyone who had used energy conservation across the impact would have obtained a bullet speed of about 28 m s⁻¹, low by a factor of fourteen.

Now you. An 8.0 g bullet is fired into a 1.50 kg block, which rises 0.176 m. What was the bullet's speed?

What is still assumed

Every body in this lesson has been treated as a point. A car has been a point, a block has been a point, and the collision has been an event with no spatial extent. That was never justified, and it is plainly false: real bodies spin, deform, and have their mass spread over metres.

The justification exists and is worth having, because it is what licenses eight lessons of point particles retroactively. For any system of particles, however complicated, there is one special point that moves exactly as a single particle would under the external forces alone. Finding it, proving that claim and then using it on a body that throws its own mass away is the next lesson.

Every calculation so far has treated cars, blocks and planets as points, which is obviously false, and this lesson supplies the theorem that makes it legitimate anyway.

Carried in: the second law in the form $\sum \overrightarrow{F}=d\overrightarrow{p}/dt$, the third law, and the fact that the total momentum of a system is unchanged by internal forces.

A system of particles

Take $N$ particles with masses $m_i$ at positions ${\overrightarrow{r}}_i$, total mass $M=\sum m_i$. Define the centre of mass as the mass weighted average position:

In components, $X=\sum m_i{x}_i/M$ and likewise for $Y$. Nothing physical has been claimed yet; this is a definition, and its justification is what follows from it.

Differentiate twice with respect to time. Since the masses are constant,

The last sum is over every force on every particle, and it splits into internal forces, which particles exert on each other, and external ones. By the third law the internal forces occur in equal and opposite pairs, so they cancel exactly, leaving

This is the result the course has been assuming since its third lesson. The centre of mass of any system moves exactly as a single particle of the total mass would, driven by the external forces alone. The system may be a spinning wrench, an exploding shell, a crowd of people or a galaxy; the internal complexity is irrelevant to that one point.

It also says $\overrightarrow{P}=M{\overrightarrow{V}}_{\text{cm}}$, so the momentum of a whole system is the total mass times the velocity of its centre of mass, which is why momentum conservation is the statement that the centre of mass keeps moving uniformly when no external force acts.

Finding it

For a few discrete masses the definition is arithmetic. For a continuous body the sum becomes an integral,

and the work is in expressing $dm$ in terms of $dx$. For a uniform rod of length $L$ and mass $M$, $dm=(M/L)dx$ and the integral gives $X=L/2$, as symmetry already said. For a rod whose density grows linearly along its length, $dm=cxdx$, and

Symmetry is the shortcut worth taking whenever it is available: the centre of mass of a uniform body lies on every plane of symmetry it has, which locates it instantly for a sphere, a cube or a cylinder. It need not lie inside the body at all. The centre of mass of a ring is at its centre, where there is no material, and the centre of mass of a high jumper arched over the bar in a Fosbury flop passes underneath the bar while every part of the jumper passes over it, which is the entire reason the technique won the 1968 Olympics.

Example. Three particles lie in a plane: 2.0 kg at the origin, 3.0 kg at (4.0, 0) and 5.0 kg at (6.0, 3.0), in metres. Where is the centre of mass?

The total mass is 10.0 kg. Then $X=[2(0)+3(4)+5(6)]/10=42/10=4.20$ m and $Y=[2(0)+3(0)+5(3)]/10=15/10=1.50$ m. The centre of mass is at (4.20, 1.50) m, which is not at any particle and is pulled towards the 5.0 kg mass, as a weighted average should be.

Now you. Three particles: 1.0 kg at the origin, 2.0 kg at (3.0, 4.0) and 4.0 kg at (5.0, 0), in metres. Find the centre of mass.

What the theorem buys

Three familiar facts become one line each.

A hammer thrown spinning across a room follows a complicated path in every part except one: its centre of mass traces a clean parabola, because gravity is the only external force. Photographs of this are the standard demonstration, and they show a point moving smoothly through a body that is tumbling wildly around it.

A shell fired on a parabola and exploding in flight has fragments flying in all directions, but the explosion is internal, so the centre of mass of the fragments carries on along the original parabola until the first piece lands.

And the Earth does not orbit the Sun. Both orbit their common centre of mass, and the same is true of the Earth and the Moon: with $m_M/(m_E+m_M)=7.342\times {10}^{22}/6.045\times {10}^{24}$ and a separation of $3.844\times {10}^8$ m, the barycentre is $4.67\times {10}^6$ m from the Earth's centre, which is 1700 km beneath the surface. The Earth wobbles about that point once a month, and the same wobble applied to stars is one of the ways planets around other stars are detected.

Example. A shell is fired from level ground at 80 m s⁻¹ at 50° and explodes into two equal fragments at the top of its flight. One fragment drops vertically from rest and lands directly below the burst. Where does the other land?

The undisturbed range would be $R=v_0^2\sin 2\theta /g=6400\sin {100}^{\circ }/9.81=642.5$ m, and the burst is at the halfway point horizontally, 321.2 m from the launch. The explosion is internal, so the centre of mass continues on the original parabola and reaches 642.5 m at the moment the pieces land, both of which land together since they fall from the same height. With equal masses, the centre of mass is midway between them, so $(321.2+x)/2=642.5$, giving $x=963.7$ m.

Now you. The same experiment with a shell fired at 60 m s⁻¹ at 45°. Where does the second fragment land?