Unit 4 - Linear Momentum

1. Linear Momentum

Title

AP Physics C: Mechanics — Unit 4

Impulse, conservation, collisions, and the centre of mass — solving problems where the forces are unknown

2. What this deck buys you

Objectives

Energy came from integrating the second law over position. Momentum comes from integrating it over time, and the two tools answer complementary questions.

That difference matters most in a collision, where the forces are enormous, last a few milliseconds, and are completely unknown. Momentum lets you answer the question anyway.

• Use impulse as the area under a force-time graph, and as the change in momentum
• Know exactly when momentum is conserved — and that it does not require the collision to be elastic
• Classify and solve elastic, inelastic and perfectly inelastic collisions
• Handle two-dimensional collisions by conserving each component separately
• Locate a centre of mass by summation and by integration, and use its motion as a shortcut
• Recognise the variable-mass situations where naive use of F = ma fails

The organising idea of the deck: momentum is a vector and energy is a scalar. Momentum problems have components and signs; energy problems do not. Nearly every collision problem needs both, used for different parts of the question.

There is a neat symmetry worth noticing at the outset. Energy came from integrating Newton's second law over position. Momentum comes from integrating the same law over time. Two integrals of one equation, giving two tools that answer different questions.

The question momentum is uniquely good at is the collision. In a collision the forces are enormous, last a few milliseconds, and vary in a way nobody can measure. Force methods are hopeless. Energy methods often fail too, because energy disappears into heat and deformation.

Momentum survives all of that. It is conserved whether or not energy is, and it needs no knowledge of the forces at all — only the masses and the velocities before and after.

The organising contrast for the whole deck: momentum is a vector and energy is a scalar. That means momentum problems have components and signs, which is more bookkeeping, and it means momentum carries directional information that energy cannot. Most collision problems use both, for different parts of the question.

3. Momentum and Impulse

Section

Section 1

4. Momentum is mass times velocity — and it is a vector

Concept

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

Momentum points the same way the velocity does, and it carries the same sign. A ball moving left has negative momentum on a rightward axis, and that minus sign is not optional — it does all the work in collision problems.

Those two objects have the same momentum and very different kinetic energies — 12 J against 36 J. That asymmetry is exactly why momentum and energy give you two independent equations about a collision, and why you often need both.

The reason is structural: momentum is linear in v and kinetic energy is quadratic. Doubling the speed doubles one and quadruples the other, so they can never carry the same information.

The definition is trivially simple, and the word that matters is vector. Momentum points where the velocity points and carries the same sign, and in one-dimensional problems that sign does all the work.

The figure makes a point worth dwelling on. A six kilogram object at two metres per second and a two kilogram object at six metres per second have identical momentum — and kinetic energies of twelve and thirty-six joules respectively. Same momentum, three times the energy.

That asymmetry is not a curiosity; it is the reason both quantities are useful. Momentum is linear in velocity and kinetic energy is quadratic, so they can never encode the same information. Doubling the speed doubles one and quadruples the other.

Practically, this is why an elastic collision gives you two independent equations. If momentum and energy said the same thing, you would have one equation and could not solve for two unknown final velocities.

5. Newton actually wrote the second law this way

Concept

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

This is the more general form, and F = ma is the special case where the mass is constant.

\[ \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} = m\vec{a} + \vec{v}\frac{dm}{dt} \]

When the mass is constant the second term vanishes and you recover F = ma. When it is not — a rocket burning fuel, a hopper loading a moving truck, a raindrop growing as it falls — that second term is real and F = ma gives wrong answers.

So this is not a rewriting for elegance. It is the version that survives when mass changes, and the last section of this deck is about exactly those cases.

The form F equals ma that you have used for two decks is a special case, and it is worth knowing which case.

Expand the derivative of mv with the product rule and you get two terms: m dv/dt, which is ma, plus v dm/dt. The second term is zero whenever the mass is constant, which is nearly always, which is why F equals ma works so well.

But it is not always. A rocket burns fuel; a hopper drops sand onto a moving truck; a raindrop grows as it falls through cloud. In all of those, the mass genuinely changes and the second term is real.

In those cases F equals ma does not give a slightly wrong answer — it gives a wrong one. The last section of this deck is about handling them correctly, and this slide is why that section has to exist.

It is also worth knowing historically: this is closer to what Newton actually wrote. He spoke of the change in the quantity of motion, not of mass times acceleration.

6. Reading the impulse-momentum theorem

Notation

Derive it in one line from the second law: if F is dp/dt, then F dt is dp, and integrating both sides gives impulse equals change in momentum. There is nothing else in it.

The practical use is that you never need to know the force in detail. A collision force spikes to some enormous unknown value for a few milliseconds; the impulse — the whole area under that spike — is exactly the change in momentum, which you can measure from the before and after velocities.

The derivation fits on one line. If force is dp/dt, then F dt is dp, and integrating both sides gives the theorem. Nothing else is in it.

The middle annotation is the one to hold onto, because it is the whole architecture of the last two decks. Work integrates force over distance and gives energy. Impulse integrates force over time and gives momentum. Same law, two integrals, two complementary tools.

The practical value is that you never need to know the force in detail. During a collision, the force spikes to some enormous unknown value for a few milliseconds. You cannot measure it. But you can measure the velocities before and after, which gives you the change in momentum, which is the entire area under that unknown spike.

Watch the vector nature in the delta p. A ball that bounces back has a much larger momentum change than one that merely stops, because the final momentum is negative rather than zero. That factor of two appears in a prediction slide shortly.

7. Impulse is the area under a force-time graph

Concept

\[ J = \int F\,dt \quad \text{and, for a constant force,} \quad J = F\Delta t \]

The two graphs deliver the same impulse — the same change in momentum — with wildly different peak forces. That single observation is the physics behind airbags, crumple zones, helmets, gymnastics mats and catching a ball with your hands moving backwards.

The logic runs one way: the change in momentum is fixed by the situation — you are going from 30 m/s to zero either way. What you can control is the time, and stretching the time is what lowers the peak force.

\[ F_{\text{avg}} = \frac{\Delta p}{\Delta t} \]

The two graphs on this slide have the same area and therefore deliver the same impulse — the same change in momentum — with peak forces differing by a factor of four.

That single observation is the physics behind airbags, crumple zones, helmets, gymnastics mats, boxing gloves, and the way a fielder pulls their hands back while catching a ball.

The causal direction matters and is easy to get backwards. The change in momentum is fixed by the situation: you are going from thirty metres per second to zero either way, and nothing can alter that. What you can control is the time over which it happens, and stretching the time is what lowers the peak force.

So safety engineering never reduces the impulse. It cannot. It extends the duration, and the force falls because the impulse was already fixed.

The average-force relation at the bottom is the working form of this. Change in momentum divided by contact time gives the average force, and that is usually what a question is asking for.

8. Impulse from a force-time graph

Worked example

A 0.15 kg ball moving at 20 m/s strikes a wall. The force during contact rises linearly to 900 N over 3.0 ms, then falls linearly to zero over the next 4.0 ms. Find the ball's rebound speed.

\[ J = \tfrac{1}{2}(7.0\times 10^{-3})(900) = 3.15\,\text{N s} \]

\[ J = mv_f - mv_i \;\Rightarrow\; -3.15 = (0.15)v_f - (0.15)(20) \]

\[ v_f = \frac{-3.15 + 3.0}{0.15} = -1.0\,\text{m/s} \]

Notice what the calculation never needed: the shape of the force curve at any particular instant. Only the area mattered.

Two things make this problem representative. The first is that the impulse is just the area of a triangle — no calculus needed, because the graph is made of straight lines.

The second is the signs. The ball arrives moving in the positive direction and the wall pushes it back, so the impulse is negative. Getting that minus sign in is what produces a rebound rather than an absurd acceleration through the wall.

The result — rebounding at one metre per second after arriving at twenty — tells you this was a highly inelastic impact. Very little of the ball's speed survived. That is a physical conclusion drawn from the numbers, and it is worth stating rather than leaving implicit.

The check at the end is a nice one. For a triangular pulse, the average force is always exactly half the peak, because that is what the area of a triangle means. Getting 450 against a peak of 900 confirms the arithmetic in one step.

And notice what the calculation never needed: the force at any particular instant. Only the total area mattered, which is the whole point of working with impulse.

9. Impulse from a force that varies with time

Worked example

A 0.40 kg object at rest is acted on along the x-axis by a force F(t) = 12t - 3t^2 newtons, from t = 0 until the force returns to zero. Find its final speed.

\[ 12t - 3t^2 = 3t(4 - t) = 0 \;\Rightarrow\; t = 0 \;\text{and}\; t = 4.0\,\text{s} \]

\[ J = \int_0^{4} (12t - 3t^2)\,dt = \Big[6t^2 - t^3\Big]_0^{4} = 96 - 64 = 32\,\text{N s} \]

\[ v_f = \frac{J}{m} = \frac{32}{0.40} = 80\,\text{m/s} \]

The general habit: integrate the force first, divide by the mass last. The impulse is the physically meaningful intermediate quantity, and keeping it intact makes the arithmetic cleaner.

This is the calculus version of the previous problem, and the structure is identical: find the area, then divide by the mass.

The first step is easy to overlook. You have to find when the force switches off, and factoring the expression makes both roots visible immediately. Integrating over the wrong interval is the most likely error here.

The integration itself is elementary. What is worth noticing is the order of operations at the end: integrate the force first, divide by the mass last.

The alternative — dividing by the mass first to get the acceleration, then integrating that — gives the same answer and takes an extra step for no benefit. The impulse is the physically meaningful intermediate quantity, and keeping it intact keeps the arithmetic clean.

The sanity check with the average force is worth building as a habit. An average of eight against a peak of twelve, for a smooth humped curve, is entirely reasonable. An average larger than the peak would signal an error immediately.

10. A bigger impulse means a bigger force

Trap

11. Why an airbag helps and a brick wall does not

Real world

A driver at 15 m/s must reach zero. That change in momentum is fixed, whatever the interior of the car is made of.

This is worth working through with actual numbers, because the size of the effect is what makes it convincing.

A seventy kilogram driver going from fifteen metres per second to rest requires an impulse of one thousand and fifty newton-seconds. That number is fixed by physics and no design decision can change it.

Hitting a rigid dashboard stops the driver in perhaps a hundredth of a second, giving an average force above a hundred kilonewtons. An airbag stretches that to a tenth of a second, giving about ten. A factor of ten in force, which is the difference between a survivable impact and a fatal one.

The counter-intuitive part is the second question. A stronger, stiffer dashboard would be worse, because strength here means less deformation, which means a shorter stopping time, which means a larger force.

That is why crash safety is entirely about designing things to fail slowly rather than to hold firm. A crumple zone is a part of the car deliberately built to be weak, and understanding why requires exactly the reasoning on this slide.

12. Two ways to catch a ball

Prediction

This has a counter-intuitive practical consequence. A bouncy object exerts a larger force on what it hits than a squishy one that sticks. It is why hail damages a roof more than rain of the same mass and speed, and why a superball hurts more than a beanbag.

The factor of two here is worth being able to produce instantly, because it appears constantly.

Catching changes the momentum from p to zero, a change of magnitude p. Bouncing changes it from p to minus p, a change of magnitude two p. The reversal doubles the impulse, because the surface must first stop the ball and then send it back just as fast.

The practical consequence is genuinely surprising: a bouncy object exerts a larger force on what it hits than a squishy one that sticks.

That explains why hail damages a roof far more than rain of the same mass and speed — hail bounces, rain splats. It is why a superball hurts more than a beanbag of the same weight. And it is why hammer heads are hard and hammer handles are not.

It also matters in reverse. If you want to maximise the force you deliver — driving a nail, hitting a golf ball — you want the collision to be as elastic as possible. If you want to minimise it, you want the impact to stick.

13. Conservation of Momentum

Section

Section 2

14. Why momentum is conserved: the third law, integrated

Concept

\[ \vec{F}_{12} = -\vec{F}_{21} \;\Rightarrow\; \frac{d\vec{p}_1}{dt} = -\frac{d\vec{p}_2}{dt} \]

\[ \frac{d}{dt}(\vec{p}_1 + \vec{p}_2) = 0 \quad \Longrightarrow \quad \vec{p}_{\text{total}} = \text{constant} \]

So conservation of momentum is not an extra law. It is Newton's third law with the time integral taken, and it holds exactly whenever the only forces are internal to the system you chose.

\[ \vec{p}_{\text{total, before}} = \vec{p}_{\text{total, after}} \qquad \text{if} \quad \sum \vec{F}_{\text{external}} = 0 \]

The condition to watch is the external one. Internal forces — however violent — always cancel in pairs. Only external forces can change a system's total momentum.

Conservation of momentum is often presented as a separate principle, and it is not. It is Newton's third law with the time integral taken, and this derivation is two lines.

The forces two objects exert on each other are equal and opposite at every instant. So their rates of momentum change are equal and opposite at every instant. Add them and you get zero, which means the total momentum does not change.

Notice that nothing in that argument mentions energy, or whether the objects stick, or how hard they hit. That is why momentum conservation is so much more robust than energy conservation in collisions, and it is the point of the trap slide later in this deck.

The condition to watch is the external one. Internal forces always cancel in pairs, however violent. Only a force from outside the chosen system can change the total.

So conservation of momentum is a conditional statement, and the condition is about the system boundary — which is the subject of the very next slide.

15. Choosing the system is the whole skill

Concept

Whether momentum is conserved is not a property of the situation. It is a property of the boundary you drew, and drawing it well is what makes a problem easy.

The last two rows are worth reading twice. A falling ball's momentum is emphatically not conserved — it grows every second. Include the Earth and it is conserved exactly, because the Earth gains an equal and opposite momentum. It just has so much mass that its velocity change is unmeasurable.

Whether momentum is conserved is not a fact about the situation. It is a fact about the boundary you drew, and the same physical event can give either answer depending on where you draw it.

The figure shows this directly. Draw the boundary around both colliding objects and the collision forces are internal, so momentum is conserved. Draw it around one of them and that same force now crosses the boundary, so it is not.

The last two rows of the table are the ones worth reading twice. A falling ball's momentum is emphatically not conserved — it grows every second. Include the Earth in the system and it is conserved exactly: the Earth gains an equal and opposite momentum. It simply has so much mass that its velocity change is far too small to measure.

That is a good example of conservation laws being about the whole system rather than about the convenient part of it.

The strategy that follows is: draw the boundary so that the forces you do not know become internal. If a problem involves an unknown collision force, an explosion, or a push you cannot measure, put both objects inside and that force disappears from your equations.

16. Is momentum conserved in these situations?

Discrimination

For each scenario, decide whether the total momentum of the stated system is conserved during the stated interval.

Item e is the one worth internalising. During a collision or explosion, the internal forces are so enormous compared with gravity or friction that momentum is treated as conserved even when small external forces exist. The approximation is excellent and universally used.

Sorting these is quick. The value is in item e, which is the approximation every collision problem quietly relies on.

A firework exploding at the top of its arc has gravity acting on it, which is an external force. Strictly, momentum is not conserved. But the explosion lasts a few milliseconds, and gravity's impulse over a few milliseconds is utterly negligible compared with the impulse of the explosive.

So we treat momentum as conserved during the explosion, and it is an excellent approximation. The same reasoning licenses ignoring friction during a collision on a rough table, and ignoring air resistance during an impact.

The rule of thumb: during a collision or explosion, internal forces are enormous and the interval is tiny, so external forces contribute essentially nothing. Momentum is conserved across the collision itself even when it is not conserved over longer times.

The items in the no bucket all share a feature: an external force is genuinely acting over a substantial time. Enlarging the system to include the other object would restore conservation in every case.

17. In a crash, the heavier vehicle exerts the larger force

Trap

18. Momentum as a currency that cannot be printed

Picture it

The bookkeeping picture from the energy deck has a momentum version, and it makes the conservation rule feel less like a formula.

Momentum behaves like a currency inside a closed system. Objects hand it back and forth in exactly matched amounts — that is the third law — so the total inside the boundary never changes.

An external force is the only thing that can add currency to the system or take it out. That is why choosing the boundary is the whole skill: draw it so the transactions you cannot measure are all internal.

Where the analogy improves on the energy version is the direction. Momentum has a sign, so handing momentum to the left is the same as receiving momentum to the right. Energy has no such structure — which is why energy can be created inside a system, by an explosion, while momentum cannot.

The energy deck used a bank-balance analogy, and momentum has one too, with one important difference.

Inside a closed system, objects hand momentum back and forth in exactly matched amounts. That matching is the third law. So the total inside the boundary never changes, and only a force crossing the boundary can add or remove any.

Where the analogy improves on the energy version is that momentum has a sign. Handing momentum to the left is the same transaction as receiving momentum to the right, so the bookkeeping is genuinely double-entry.

And where the two differ is the crucial part. Energy can be created inside a system — an explosion converts chemical energy into kinetic energy without anything crossing the boundary. Momentum cannot. There is no internal process whatever that increases the total momentum of a system.

That is a much stronger statement than energy conservation, and it is why momentum is the reliable tool when you do not know what is going on inside.

19. The momentum recipe

Pattern

1. Draw before and after, as two separate sketches. This is the momentum equivalent of the free-body diagram and it is just as non-optional.
2. Choose the system so the forces you do not know become internal.
3. Choose an axis and mark the positive direction. Every velocity gets a sign from it.
4. Write the momentum sum before and after, and set them equal — once per dimension.
5. Check whether energy is also conserved. If the collision is elastic you get a second equation; if not, you do not, and you must not pretend otherwise.
6. Solve, then check the signs against the picture. Objects should not pass through each other.

\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]

Step three is where nearly every error in this unit lives. Momentum is a vector, and in one dimension that shows up entirely as signs. An object moving left gets a negative velocity, before and after.

Step six catches a specific class of error. If your answer has the trailing object moving faster than the one in front, they have passed through each other, and something is wrong.

Six steps, and the first and third are where the marks live.

Draw before and after as two separate sketches. This is the momentum equivalent of the free-body diagram, and it is just as non-optional. Trying to hold both states in your head is how velocities get mixed up.

Step three is where nearly every error in this unit originates. Momentum is a vector, and in one dimension that shows up entirely as signs. An object moving left gets a negative velocity — before and after — and the axis you chose decides which direction that is.

Step five is the branch point of the whole deck. Momentum is always available. The energy equation is available only if the collision is elastic, and using it when it is not will produce a confidently wrong answer.

Step six catches a specific and common failure. If your answer has the trailing object moving faster than the one in front, they have passed through each other, which is not something objects do. Check the signs before writing anything down.

20. Which system makes this easy?

Check

A 60 kg skater throws a 3.0 kg ball horizontally while standing at rest on frictionless ice. You want the skater's recoil speed.

The phrase to notice is without knowing the throwing force. That is the question telling you which system to choose.

Put both the skater and the ball inside the boundary and the throwing force becomes internal, cancelling with its third-law partner. The total momentum starts at zero and must remain zero, which gives the recoil speed immediately.

You never learn how hard the throw was, or how long it lasted, and you do not need to. That is the entire reason for choosing a system this way.

Option D is worth a comment because it is not wrong, merely wasteful. Adding the frictionless ice changes nothing, because the ice exerts no horizontal force. But adding irrelevant objects to a system is a reliable way to confuse yourself, so keep the boundary as tight as it can be while still enclosing the unknown forces.

21. Recoil: a rifle and a bullet

Worked example

A 4.0 kg rifle fires a 12 g bullet at 380 m/s. Find the recoil speed of the rifle, and compare the kinetic energies.

\[ 0 = m_b v_b + m_r v_r \]

\[ v_r = -\frac{m_b v_b}{m_r} = -\frac{(0.012)(380)}{4.0} = -1.14\,\text{m/s} \]

\[ K_b = \tfrac{1}{2}(0.012)(380)^2 = 866\,\text{J} \qquad K_r = \tfrac{1}{2}(4.0)(1.14)^2 = 2.6\,\text{J} \]

\[ K = \frac{p^2}{2m} \]

That relation is worth remembering in its own right. For a given momentum, lighter means more energetic. It is why the bullet does the damage and the shooter merely feels a shove, and why the chemical energy of the propellant goes overwhelmingly into the bullet.

The momentum part of this problem is two lines. The energy comparison at the end is the part worth remembering.

The bullet and the rifle have exactly equal and opposite momenta — that is what conservation from rest means. Their kinetic energies differ by a factor of three hundred and thirty-three.

The relation behind that is K equals p squared over 2m, which is worth deriving once and then keeping. For a given momentum, kinetic energy is inversely proportional to mass. Lighter means more energetic.

That is why the bullet does the damage and the shooter merely feels a shove, and why almost all the chemical energy of the propellant ends up in the bullet rather than the rifle.

The same relation explains a lot elsewhere. It is why a small fragment of an explosion is more dangerous than a large one, and why particle physicists accelerate light particles rather than heavy ones when they want high energies.

22. A one-dimensional collision with signs

Worked example

A 2.0 kg cart moving right at 5.0 m/s collides with a 3.0 kg cart moving left at 2.0 m/s. After the collision the 2.0 kg cart moves left at 1.6 m/s. Find the velocity of the other cart.

\[ (2.0)(5.0) + (3.0)(-2.0) = (2.0)(-1.6) + (3.0)v_{2f} \]

\[ 4.0 = -3.2 + 3.0v_{2f} \]

\[ v_{2f} = \frac{7.2}{3.0} = +2.4\,\text{m/s} \]

That last check is worth doing on every collision problem. Kinetic energy may stay the same or fall, never rise — unless something inside the system released stored energy, like an explosion or a compressed spring.

This problem is entirely about bookkeeping, and the bookkeeping is where it goes wrong.

Fix the axis first, then translate every stated speed into a signed velocity. Moving left at two metres per second becomes negative two. Doing this as an explicit step, written down, prevents most errors in this unit.

The result — both carts reversing — is physically sensible for a genuine collision, and worth checking. The lighter cart bounces back and the heavier one is turned around, which fits: lighter things are easier to reverse.

The energy check at the end is the habit to adopt. Kinetic energy fell from thirty-one joules to eleven, which is fine. Had it risen, the answer would have been impossible, because no collision creates kinetic energy unless something inside released stored energy.

That check costs about fifteen seconds and catches sign errors that produce plausible-looking but impossible answers. Use it on every collision problem.

23. Collisions

Section

Section 3

24. Three kinds of collision

Concept

Momentum is conserved in all three. That is the single most important row of the table, and the misconception it corrects is the subject of a trap slide two pages from here.

What distinguishes them is kinetic energy. Elastic collisions are the idealisation — genuinely exact for billiard balls to a good approximation, and exact for atomic and subatomic particles.

A perfectly inelastic collision loses the maximum possible energy consistent with conserving momentum. It cannot lose all of it, because the combined object still has to carry the total momentum, and carrying momentum requires motion.

Practical translation: 'stick together' in a problem statement means you get one equation, momentum, and the final velocity comes straight out of it. 'Elastic' means you get two equations and can solve for two unknowns.

The most important row of this table is the momentum column, because every entry in it says the same thing: conserved.

Momentum is conserved in all three kinds of collision. What distinguishes them is entirely what happens to the kinetic energy, and that is a separate question with a separate answer.

Elastic collisions are an idealisation, and a good one for billiard balls and an exact one for atomic and subatomic particles. Nothing macroscopic is perfectly elastic.

A perfectly inelastic collision loses the maximum energy consistent with conserving momentum. Notice that it cannot lose all of it — the combined object still has to carry the total momentum, and carrying momentum requires motion, which requires kinetic energy.

For problem-solving, the translation is: stick together means one equation and the answer falls out. Elastic means two equations and two unknowns. Anything else means momentum only, and the problem must give you enough other information.

25. Momentum is only conserved in elastic collisions

Trap

26. A perfectly inelastic collision

Worked example

A 1200 kg car at 18 m/s rear-ends a stationary 900 kg car; the two lock together. Find their common speed and the energy lost.

\[ m_1 v_1 = (m_1 + m_2)v_f \]

\[ v_f = \frac{(1200)(18)}{2100} = 10.3\,\text{m/s} \]

\[ K_i = \tfrac{1}{2}(1200)(18)^2 = 194\,\text{kJ} \qquad K_f = \tfrac{1}{2}(2100)(10.3)^2 = 111\,\text{kJ} \]

\[ \frac{\Delta K}{K_i} = \frac{83}{194} = 0.43 \]

That general fraction is worth knowing. Hit something much lighter and you keep almost all your energy; hit something much heavier and you lose almost all of it — which is why a car hitting a wall (effectively infinite mass) loses essentially everything.

The phrase lock together is the whole setup. It means one final velocity instead of two, which means momentum alone is sufficient — and it means no energy equation is available, because a great deal of energy has just been destroyed.

The calculation is one line. The interesting part is the energy accounting afterwards: forty-three percent of the kinetic energy is gone, into heat, sound and crumpled metal.

The general result at the end is worth keeping. For a mass hitting a stationary one and sticking, the fraction of kinetic energy retained is m1 over m1 plus m2.

Read that formula at its extremes. Hit something much lighter and you keep almost all your energy — the fraction approaches one. Hit something much heavier and you lose almost all of it.

A car hitting a wall is the extreme case: the wall is effectively infinitely massive, so the retained fraction goes to zero and essentially all the kinetic energy goes into destroying the car. That is why a head-on impact with a solid barrier is so much worse than a rear-end collision at the same speed.

27. An elastic collision, derived properly

Worked example

Object 1 of mass m1 at speed v1 strikes a stationary object of mass m2 elastically. Find both final velocities.

\[ m_1v_1 = m_1v_{1f} + m_2v_{2f} \qquad \tfrac{1}{2}m_1v_1^2 = \tfrac{1}{2}m_1v_{1f}^2 + \tfrac{1}{2}m_2v_{2f}^2 \]

\[ m_1(v_1 - v_{1f}) = m_2v_{2f} \qquad m_1(v_1^2 - v_{1f}^2) = m_2v_{2f}^2 \]

\[ \frac{(v_1 - v_{1f})(v_1 + v_{1f})}{v_1 - v_{1f}} = v_{2f} \;\Rightarrow\; v_1 + v_{1f} = v_{2f} \]

\[ v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_1 \qquad v_{2f} = \frac{2m_1}{m_1 + m_2}v_1 \]

The quadratic never had to be solved, which is the point of the divide-the-equations trick. Doing it the brute-force way works but is much longer and much easier to get wrong.

The brute-force approach here is to solve the quadratic energy equation simultaneously with the linear momentum equation. It works and it is unpleasant.

The trick is to rearrange both equations so the mass-two terms are alone on the right, then divide one by the other. The difference of two squares factorises, one factor cancels, and what is left is linear.

That single manoeuvre converts a quadratic problem into a linear one, and it is why the relative-velocity rule on the next slide exists at all.

The verification with equal masses is worth doing every time you use this result. Set the masses equal and the first object stops dead while the second leaves with the full speed — which is exactly what a head-on billiard shot looks like.

Learning the derivation matters more than memorising the two final formulas, because the divide-the-equations trick recurs. It is the same idea as dividing the two component equations to eliminate the tension in a conical pendulum.

28. The relative-velocity rule for elastic collisions

Concept

\[ v_{1i} - v_{2i} = -(v_{1f} - v_{2f}) \]

In an elastic collision the relative velocity reverses but keeps its magnitude. They approach at some speed and separate at exactly that speed.

This is a direct consequence of the two conservation laws — it falls out of the algebra on the previous slide — but it is far easier to use, because it is linear. Two linear equations beat one linear and one quadratic every time.

The practical recipe: for an elastic collision, use momentum conservation plus the relative-velocity rule. Never use the energy equation directly unless you are asked to prove something — it is the same information in a harder form.

This is the most useful single result about elastic collisions, and it is far easier to work with than the energy equation.

It says the relative velocity reverses without changing size. They approach at some speed and separate at exactly that speed, whatever the masses are.

The reason it is so much better than the energy equation is that it is linear. Two linear equations solve in a couple of lines; one linear plus one quadratic does not.

So the working recipe for any elastic collision is: momentum conservation plus this rule. Do not use the raw kinetic-energy equation unless a question specifically asks you to demonstrate energy conservation — it contains the same information in a much harder form.

It is worth checking the rule against the equal-mass case you already know. They approach at v and separate at v, with the roles swapped. Consistent.

29. Equal masses, head-on and elastic

Prediction

Option A is worth dismantling because it is the most tempting. Two balls at half speed do carry the right total momentum — but their combined kinetic energy is only half the original, since energy goes as the square. Momentum alone cannot decide between the options; you need the energy condition too.

This is the Newton's cradle result, and it is why exactly one ball swings out when one ball is released. Two balls swinging out at half speed would violate energy conservation, and three at a third speed would violate it further.

Option A is the tempting answer and is worth dismantling carefully, because the reasoning behind it is half right.

Two balls each moving at half the original speed do carry exactly the right total momentum. Momentum conservation alone cannot rule it out.

What rules it out is energy. Kinetic energy goes as the square of the speed, so two balls at half speed carry only half the original kinetic energy. The other half would have to have gone somewhere, and in an elastic collision it does not.

So this is a good demonstration that in an elastic collision you genuinely need both laws. Momentum alone is not enough to determine the outcome.

This is also the Newton's cradle result, which is a nice thing to be able to explain. One ball in, one ball out, at the same speed. Two balls out at half speed would violate energy conservation, and three at a third speed would violate it further.

30. Sweep the mass ratio in an elastic collision

Tweak it

The doubling in the light-target limit is used deliberately in physics and engineering. It is how a golf club transfers speed to a ball, and the same principle scaled up is the gravitational slingshot that sends spacecraft to the outer planets.

The turning point is at equal masses, and it is worth knowing which side of it you are on before doing any algebra.

Hit something lighter than yourself and you continue forward, slowed. Hit something heavier and you bounce back. At exactly equal masses you stop dead.

The extreme cases are the ones to remember. Hitting a wall — effectively infinite mass — returns you at nearly your original speed, which is why balls bounce. Hitting a speck barely slows you at all.

The top row of the table contains the result people find hardest to believe: a very light target can leave at up to twice the incoming speed. That violates nothing. It carries very little momentum and very little energy precisely because its mass is so small.

That doubling is used deliberately. It is roughly how a golf club transfers speed to a ball, and scaled up enormously it is the gravitational slingshot that sends spacecraft to the outer planets — a probe bouncing elastically off a planet moving toward it can leave with far more speed than it arrived with.

31. Match each description to its collision type

Matching

Item e is the case worth keeping in view. An explosion is a collision run backwards in the energy accounting: the total kinetic energy rises. Momentum is conserved anyway, because the third law does not care where the energy came from.

Sort these by asking what happened to the kinetic energy, since that is the only thing that distinguishes the categories.

Objects that end up moving together have lost the maximum possible amount and are perfectly inelastic. Objects that bounce with almost no loss are elastic. Almost everything real is somewhere in between.

Item e is the case worth keeping in view, because it is the one that runs the accounting backwards. An explosion increases the total kinetic energy, drawing on stored chemical energy.

And momentum is conserved anyway. The third law does not care where the energy came from, so an explosion conserves momentum exactly as a collision does. That is why an explosion problem and a collision problem are solved with identical algebra.

It is worth noting that a collision and an explosion are, mathematically, the same event viewed with the before and after swapped. A perfectly inelastic collision run backwards is an explosion.

32. The coefficient of restitution: a sliding scale

Concept

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

Real collisions are almost never exactly elastic or exactly perfectly inelastic. The coefficient of restitution measures where a collision sits between those two extremes.

Notice that e = 1 is exactly the relative-velocity rule from the previous slide, and e = 0 says the separation speed is zero — which is what sticking together means. The two cases you know are the endpoints of one continuous scale.

A convenient consequence for the lab: a ball dropped from height h and bouncing to height h' has e equal to the square root of h' over h. That gives you a way to measure the coefficient with nothing but a metre rule, and it is a standard AP experiment.

The AP exam does not require e by name, but it does ask about collisions that are neither ideal, and knowing that they lie on a continuum makes those questions far less alarming.

Real collisions are almost never exactly elastic or exactly perfectly inelastic, and it helps to have a way of talking about the space in between.

The coefficient of restitution is the ratio of separation speed to approach speed. Look at what its extreme values say: e equal to one is exactly the relative-velocity rule from two slides ago, and e equal to zero says the separation speed is zero, which is what sticking together means.

So the two cases you already know are the endpoints of a single continuous scale, and everything real lives somewhere along it.

The lab consequence is convenient. A ball dropped from height h and bouncing to h prime has e equal to the square root of h prime over h. That gives you a measurement with nothing but a metre rule, and it is a standard AP experiment.

The exam does not require the coefficient by name. It does ask about collisions that are neither ideal, and knowing they sit on a continuum makes those questions much less alarming than they look.

33. Does a collision look different from a moving train?

Socratic

This question is worth thinking through because the answer reveals something structural about conservation laws.

Both observers find momentum conserved. Adding a constant velocity to everything adds the same constant amount to the total on both sides of the equation, so the conservation statement survives unchanged.

More deeply, it has to. Momentum conservation follows from the third law, and the third law holds in every inertial frame. Anything derived from it is frame-independent.

The surprising part is that both observers also agree on how much kinetic energy was lost, even though they disagree about every individual kinetic energy. The extra terms introduced by the frame shift depend only on the total momentum, which is unchanged, so they cancel in the subtraction.

What the two observers genuinely disagree about is which object was moving. In one frame a cart may be at rest beforehand; in the other it never is. That is why the moving one is never a physically meaningful description.

And there is a frame worth knowing about: the one moving with the centre of mass, in which the total momentum is zero. In that frame the two objects always approach with equal and opposite momenta and leave the same way, which makes many hard collision problems nearly trivial.

34. The ballistic pendulum

Worked example

A 9.0 g bullet embeds in a 2.5 kg block hanging from a string. The block swings up 8.0 cm. Find the bullet's speed.

\[ mv = (m + M)V \;\Rightarrow\; V = \frac{mv}{m+M} \]

\[ \tfrac{1}{2}(m+M)V^2 = (m+M)gh \;\Rightarrow\; V = \sqrt{2gh} \]

\[ V = \sqrt{2(9.8)(0.080)} = 1.25\,\text{m/s} \]

\[ v = \frac{(m+M)V}{m} = \frac{(2.509)(1.25)}{0.009} = 348\,\text{m/s} \]

This is the classic two-phase problem, and it is on AP exams regularly because it tests whether you know which conservation law applies where.

Phase one, the embedding, is perfectly inelastic. Over ninety-nine percent of the kinetic energy is destroyed. Momentum is conserved; energy emphatically is not.

Phase two, the swing, is smooth and dissipationless, so mechanical energy is conserved. But momentum is not, because the string tension and gravity are external forces acting for a substantial time.

So each phase has exactly one applicable conservation law, and they are different laws. Mixing them is the standard error, and the next slide is devoted to what happens if you do.

The two phases meet at V, the common speed immediately after embedding. Work phase two backwards from the measured height to find V, then use phase one to find the bullet speed. That is the structure.

The energy check at the end quantifies why the split matters: 545 joules before, under two joules after. Assuming energy conservation across phase one would throw away 99.6 percent of the physics.

35. Find the mistake in this ballistic-pendulum attempt

Error analysis

A student solves the previous problem in one step, using energy conservation from the bullet's arrival to the top of the swing.

The general rule this illustrates is worth stating on its own: an inelastic collision destroys kinetic energy, so no energy equation may be written across it. Momentum is the only law that survives.

And the reverse: during the swing, the string tension and gravity are external forces, so momentum is not conserved there. Each phase has exactly one applicable law, and identifying which is the skill.

A good habit for any multi-phase problem: draw a vertical line at each phase boundary and label which conservation law applies on each side of it before writing anything.

This student's equation is not carelessly written — it is a perfectly reasonable-looking application of energy conservation. It is simply applied across a phase where energy is not conserved.

The result is wrong by a factor of seventeen, and helpfully wrong in a way you could catch. It gives a bullet speed of about twenty-one metres per second, which is roughly the speed of a thrown cricket ball. Bullets do not travel at that speed, so the answer fails a plausibility check even without knowing the correct method.

The rule to extract: an inelastic collision destroys kinetic energy, so no energy equation may be written across it. Momentum is the only law that survives.

And the reverse holds for the swing: momentum is not conserved there, because external forces act. Neither law works for both phases.

The habit that prevents this is mechanical. In any multi-phase problem, draw a vertical line at each phase boundary and label which conservation law applies on each side before writing a single equation. It takes ten seconds and it makes the error structurally impossible.

36. Collisions in Two Dimensions

Section

Section 4

37. Conserve each component separately

Concept

\[ \sum p_{x,i} = \sum p_{x,f} \qquad \sum p_{y,i} = \sum p_{y,f} \]

Momentum is a vector, so conserving it means conserving each component independently. In two dimensions that is two equations, and they do not talk to each other.

The figure shows the most common exam setup: one object incoming along the x-axis, nothing moving in y beforehand. That makes the initial y-momentum zero, which forces the two outgoing y-components to cancel exactly.

That is often the more useful of the two equations, because a zero on one side removes a term. Look for it: any direction with no initial motion gives you a sum-to-zero condition.

Momentum is a vector, so conserving it means conserving each component independently. In two dimensions that is two equations, and they do not interact — the same independence you met with projectiles.

The figure shows the standard exam configuration: one object incoming along the x-axis with nothing moving in y beforehand.

That makes the initial y-momentum zero, and it must remain zero. So the two outgoing y-components have to cancel exactly. That is often the more useful of the two equations, because a zero on one side eliminates a term.

The general habit worth building: look for any direction in which nothing was moving initially. That direction gives you a sum-to-zero condition, which is the easiest kind of equation to use.

It also gives you a quick qualitative check on any two-dimensional collision answer. If nothing was moving north or south before, the outgoing objects must go one each way, and if your answer sends both the same way, something is wrong.

38. A two-dimensional collision

Worked example

A 3.0 kg puck moving east at 4.0 m/s strikes a stationary 2.0 kg puck. Afterwards the 3.0 kg puck moves at 2.0 m/s at 30 degrees north of east. Find the velocity of the 2.0 kg puck.

\[ p_{1fx} = (3.0)(2.0)\cos 30^\circ = 5.20 \qquad p_{1fy} = (3.0)(2.0)\sin 30^\circ = 3.00 \]

\[ (3.0)(4.0) = 5.20 + p_{2fx} \;\Rightarrow\; p_{2fx} = 6.80\,\text{kg m/s} \]

\[ 0 = 3.00 + p_{2fy} \;\Rightarrow\; p_{2fy} = -3.00\,\text{kg m/s} \]

\[ v_{2fx} = 3.40 \qquad v_{2fy} = -1.50 \qquad v_{2f} = \sqrt{3.40^2 + 1.50^2} = 3.72\,\text{m/s} \]

\[ \theta = \arctan\frac{1.50}{3.40} = 23.8^\circ \text{ south of east} \]

The procedure is entirely mechanical once the axes are set: resolve, write two equations, solve, recombine.

Notice that only one velocity needed resolving, because the incoming puck moved along an axis. Choosing axes to align with a known velocity is worth doing whenever the problem allows it.

The y-equation is the easy one and is worth doing first. Nothing was moving north or south initially, so the two y-momenta must cancel, and the second puck's y-momentum follows immediately with a minus sign.

Converting back from momentum components to a velocity requires dividing by the second puck's mass, which is easy to forget when you have been working in momentum for several lines. Keep track of which quantity you are holding.

The energy check at the end tells you something the question did not: this collision was inelastic, losing about eighteen percent of the kinetic energy. That is a legitimate physical conclusion and it also confirms the answer is at least possible.

39. An explosion into three fragments

Worked example

A stationary 9.0 kg shell bursts into three pieces. A 2.0 kg piece flies east at 30 m/s; a 3.0 kg piece flies north at 20 m/s. Find the velocity of the third piece.

\[ \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \]

\[ \vec{p}_1 = (60)\,\hat{\imath} \qquad \vec{p}_2 = (60)\,\hat{\jmath} \]

\[ \vec{p}_3 = -(60\,\hat{\imath} + 60\,\hat{\jmath}) \]

\[ |\vec{p}_3| = \sqrt{60^2 + 60^2} = 84.9 \;\Rightarrow\; v_3 = \frac{84.9}{4.0} = 21.2\,\text{m/s} \]

The setup line here does all the work: the shell was at rest, so the total momentum is zero and must remain zero.

That converts the problem into a vector addition. The third fragment's momentum must be exactly the negative of the other two combined, which fixes both its magnitude and its direction with no further physics.

The mass of the third piece is whatever is left over. That is a small bookkeeping step that is easy to skip, and it matters because momentum divided by the wrong mass gives the wrong velocity.

The energy line at the end is the point worth extracting. Two thousand four hundred joules of kinetic energy now exist where there were none, all of it converted from chemical energy in the explosive.

And the momentum stayed at zero throughout. That contrast is the cleanest statement of the difference between the two conservation laws: internal processes can create kinetic energy freely, and can never create momentum.

40. A two-dimensional sum-to-zero

Check

A stationary object explodes into exactly two fragments. One flies off due north at 6.0 m/s.

Option C is the one to think about, because it is right about direction and wrong about magnitude.

What must be equal and opposite is the momentum, not the velocity. If the two fragments have different masses, they have different speeds, and only their momenta match.

The direction, though, is completely determined regardless of the masses. Two fragments from a stationary object must fly apart along the same line in opposite directions, because there is no way to make two non-collinear vectors sum to zero.

That is worth remembering as a general result about two-body explosions. Three or more fragments can go in genuinely different directions; exactly two cannot.

Option D is worth dismissing explicitly. The explosive energy sets how much total kinetic energy there is, and therefore how fast the fragments go. It has no bearing at all on the directions, which momentum conservation fixes by itself.

41. The Centre of Mass

Section

Section 5

42. The mass-weighted average position

Concept

\[ x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i} \qquad \vec{r}_{\text{cm}} = \frac{1}{M}\sum m_i\vec{r}_i \]

The centre of mass is the point that behaves as though all the mass were concentrated there. It is a weighted average, with each position weighted by its own mass.

The picture gives you a shortcut worth having: the centre of mass divides the line between two objects in inverse proportion to their masses. Three times the mass means one third of the distance.

It need not lie inside any object. The centre of mass of a doughnut is in the hole, and the centre of mass of the Earth-Moon system is inside the Earth but well off-centre.

The centre of mass is a weighted average, with each position weighted by its own mass. That is the whole definition, and everything else follows from it.

The figure gives a shortcut worth having: the centre of mass divides the line between two objects in inverse proportion to their masses. Three times the mass means one third of the distance from it.

That lets you locate a two-body centre of mass by inspection in most cases, and it gives you a check on any calculation — the answer must be closer to the heavier object.

It need not lie inside any object. The centre of mass of a doughnut is in the hole; of a boomerang, in the air beside it; of the Earth-Moon system, inside the Earth but about three-quarters of the way out from the centre.

That is not a paradox. The centre of mass is a location, not a piece of matter, and there is no requirement for anything to be there.

43. Centre of mass of a set of particles

Worked example

Three masses sit in the xy-plane: 2.0 kg at (0, 0), 3.0 kg at (4.0, 0), and 5.0 kg at (2.0, 3.0). Find the centre of mass.

\[ x_{\text{cm}} = \frac{(2)(0) + (3)(4) + (5)(2)}{10} = \frac{22}{10} = 2.2 \]

\[ y_{\text{cm}} = \frac{(2)(0) + (3)(0) + (5)(3)}{10} = \frac{15}{10} = 1.5 \]

Notice that the centre of mass is at a point where there is no mass at all. That is entirely normal and is not a problem: it is a location, not an object.

The mechanics of this are simple: each coordinate is its own weighted average, and the two calculations are completely independent, exactly as x and y were for projectiles.

The one thing to get right is the denominator. It must be the total mass of the system, not the number of particles. Dividing by three here instead of by ten kilograms is the standard slip and produces an answer that is not even dimensionally sensible.

The check against intuition is worth doing. The heaviest mass sits at the top of the arrangement, so the centre of mass should be pulled upward — and it is, sitting at y equal to 1.5 even though two of the three masses are on the axis.

Notice that the result is a point where no mass exists. That is entirely ordinary, and it will be true again in almost every extended-body problem you meet.

44. For a continuous body, the sum becomes an integral

Concept

\[ x_{\text{cm}} = \frac{1}{M}\int x\,dm \qquad \text{with} \quad dm = \lambda\,dx \;\text{(rod)}, \quad dm = \sigma\,dA \;\text{(sheet)} \]

The structure is identical to the discrete case — a weighted average — with the sum replaced by an integral and the individual masses by mass elements.

The step that actually requires thought is expressing dm in terms of the integration variable. For a rod that means the linear density lambda, which may itself depend on position.

Use symmetry whenever it is available. For any object with a plane of symmetry and uniform density, the centre of mass lies on that plane — which usually settles one coordinate for free and sometimes all of them.

The structure of the continuous case is identical to the discrete one — still a weighted average — with a sum replaced by an integral and individual masses replaced by mass elements.

The step that actually requires thought is expressing dm in terms of the integration variable. For a rod that means the linear density, which may itself vary with position. Getting that substitution right is the whole difficulty of these problems.

The table lists the common cases. Note the middle row: a density proportional to x means dm is proportional to x dx, and the resulting integral is one power higher than you might expect.

The last line is the practical advice. Use symmetry whenever it is available. Any uniform object with a plane of symmetry has its centre of mass on that plane, which often settles one coordinate for free and sometimes all of them.

A uniform sphere, cube or rod needs no integration at all — the centre of mass is at the geometric centre, by symmetry. Integrate only when the density varies or the shape is genuinely irregular.

45. Centre of mass of a rod with varying density

Worked example

A rod of length L lies along the x-axis from 0 to L, with linear density lambda(x) = kx. Find its total mass and its centre of mass.

\[ M = \int_0^L kx\,dx = \frac{kL^2}{2} \]

\[ \int_0^L x\,dm = \int_0^L x(kx)\,dx = \int_0^L kx^2\,dx = \frac{kL^3}{3} \]

\[ x_{\text{cm}} = \frac{kL^3/3}{kL^2/2} = \frac{2L}{3} \]

This is the standard AP integration problem for centre of mass, and it comes in two parts that are easy to conflate.

First find the total mass by integrating dm. You need it as the denominator, and it is not simply lambda times L, because lambda varies.

Then compute the weighted integral, multiplying each mass element by its position. Notice that this integrand is one power of x higher than the mass integrand, which is what makes the answer come out at two thirds rather than one half.

The constant k cancels in the division, and it must. The shape of the density profile determines where the centre of mass is; the overall scale of the density does not.

Two checks are worth doing. The answer must lie past the midpoint, since the rod is denser toward the far end — two thirds of L qualifies. And running the same procedure with a constant density must give L over two, which it does.

46. The system moves as though all its mass were at the centre of mass

Concept

\[ M\vec{a}_{\text{cm}} = \sum \vec{F}_{\text{external}} \qquad \vec{p}_{\text{total}} = M\vec{v}_{\text{cm}} \]

This is the deepest result in the deck. However complicated a system's internal motion — spinning, colliding, exploding — its centre of mass obeys Newton's second law as a single particle of the total mass.

Internal forces cannot move the centre of mass, because they cancel in pairs. Only external forces can accelerate it.

Two consequences worth having ready. A system with no external force has a centre of mass that moves at constant velocity, no matter what its parts are doing. And conservation of momentum is exactly the statement that v_cm is constant.

This is also why you were allowed to treat blocks, cars and planets as point particles for three decks. You were, all along, describing the motion of their centres of mass.

This is the deepest result in the deck, and it is worth stating carefully because it is easy to under-appreciate.

However complicated the internal motion of a system — parts spinning, colliding, exploding, running about — the centre of mass obeys Newton's second law as though it were a single particle carrying the total mass, driven by the external forces alone.

Internal forces cannot move it, because they cancel in pairs. That is the third law again, in its most useful form.

Two consequences to have ready. A system with no external force has a centre of mass moving at constant velocity, no matter what its parts do. And conservation of momentum is precisely the statement that the centre-of-mass velocity is constant — the two are the same fact.

This is also, retrospectively, why you were allowed to treat blocks, cars and planets as point particles for three decks. You were describing their centres of mass the whole time, and this theorem is what made that legitimate.

47. Where is the centre of mass?

Check

A 2.0 kg mass sits at x = 0 and a 6.0 kg mass sits at x = 8.0 m.

This is worth doing by both routes, because each catches a different error.

The formula route: weighted average, forty-eight divided by eight, six metres. The proportion route: the six kilogram mass is three times heavier, so it sits one third as far from the centre of mass, which puts the centre two metres from it and six from the other.

Option A is the geometric midpoint, which would be right only for equal masses. It is the answer you get by forgetting to weight at all.

Option C is on the wrong side entirely — closer to the lighter mass. Any answer on the light side of the midpoint is immediately wrong, which makes this a good quick check on any centre-of-mass calculation.

The habit worth building: before computing, say out loud which object the answer should be closer to. It takes a second and it eliminates half the possible mistakes.

48. The projectile that explodes in mid-air

Anomaly

A firework is launched and follows a parabola. At the top of its arc it explodes into many fragments flying in all directions.

Air resistance eventually spoils this for real fireworks, since drag is external and affects small fragments strongly. But for the first moments after the burst it is startlingly accurate.

This is the most striking demonstration of the centre-of-mass theorem, and it is worth being sure why it works.

The explosion is entirely internal. Internal forces cancel in pairs, so they cannot change the centre-of-mass motion at all — not its position, not its velocity, not its acceleration.

The only external force is still gravity, which was the only external force before the explosion too. So the centre of mass continues to accelerate downward at g, along exactly the parabola the unexploded firework would have followed, right down to where it would have landed.

The fragments meanwhile fly in all directions, some upward, some backward, some at high speed. Their weighted average position is completely undisturbed by any of it.

Air resistance eventually spoils the effect for real fireworks, because drag is external and affects small fragments strongly. But for the first moments after the burst it is startlingly accurate, and high-speed footage shows it clearly.

49. Two people walking on a boat

Worked example

A 4.0 m boat of mass 120 kg floats at rest on still water. A 60 kg person walks from one end to the other. Ignoring water resistance, how far does the boat move?

\[ m_p(4.0 - d) = m_b \,d \]

\[ 60(4.0 - d) = 120d \;\Rightarrow\; 240 = 180d \;\Rightarrow\; d = 1.33\,\text{m} \]

This is not a curiosity. It is why stepping off a small boat onto a dock is difficult, and why an astronaut who throws a tool drifts the other way — with no external force available, the only way to move yourself is to move something else the opposite way.

The constraint that solves this problem is stated in one line: there are no external horizontal forces and everything starts at rest, so the centre of mass cannot move at all.

Everything else is bookkeeping. The person and the boat must move in opposite directions by distances in inverse proportion to their masses, so that the weighted average stays exactly where it was.

The check confirms it: the person moves 2.67 metres and the boat 1.33, a ratio of two to one, which is the inverse of the sixty to one hundred and twenty mass ratio.

It is worth seeing that this is momentum conservation integrated over time. At every instant the person's forward momentum equals the boat's backward momentum, so the total is always zero — and integrating zero momentum over time gives zero displacement of the centre of mass.

The practical versions are everywhere. Stepping off a small boat pushes it away beneath you. An astronaut who throws a tool drifts the other way. With no external force available, the only way to move yourself is to move something else in the opposite direction.

50. Why can internal forces never move the centre of mass?

Explain it to yourself

Being able to state this argument in your own words is worth more than being able to use the formula, because the argument generalises and the formula does not.

Every internal force comes in a third-law pair, equal and opposite, acting on two members of the system. When you sum all the forces to find the centre-of-mass acceleration, those pairs cancel exactly. Only external forces survive the sum.

So the centre of mass is deaf to internal activity, however violent. It carries on exactly as the external forces alone dictate.

The boat question has a clean answer: no, if the dock is out of reach. Walking moves the boat, but only by moving yourself the opposite way, and when you stop walking the boat stops too. The centre of mass never moved.

But the loophole is the interesting part. You can throw something. Throwing a heavy object away from the dock ejects it from the system, and what remains genuinely does move toward the dock. The centre of mass of everything is still fixed — the thrown object has carried the compensating momentum away with it.

That loophole is the entire principle of rocket propulsion, which is where this deck goes next.

51. When the Mass Changes

Section

Section 6

52. Why F = ma fails for a rocket

Concept

A rocket's mass falls as it burns fuel, sometimes to a small fraction of what it started with. The second law in the form F = ma assumes constant mass, so it simply does not apply.

\[ \vec{F}_{\text{ext}} = \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} \]

The correct approach is to go back to momentum conservation for the whole system, rocket and expelled gas together. Nothing external needs to act at all — the gas going one way is what sends the rocket the other.

This is the answer to the common question of what a rocket pushes against in space. Nothing. It pushes against its own exhaust, by the third law, and it works better in vacuum than in atmosphere.

A rocket's mass can fall to a small fraction of its launch value, so the constant-mass assumption behind F equals ma is badly violated.

The correct approach is to go back to momentum conservation for the whole system — rocket plus expelled gas together. Nothing external needs to act at all. The gas going backward is exactly what sends the rocket forward.

This answers a question people ask constantly: what does a rocket push against in space? Nothing. It pushes against its own exhaust, by the third law.

And it works better in vacuum than in atmosphere, which is the opposite of what the push-against-something intuition predicts. Air resistance and back pressure both reduce a rocket's performance.

The picture to keep is the momentum ledger. The total momentum of rocket plus exhaust is unchanged; the rocket gains exactly what the gas loses, in the opposite direction.

53. The rocket equation

Worked example

A rocket of instantaneous mass m expels fuel at a constant speed u relative to itself. Derive how its speed changes as its mass falls.

\[ mv = (m + dm)(v + dv) + (-dm)(v - u) \]

\[ 0 = m\,dv + u\,dm \]

\[ \int_{v_0}^{v} dv' = -u\int_{m_0}^{m}\frac{dm'}{m'} \]

\[ v - v_0 = u\ln\frac{m_0}{m} \]

This is Tsiolkovsky's rocket equation, and the logarithm is the reason spaceflight is hard. A rocket needing a speed change of three times its exhaust speed must start out about twenty times as massive as it finishes — which is why launch vehicles are staged.

The derivation is short and the key step is the one that looks like cheating: discarding the product of two infinitesimals.

That is legitimate because dm and dv are both going to zero, so their product goes to zero much faster than either alone. It is the same reasoning that lets you ignore second-order terms in any differentiation.

What remains separates cleanly, and the integral of dm over m gives a logarithm. That logarithm is the entire character of the result.

Read what it implies. To double the speed you gain, you must square the mass ratio. A rocket needing a speed change of three times its exhaust speed must start out about twenty times as massive as it finishes.

That is why spaceflight is hard, and why launch vehicles are staged. Once a tank is empty it is dead weight sitting in the denominator, so you throw it away and start a fresh mass ratio with what remains.

This is Tsiolkovsky's equation, derived in 1903, and it still sets the fundamental limits on what chemical rockets can do.

54. Sand falling onto a moving conveyor belt

Worked example

Sand falls vertically at a rate of 25 kg/s onto a belt moving horizontally at 2.0 m/s. What horizontal force must the motor supply to keep the belt at constant speed?

\[ F = \frac{dp}{dt} = v\frac{dm}{dt} \]

\[ F = (2.0)(25) = 50\,\text{N} \]

\[ P_{\text{supplied}} = 100\,\text{W} \qquad \frac{dK}{dt} = 50\,\text{W} \]

The lesson is that in variable-mass problems, energy is not conserved even when nothing looks lossy. The mass being added arrives at the wrong speed, and bringing it up to speed always dissipates something.

This problem is a clean example of the v dm/dt term doing all the work, with no ma term at all.

The belt moves at constant speed, so there is no acceleration. Yet a force is required, because newly arrived sand keeps having to be brought up to belt speed. That force is entirely the mass-changing term.

The energy accounting at the end is the interesting part and is worth working through. The motor supplies one hundred watts. The sand's kinetic energy grows at only fifty watts. Half the input is unaccounted for.

It goes into heat, through the sliding friction as each grain is accelerated from rest to belt speed. This factor of exactly one half is a famous result for this class of problem, and it is not an approximation.

The general lesson: in variable-mass problems, energy is not conserved even when nothing looks lossy. Mass that joins the system arrives at the wrong speed, and bringing it up to speed always dissipates something.

55. Push the exhaust speed and the mass ratio

Edge cases

The two parameters in the rocket equation enter very differently, and knowing which to improve is a real engineering question.

Mass ratio enters logarithmically, so every doubling adds a fixed amount rather than a fixed factor. Doubling from two to four buys you the same delta-v as doubling from twenty to forty. The returns are brutally diminishing.

Exhaust speed enters linearly, so doubling it doubles the delta-v outright. That is why ion drives, with exhaust speeds ten times higher than chemical rockets, are used for deep-space missions despite their almost comically small thrust.

And the mass-ratio route hits a hard physical wall. A ratio of forty means the vehicle is 97.5 percent propellant, leaving almost nothing for structure, engines and payload. You cannot build that.

Which is exactly why rockets are staged. Discarding empty tanks lets you restart the mass ratio partway up, and multi-stage vehicles achieve delta-v that no single stage could.

56. Putting It Together

Section

Section 7

57. Momentum versus energy, side by side

Comparison

The last row is the one that catches people. Internal forces cannot change momentum but absolutely can change kinetic energy — an explosion creates kinetic energy from stored chemical energy while leaving the total momentum untouched.

That asymmetry has a clean explanation. Momentum conservation follows from the third law, which is about forces coming in equal and opposite pairs. Energy conservation is about whether anything is stored or released internally, which the third law says nothing about.

The last row of this table is the one that catches people, and it is worth being clear about.

Internal forces cannot change momentum — that is the third law. But internal forces absolutely can change kinetic energy: an explosion creates kinetic energy out of stored chemical energy while leaving the total momentum untouched at zero.

The asymmetry has a clean explanation. Momentum conservation follows from forces coming in equal and opposite pairs, which is a statement about direction. Energy conservation is about whether anything is stored or released internally, which the third law says nothing about.

The third row is the architectural one. Momentum comes from integrating force over time; energy comes from integrating it over position. Everything else in the table follows from that difference.

It is worth reading this table before any collision problem. It tells you exactly which law you may write down and which you may not.

58. Deciding what to conserve

Pattern

1. Is anything external acting along the direction you care about? If not, momentum is conserved in that direction — even if other directions are spoiled.
2. Is the interval very short? Then gravity and friction contribute negligible impulse, and momentum is conserved to an excellent approximation.
3. Do the objects stick together? Perfectly inelastic. One equation, momentum, and the final velocity falls out.
4. Does the problem say elastic? Two equations — use momentum plus the relative-velocity rule, never the raw energy equation.
5. Are there several phases? Draw a line at each boundary and decide separately which law applies on each side.
6. Is the mass changing? Go back to F = dp/dt and keep the v dm/dt term.

Point five is the one that separates strong answers from weak ones. The ballistic pendulum is the canonical example: momentum for the embedding, energy for the swing, and mixing them destroys the answer.

Point one deserves emphasis too. Momentum can be conserved in one direction and not another. A ball bouncing off the ground has its vertical momentum changed by the floor while its horizontal momentum is untouched — and the horizontal equation is still available to you.

Six questions, and the fifth is the one that separates strong answers from weak ones.

Multi-phase problems need a separate decision for each phase. The ballistic pendulum is the canonical case: momentum for the embedding, energy for the swing, and mixing them destroys the answer completely.

Point one deserves emphasis too, because it is underused. Momentum can be conserved in one direction and not another. A ball bouncing off the ground has its vertical momentum changed by the floor while its horizontal momentum is completely untouched — and that horizontal equation is still available to you.

Point two is the approximation that makes collision problems possible at all. If the interval is very short, gravity and friction contribute negligible impulse and can be ignored.

Point four is worth repeating: when a problem says elastic, use momentum plus the relative-velocity rule, never the raw energy equation. Same information, far less algebra.

59. How hard does a tennis ball hit a racket?

Estimation

The reversal is the point worth extracting. Because the ball comes back, the speeds add in the momentum change. Forgetting that and computing 40 minus 30 gives 116 N, wrong by a factor of seven.

This is also why the contact time matters so much to players. A softer string bed extends the contact and lowers the peak force on the arm — the same impulse spread over more time, exactly as with the airbag.

The step this question is testing is the reversal, and it is the step most people get wrong under time pressure.

The ball arrives at thirty and leaves at forty in the opposite direction. Those speeds add in the momentum change, giving seventy metres per second of velocity change rather than ten.

Computing forty minus thirty gives one hundred and sixteen newtons instead of eight hundred and ten — wrong by a factor of seven, and wrong in a way that produces a plausible-looking number.

The habit that prevents it is to convert speeds to signed velocities explicitly before subtracting. Plus forty minus minus thirty is visibly seventy; forty minus thirty is not.

The result also explains something players care about. A softer string bed extends the contact time and lowers the peak force transmitted to the arm — the same impulse spread over more time, which is the airbag principle applied to tennis elbow.

60. Four questions before you close the deck

Exit ticket

Answer these from memory first. All four target the boundaries of the concepts rather than calculation, which is where exam questions live.

The first tests system choice. Not conserved for the ball alone, because the floor pushes it. Conserved exactly if you include the Earth — whose velocity change is real but unmeasurably small.

The second tests the reversal factor. Bouncing needs twice the impulse of catching, because the momentum goes from p to minus p rather than from p to zero.

The third tests the centre-of-mass theorem. The explosion is internal and cannot affect it, so the centre of mass carries on along the original parabola.

The fourth tests whether you know the efficient route through an elastic collision. Momentum plus the relative-velocity rule, both linear, rather than momentum plus the quadratic energy equation.

Any hesitation points at a specific slide rather than at the whole deck, which is the more efficient thing to revisit.

61. What you can do now

Recap

• Use impulse as a time integral, as the area under a force-time graph, and as the change in momentum
• Explain why extending the contact time reduces the force without changing the impulse
• Choose a system so the unknown forces become internal, and state exactly when momentum is conserved
• Solve elastic collisions with momentum plus the relative-velocity rule instead of a quadratic
• Split a multi-phase problem and apply the right conservation law to each phase
• Conserve momentum component by component in two dimensions
• Locate a centre of mass by summing or integrating, and use its motion as a shortcut
• Handle variable-mass problems with F = dp/dt rather than F = ma

Next deck: rotation. Every idea in this course so far has a rotational counterpart — torque for force, moment of inertia for mass, angular momentum for momentum — and the conservation argument you just met for linear momentum runs again, almost word for word, for angular momentum.

You now have three ways of attacking a mechanics problem — forces, energy, and momentum — and the skill that matters most is choosing between them quickly.

The organising idea of this deck is that momentum comes from integrating the second law over time, which is why it knows about durations and forces and nothing about distances. Energy came from integrating over position, which is why it is the reverse.

The results with the longest reach are the conditional ones. Momentum is conserved when no external force acts, which makes system choice the central skill. All collisions conserve momentum and only elastic ones conserve energy, which is what makes the momentum equation always safe to write first.

The centre-of-mass theorem is the most powerful single statement here. A system's centre of mass responds only to external forces, which means it ignores collisions, explosions and internal rearrangement entirely.

Next comes rotation, and the correspondence is remarkably close. Torque plays the role of force, moment of inertia the role of mass, and angular momentum the role of momentum. The conservation argument you just met runs again for angular momentum, almost word for word — and the reason is the same third law, applied to torques instead of forces.