7 concepts18 worked examples32 exercises5 exam-level7 figures
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11Linear Momentum
A bat meets a 0.145 kg ball that is coming in at 40.0 m/s and sends it back the other way at 45.0 m/s. The whole meeting lasts about 1.5 thousandths of a second. Ask for the force the bat applied and every tool from the last ten weeks fails: the force is nowhere near constant, so there is no single acceleration to put into Newton's second law, and the kinetic energy went up rather than staying put, so the conserved total is no help either.
By the end of this section you can turn that 1.5 ms into a number in newtons in two lines, predict the speeds of two bodies after they hit each other without knowing anything about the force between them, and say which of the two conserved quantities you are allowed to use before you write a single equation.
In 60 seconds
Mass times velocity is the quantity a force changes at a rate, so a force acting for a time changes it by exactly $F\Delta t$; and because the two bodies in a collision push on each other with equal and opposite forces for exactly the same time, whatever one gains the other loses and the total for the pair comes out of the crash unchanged even though the kinetic energy usually does not.
Momentum, and the law that changes it
$$\vec p = m\vec v,\qquad \sum \vec F = \frac{d\vec p}{dt}$$
always; it is a vector along the velocity, and a net force is the rate at which it changes
Impulse, and the momentum change it produces
$$\vec J = \bar{\vec F}\,\Delta t = \Delta \vec p$$
a short violent contact: a bat, a kick, a landing, a crash, an airbag
the bodies stick, or the question says elastic; the rest of the formula card carries the special cases
Three most common mistakes
Dropping the sign of a velocity that reverses. A 0.145 kg ball arriving at 40.0 m/s and leaving at 45.0 m/s the other way changes its momentum by $0.145(-45.0-40.0) = -12.3$ kg m/s, not by $0.145(45.0-40.0)=0.725$ kg m/s. The wrong version is seventeen times too small.
Conserving kinetic energy in a collision that never said elastic. Unless the question uses that word, or says the bodies bounce apart perfectly, the kinetic energy after is smaller than before and setting the two equal gives a confident wrong answer.
Adding momenta as numbers when they point in different directions. Two cars, one carrying 24000 kg m/s east and one 21600 kg m/s north, leave a wreck carrying 32300 kg m/s, not 45600 kg m/s; perpendicular vectors add by the theorem of Pythagoras.
The assessment table gives 20% to each midterm, 25% to the final, 10% to the quizzes in total, 5% to homework and 20% to the laboratory. It says nothing about which topic appears on which paper, so no claim is made here about that. What can be said is that a momentum question usually arrives welded to an energy question: the collision is one line, and the ramp or spring or friction on either side of it is last week's work.
How much time do you have?
10 minutes
You leave with the two equations that do almost everything, the impulse relation and the conservation of the total, plus the one word in a question that decides whether the kinetic energy equation may also be used.
In 60 seconds · Formula card · Impulse: a force that acts for a time · Conservation of momentum: what the pair keeps · Mistake ledger
45 minutes
You add the two collision types that carry most of the marks, sticking and elastic, the method box that tells them apart, and the fading ladder that makes you write the reasoning yourself instead of reading it.
In 60 seconds · Conventions used here · Momentum: the quantity a force changes at a rate · Impulse: a force that acts for a time · Conservation of momentum: what the pair keeps · Sticking together: the perfectly inelastic collision · Elastic collisions on a line · Method boxes · Scaffolding comes off · Practice B (computation) · Check yourself
full read
Everything above plus collisions that leave the line, which is where the two-equation bookkeeping earns its keep, the centre of mass and the reason it refuses to notice an explosion, the exam-style question in full, and the interleaved set that makes you choose the tool before using it.
The opening pages · Recall first · Try it yourself first · Notation · All seven concept blocks · Method boxes · Contrast pairs · Scaffolding comes off · Full exam-style question · Practice A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the momentum of a body and the change in that momentum when the velocity reverses, keeping the sign that the reversal puts into the answer.
Find an from a momentum change and a contact time, and explain why lengthening the contact time lowers the force without changing the impulse.
Decide whether the momentum of a stated system is conserved by naming the external forces and comparing their impulse with the internal one, then write the conservation line with correct signs.
Solve a collision in which the bodies stick together, find the kinetic energy that disappeared, and combine the collision with an energy calculation on either side of it.
Apply the reversal of relative velocity to an elastic collision on a line and predict what happens for equal, heavy and light masses without solving anything.
Resolve a collision or an explosion into components and solve the two conservation equations for a speed and a direction.
Locate the centre of mass of a set of particles and use the fact that it answers only to external forces to solve a problem no other method reaches.
Syllabus coverage
Momentum and its relation to force, impulse, conservation of momentum for a system, collisions in one dimension both elastic and perfectly inelastic, collisions and explosions in two dimensions, and the centre of mass with the way it moves
The week line carries no chapter numbers, so no chapter number is quoted anywhere in this section. The scope taken is the standard content of that title in the set textbook, spread over the seven blocks named here.
covered
Centre of mass and the motion of a system of particles
The mass weighted average position, and the result that external forces alone decide how it moves
Listed separately because a student scanning the week line for the word collision would not expect it. It belongs here twice over: the course catalogue names the dynamics of a system of particles, and the same textbook chapter carries it.
covered
Systems whose mass changes, such as a rocket
What happens to the momentum bookkeeping when the body throws part of itself away
Named and not developed. It is the closing topic of the same textbook chapter and it follows from the force form of the law, but the week line does not ask for it and no question here needs it. If it appears in your own lecture, the entry point is the relation between force and the rate of change of momentum in the first block.
off_syllabus
Angular momentum and collisions that make things spin
The rotational partner of everything on this page
Deferred to the rotational sections later in the course. Every body on this page is treated as a point, and every collision here is arranged so that nothing is left spinning; a bullet that hits a door off centre is a question for that section, not this one.
deferred
The restitution coefficient
The single number that grades a collision between perfectly elastic and perfectly sticking
Three sentences at the end of the elastic block, because students meet the ratio of separation speed to approach speed in problem sets and it explains why the two extremes on this page are the two ends of one scale. No question here asks for it.
off_syllabus
Recall first
Newton's third law
When body A pushes on body B, body B pushes back on A with a force of the same size in the opposite direction, at every instant of the contact. The two forces act on different bodies, which is why they never cancel in the equation for one body.
This is the whole reason a collision conserves anything: the two forces are equal and opposite and last exactly the same time, so the two momentum changes are equal and opposite.
Newton's second law for a single body
$\sum \vec F = m\vec a$, applied to one named body, with the forces resolved along chosen axes.
The momentum form on this page is this law rewritten, and every worked example that needs a force between the collision and the next stage goes back to it.
Kinetic energy and the work-energy principle
$KE = \tfrac12 mv^{2}$, a scalar, never negative, and $W_{\rm net} = \Delta KE$ for the work done by all forces together.
Every collision here is graded by how much kinetic energy survived it, and half the questions attach an energy calculation to the collision.
Conservation of mechanical energy
$\tfrac12 mv_1^{2} + U_1 = \tfrac12 mv_2^{2} + U_2$ when only conservative forces do work, with $U = mgy$ near the ground and $U = \tfrac12 kx^{2}$ for a spring.
A , a block that runs into a spring and a body that arrives down a smooth ramp are all two-stage problems: this equation for one stage, momentum for the other.
The energy ledger with friction
$\tfrac12 mv_1^{2} + U_1 = \tfrac12 mv_2^{2} + U_2 + f_k d$, where $f_k = \mu_k N$ and $d$ is the distance travelled along the surface.
The most common exam question in this section ends with the wreckage sliding to a stop, and this is the line that finds how far.
Resolving a vector into components
A vector of magnitude $A$ at an angle $\theta$ to the $x$ axis has components $A\cos\theta$ and $A\sin\theta$; a vector is rebuilt from its components by $A=\sqrt{A_x^{2}+A_y^{2}}$ and $\tan\theta = A_y/A_x$.
Momentum is a vector, so a collision that leaves the line becomes two independent equations, one for each component.
Try it yourself first (3 questions)
1§04 recall — forces in a head-on crash●●○○○
Three short questions from the sections behind this one, to find what needs rereading before you start. Getting one wrong costs nothing. First: a 20000 kg lorry runs head-on into a 1000 kg car, and the car is wrecked while the lorry is dented.
Given
lorry $2.00\times10^{4}$ kg, car $1.00\times10^{3}$ kg
the two are in contact for the same interval
the damage to the car is far worse
Find
(a) During the contact, which vehicle exerts the larger force on the other?
Hint 1/4
The question asks about the two forces, not about the two accelerations. Separate them before you answer, because the damage you can see is about the second pair, not the first.
Hint 2/4
Newton's third law: the force of A on B and the force of B on A are equal in size and opposite in direction, whatever the masses are.
Hint 3/4
With a force $F$ acting on each, the accelerations are $F/1.00\times10^{3}$ for the car and $F/2.00\times10^{4}$ for the lorry: the same force, twenty times the acceleration.
Hint 4/4
The two forces are equal. The car is wrecked because it has less mass to resist the same force, not because it is pushed harder.
Show solution
Answering with the third law first and the second law second keeps the force question and the damage question apart, which is exactly where the intuition fails.
Test the alternative: if the lorry really pushed harder, the pair of forces would not sum to zero and the two vehicles taken together would gain momentum with nothing outside pushing them, which no crash has ever done.
2§03 recall — combining two perpendicular vectors●●○○○
Second question, and it is pure vector work. One vector of size 24000 units points east and a second of size 21600 units points north.
Given
east component $2.40\times10^{4}$
north component $2.16\times10^{4}$
the two are at right angles
Find
(a) Find the size of the sum of the two vectors.
Hint 1/4
Draw them nose to tail. The sum is the hypotenuse of a right-angled triangle, so it must be longer than either one and shorter than their arithmetic total.
Hint 2/4
For perpendicular components, $A = \sqrt{A_x^{2}+A_y^{2}}$.
Hint 3/4
Here $A_x = 2.40\times10^{4}$ and $A_y = 2.16\times10^{4}$, so the squares are $5.76\times10^{8}$ and $4.67\times10^{8}$.
Hint 4/4
The sum has size $\sqrt{1.043\times10^{9}} = 3.23\times10^{4}$ units, and it points at $42.0^{\circ}$ north of east.
Show solution
Squaring first is safer than drawing to scale, and the bracket between the longer side and the plain sum catches the two usual errors at once.
measured from the east direction because the east component was put on the bottom
Answer $$\boxed{\;A = 3.23\times10^{4}\ \text{units at }42.0^{\circ}\ \text{north of east}\;}$$
Check
Bracket check: the answer must lie between the longer component, 24000, and the plain sum, 45600, and 32300 sits between them. The angle is a little under 45 degrees, as it should be when the east component is the larger.
3§09 recall — how far a sliding block travels●●○○○
Third question, and this one is the second half of half the problems in this section. A block is moving at 4.00 m/s along a horizontal floor when it meets a rough patch with a coefficient of kinetic friction of 0.250, and it slides to a stop.
Given
$v = 4.00\ \mathrm{m/s}$
$\mu_k = 0.250$
$g = 9.80\ \mathrm{m/s^{2}}$
horizontal floor, no other horizontal force
Find
(a) Find the distance the block slides before stopping.
Hint 1/4
The block arrives with a store of kinetic energy and leaves with none. Something removed it at a fixed rate per metre; find that rate first.
Hint 2/4
The energy ledger with friction: $\tfrac12 mv^{2} = f_k d$ with $f_k = \mu_k mg$ on a horizontal floor.
Hint 3/4
The mass cancels, leaving $d = v^{2}/(2\mu_k g)$ with $v = 4.00$ m/s, $\mu_k = 0.250$ and $g = 9.80$ m/s$^{2}$.
Hint 4/4
$d = 16.0/(2\times0.250\times9.80) = 16.0/4.90 = 3.27$ m.
Show solution
The energy route is chosen because the question mentions no time; the kinematic route is kept for the check.
Write the ledger for the slide
$$\tfrac12 mv^{2} = \mu_k m g\,d$$
the whole kinetic energy is removed by friction and there is no height change to store any of it
the mass cancels from both sides because friction here is proportional to the weight
Answer $$\boxed{\;d = 3.27\ \mathrm{m}\;}$$
Check
Second route with kinematics, which is legal here because the friction force is constant: $a = -\mu_k g = -2.45$ m/s$^{2}$, and $v^{2}=2ad$ gives $d = 16.0/4.90 = 3.27$ m. Order of magnitude: a shove across a rough floor stopping in about three metres is what the number claims, and that is believable.
Notation
symbol
reads as
means
watch out
$\vec p$
pee vector
the linear momentum of a body, mass times velocity, in kilogram metres per second
A vector. It points wherever the velocity points, and in one dimension its sign is the direction.
$\Delta \vec p$
delta pee
the change in momentum, final minus initial
Final minus initial, never the other way round, and never the difference of the two speeds ignoring direction.
$\vec J$
jay
impulse, the effect of a force acting for a time, in
One newton second is one kilogram metre per second; the two units are the same thing wearing different clothes.
$\bar{F}$
eff bar
the average force during a contact, the constant force that would deliver the same impulse in the same time
Not the peak force. In a real collision the peak is roughly twice the average.
$v_1, v_2$
vee one, vee two
the velocities of the two bodies before the collision
Both carry signs from the direction chosen as positive, and both are velocities of the same instant.
$v_1', v_2'$
vee one prime, vee two prime
the velocities of the same two bodies after the collision
The prime is the only thing separating before from after in an equation, so it is worth writing clearly.
$v'$
vee prime
the single common velocity when two bodies stick together
One symbol without a body number means exactly that: there is only one velocity left.
$M$
big em
the total mass of the system, the sum of the parts
In a sticking collision the denominator is this total, not the mass of one body.
$x_{\rm cm}$
ex see em
the position of the centre of mass along the chosen axis
A mass weighted average, not a midpoint, and it need not sit inside any of the bodies.
$\vec v_{\rm cm}$
vee see em
the velocity of the centre of mass, the total momentum divided by the total mass
In an isolated system this is a constant, and it is the same before, during and after a collision.
Conventions used here
The positive direction, declared before any momentum line
Every problem here names one direction as positive and keeps it for the whole problem: to the right, or east, or the direction the first body was moving. A velocity against that direction goes into the equation with a minus sign, and a final answer that comes out negative means the body moves the other way. That last sentence is a result, not an error.
Momentum is a vector and this section adds momenta constantly; a single unsigned speed in a sum is the most common wrong line on the page.
Naming the system before writing the conservation line
Before the words momentum is conserved appear, the solution says which bodies are in the system. A force is internal or external only with respect to a stated system, and the same force can be either: the push between two skaters is internal to the pair and external to one skater.
Conservation is a property of a system, not of a body, and half the wrong answers in this section come from a system that was never named.
What during the collision means for gravity and friction
When a collision is described as brief, the external forces are ignored across it. Gravity acting on a 1.2 kg block for 2 ms delivers an impulse of about 0.024 kg m/s, while the internal impulse in the same 2 ms is of the order of tens of kilogram metres per second. Momentum is therefore taken as conserved through the contact even when the system is falling, and the external forces are put back for the motion before and after.
It is an approximation, not a law, and a student who is not shown the two numbers cannot tell when it stops being safe.
The force from a surface against the net force on the body
A question that asks for the force from the ground, the floor or the wall wants the contact force itself, and that is the net force plus whatever else acts during the contact. In a vertical landing or bounce the extra term is the weight: $\bar N = \Delta p/\Delta t + mg$. In a violent stop the weight is a fraction of a per cent of the answer and is often dropped, but in a soft stop it is a quarter of it, so it is carried everywhere on this page.
The impulse equation delivers the net force and nothing else, and a solution that boxes it under the label force from the ground is wrong by exactly one weight.
Constants and data fixed for every calculation here
$g = 9.80\ \mathrm{m/s^{2}}$ near the ground, as in every earlier section, and no block quietly rounds it to 10. Masses and speeds are read from the problem; nothing is imported from outside the question.
A constant that drifts between blocks makes two correct routes disagree in the third digit and destroys every check you try to run.
Digits kept in a momentum answer
Answers are quoted to three significant figures and the intermediate values are carried at more, so a printed 7.14 m/s may be 7.1429 m/s inside the next line. Where a squared quantity follows, the unrounded value is used; a number rounded to two digits and then squared is how a check that should agree ends up disagreeing.
This section chains a momentum step into an energy step constantly, and the energy step squares whatever the momentum step produced.
What is idealised away in every collision on this page
Bodies are treated as points, so nothing is left spinning and no energy goes into rotation. Surfaces called frictionless have no friction, air resistance is absent unless the question puts it there, and strings and springs have no mass of their own. Where one of these would visibly change the answer it is said in the example rather than left for you to find.
Spinning is real and is the first thing a real collision produces; a student who is not told it was dropped cannot tell an idealisation from an error.
11.1Momentum: the quantity a force changes at a rate
Mass times velocity is what a force actually changes, and the rate of that change is the force itself.
Two sections asked what a force does over a distance; this one asks what it does over a time.
Solvable with what we have
Find the acceleration of a crate under a known constant tension.
Find the speed at the foot of a smooth ramp of known height.
Find how far a block slides against a known friction force.
Not solvable yet
Find the force a bat puts on a ball during 1.5 ms of contact.
Find the speed of two wagons after they couple, the coupling force unknown.
Find where the pieces of a shell go, the explosive force unknown.
The second law first. The ball arrives at $+40.0$ m/s and leaves at $-45.0$ m/s, so pretending the acceleration is constant, $a = (-45.0-40.0)/0.0015 = -5.67\times10^{4}$ m/s$^{2}$ and $F = ma = -8.2\times10^{3}$ N. Now the energy route: $\tfrac12(0.145)(45.0)^{2}-\tfrac12(0.145)(40.0)^{2} = +30.8$ J, which divided by an unknown contact distance gives a force we cannot evaluate.
Why it fails
The first line is not wrong so much as unjustified: the force in a bat strike rises and dies inside a millisecond, so there is no constant acceleration and $a = \Delta v/\Delta t$ is an average nobody has justified. The second fails outright: energy needs the distance the force acted over, and the question gives a time. Both circle the same product $m\Delta v$, which is worth a name.
DefinitionDefinition 11.1: linear momentum, and the law it satisfies
Conditions
The momentum of a body is defined for any velocity, constant or not
The force form holds for a single body or for a whole system, as long as the same body or system is meant on both sides
The reduction to $m\vec a$ needs the mass to be constant, which is true for every body in this section
$$\boxed{\;\vec p = m\vec v\;;\qquad \sum \vec F = \frac{d\vec p}{dt}\;}$$
The momentum of a body is its mass times its velocity: a vector along the motion, and doubling either factor doubles it. The second line says a net force is nothing but the rate at which that momentum changes, so one newton for one second changes it by one kilogram metre per second.
Why this is the same second law you already have
Differentiate the definition with the mass held constant: $d\vec p/dt = d(m\vec v)/dt = m\,d\vec v/dt = m\vec a$. So for a body of fixed mass the new statement and $\sum\vec F = m\vec a$ say the same thing. The point is not new physics but a new shape: written this way, the law survives when the mass changes, and it integrates over time into something useful.
The ball at the bat, on a momentum axis. It arrives $\textcolor{#1f6feb}{+5.80}$ and leaves $\textcolor{#d1690a}{-6.53}$ kg m/s, so the $\textcolor{#116329}{\text{change}}$ is $-12.3$ kg m/s: longer than either arrow, because the two point opposite ways.
Looks like this, but is not
Momentum is another name for kinetic energy, so more of one means more of the other.
Take a 0.0100 kg bullet at 400 m/s and a 4.00 kg medicine ball at 1.00 m/s. Both carry $p = 4.00$ kg m/s, while their kinetic energies, $800$ J and $2.00$ J, are four hundred times apart. Equal momentum means the same impulse stops either; equal energy would mean the same work does. Speed enters them once and twice, which is why this section cannot be done with energy.
The lorry and the sports car that are equally hard to stop
A 1200 kg car travels at 10.0 m/s and a 20000 kg lorry crawls through a yard at 0.600 m/s. Find the momentum and the kinetic energy of each, and say what each pair of numbers means for the driver of a vehicle that has to stop them.
Ratio check without arithmetic: momentum goes as $mv$ and the two products are $1200\times10.0$ and $20000\times0.600$, both $1.20\times10^{4}$; energy goes as $mv^{2}$, so the car beats the lorry by the extra factor of $10.0/0.600 = 16.7$, and $6.00\times10^{4}/3.60\times10^{3}$ is indeed 16.7.
Four multiplications, no algebra.
Momentum measures how much force-time it takes to stop something; kinetic energy measures how much force-distance. A question that gives you a time wants the first, and a question that gives you a distance wants the second.
Braking force on a car from the rate its momentum falls
A 1000 kg car slows steadily from 25.0 m/s to rest in 8.00 s on a straight level road. Find the average net force on it, working from the momentum rather than from the acceleration.
Given
$m = 1000\ \mathrm{kg}$
$v_1 = 25.0\ \mathrm{m/s}$, $v_2 = 0$
$\Delta t = 8.00\ \mathrm{s}$
direction of travel positive
Find
the average net force on the car
Solution
The momentum route is used because it needs only the two velocities and the time, and it does not care whether the deceleration was steady.
the state before, with the sign of the direction of travel
$$p_2 = 0$$
at rest, so nothing is left to carry
Divide the change by the time
$$\bar F = \frac{\Delta p}{\Delta t} = \frac{0-2.50\times10^{4}}{8.00}$$
the force form of the second law, used with an average because the exact shape of the braking is unknown
$$\bar F = -3.13\times10^{3}\ \mathrm{N}$$
negative, meaning backwards along the motion, which is what a brake does
Answer $$\boxed{\;\bar F = -3.13\times10^{3}\ \mathrm{N}\;}$$
Check
Independent route through the acceleration: $a = (0-25.0)/8.00 = -3.125$ m/s$^{2}$ and $F = ma = -3.13\times10^{3}$ N, the same number. Order of magnitude: 3130 N is about a third of the 9800 N weight of the car, so the deceleration is about a third of $g$, which is comfortable braking rather than an emergency stop.
Two lines.
For a body of fixed mass the two routes always agree, so use whichever the question feeds you. The momentum route starts to earn its keep the moment the force stops being constant.
Checkpoint
§11.1 — momentum against kinetic energy●○○○○
Thirty seconds, no calculator needed. A 2.00 kg ball moving at 3.00 m/s to the right and a 6.00 kg ball moving at 1.00 m/s to the right are compared.
Given
ball A: 2.00 kg at 3.00 m/s right
ball B: 6.00 kg at 1.00 m/s right
Find
(a) Which statement about the pair is correct?
Hint 1/4
Two quantities are being compared, and they are built from the same two numbers in different ways. Work out both for each ball before choosing.
Hint 2/4
$p = mv$ and $KE = \tfrac12 mv^{2}$: the speed appears once in the first and twice in the second.
Hint 3/4
Ball A: $p = 6.00$ kg m/s and $KE = 9.00$ J. Ball B: $p = 6.00$ kg m/s and $KE = 3.00$ J.
Hint 4/4
The momenta are equal and the kinetic energy of the lighter, faster ball is three times the other.
Show solution
Computing both quantities for both balls is four multiplications and removes all guesswork about which one the question is testing.
Ratio argument with no numbers: the two balls have equal momenta by construction, and for equal momenta the kinetic energy is $p^{2}/2m$, which is larger for the smaller mass. Ball A has one third of the mass, so three times the energy.
⚠ Treating momentum as a size with no direction
Kinetic energy has no direction and was the quantity in use for two whole sections, so the habit carries over.
⚠ Assuming that more momentum means more kinetic energy
Both grow with mass and with speed, so they feel like the same idea measured in different units.
wrong$$p_A > p_B \;\Rightarrow\; KE_A > KE_B$$
right$$KE = \frac{p^{2}}{2m}\;\text{: at equal }p\text{, the smaller mass carries the greater }KE$$
11.2Impulse: a force that acts for a time
Multiply a force by how long it acts and you get the momentum change it produces, whatever shape the force had.
The force form of the law says how fast momentum changes; multiply both sides by a time interval and it says how much it changed, which is the question a collision actually asks.
TheoremTheorem 11.2: the impulse-momentum principle
Conditions
The force is the net force on the body during the interval
The interval runs from a named instant to a named instant, and the momenta are taken at those two instants
Nothing is assumed about the shape of the force: it may spike, wobble or vanish inside the interval
The average force means the constant force that would produce the same impulse in the same time, not the peak
$$\boxed{\;\vec J = \int_{t_1}^{t_2}\!\sum\vec F\,dt = \bar{\vec F}\,\Delta t = \Delta\vec p = m\vec v_2 - m\vec v_1\;}$$
The area under the graph of net force against time, over any interval you like, equals the momentum the body gained during that interval. If the force is complicated, replace it by the flat force with the same area, and that flat value is what we call the average force. Read backwards it is a recipe: measure how much the momentum changed and how long the contact lasted, divide, and you have the average force, without ever knowing what the real force was doing inside the millisecond.
One integration
Start from $\sum\vec F = d\vec p/dt$ and integrate both sides over the contact: $\int_{t_1}^{t_2}\sum\vec F\,dt = \int_{p_1}^{p_2} d\vec p = \vec p_2-\vec p_1$. The right-hand side needs only the two end states, which is why the messy middle never has to be known. The average force is then defined so that $\bar{\vec F}\Delta t$ reproduces the same integral, which makes the definition useful rather than arbitrary.
The bat strike as a graph. The $\textcolor{#1f6feb}{\text{real force}}$ rises to about 12.9 kN and dies away in 1.5 ms; the $\textcolor{#d1690a}{\text{dashed rectangle}}$ has the same area, so its height 8.22 kN is the average force. Same area means same impulse means same final speed.
Looks like this, but is not
Landing with bent knees is gentler, so the impulse on the body is smaller.
The impulse is fixed before the landing begins. A 65.0 kg person arriving at 6.00 m/s and ending at rest has $|\Delta p| = 390$ kg m/s, and no amount of bending changes either of those two numbers. What bending changes is the time over which the 390 is delivered: about 0.010 s with locked legs, about 0.20 s with a deep bend. The impulse is the same, the average net force is 39.0 kN in the first case and 1.95 kN in the second, twenty times smaller. Everything protective in engineering, from airbags to crash barriers to landing on your feet, works this way: it cannot touch the impulse, so it stretches the time.
Force of the bat on the ball, from 1.5 milliseconds
A 0.145 kg ball arrives at the bat at 40.0 m/s, and leaves in the opposite direction at 45.0 m/s. The contact lasts 1.5 ms. Find the impulse the bat delivered and the average force it exerted, and check whether ignoring the weight of the ball during the contact was fair.
Given
$m = 0.145\ \mathrm{kg}$
$v_1 = +40.0\ \mathrm{m/s}$ towards the bat, taken as positive
$v_2 = -45.0\ \mathrm{m/s}$, back the way it came
$\Delta t = 1.5\times10^{-3}\ \mathrm{s}$
Find
the impulse and the average force, and whether the weight matters
Solution
The impulse route is chosen because the question supplies a time and no distance; an energy calculation here would need the compression of the ball, which nobody measured.
the bat force is nearly six thousand times the weight, so leaving the weight out changes nothing at three digits
Answer $$\boxed{\;J = -12.3\ \mathrm{kg\,m/s},\qquad \bar F = -8.2\times10^{3}\ \mathrm{N}\;}$$
Check
Independent check by the size of the answer: 8200 N is the weight of an 840 kg mass, roughly a small car resting on the ball for the millisecond of contact, which matches the fact that a struck ball deforms visibly. A second check on the sign: the ball ends up moving in the negative direction, so the net force on it had to be negative.
Three lines, plus one line of scepticism about the weight.
Whenever a question gives you a contact time, the answer is the momentum change divided by that time, and the hardest part of the whole calculation is remembering that the incoming and outgoing velocities have opposite signs.
Landing on stiff legs against landing with a deep bend
A 65.0 kg person drops from a wall and reaches the ground at 6.00 m/s. Find the average force from the ground if the body is brought to rest in 0.010 s with locked knees, and again if a deep bend stretches the stop to 0.20 s. Compare each with the weight of the person.
Given
$m = 65.0\ \mathrm{kg}$
arrival speed $6.00\ \mathrm{m/s}$ downwards
final speed zero
stopping times $0.010\ \mathrm{s}$ and $0.20\ \mathrm{s}$
downwards is negative here
Find
the average force the ground exerts in each case, and its size compared with the weight
Solution
The two cases are run with the same impulse and different times on purpose, because that is the entire content of the safety argument.
The impulse, and what the ground has to supply
$$\Delta p = m(0) - m(-6.00) = +390\ \mathrm{kg\,m/s}$$
positive because the ground has to push the person upwards to stop a downward motion
the force from the ground is not the net force: gravity keeps pulling all through the stop, so the ground supplies the deceleration and the weight together
the weight adds under two per cent here, which is why it is so often dropped in the stiff case; sixty-two times body weight is the region where bones break
the same 637 N is now a quarter of the answer, so leaving it out would report a force 25 per cent too small: the softer the landing, the less the weight may be ignored
Second route through distance instead of time: at an average speed of 3.00 m/s the stiff landing lasts 0.010 s and so compresses through 0.030 m. Over that distance the ground removes the kinetic energy while gravity keeps adding to it, so $\bar N = KE/d + mg = 1170/0.030 + 637 = 3.96\times10^{4}$ N, the same figure by a route that never mentions time.
One impulse, two divisions, one weight added twice.
The design rule behind airbags, crash barriers, packaging and gymnastics mats is one line long: you cannot change the impulse, so buy time.
Checkpoint
§11.2 — what a soft landing changes●●○○○
Thirty seconds. The same egg is dropped from the same height twice, once onto a stone floor where it breaks and once onto a thick cushion where it survives, and in both cases it ends at rest.
Given
same egg, same drop height, so the same arrival speed
final state at rest in both cases
stone floor in one trial, thick cushion in the other
Find
(a) Which quantity is the same in the two trials?
Hint 1/4
Write down what you know about the two ends of the motion in each trial, and notice that the two ends are identical in the two trials.
Hint 2/4
The impulse equals the momentum change, and the momentum change is fixed by the two end states alone.
Hint 3/4
Same egg and same arrival speed mean the same momentum before, and rest means zero momentum after, in both trials.
Hint 4/4
The impulse is identical; the cushion stretches the time, so the average force falls, and that is why the egg survives.
Show solution
The question is settled by comparing end states rather than by imagining the landing, which is what makes it a thirty second question.
Compare the ends, not the middles
$$\Delta p = 0 - m v_{\rm arrival}$$
identical in the two trials because both the mass and the arrival speed are identical
$$J = \Delta p \ \text{is therefore identical}$$
the impulse depends on nothing except the two end states
the same numerator over a much larger denominator gives a much smaller force
Answer $$\boxed{\;J\ \text{is the same; } \bar F\ \text{is not}\;}$$
Check
Limit test: if the cushion changed the impulse, an infinitely soft cushion would deliver no impulse at all, and the egg would never stop. It does stop, so the impulse survives and only the time is negotiable.
⚠ Using the change of speed instead of the change of momentum
The word impulse sounds like a property of the motion, and the mass is easy to leave in the problem statement.
wrong$$J = \Delta v\,\Delta t$$
right$$J = m\,\Delta v = \bar F\,\Delta t$$
⚠ Believing that a longer contact means a smaller impulse
A soft landing feels gentler in every respect, so every quantity in it feels smaller.
Add up the momentum of every body in the system, as vectors, before anything happens, and add them up again afterwards. If nothing outside the system pushed on it, the two totals are equal. The bodies inside may have hit each other, stuck, exploded or bounced; those forces are internal and they only move momentum from one member of the system to another.
The third law, integrated over the contact
During the contact, cart A feels $\vec F_{BA}$ and cart B feels $\vec F_{AB} = -\vec F_{BA}$, at every instant. Integrating each over the same interval gives $\Delta\vec p_A = \int \vec F_{BA}\,dt$ and $\Delta\vec p_B = -\int\vec F_{BA}\,dt$, so $\Delta\vec p_A + \Delta\vec p_B = 0$ no matter what the force did in between. The two curves in the figure are that statement drawn: the same shape, mirrored, so the two areas cancel.
Why anything is conserved at all. The $\textcolor{#1f6feb}{\text{force on B}}$ and the $\textcolor{#d1690a}{\text{force on A}}$ are equal and opposite at every instant and last exactly the same time, so the two shaded areas, which are the two impulses, are equal and opposite: whatever B gains, A loses. The numbers are one laboratory run, cart A of 4.10 kg dropping from 6.00 to 3.00 m/s against cart B in a contact of 1.5 ms, so each area is $(4.10)(3.00) = 12.3$ N s and the peak force is about 13 kN.
Looks like this, but is not
Momentum is conserved in every collision, so it can be applied to a ball bouncing off a wall.
Take the ball alone as the system. It arrives at $+8.00$ m/s and leaves at $-8.00$ m/s, so its momentum has changed sign and certainly is not constant. Nothing is wrong with the law: the wall pushed on the ball, and the wall is outside the chosen system, so an external force acted. Enlarge the system to ball plus wall plus building plus planet and the total is conserved again; the Earth picks up the missing momentum and, with a mass of $6\times10^{24}$ kg, moves at a speed no instrument will ever record. The lesson is that conservation is a statement about a system, and if you have not said what the system is, you have not said anything.
Throwing a ball while standing on ice
A 60.0 kg skater stands at rest on frictionless ice holding a 2.00 kg ball, and throws it horizontally at 8.00 m/s. Find the velocity of the skater afterwards, and find where the kinetic energy came from.
Given
skater $60.0\ \mathrm{kg}$, ball $2.00\ \mathrm{kg}$, both at rest at the start
ball leaves at $8.00\ \mathrm{m/s}$ horizontally
frictionless ice, so no horizontal external force
direction of the throw taken as positive
Find
the velocity of the skater, and the source of the kinetic energy
Solution
Momentum is the only tool available: the force of the arm on the ball is unknown, and the energy is not conserved because the skater is a source.
the total after the throw, with almost all of it in the ball
$$KE_{\rm before} = 0 \;\Rightarrow\; 66.1\ \mathrm{J}\ \text{came from the skater}$$
chemical energy in muscle, which is outside the mechanical account and is why energy is not conserved here while momentum is
Answer $$\boxed{\;v_s' = -0.267\ \mathrm{m/s}\ \text{(backwards)},\qquad 66.1\ \mathrm{J}\ \text{supplied by the skater}\;}$$
Check
Momentum check by substitution: $(2.00)(8.00) + (60.0)(-0.267) = 16.0 - 16.0 = 0$, the value it started with. Order of magnitude: 0.267 m/s is a slow walk backwards, and a 60 kg adult throwing a 2 kg ball hard would indeed slide gently on ice rather than be thrown off their feet.
One conservation line and one division.
Every recoil problem, from a rifle to a rocket to a person stepping off a boat, is this calculation with different numbers: total momentum zero before, so the two afterwards are equal in size and opposite in direction.
The second cart, from the momentum the first one lost
On a horizontal air track a 0.500 kg cart moving at 2.40 m/s runs into a 0.300 kg cart at rest. After the collision the 0.500 kg cart is still moving forwards at 0.900 m/s. Find the velocity of the second cart, and decide whether the collision was elastic.
the velocity of the second cart and the fate of the kinetic energy
Solution
The momentum equation is written before any thought is given to energy, because it holds whether or not the collision is elastic; the energy is then a diagnosis rather than an assumption.
Conservation line for the pair
$$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$$
the air track removes the only external horizontal force, so the total is unchanged
Independent check with relative speeds: the carts approached at $2.40$ m/s and separate at $2.50-0.900 = 1.60$ m/s. A separation speed smaller than the approach speed is exactly what a collision that loses energy must give, and a value larger than the approach speed would have meant an arithmetic error.
One conservation line, then two kinetic energies to grade it.
Momentum tells you what happened; energy tells you what kind of collision it was. Never use the second to find the first unless the question uses the word elastic.
Checkpoint
§11.3 — choosing a system whose momentum is conserved●●○○○
Thirty seconds, no arithmetic. Four situations are described, and in only one of them is the total momentum of the stated system constant during the event.
Given
each situation names its own system
each event lasts a short time
the question is about the total momentum of the named system
Find
(a) In which case is the total momentum of the stated system conserved?
Hint 1/4
For each case, list the forces that come from outside the named system. Do not think about the collision itself at all.
Hint 2/4
The total is constant when the external forces sum to zero, or when their impulse over the short interval is negligible beside the internal one.
Hint 3/4
A wall or the ground pushing horizontally is external and fatal; gravity acting for two milliseconds delivers an impulse thousands of times smaller than a collision does.
Hint 4/4
The two carts on the air track qualify: the only external forces are vertical and they cancel, so nothing acts along the line of motion.
Show solution
Listing external forces is faster and more reliable than picturing the event, because a violent event can still have a very short external force list.
Apply one test to each case
$$\text{list } \vec F_{\rm ext}\ \text{on the named system}$$
the same force can be internal or external, so the system decides the answer
$$\sum \vec F_{\rm ext} = 0 \ \text{along the line of motion?}$$
only the component along the motion can change the total that the question asks about
Run the test
$$\text{carts on air track: only vertical forces, and they cancel}$$
so the horizontal total is constant, which is what is being asked
$$\text{ball and wall, sliding crate, falling body: an external push acts}$$
in each of those a body outside the system supplies a horizontal impulse
Answer $$\boxed{\;\text{the two carts on the air track}\;}$$
Check
Cross-check on the rejected cases: in each of them the momentum of the named system is visibly different afterwards, and in each of them you can point at the outside body that took the difference.
⚠ Writing the conservation line without naming the system
The equation looks the same in every problem, so it gets copied before the question of what is inside it is asked.
wrong$$\text{ball on a wall: } mv = mv'\;\Rightarrow\; v' = v$$
right$$\text{ball alone is not isolated: } \Delta p_{\rm ball} = J_{\rm wall\ on\ ball} \ne 0$$
⚠ Adding momenta without signs when the bodies approach each other
The word total suggests adding sizes, and speeds are quoted in questions as positive numbers.
right$$p_{\rm tot} = (3)(+4)+(2)(-5) = +2\ \mathrm{kg\,m/s}\ \text{when they approach}$$
11.4Sticking together: the perfectly inelastic collision
When the bodies leave as one object there is only one unknown velocity, and momentum alone is enough to find it.
The conservation line has two unknown velocities in it and only one equation, which is one short; the simplest way to close the gap is a collision in which the two velocities are forced to be equal.
RuleRule 11.4: bodies that stay together after the collision
Conditions
The bodies stick, couple, embed, lock or are caught: one velocity afterwards, not two
Momentum is conserved through the contact, so the system must be isolated or the contact brief
Kinetic energy is not conserved and must never be assumed to be
The velocities on the right carry their signs; a head-on pair gives a small or even a reversed result
$$\boxed{\;v' = \frac{m_1v_1+m_2v_2}{m_1+m_2}\;;\qquad \Delta KE = -\tfrac12\,\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^{2}\;}$$
The common velocity after a sticking collision is the average of the two velocities before it, weighted by the masses, which is also the velocity of the centre of mass of the pair and always was. The second expression says how much kinetic energy the collision destroyed: it depends on the and never on how fast the pair was travelling as a whole, which is why a collision looks the same to a passenger in either vehicle.
One equation, one unknown, and then the energy that went missing
Conservation gives $m_1v_1+m_2v_2 = (m_1+m_2)v'$ because the two bodies share one velocity afterwards, and dividing by the total mass gives the first result. For the second, put that $v'$ back into $\tfrac12(m_1+m_2)v'^2$ and subtract it from $\tfrac12 m_1v_1^{2}+\tfrac12 m_2v_2^{2}$; the algebra collapses to the product of masses over their sum, times the square of the approach speed. The combination $m_1m_2/(m_1+m_2)$ is smaller than either mass, and it is the factor that decides how expensive a crash is.
Two 10 000 kg wagons couple. The $\textcolor{#116329}{\text{momentum}}$ bar is the same height before and after, which is the law; the $\textcolor{#8250df}{\text{kinetic energy}}$ bar loses exactly half, which is the price of sticking. Nothing here is a contradiction: one is linear in the speed, the other quadratic.
Looks like this, but is not
Kinetic energy disappeared, so momentum must have disappeared with it.
In the coupling above, $2.88$ MJ becomes $1.44$ MJ and half the energy is gone, into crumpled metal, sound and heat. The momentum is $2.40\times10^{5}$ kg m/s before and $(2.00\times10^{4})(12.0) = 2.40\times10^{5}$ kg m/s after, unchanged to the last digit. The reason is that the two quantities have different escape routes. Energy can hide in forms that are not motion, so it can leave the mechanical account without going anywhere in particular. Momentum is a vector tied to mass and velocity; for the total to fall, something outside the system has to push, and the couplings are inside it.
Two railway wagons coupling, and the half that vanishes
A 10 000 kg railway wagon rolls at 24.0 m/s along a level track into an identical wagon standing at rest, and the two couple automatically. Find the speed of the pair afterwards and the kinetic energy the coupling destroyed.
Given
$m_1 = m_2 = 1.00\times10^{4}\ \mathrm{kg}$
$v_1 = 24.0\ \mathrm{m/s}$, $v_2 = 0$
the wagons couple, so one velocity afterwards
level track, brief coupling
Find
the common speed afterwards and the kinetic energy lost
Solution
Momentum is used for the speed because the coupling force is unknown; the energy is computed afterwards as a diagnosis, never as an assumption.
Conservation with one unknown
$$m_1v_1 + m_2v_2 = (m_1+m_2)v'$$
the wagons leave as a single 20 000 kg object, so there is only one velocity to find
twice the mass but a quarter of the squared speed, so half the energy
$$\Delta KE = -1.44\times10^{6}\ \mathrm{J}$$
exactly half, and it went into the buffers, the noise and the warmth of the couplings
Answer $$\boxed{\;v' = 12.0\ \mathrm{m/s},\qquad \Delta KE = -1.44\times10^{6}\ \mathrm{J}\;}$$
Check
Independent check with the loss formula: $\tfrac12\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^{2} = \tfrac12(5.00\times10^{3})(24.0)^{2} = 1.44\times10^{6}$ J, which agrees with the difference of the two kinetic energies although it never computed either of them.
One line for the speed, two for the energy.
Equal masses, one at rest, sticking: half the energy always goes, whatever the numbers. It is worth remembering as a check rather than as a formula.
Speed of a bullet from how high the block swings
A 0.0100 kg bullet is fired into a 2.00 kg wooden block hanging at rest from long light strings, and stays inside it. The block with the bullet in it swings up to a height of 0.100 m above its lowest point. Find the speed of the bullet just before impact.
$g = 9.80\ \mathrm{m/s^{2}}$, strings light, air resistance ignored
Find
the speed of the bullet before the collision
Solution
Working backwards from the height is chosen because the swing is the only stage in which anything was measured; going forwards would need the unknown bullet speed as an input.
Split the problem at the instant the bullet stops moving through the wood
each stage has its own conserved quantity, and mixing them is the classic error here
$$\text{momentum for stage 1},\quad \text{mechanical energy for stage 2}$$
the collision destroys energy but is too brief for gravity to matter; the swing destroys no energy but lasts long enough that momentum is not conserved
Work backwards through the swing
$$\tfrac12 (M+m)v'^{2} = (M+m)gh$$
after the collision only gravity and the string act, and the string does no work
the factor of 201 between the masses is what turns a slow swing into a fast bullet
Answer $$\boxed{\;v = 281\ \mathrm{m/s}\;}$$
Check
Energy audit as an independent check: the bullet arrived with $\tfrac12(0.0100)(281.4)^{2} = 396$ J and the pair left the collision with $\tfrac12(2.010)(1.40)^{2} = 1.97$ J, a surviving fraction of 0.00497, which is exactly $m/(M+m) = 0.0100/2.010$ as the theory of a sticking collision demands. Order of magnitude: 281 m/s is about 1000 km/h, the right region for a handgun.
Two conservation statements, applied to two different stages, in the right order.
Any question in which something fast hits something that then swings, slides or compresses a spring is this two-stage pattern. Draw the dividing line first and label each side with the law that survives there.
Checkpoint
§11.4 — which law belongs to which stage●●○○○
Thirty seconds, no arithmetic. A bullet embeds itself in a hanging block and the block then swings upwards, and a student wants to solve it in one line.
Given
stage one: the bullet embeds in the block, in about a millisecond
stage two: the block with the bullet inside swings up on light strings
the strings are long and light and air resistance is ignored
Find
(a) Which combination of laws is correct for the two stages?
Hint 1/4
Ask, for each stage separately, whether energy escapes and whether an external force acts for long enough to matter.
Hint 2/4
Momentum survives a brief violent contact; mechanical energy survives a smooth slow motion with no friction.
Hint 3/4
In stage one wood is splintered and heated, so energy leaves. In stage two gravity acts for a large fraction of a second, so momentum changes continuously.
Hint 4/4
Momentum for the collision, mechanical energy for the swing, and neither law is allowed to cross the boundary.
Show solution
The stages are examined separately and by the same two questions each time, which is quicker than remembering a rule for pendulums.
Interrogate stage one
$$\Delta t \approx 10^{-3}\ \mathrm{s} \Rightarrow J_{\rm ext} \approx (M+m)g\Delta t \approx 0.02\ \mathrm{kg\,m/s}$$
negligible beside the internal impulse of a few kilogram metres per second, so momentum is conserved
$$\text{wood splinters and heats} \Rightarrow KE\ \text{not conserved}$$
energy leaves the mechanical account in a form no equation on this page tracks
the pair is brought to rest at the top, so its momentum certainly does not stay constant
$$\text{no friction, string does no work} \Rightarrow KE+U\ \text{conserved}$$
which is exactly the condition for the energy line to be legal
Answer $$\boxed{\;\text{momentum for the impact, mechanical energy for the swing}\;}$$
Check
Test by contradiction: assume energy is conserved through the impact as well. Then the pair would leave with 396 J and rise by more than 20 m rather than 0.100 m, which no laboratory pendulum does.
⚠ Conserving kinetic energy through a collision that was not elastic
The previous section conserved energy in every problem, and the habit is only two weeks old.
wrong$$\tfrac12 mv^{2} = \tfrac12 (M+m)v'^{2}\ \text{for a bullet embedding}$$
right$$mv = (M+m)v'\ \text{for the collision, and the energy falls by the factor } \frac{m}{M+m}$$
⚠ Leaving the embedded body out of the mass afterwards
The bullet is small, so it feels safe to ignore, and it is written on the other side of the equation.
wrong$$v' = \frac{mv}{M}$$
right$$v' = \frac{mv}{M+m}$$
11.5Elastic collisions on a line
When the kinetic energy also survives, the speed at which the bodies separate equals the speed at which they approached.
Sticking closed the gap by forcing the two final velocities to be equal; the other way to close it is to add a second conservation law, and that is what the word elastic buys you.
TheoremTheorem 11.5: elastic collision in one dimension
Conditions
The collision is elastic: the total kinetic energy after equals the total before
Momentum is conserved as well, so the system is isolated or the contact is brief
Everything happens along one line, so the velocities are signed numbers rather than vectors with two components
The second boxed pair holds only when the target starts at rest; the first line holds always
In an elastic collision the two bodies separate exactly as fast as they approached, with the direction reversed: if they closed on each other at three metres per second, they part at three metres per second. That statement replaces the kinetic energy equation and it is linear, so a pair of linear equations is all you ever have to solve. The second pair is the answer already worked out for the common case of a target sitting at rest, and the mass ratio in front is the whole story.
Where the relative velocity rule comes from
Write the two conservation laws as $m_1(v_1-v_1') = m_2(v_2'-v_2)$ and $m_1(v_1^{2}-v_1'^{2}) = m_2(v_2'^{2}-v_2^{2})$. Factor the second as a difference of squares and divide it by the first; the masses and the common factors cancel and what is left is $v_1+v_1' = v_2+v_2'$, which rearranges to the relative velocity rule. The division is legal as long as the collision actually happens, that is as long as $v_1 \ne v_1'$. Solving the rule together with the momentum equation, with $v_2=0$, gives the second boxed pair in three lines of algebra.
Three elastic collisions with the target at rest, $\textcolor{#1f6feb}{\text{before}}$ on the left and $\textcolor{#d1690a}{\text{after}}$ on the right. Only the mass ratio changes between the rows, and it decides everything: a swap, a near miss that launches the small one, or a bounce straight back.
Looks like this, but is not
In an elastic collision the two bodies exchange velocities.
True for equal masses and false for every other case. Put $m_1=m_2$ into the boxed pair and the fractions become $0$ and $1$, so the mover stops and the target leaves with the whole of the incoming speed, which is the exchange. Now take $m_1 = 0.500$ kg onto $m_2 = 0.300$ kg at rest with $v_1 = 3.00$ m/s: the formulas give $0.750$ m/s and $3.750$ m/s, and neither body has the other's old velocity. The exchange is a special case of the rule, not the rule itself, and it fails as soon as the masses differ by anything at all.
A 0.500 kg glider striking a 0.300 kg glider at rest
On a frictionless track a 0.500 kg glider moving at 3.00 m/s hits a 0.300 kg glider at rest. The bumpers are magnetic and the collision is elastic. Find both velocities afterwards.
elastic collision, frictionless track, motion along one line
Find
the two velocities after the collision
Solution
The ready-made formulas are used because the target really is at rest; if it were not, they would be wrong and the two-equation route below would be needed.
Use the ready-made pair, since the target is at rest
Two independent checks. Momentum: $(0.500)(0.750)+(0.300)(3.75) = 0.375+1.125 = 1.50$ kg m/s, matching $(0.500)(3.00)$. Kinetic energy: $\tfrac12(0.500)(0.750)^{2}+\tfrac12(0.300)(3.75)^{2} = 0.141+2.109 = 2.25$ J, matching $\tfrac12(0.500)(3.00)^{2}$. Separation speed $3.75-0.750 = 3.00$ m/s equals the approach speed, as the rule demands.
Two substitutions, no algebra.
Notice that the struck body can come out faster than the striker ever was. Nothing is violated: it carries less mass, so its momentum and its energy are both smaller than the totals.
Head-on elastic collision with both bodies moving
A 0.400 kg ball moving to the right at 2.00 m/s meets a 0.600 kg ball moving to the left at 1.00 m/s in a head-on elastic collision. Find both velocities afterwards.
The relative velocity rule is preferred to the energy equation because it is linear: the energy route ends in a quadratic with a spurious root that describes the two bodies passing through each other untouched.
Kinetic energy, which was never used in the solution and is therefore a genuine test: before, $\tfrac12(0.400)(4.00)+\tfrac12(0.600)(1.00) = 1.10$ J; after, $\tfrac12(0.400)(2.56)+\tfrac12(0.600)(1.96) = 0.512+0.588 = 1.10$ J. Momentum after is $(0.400)(-1.60)+(0.600)(1.40) = +0.200$ kg m/s, the value it started with.
Two linear equations and one substitution.
Real collisions sit between the two extremes on this page, and the ratio of separation speed to approach speed is the number that grades them: one for elastic, zero for sticking. It is not needed for anything here, but it is why these two cases are the ends of a single scale rather than two unrelated tricks.
Checkpoint
§11.5 — equal masses, elastic, target at rest●○○○○
Thirty seconds. A snooker ball hits an identical ball squarely, dead centre, and the collision is very nearly elastic.
Given
equal masses
the struck ball is at rest
elastic, head-on
Find
(a) What happens immediately after the impact?
Hint 1/4
Put equal masses into the two conditions you have and see what they force, rather than recalling what snooker balls look like.
Hint 2/4
With $v_2=0$ the first ball leaves at $\frac{m_1-m_2}{m_1+m_2}v_1$ and the second at $\frac{2m_1}{m_1+m_2}v_1$.
Hint 3/4
Equal masses make the first fraction $0/2m$ and the second $2m/2m$.
Hint 4/4
The striker stops dead and the struck ball moves off with the whole of the incoming speed.
Show solution
Substituting into the general result is safer than recalling the snooker fact, because the fact is only true for equal masses.
Substitute equal masses
$$v_1' = \frac{m-m}{2m}v_1 = 0$$
the numerator vanishes, so the striker is left with nothing
$$v_2' = \frac{2m}{2m}v_1 = v_1$$
the whole of the incoming velocity is handed over
Answer $$\boxed{\;v_1' = 0,\qquad v_2' = v_1\;}$$
Check
Both laws hold on their own: momentum $mv_1$ before and $mv_1$ after, kinetic energy $\tfrac12 mv_1^{2}$ before and after. Any other split of the speed between the two balls would satisfy one law and break the other.
⚠ Using the kinetic energy equation and picking the wrong root
The energy equation is the one that was quoted in the definition of elastic, so it feels like the one to solve.
wrong$$v_1' = v_1,\ v_2' = v_2\ \text{(a root of the quadratic)}$$
right$$v_2'-v_1' = v_1-v_2\ \text{, the linear rule, which discards the no-collision root}$$
⚠ Using the target-at-rest formulas when the target is moving
The boxed pair is short and memorable and its condition is written in small print beside it.
wrong$$v_1' = \frac{m_1-m_2}{m_1+m_2}v_1\ \text{ with } v_2 \ne 0$$
right$$\text{solve } m_1v_1+m_2v_2 = m_1v_1'+m_2v_2'\ \text{ together with } v_2'-v_1' = v_1-v_2$$
11.6Collisions that leave the line
A vector law becomes two scalar laws, one for each direction, and the two are solved together.
Every collision so far has been arranged so that everything happens along one line; take that away and nothing new has to be learned, because a vector equation is simply two equations wearing one hat.
RuleRule 11.6: momentum conservation, component by component
Conditions
Axes are chosen and named before any number is written, usually along the initial velocity of one body
The system is isolated in each direction separately: the $x$ sum may be conserved while the $y$ sum is not
Kinetic energy gives a third equation, but only when the collision is stated to be elastic
Angles are measured from a stated axis, and a velocity below that axis carries a negative $y$ component
Resolve every velocity before and after into its component along the first axis and its component along the second. The sum of the mass-times-velocity products along the first axis is the same before and after, and separately the same is true along the second axis. Two unknowns can therefore be found from a two dimensional collision with no extra information, and a third unknown needs the word elastic.
Nothing new: a vector equation is a pair of scalar ones
The conservation law was derived as a vector statement, $\sum\vec p$ constant, and two vectors are equal exactly when each pair of components is equal. Writing the same law twice, once for $x$ and once for $y$, is therefore not an extra assumption. The one thing to be careful about is that the licence can hold in one direction and fail in the other: a ball bouncing off a smooth floor keeps its horizontal momentum, because the floor pushes only upwards, while its vertical momentum reverses.
An intersection collision as a vector sum. $\textcolor{#1f6feb}{\text{Car A}}$ brings $2.25$ and $\textcolor{#d1690a}{\text{car B}}$ brings $2.00$, in units of $10^{4}$ kg m/s, and the $\textcolor{#116329}{\text{total}}$ is the diagonal of the rectangle: $3.01$ at $41.6^{\circ}$. The wreck must slide along that diagonal, which is how the direction of the skid marks becomes evidence.
Looks like this, but is not
The total momentum is the sum of the two sizes, so the wreck carries 4.25 units.
Sizes add only when the vectors point the same way. Here they are at right angles, so the total is $\sqrt{2.25^{2}+2.00^{2}} = 3.01$ units, not $4.25$: the naive answer is 41% too large and would put the speed of the wreck at 17.0 m/s instead of 12.0 m/s. The way to keep this straight is never to add momenta at all, but to add their components, which are ordinary signed numbers and behave the way arithmetic expects. The magnitude is computed once, at the end, from the components that survived.
The break on a snooker table, with the angles measured
A 0.160 kg ball moving east at 2.00 m/s strikes an identical stationary ball off centre. After the elastic collision the first ball moves at $30.0^{\circ}$ north of east and the second at $60.0^{\circ}$ south of east. Find both speeds, and check the collision really was elastic.
Given
$m_1 = m_2 = 0.160\ \mathrm{kg}$
$v_1 = 2.00\ \mathrm{m/s}$ east, second ball at rest
after: ball 1 at $30.0^{\circ}$ north of east, ball 2 at $60.0^{\circ}$ south of east
The energy check above was not used in the solution and therefore counts as independent. A second check: the two outgoing directions are $30.0^{\circ}$ and $-60.0^{\circ}$, separated by $90.0^{\circ}$, and a right angle between the paths is what equal masses in an elastic collision with a stationary target always produce, which is why a snooker player can aim the second ball by aiming the first.
Two component equations and one substitution.
In any two dimensional collision, write both equations before solving either. The one with fewer known numbers in it usually gives the cleanest relation between the unknowns.
Which car was speeding, from the direction of the skid
A 1500 kg car travelling east at 15.0 m/s collides at an intersection with a 1000 kg car travelling north at 20.0 m/s, and the two lock together. Find the velocity of the wreck immediately after the impact, as a speed and a direction.
Given
car A: $1500\ \mathrm{kg}$ east at $15.0\ \mathrm{m/s}$
car B: $1000\ \mathrm{kg}$ north at $20.0\ \mathrm{m/s}$
the cars lock together
east taken as $x$, north as $y$
Find
the speed and direction of the wreck just after the collision
Solution
Components are used rather than a scale drawing because the answer has to be good to three digits and because the same components carry straight on into the next stage of a typical question, the slide of the wreck.
the components are perpendicular, so the magnitude is the hypotenuse
$$\theta = \arctan\frac{8.00}{9.00} = 41.6^{\circ}\ \text{north of east}$$
measured from east because the eastward component was put in the denominator
Answer $$\boxed{\;v' = 12.0\ \mathrm{m/s}\ \text{at}\ 41.6^{\circ}\ \text{north of east}\;}$$
Check
Independent route through the total momentum: $|\vec p| = \sqrt{(2.25\times10^{4})^{2}+(2.00\times10^{4})^{2}} = 3.01\times10^{4}$ kg m/s, and dividing by 2500 kg gives 12.0 m/s, the same answer without ever forming the components of the velocity. The direction is under $45^{\circ}$, which it must be because the eastward momentum is the larger of the two. Kinetic energy falls from $3.69\times10^{5}$ J to $1.81\times10^{5}$ J, which is a crash, not a bounce.
Two components in, two components out, one Pythagoras.
The direction of the wreck depends on the two momenta, not the two speeds, so a heavy slow vehicle can dominate the outcome. This is why investigators can work backwards from skid marks to a speed nobody witnessed.
Checkpoint
§11.6 — adding two perpendicular momenta●○○○○
Thirty seconds. In an isolated system one body carries 3.00 kg m/s due east and a second carries 4.00 kg m/s due north.
Given
3.00 kg m/s east
4.00 kg m/s north
isolated system
Find
(a) What is the total momentum of the system?
Hint 1/4
Sketch the two arrows nose to tail. The total is the closing arrow, and it is shorter than the two laid end to end in a straight line.
Hint 2/4
Add components, then rebuild: $p = \sqrt{p_x^{2}+p_y^{2}}$ for perpendicular contributions.
Hint 3/4
Here $p_x = 3.00$ and $p_y = 4.00$, so the squares are 9.00 and 16.0.
Hint 4/4
The total has magnitude $5.00$ kg m/s, at $53.1^{\circ}$ north of east.
Show solution
Components are kept separate until the last line, which is the habit that makes every two dimensional problem in this section routine.
Keep the components apart, then combine once
$$p_x = 3.00,\qquad p_y = 4.00$$
each body contributes to one axis only, which is what perpendicular means
$$p = \sqrt{9.00+16.0} = 5.00\ \mathrm{kg\,m/s}$$
the magnitude is built at the end, from components that were never mixed
Answer $$\boxed{\;p = 5.00\ \mathrm{kg\,m/s}\ \text{at}\ 53.1^{\circ}\ \text{north of east}\;}$$
Check
Bracket check: the total must be larger than either contribution and smaller than their plain sum, so it lies between 4.00 and 7.00, and 5.00 does.
⚠ Adding the magnitudes of two perpendicular momenta
The word total invites addition, and the two numbers are sitting side by side in the question.
⚠ Writing one conservation equation for a two dimensional collision
One equation was enough on every previous page, and the habit is stronger than the diagram.
wrong$$m_1v_1+m_2v_2 = (m_1+m_2)v'\ \text{with speeds from two directions}$$
right$$\text{one equation for } x,\ \text{a second for } y,\ \text{then } v' = \sqrt{v_x'^{2}+v_y'^{2}}$$
11.7The centre of mass, and why it ignores the collision
One point represents the whole system, and only outside forces can change how that point moves.
Every conservation line written so far has quietly been a statement about a single point, and giving that point a name turns several awkward problems into one-line problems.
DefinitionDefinition 11.7: the centre of mass and the law it obeys
Conditions
The positions are measured from an origin you choose, and the answer moves with that choice while the physics does not
For bodies rather than particles, the same formula holds with each body replaced by a particle at its own centre of mass
The second statement holds for any system whatsoever, rigid or flying apart
Internal forces never appear in it, however violent they are
The centre of mass is the average position of the material, with each position counted in proportion to how much mass sits there, so it lies closer to the heavy end and it need not be inside any body at all. The second statement says that this one point moves exactly as a single particle of the total mass would move under the external forces alone. A shell that bursts in flight has a centre of mass that carries on along the original path as though nothing had happened, because the explosion was internal.
Why the total momentum is the momentum of one point
Differentiate the definition: $M\vec v_{\rm cm} = \sum_i m_i\vec v_i = \vec p_{\rm total}$, so the velocity of the centre of mass is the total momentum divided by the total mass. Differentiate once more and use the force law on each body: $M\vec a_{\rm cm} = \sum_i \vec F_i$, and every internal force in that sum is cancelled by its third-law partner, leaving only the external ones. Conservation of momentum is therefore the special case of the second boxed statement in which the external forces add to zero and $\vec v_{\rm cm}$ is constant.
Position against time for a 1.0 kg body $\textcolor{#1f6feb}{\text{A}}$ that runs into a 3.0 kg body $\textcolor{#d1690a}{\text{B}}$ at rest and sticks to it. Both paths bend at the collision; the $\textcolor{#116329}{\text{centre of mass}}$ does not, because the collision was internal to the pair. After the impact all three lines are the same line.
Looks like this, but is not
The centre of mass is a point of the object, so it is always somewhere in the material.
Take a uniform ring. By symmetry its centre of mass is at the geometric centre, which is a hole. The same happens for a horseshoe, a boomerang and a high jumper arched over the bar, whose centre of mass can pass under the bar while every part of the athlete passes over it. The formula is an average of positions, and an average of two values need not be one of the values: the average of 1 and 9 is 5, which is neither. What the point does have is the property in the second boxed line, and that is a statement about motion rather than about material.
Centre of mass of three bolts on a plate
Three bolts are fixed to a light plate: 1.00 kg at the origin, 2.00 kg at $x = 2.00$ m on the $x$ axis, and 3.00 kg at $y = 3.00$ m on the $y$ axis. Find the coordinates of the centre of mass of the three.
Given
$m_1 = 1.00\ \mathrm{kg}$ at $(0,\,0)$
$m_2 = 2.00\ \mathrm{kg}$ at $(2.00,\,0)$
$m_3 = 3.00\ \mathrm{kg}$ at $(0,\,3.00)$
the plate itself has negligible mass
Find
the coordinates of the centre of mass
Solution
Coordinates are handled separately because the definition treats them separately; there is no shortcut that combines them.
Plausibility check by geometry: the point must lie inside the triangle whose corners are the three bolts, and each coordinate must sit at the fraction of the way across that the masses demand. Both hold: $y_{\rm cm}/3.00 = 0.50$, half way up towards the 3.00 kg bolt because that bolt carries half of the 6.00 kg total, and $x_{\rm cm}/2.00 = 0.333$, a third of the way across towards the 2.00 kg bolt because that bolt carries a third of it.
Two weighted averages.
Moving the origin moves the answer but not the physics: put the origin at the 3.00 kg bolt and the coordinates change, while the distance from that bolt to the centre of mass does not.
Walking to the front of a boat and pushing it backwards
A 60.0 kg person stands at the back of a 40.0 kg boat, both at rest on still water, and walks 3.00 m forward along the boat. Ignoring the resistance of the water, find how far the boat moves and in which direction.
Given
person $60.0\ \mathrm{kg}$, boat $40.0\ \mathrm{kg}$
both at rest at the start
the person moves $3.00\ \mathrm{m}$ relative to the boat, forwards
no horizontal external force from the water
Find
the displacement of the boat relative to the water
Solution
The centre of mass route is chosen because the forces between the feet and the deck are unknown and irrelevant; no force calculation could produce this answer in fewer than several lines.
Substitute back into the definition: $(60.0)(+1.20)+(40.0)(-1.80) = 72.0-72.0 = 0$, so the centre of mass really has not moved. Limit check: make the boat enormously heavy and the formula gives almost no boat motion and a person displacement of nearly the full 3.00 m, which is what walking along a pier feels like.
Two lines once the relative displacement is written correctly.
Whenever a system starts at rest and rearranges itself, the fixed centre of mass gives one equation for free. The only trap is the difference between a displacement measured relative to the moving body and one measured relative to the ground.
Checkpoint
§11.7 — the centre of mass of an exploding shell●●○○○
Thirty seconds. A firework shell is fired and bursts at the top of its flight into many fragments that fly off in all directions, and air resistance is ignored throughout.
Given
the shell follows a projectile path before the burst
the burst is caused by an explosive inside the shell
air resistance is ignored
Find
(a) What does the centre of mass of the fragments do after the burst?
Hint 1/4
Ask which forces on the system of fragments come from outside it. The explosion is not one of them.
Hint 2/4
$\sum\vec F_{\rm ext} = M\vec a_{\rm cm}$, and internal forces cancel in pairs however violent they are.
Hint 3/4
The only external force here is gravity, exactly as before the burst, so the acceleration of the centre of mass is still $g$ downwards.
Hint 4/4
It carries on along the same parabola the unexploded shell would have followed, until the first fragment lands and the ground starts pushing.
Show solution
Listing external forces settles the question in one line, whereas following the fragments would need every fragment velocity.
List the external forces on the system of fragments
$$\sum \vec F_{\rm ext} = M\vec g$$
gravity is the only force from outside; the explosive is inside the system
$$\vec a_{\rm cm} = \vec g$$
unchanged by the burst, so the centre of mass does not even notice it
Answer $$\boxed{\;\text{the same parabola as before the burst}\;}$$
Check
Test with a symmetric case that can be checked by hand: a shell that splits into two equal fragments thrown apart horizontally with equal and opposite velocities. Their average position is the midpoint, and the midpoint follows the original path exactly.
⚠ Taking the midpoint instead of the mass weighted average
The word centre suggests geometry, and for equal masses the two really do agree.
⚠ Expecting the centre of mass to react to an internal event
Explosions and collisions are dramatic, and it is hard to believe that any quantity ignores them.
wrong$$\vec a_{\rm cm}\ \text{changes when the shell bursts}$$
right$$M\vec a_{\rm cm} = \sum\vec F_{\rm ext}\ \text{only, so it does not change}$$
Solving any collision or explosion
Two or more bodies interact for a short time and the question asks for a velocity afterwards, or for a velocity before that would explain what was seen afterwards.
Name the system and the two instants.
Write down which bodies are inside the system, and what just before and just after mean. Most wrong solutions are wrong here rather than in the algebra.
Declare a positive direction, or a pair of axes.
One line: rightwards positive, or east is $x$ and north is $y$. Every velocity in the problem then gets a sign, including the ones the question quoted as positive numbers.
Check the licence for conservation.
List the external forces on the system. If they cancel, or if the contact is brief enough that their impulse is negligible, momentum is conserved. If a wall, the ground or a hand is involved and is not inside the system, it is not.
Write the conservation line in full, then delete the zeros.
$m_1v_1+m_2v_2 = m_1v_1'+m_2v_2'$, one line per axis. Deleting terms that are zero is safer than trying to remember which terms belong.
Count the unknowns, and find the second equation if you need one.
One equation per axis. If there is one unknown too many, the question must have supplied either the word elastic, giving the relative velocity rule, or the fact that the bodies stick, forcing the two final velocities to be equal.
Solve, then check with the other quantity.
Compute the kinetic energy before and after. It must fall for a sticking collision, stay equal for an elastic one, and rise only if something inside the system supplied energy, as in an explosion.
Where it goes wrong
Speeds added without signs when the bodies approach each other.
Kinetic energy conserved in a collision that never used the word elastic.
The struck body left out of the total mass after the two stick together.
A two dimensional collision squeezed into a single equation.
Getting a force out of a contact that lasted milliseconds
The question gives a contact time, or a stopping time, and asks for a force; or it gives a force and asks how long the contact must have lasted.
Write the velocity before and after, with signs.
A rebound reverses the sign, and that single minus sign roughly doubles the answer. It is the most valuable character in the calculation.
Form the momentum change.
$\Delta p = mv_2-mv_1$, final minus initial, in kilogram metres per second.
Divide by the time, or multiply, depending on what is missing.
$\bar F = \Delta p/\Delta t$ for the force, $\Delta t = \Delta p/\bar F$ for the time. There is nothing else in this method.
Test whether the external forces were fairly ignored.
Compare the answer with the weight of the body. If the contact force is hundreds of times larger, ignoring gravity during the contact was safe; if it is comparable, it was not.
Say what the answer means.
Quote it as a multiple of the weight, or compare it with a familiar force. A force of 39 kN on a person means nothing until it is called sixty times body weight.
Where it goes wrong
The reversal of direction ignored, giving an answer many times too small.
The mass left out, so a change of velocity is reported as an impulse.
The peak force and the average force treated as the same number.
Milliseconds not converted to seconds, giving an answer a thousand times out.
Problems that need momentum for one stage and energy for the other
Something fast hits something that afterwards swings, slides, rises or squashes a spring; or a body arrives down a ramp and then collides.
Draw the dividing line and label the instant on it.
The line sits at the moment the contact ends. Call the velocity there $v'$ and treat it as the answer to one stage and the data for the other.
Give each side its own law.
Collision side: momentum, because the contact force is unknown and the interval is short. Smooth side: mechanical energy, because gravity or a spring acts over a distance and does known work.
Start from the stage where something was actually measured.
In a ballistic pendulum that is the height, so the swing is solved first and the collision second; in a bullet into a block on a rough floor it is the sliding distance.
Never let a law cross the line.
Energy is not conserved through the collision and momentum is not conserved through the swing. Writing one equation from the top of the swing to the moment before impact is the standard way to lose all the marks.
Check the size of the energy that was destroyed.
For a body of mass $m$ embedding in a mass $M$ at rest, the surviving fraction of the kinetic energy is $m/(M+m)$. If your numbers disagree with that, one of the two stages has been solved wrongly.
Where it goes wrong
Conservation of energy applied across the impact, giving a bullet speed far too low.
Momentum applied across the swing, giving a height that is far too large.
The two stages solved in the wrong order, so the unknown never cancels.
The mass after the collision taken as the struck body alone.
The same two gliders, elastic
A 0.500 kg glider at 3.00 m/s hits a 0.300 kg glider at rest on a frictionless track. The bumpers are magnetic, so the collision is elastic. Find the two velocities afterwards and the kinetic energy of the pair.
Given
$m_1 = 0.500$ kg at 3.00 m/s, $m_2 = 0.300$ kg at rest
Separation speed is $3.75-0.750 = 3.00$ m/s, equal to the approach speed, as an elastic collision requires.
Two substitutions.
The same two gliders, sticking
The same 0.500 kg glider at 3.00 m/s hits the same 0.300 kg glider at rest, but this time the bumpers carry putty and the gliders stay together. Find the velocity afterwards and the kinetic energy of the pair.
Given
$m_1 = 0.500$ kg at 3.00 m/s, $m_2 = 0.300$ kg at rest
The loss formula gives $\tfrac12\frac{(0.500)(0.300)}{0.800}(3.00)^{2} = 0.844$ J, matching the difference $2.25-1.41$ J.
One division.
Identical masses, identical approach speed, identical conservation of momentum, and yet one collision leaves the pair with 2.25 J and the other with 1.41 J, because in the second the two bodies are forced to end with the same velocity and that costs energy.
How to tell them apart
Read the sentence that describes the bumpers. Magnetic, springy, superball, elastic: use the relative velocity rule. Putty, coupling, embedding, catching, locking: use one velocity for both. If the question says neither, you have been given a final velocity somewhere and you should use momentum alone.
Scaffolding comes off
The common skeleton
Name the system and the two instants, and say which stage of the problem you are in.
Declare the positive direction, or the pair of axes.
Check that no external force spoils the conservation over that interval.
Write the conservation line in full and delete the terms that are zero.
Solve for the one unknown; if there is more than one, find the second condition the question gave you.
Check the answer with the quantity you did not use, or with a limiting case.
1 · fully worked
A bullet embedded in a block, and how far the pair slides
A 0.0250 kg bullet travelling at 350 m/s embeds itself in a 1.20 kg wooden block resting on a rough horizontal floor with a coefficient of kinetic friction of 0.400. Find how far the block slides before stopping.
Given
bullet $0.0250\ \mathrm{kg}$ at $350\ \mathrm{m/s}$
block $1.20\ \mathrm{kg}$, initially at rest
$\mu_k = 0.400$ between block and floor
$g = 9.80\ \mathrm{m/s^{2}}$; the bullet stays inside the block
Find
the sliding distance of the block with the bullet in it
Solution
The stages are solved in the order collision then slide because the data runs that way; a question that gave the sliding distance instead of the bullet speed would be run backwards through the same two steps.
Divide the problem at the instant the bullet stops moving inside the wood
the total mass cancels, which is why the mass of the bullet never appears in this stage
$$d = 6.51\ \mathrm{m}$$
a long slide, because 7.14 m/s is a running speed and the floor is not very rough
Answer $$\boxed{\;d = 6.51\ \mathrm{m}\;}$$
Check
Two independent checks. Energy audit: the bullet arrived with $\tfrac12(0.0250)(350)^{2} = 1531$ J and the pair left the impact with $\tfrac12(1.225)(7.143)^{2} = 31.2$ J, a fraction of 0.0204, which is exactly $m/(m+M) = 0.0250/1.225$. Kinematics: the deceleration during the slide is $\mu_k g = 3.92$ m/s$^{2}$, and $v'^{2} = 2ad$ gives $d = 51.02/7.84 = 6.51$ m.
Two conservation statements, one per stage.
Ninety-eight per cent of the energy of the bullet is destroyed in the wood, and the two per cent that survives is what pushes the block six and a half metres. That ratio is a property of the mass ratio alone.
2 · you write the reasoning
Now an easier problem with the reasoning stripped out. A 2.00 kg trolley moving at 3.00 m/s on a level frictionless track runs into a 4.00 kg trolley at rest, and the two lock together. The steps are given; write the reason beside each one before opening the model answers. The physics is deliberately easier than rung 1, because here the work is the explaining.
System: the two trolleys. State 1: just before contact. State 2: just after they lock. Rightwards positive.
reasoning
Both instants have to exist on paper before an equation can compare them, and the positive direction has to be fixed before any velocity is written down, because a velocity without a declared axis is not yet a number.
No external horizontal force acts on the pair, so the total momentum before equals the total momentum after.
reasoning
The licence is checked, not assumed: gravity and the normal force are vertical and cancel, the track is frictionless, and nothing else touches the pair, so the horizontal total cannot change.
The line becomes $(2.00)(3.00) + 0 = (6.00)v'$.
reasoning
Two terms are deleted for two different reasons: the second trolley is at rest, and the two bodies share one velocity afterwards because they lock, which is what turns two unknowns into one.
So $v' = 1.00$ m/s, and the kinetic energy falls from 9.00 J to 3.00 J.
reasoning
The energy is computed afterwards as a diagnosis rather than used as an assumption; two thirds of it is destroyed in the coupling, which is what the mass ratio predicts and is not an error.
3 · find the buried error
Harder than rung 2, and this solution contains exactly two errors. A 1400 kg car travelling east at 18.0 m/s collides at an intersection with a 2000 kg van travelling north at 12.0 m/s. The two lock together. Find the speed of the wreck immediately after the impact. Find the two wrong steps.
Step 1. East is $x$ and north is $y$. The car carries $p_x = (1400)(18.0) = 2.52\times10^{4}$ kg m/s.
Step 2. The van carries $p_y = (2000)(12.0) = 2.40\times10^{4}$ kg m/s northwards.
Step 3. The total momentum is therefore $2.52\times10^{4} + 2.40\times10^{4} = 4.92\times10^{4}$ kg m/s.
Step 4. Dividing by the mass of the van, $v' = 4.92\times10^{4}/2000 = 24.6$ m/s.
the two buried errors (2)
⚠ step 3
The two momenta are perpendicular, so they cannot be added as numbers. The total is $\sqrt{(2.52\times10^{4})^{2}+(2.40\times10^{4})^{2}} = 3.48\times10^{4}$ kg m/s, not $4.92\times10^{4}$, and it points $43.6^{\circ}$ north of east.
Both quantities are called momentum and both are written as positive numbers on the same page, so the plus sign is written without noticing that one is east and the other north.
right
Keep the components apart and combine only at the end: $p = \sqrt{p_x^{2}+p_y^{2}} = 3.48\times10^{4}$ kg m/s.
⚠ step 4
After the impact the two vehicles move as one body of $1400+2000 = 3400$ kg, so the division must be by 3400 and not by 2000. With the corrected total, $v' = 3.48\times10^{4}/3400 = 10.2$ m/s.
The last mass written down is the one still in mind, and the phrase they lock together is easy to read as a description rather than as an instruction about the mass.
right
$v' = p_{\rm tot}/(m_1+m_2) = 3.48\times10^{4}/3400 = 10.2\ \mathrm{m/s}$ at $43.6^{\circ}$ north of east.
4 · the bare problem
§11.4 — a sticking collision with no scaffolding●●●○○
No steps this time, and no hints until you ask for them. Two trolleys meet on a level frictionless track and their magnetic couplers lock on contact.
Given
trolley 1: $0.400\ \mathrm{kg}$ at $5.00\ \mathrm{m/s}$
trolley 2: $0.600\ \mathrm{kg}$, at rest
the two lock together on contact
level frictionless track
Find
(a) Find the speed of the pair immediately after the collision.
(b) Find the kinetic energy destroyed in the coupling.
Hint 1/4
Two bodies go in and one body comes out, so there is a single unknown velocity. Decide which conserved quantity survives the contact before you write anything.
Hint 2/4
$m_1v_1+m_2v_2 = (m_1+m_2)v'$ for a sticking collision on a frictionless track, and the kinetic energy is computed afterwards rather than conserved.
Hint 3/4
With $m_1 = 0.400$ kg at $5.00$ m/s and $m_2 = 0.600$ kg at rest, the left side is $2.00$ kg m/s and the total mass is $1.00$ kg.
Hint 4/4
$v' = 2.00$ m/s, and the kinetic energy falls from $5.00$ J to $2.00$ J, so $3.00$ J was destroyed.
Show solution
Momentum first, always: it is the only law that is valid across the contact, and the energy question can only be answered once the final speed is known.
Momentum through the contact
$$(0.400)(5.00) + 0 = (1.00)v'$$
one velocity afterwards, because the couplers lock, and no external horizontal force acts
$$v' = 2.00\ \mathrm{m/s}$$
the same 2.00 kg m/s now carried by two and a half times the mass
so 3.00 J went into the couplers as heat and sound
Answer $$\boxed{\;v' = 2.00\ \mathrm{m/s},\qquad \Delta KE = -3.00\ \mathrm{J}\;}$$
Check
Independent check with the loss formula: $\tfrac12\frac{(0.400)(0.600)}{1.00}(5.00)^{2} = \tfrac12(0.240)(25.0) = 3.00$ J, which agrees without computing either kinetic energy.
Full exam-style question
Bullet into a block against a spring, in four partsexam format
A 0.0300 kg bullet is fired horizontally at 400 m/s into a 2.97 kg block resting on a frictionless horizontal surface. The block is in contact with a spring of stiffness 800 N/m whose other end is fixed to a wall. The bullet embeds itself in the block. (a) Find the speed of the block with the bullet inside it immediately after the impact. (b) Find the maximum compression of the spring. (c) Find the percentage of the kinetic energy of the bullet that survives the impact. (d) If the bullet is brought to the common speed in 1.20 ms, find the average force the block exerted on it.
Given
bullet $0.0300\ \mathrm{kg}$ at $400\ \mathrm{m/s}$
block $2.97\ \mathrm{kg}$, at rest, on a frictionless surface
spring stiffness $k = 800\ \mathrm{N/m}$, other end fixed
the bullet embeds; the embedding takes $1.20\ \mathrm{ms}$
Find
the speed after impact, the spring compression, the surviving energy fraction, and the average force on the bullet
Solution
Each part is given the law that survives its own interval, and the parts are done in order because each one feeds the next; attempting (b) before (a) would leave two unknowns in one equation.
(a) The collision, by momentum
$$mv = (m+M)v'$$
the bullet stays inside, so one velocity afterwards; the spring is far too slow to matter in 1.20 ms
the total mass is a round 3.00 kg, which is why the numbers were chosen this way
(b) The compression, by energy
$$\tfrac12 (m+M)v'^{2} = \tfrac12 kx^{2}$$
after the impact the surface is frictionless and the spring is ideal, so mechanical energy is conserved from the impact to the moment of maximum squash
the bullet is the body being asked about, so only its own momentum change is used
$$\bar F = \frac{-11.88}{1.20\times10^{-3}} = -9.90\times10^{3}\ \mathrm{N}$$
negative, meaning backwards along the flight, which is the direction the wood resists
Answer $$\boxed{\;v' = 4.00\ \mathrm{m/s};\quad x = 0.245\ \mathrm{m};\quad 1.00\%\ \text{survives};\quad \bar F = 9.90\times10^{3}\ \mathrm{N}\;}$$
Check
Part (d) checked from the other body, which is a genuinely independent route: the block gains $(2.97)(4.00) = 11.88$ kg m/s in the same 1.20 ms, so the force on it is $+9.90\times10^{3}$ N and the third law is satisfied to the last digit. Part (b) checked for size: 9900 N is about the weight of a tonne, which is the sort of force that stops a bullet, and 0.245 m of compression against 800 N/m needs 196 N, a force a person could apply by leaning on it.
Four parts, three different laws, one shared intermediate value.
Look at how much work the single number 4.00 m/s does: it is the answer to (a), the input to (b), the numerator of (c) and one end of the momentum change in (d). Exam questions are built this way on purpose, which is why an error in part (a) is expensive and why part (a) deserves the check.
Practice
A · concept 4 questions
1§11.1 — equal energy does not mean equal momentum●●○○○
A statement about two bodies is offered for judgement. Decide whether it is always true, and give the reason rather than the verdict alone.
Given
the two bodies have different masses
they have the same kinetic energy
Find
(a) True or false: two bodies with the same kinetic energy must carry the same momentum. Justify.
Hint 1/4
Do not argue in words. Pick two masses that differ a lot, force their kinetic energies to be equal, and compare the momenta you get.
Hint 2/4
$KE = \tfrac12 mv^{2}$ and $p = mv$, so $p = \sqrt{2mKE}$ for a given kinetic energy.
Hint 3/4
With $KE = 100$ J: a 2.00 kg body needs $v = 10.0$ m/s and carries $p = 20.0$; a 50.0 kg body needs $v = 2.00$ m/s and carries $p = 100$.
Hint 4/4
False. At equal kinetic energy the momentum goes as $\sqrt{m}$, so the heavier body always carries more.
Show solution
A single explicit counterexample is enough to demolish a universal claim, and it is faster than any general argument.
Express one quantity in terms of the other
$$KE = \frac{p^{2}}{2m} \;\Longrightarrow\; p = \sqrt{2mKE}$$
eliminating the speed leaves a relation containing the mass, which is the source of the difference
Build the counterexample
$$m = 2.00\ \mathrm{kg},\ KE = 100\ \mathrm{J} \Rightarrow p = 20.0\ \mathrm{kg\,m/s}$$
a light fast body
$$m = 50.0\ \mathrm{kg},\ KE = 100\ \mathrm{J} \Rightarrow p = 100\ \mathrm{kg\,m/s}$$
a heavy slow body with the same energy and five times the momentum
Answer $$\boxed{\;\text{False: } p = \sqrt{2mKE}\;}$$
Check
Limiting case: let the mass grow without bound at fixed energy. The speed falls as one over the root of the mass while the momentum grows as the root of it, so the two quantities move in opposite directions and cannot both stay equal.
2§11.3 — lost energy against lost momentum●●○○○
A second statement for judgement, this time about what a collision destroys.
Given
a collision in which kinetic energy is definitely lost
the system is isolated
Find
(a) True or false: if kinetic energy is lost in a collision, some momentum must be lost too. Justify.
Hint 1/4
Ask what each of the two quantities would need in order to leave the system, and whether a collision provides it.
Hint 2/4
Momentum leaves a system only through an external force; kinetic energy leaves it whenever the motion turns into heat, sound or permanent deformation.
Hint 3/4
Two 10 000 kg wagons coupling at 24.0 m/s: energy falls from $2.88\times10^{6}$ J to $1.44\times10^{6}$ J while the momentum stays $2.40\times10^{5}$ kg m/s.
Hint 4/4
False, and the two quantities have entirely different escape routes, which is why the section works at all.
Show solution
The two quantities are compared by asking how each could leave the system, which is a question with a short answer, rather than by inspecting the collision.
Ask how each quantity could leave
$$\Delta\vec p_{\rm total} = \vec J_{\rm ext}$$
the only way the total momentum changes is an impulse from outside, and there is none
$$\Delta KE = -Q\ \text{(heat, sound, deformation)}$$
energy has forms that are not motion, and it can leave the mechanical account without leaving the system
Test the converse for consistency: an explosion increases the kinetic energy of an isolated system while its total momentum stays exactly where it was, so the two really are independent in both directions.
3§11.2 — bouncing against sticking●●●○○
Two balls of equal mass are thrown at a wall with equal speeds. One is made of putty and sticks to the wall; the other is a superball and rebounds with almost its original speed.
Given
equal masses and equal incoming speeds
the putty ball ends at rest on the wall
the superball rebounds at nearly the incoming speed
Find
(a) Which ball delivers the larger impulse to the wall, and roughly by what factor?
Hint 1/4
Work out the momentum change of each ball. The impulse on the wall is equal and opposite to the impulse on the ball, so the larger change wins.
Hint 2/4
$J = mv_2-mv_1$, with the incoming direction positive and the rebound direction negative.
Limiting check: a ball that rebounds at zero speed is the putty case and the factor becomes one; a ball that rebounds at its full speed gives exactly two. Every real ball sits between, which brackets the answer.
4§11.7 — what an internal push can do●●○○○
A last statement for judgement, about a system left entirely to itself in deep space.
Given
an isolated system of several bodies
the bodies push, pull and collide with each other
no force acts from outside
Find
(a) True or false: if the parts push hard enough on each other, the centre of mass of the system can be made to accelerate. Justify.
Hint 1/4
Write the law that governs the centre of mass and see which forces appear in it and which do not.
Hint 2/4
$\sum\vec F_{\rm ext} = M\vec a_{\rm cm}$, in which every internal force has been cancelled by its third-law partner.
Hint 3/4
With no external force the right side gives zero acceleration, however violent the internal forces are.
Hint 4/4
False. No internal arrangement can move the centre of mass of an isolated system, which is why a rocket must throw mass away rather than push against itself.
Show solution
Sorting the forces into internal and external answers the question in two lines and makes clear which half of the sum is doing the work.
no external force means no acceleration of the centre of mass, whatever happens inside
Answer $$\boxed{\;\text{False}\;}$$
Check
Consistency with the conservation law: constant velocity of the centre of mass and constant total momentum are the same statement, since the total momentum is the total mass times that velocity.
B · computation 8 questions
1§11.2 — impulse on a bouncing ball●●●○○
A ball is dropped onto a hard floor and bounces back, losing some speed in the contact. The contact is filmed at high speed and timed.
Given
$m = 0.200\ \mathrm{kg}$
arrives at $8.00\ \mathrm{m/s}$ downwards
leaves at $6.00\ \mathrm{m/s}$ upwards
contact lasts $0.0120\ \mathrm{s}$
upwards is positive; $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the impulse the floor delivered to the ball.
(b) Find the average force the floor exerted on the ball during the contact.
Hint 1/4
The velocity reverses, so the two momenta have opposite signs and their difference is larger than either. Fix the positive direction before writing anything.
Hint 2/4
$J = \Delta p = mv_2-mv_1$; the average net force is $J/\Delta t$, and the floor force is the net force plus the weight it also has to hold.
Hint 3/4
With upwards positive, $v_1 = -8.00$ m/s and $v_2 = +6.00$ m/s, $m = 0.200$ kg and $\Delta t = 0.0120$ s.
Hint 4/4
$J = +2.80$ kg m/s and the floor pushes with $233 + 1.96 = 235$ N upwards.
Show solution
The weight is kept in part (b) rather than dropped, because the question asks for the force from the floor and not for the net force.
the average of everything acting, which during the contact is the floor and gravity together
$$\bar N = 233 + (0.200)(9.80) = 235\ \mathrm{N}$$
the floor also has to hold the weight, although at 1.96 N it barely changes the answer
Answer $$\boxed{\;J = +2.80\ \mathrm{kg\,m/s},\qquad \bar N = 235\ \mathrm{N}\;}$$
Check
Size check: 235 N is 120 times the 1.96 N weight of the ball, which is why the weight was almost irrelevant and why it is fair to ignore gravity during short contacts. Sign check: the ball ends up moving upwards, so the net impulse on it had to point upwards, and it does.
2§11.2 — impulse from a force that changes in steps●●○○○
A 3.00 kg trolley starts from rest on a frictionless track. A motor pushes it with a force that is held constant over each of three intervals, recorded in the table.
Given
$m = 3.00\ \mathrm{kg}$, initially at rest, frictionless track
from $0$ to $0.20$ s the force is $6.00$ N
from $0.20$ to $0.50$ s the force is $2.00$ N
from $0.50$ to $0.80$ s the force is zero
the force is always along the direction of motion
Find
(a) Find the total impulse delivered to the trolley in the 0.80 s.
(b) Find the speed of the trolley at the end of the 0.80 s.
Hint 1/4
The force is constant inside each interval, so the graph is a staircase and the area under it is a sum of rectangles.
Hint 2/4
$J = \sum \bar F_i \Delta t_i$, and $J = \Delta p = m(v_2-v_1)$ closes the problem.
Hint 3/4
The rectangles are $(6.00)(0.20)$, $(2.00)(0.30)$ and $(0)(0.30)$, with $m = 3.00$ kg starting from rest.
Hint 4/4
$J = 1.20+0.60+0 = 1.80$ N s, so $v = 1.80/3.00 = 0.600$ m/s.
Show solution
The impulse route handles the whole staircase in one line, whereas the stage-by-stage route needs a separate acceleration for each interval.
Area under the staircase
$$J_1 = (6.00)(0.20) = 1.20\ \mathrm{N\,s}$$
a rectangle, because the force is constant across that interval
the trolley started from rest, so the whole impulse shows up as final momentum
Answer $$\boxed{\;J = 1.80\ \mathrm{N\,s},\qquad v = 0.600\ \mathrm{m/s}\;}$$
Check
Second route by stages: after 0.20 s at $a = 2.00$ m/s$^{2}$ the speed is 0.400 m/s; over the next 0.30 s at $a = 0.667$ m/s$^{2}$ it gains 0.200 m/s, and 0.600 m/s is reached before the force stops. Both routes agree, and the third interval correctly changes nothing.
3§11.3 — recoil of a rifle●●○○○
A rifle is fired while held loosely, so that nothing outside the rifle and bullet acts horizontally during the shot.
(b) Find the ratio of the kinetic energy of the bullet to that of the rifle.
Hint 1/4
The system starts at rest, so its total momentum is zero and must stay zero. That single fact fixes the recoil.
Hint 2/4
$0 = m_bv_b + m_rv_r$, so the two momenta are equal in size and opposite in direction.
Hint 3/4
With $m_b = 0.0120$ kg at $620$ m/s and $m_r = 4.50$ kg, the bullet carries $7.44$ kg m/s.
Hint 4/4
$v_r = -7.44/4.50 = -1.65$ m/s, and the energy ratio is $m_r/m_b = 375$.
Show solution
The energy ratio is done through $p^{2}/2m$ rather than by computing both energies, because the momenta are known to be equal and the ratio then needs no arithmetic at all.
Recoil from a total of zero
$$0 = (0.0120)(620) + (4.50)v_r$$
nothing outside pushes horizontally, so the total stays at the zero it started with
Direct check of the ratio: $KE_b = \tfrac12(0.0120)(620)^{2} = 2306$ J and $KE_r = \tfrac12(4.50)(1.6533)^{2} = 6.15$ J, and $2306/6.15 = 375$. Order of magnitude: a 1.65 m/s recoil is a firm shove into the shoulder rather than a knock-down, which matches the experience of firing a rifle.
4§11.4 — a head-on collision that stops both bodies●●●○○
Two lumps of clay slide towards each other on a frictionless bench and stick on contact.
Given
$m_1 = 3.00\ \mathrm{kg}$ moving right at $4.00\ \mathrm{m/s}$
$m_2 = 2.00\ \mathrm{kg}$ moving left at $6.00\ \mathrm{m/s}$
they stick together
rightwards positive, frictionless bench
Find
(a) Find the velocity of the combined lump.
(b) Find the kinetic energy destroyed in the collision.
Hint 1/4
One body is moving against the chosen positive direction, so one of the two momenta enters the sum negative. Work out the total before doing anything else.
Hint 2/4
$m_1v_1+m_2v_2 = (m_1+m_2)v'$, with the signs of the velocities kept.
Hint 3/4
Here $(3.00)(+4.00) = +12.0$ and $(2.00)(-6.00) = -12.0$ kg m/s.
Hint 4/4
The total is zero, so $v' = 0$: the lump is at rest, and all $60.0$ J of kinetic energy has been destroyed.
Show solution
The signs are put in before any arithmetic, because this problem is designed so that a missing sign gives 24.0 kg m/s and a plausible looking wrong answer of 4.80 m/s.
energies are never negative, so unlike the momenta these two add rather than cancel
$$\Delta KE = -60.0\ \mathrm{J}$$
everything is destroyed, which is the most a collision can ever do
Answer $$\boxed{\;v' = 0,\qquad \Delta KE = -60.0\ \mathrm{J}\;}$$
Check
Check with the loss formula: $\tfrac12\frac{(3.00)(2.00)}{5.00}(4.00-(-6.00))^{2} = \tfrac12(1.20)(100) = 60.0$ J, the entire kinetic energy, as it must be when the pair ends at rest.
5§11.5 — elastic collision with a lighter target●●○○○
On a horizontal frictionless track a heavy glider with a magnetic bumper runs into a lighter glider standing still, and the collision is elastic.
Given
$m_1 = 4.00\ \mathrm{kg}$ at $5.00\ \mathrm{m/s}$
$m_2 = 1.00\ \mathrm{kg}$ at rest
elastic collision along one line
Find
(a) Find both velocities after the collision.
(b) Show that both the momentum and the kinetic energy come out right.
Hint 1/4
The target is at rest, so the ready-made pair of formulas applies directly and no simultaneous equations are needed.
Hint 2/4
$v_1' = \frac{m_1-m_2}{m_1+m_2}v_1$ and $v_2' = \frac{2m_1}{m_1+m_2}v_1$ when $v_2 = 0$.
Hint 3/4
With $m_1 = 4.00$ kg, $m_2 = 1.00$ kg and $v_1 = 5.00$ m/s, the fractions are $3.00/5.00$ and $8.00/5.00$.
Hint 4/4
$v_1' = 3.00$ m/s and $v_2' = 8.00$ m/s, and both conservation checks come out exactly.
Show solution
The ready-made formulas are legitimate here only because the target is at rest, and that condition is checked before they are used.
Relative velocity check, independent of both conservation lines: the gliders approached at $5.00$ m/s and separate at $8.00-3.00 = 5.00$ m/s, exactly reversed, which is the signature of an elastic collision.
6§11.6 — an explosion into three pieces●●●○○
A 3.00 kg object is at rest on a frictionless horizontal sheet of ice when an internal charge bursts it into three equal pieces that slide away across the ice.
Given
total mass $3.00\ \mathrm{kg}$, at rest before the burst
three pieces of $1.00\ \mathrm{kg}$ each
piece 1 moves east at $12.0\ \mathrm{m/s}$
piece 2 moves north at $12.0\ \mathrm{m/s}$
east is $x$, north is $y$
Find
(a) Find the velocity of the third piece, as a speed and a direction.
(b) Find the kinetic energy released by the charge.
Hint 1/4
The object was at rest, so the three momenta afterwards must add to zero as vectors. Work with the two components separately.
Hint 2/4
$\sum p_x = 0$ and $\sum p_y = 0$ separately, then rebuild the magnitude with the theorem of Pythagoras.
Hint 3/4
Pieces 1 and 2 carry $+12.0$ kg m/s east and $+12.0$ kg m/s north, so the third must carry $-12.0$ in each.
Hint 4/4
$v_3 = \sqrt{12.0^{2}+12.0^{2}} = 17.0$ m/s, pointing southwest, and the charge released $288$ J.
Show solution
Components are used rather than a scale diagram because the third momentum is not along either axis and the answer needs three digits.
Vector check by drawing: two arrows of 12.0 units at right angles and one of 17.0 units on the diagonal between them, reversed, close a triangle exactly. Energy check on the direction of the change: the total energy went up, which is allowed here because a charge inside the system supplied it, while the total momentum stayed at zero throughout.
7§11.7 — centre of mass of three masses on a line●●○○○
Three small blocks are glued at measured positions along a light metre rule, which itself has negligible mass.
Given
$2.00\ \mathrm{kg}$ at $x = 0$
$3.00\ \mathrm{kg}$ at $x = 1.50\ \mathrm{m}$
$5.00\ \mathrm{kg}$ at $x = 4.00\ \mathrm{m}$
the rule itself has no mass worth counting
Find
(a) Find the position of the centre of mass.
(b) State where the centre of mass would move if the origin were placed at the 5.00 kg block instead.
Hint 1/4
This is an average, not a midpoint. Each position is counted as many times as there are kilograms sitting at it.
Hint 2/4
$x_{\rm cm} = \sum m_ix_i / \sum m_i$.
Hint 3/4
The products are $0$, $(3.00)(1.50) = 4.50$ and $(5.00)(4.00) = 20.0$, over a total mass of $10.0$ kg.
Hint 4/4
$x_{\rm cm} = 24.5/10.0 = 2.45$ m from the first block, and moving the origin changes the number but not the point.
Show solution
Part (b) is answered by shifting rather than recomputing, which makes the point that the physics does not depend on where the ruler starts.
a mass blind average of the three positions would give 1.83 m; the weighting drags it out to 2.45 m, towards the heaviest block at 4.00 m
Shift the origin
$$x_{\rm cm}' = 2.45 - 4.00 = -1.55\ \mathrm{m}$$
every coordinate drops by 4.00 m, so the average does too; the point itself has not moved
Answer $$\boxed{\;x_{\rm cm} = 2.45\ \mathrm{m}\ \text{from the 2.00 kg block}\;}$$
Check
Bracket check: the answer has to lie between 0 and 4.00 m, and above the plain midpoint of 2.00 m because the heaviest block sits at the far end. It also has to be nearer to 4.00 m than a mass-blind average would put it, and 2.45 m satisfies both.
8§11.7 — a child walking along a raft●●●●○
A child stands at one end of a raft floating at rest on still water, and walks to the other end. The water offers no resistance worth counting.
raft length $6.00\ \mathrm{m}$, child walks from one end to the other
both at rest at the start
no horizontal force from the water
Find
(a) Find how far the raft moves and in which direction.
(b) Find how far the child moves relative to the water.
Hint 1/4
Nothing outside pushes horizontally and the system starts at rest, so there is one point in this problem that cannot move at all. Find it and use it.
Hint 2/4
$\vec v_{\rm cm} = 0$ throughout, so $m_1\Delta x_1 + m_2\Delta x_2 = 0$ for the displacements measured relative to the water.
Hint 3/4
The 6.00 m is measured along the raft, so $\Delta x_{\rm child} = \Delta x_{\rm raft} + 6.00$, with masses 45.0 kg and 75.0 kg.
Hint 4/4
$\Delta x_{\rm raft} = -2.25$ m and $\Delta x_{\rm child} = +3.75$ m, and the two mass weighted displacements cancel.
Show solution
The fixed centre of mass is used because the forces between the feet and the deck are unknown, and because it turns the whole problem into one linear equation.
Substitute back: $(45.0)(+3.75)+(75.0)(-2.25) = 168.75-168.75 = 0$, so the centre of mass really is where it started. Limit check: make the raft enormously heavy and the formula gives almost no raft motion with the child covering nearly the full 6.00 m, which is what walking on a jetty feels like.
C · exam level 5 questions
1§11.4 — ballistic pendulum, in three parts●●●●○
A laboratory ballistic pendulum is used to measure the speed of a bullet from a small rifle. The block hangs at rest from long light strings and the bullet stays inside it.
the block is at rest before the shot and the bullet embeds
the bullet arrives at $250\ \mathrm{m/s}$
$g = 9.80\ \mathrm{m/s^{2}}$, strings light, air resistance ignored
Find
(a) Find the speed of the block with the bullet in it immediately after the impact.
(b) Find the height the block rises.
(c) Find the percentage of the kinetic energy of the bullet that survives the impact.
Hint 1/4
Two stages with a boundary between them: the embedding, which is fast and violent, and the swing, which is slow and smooth. Decide which law survives each before writing anything.
Hint 2/4
Momentum for the impact, $mv = (m+M)v'$; mechanical energy for the swing, $\tfrac12 (m+M)v'^{2} = (m+M)gh$.
Hint 3/4
With $m = 0.0150$ kg at $250$ m/s and $M = 1.50$ kg, the momentum is $3.75$ kg m/s and the total mass is $1.515$ kg.
Hint 4/4
$v' = 2.48$ m/s, $h = v'^{2}/2g = 0.313$ m, and only $0.990\%$ of the energy survives.
Show solution
The stages are solved forwards here because the bullet speed is given; in the laboratory the height is what gets measured and the same two steps are run in reverse.
(a) The impact
$$(0.0150)(250) = (1.515)v'$$
momentum survives the millisecond of impact; kinetic energy does not
$$v' = 2.48\ \mathrm{m/s}$$
a walking pace, which is what makes a laboratory pendulum swing a measurable amount
(b) The swing
$$\tfrac12(1.515)v'^{2} = (1.515)gh$$
no friction and the string does no work, so mechanical energy is conserved from the impact to the top
Part (c) is also the theoretical fraction $m/(m+M) = 0.0150/1.515 = 0.00990$, computed without either energy, so the two routes agree. Size check on (b): a 31 cm rise on a laboratory pendulum is easily measured, which is the whole point of the instrument.
2§11.5 — elastic collision followed by a spring●●●●○
Two carts sit on a level frictionless track. The second one is free to run into a spring whose far end is clamped to the bench.
Given
cart 1: $0.800\ \mathrm{kg}$ at $4.00\ \mathrm{m/s}$
cart 2: $0.400\ \mathrm{kg}$, at rest
the collision between them is elastic
cart 2 then meets a spring of stiffness $200\ \mathrm{N/m}$
Find
(a) Find the velocity of each cart immediately after the collision.
(b) Find the maximum compression of the spring.
(c) Verify that the kinetic energy immediately after the collision is the same as before it.
Hint 1/4
Two separate events, and they do not overlap: first a collision between two bodies, then one body against a spring. Solve them in that order.
Hint 2/4
Elastic with a target at rest: $v_1' = \frac{m_1-m_2}{m_1+m_2}v_1$, $v_2' = \frac{2m_1}{m_1+m_2}v_1$. Then $\tfrac12 m_2v_2'^{2} = \tfrac12 kx^{2}$.
Hint 3/4
Here $m_1 = 0.800$ kg, $m_2 = 0.400$ kg, $v_1 = 4.00$ m/s and $k = 200$ N/m.
Hint 4/4
$v_1' = 1.33$ m/s, $v_2' = 5.33$ m/s, and the spring squashes by $0.239$ m.
Show solution
The unrounded 5.3333 m/s is carried into the spring calculation rather than the rounded 5.33, because the next step squares it and the rounding would show in the third digit.
Momentum check, not used in the solution: $(0.800)(1.3333)+(0.400)(5.3333) = 1.067+2.133 = 3.20$ kg m/s, equal to $(0.800)(4.00)$. Relative velocity check: approach $4.00$ m/s, separation $5.333-1.333 = 4.00$ m/s.
3§11.6 — two pucks and an unknown direction●●●●○
Two identical pucks slide on frictionless ice. One is at rest until it is struck off centre by the other, and a camera records the direction the striker takes afterwards.
Given
both pucks $0.250\ \mathrm{kg}$
puck 1 arrives at $3.00\ \mathrm{m/s}$ due east, puck 2 is at rest
after the collision puck 1 moves at $1.50\ \mathrm{m/s}$ at $60.0^{\circ}$ north of east
$v_{2x} = 2.25$ m/s and $v_{2y} = -1.30$ m/s, so $2.60$ m/s at $30.0^{\circ}$ south of east, and the kinetic energy is unchanged.
Show solution
The components are solved before any thought is given to energy, because momentum holds whether or not the collision is elastic and part (c) is then a genuine test rather than an assumption.
The right angle is the independent check: for equal masses in an elastic collision with a stationary target, the two outgoing paths must be $90^{\circ}$ apart, and $60.0^{\circ}$ north plus $30.0^{\circ}$ south is exactly that. Had the energy check failed, the angle would have come out different from $90^{\circ}$ as well.
4§11.2 — what the airbag is worth●●●○○
A 75.0 kg driver travelling at 20.0 m/s is brought to rest in a crash. With an airbag the stop takes 0.100 s; against a rigid steering column it would take 0.0100 s.
Given
$m = 75.0\ \mathrm{kg}$, from $20.0\ \mathrm{m/s}$ to rest
stopping time with the airbag $0.100\ \mathrm{s}$
the weight of the driver is $735\ \mathrm{N}$
Find
(a) Find the average force on the driver with the airbag, and say what it is as a multiple of body weight.
Hint 1/4
The impulse is fixed by the two end states and is the same with or without the airbag. Only the time differs, so work out the impulse first.
Hint 2/4
$\bar F = \Delta p/\Delta t$, and a force is best reported as a multiple of the weight $mg$.
Hint 3/4
$\Delta p = (75.0)(20.0) = 1500$ kg m/s, $\Delta t = 0.100$ s, and $mg = 735$ N.
Hint 4/4
$\bar F = 1500/0.100 = 1.50\times10^{4}$ N, which is 20.4 times body weight; without the airbag it would be ten times worse.
Show solution
The impulse is computed once and reused for both scenarios, which makes the comparison exact rather than approximate.
Impulse first, because it is the same in both scenarios
with the bag, the stop is stretched over a tenth of a second
$$\frac{1.50\times10^{4}}{735} = 20.4$$
twenty times body weight is survivable; the same calculation at 0.0100 s gives 204 times, which is not
Answer $$\boxed{\;\bar F = 1.50\times10^{4}\ \mathrm{N} = 20.4\,mg\;}$$
Check
Cross-check by distance: at an average speed of 10.0 m/s the 0.100 s stop covers 1.00 m, and the work-energy principle gives $F = KE/d = 15\,000/1.00 = 1.50\times10^{4}$ N, the same figure by a route that never uses the time.
5§11.4 — ramp, collision and rough floor●●●●●
A block is released from rest on a smooth curved ramp, hits a second block waiting at the bottom, and the pair slides away across a rough floor.
Given
block 1: $0.500\ \mathrm{kg}$, released from rest $0.800\ \mathrm{m}$ above the floor
the ramp is smooth; the floor beyond it is rough with $\mu_k = 0.300$
block 2: $1.50\ \mathrm{kg}$, at rest at the foot of the ramp
the blocks stick together on impact; $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of block 1 at the foot of the ramp.
(b) Find the speed of the pair immediately after the collision.
(c) Find how far the pair slides before stopping.
(d) Find the kinetic energy destroyed in the collision.
Hint 1/4
Three stages, each with its own law: a smooth descent, a sticking collision and a rough slide. Draw the two boundaries before writing an equation.
Hint 2/4
Energy for the ramp, $v=\sqrt{2gh}$; momentum for the impact, $m_1v = (m_1+m_2)v'$; the friction ledger for the slide, $\tfrac12 Mv'^{2} = \mu_k Mgd$.
Hint 3/4
With $h = 0.800$ m, $m_1 = 0.500$ kg, $m_2 = 1.50$ kg and $\mu_k = 0.300$.
Hint 4/4
$v = 3.96$ m/s, $v' = 0.990$ m/s, $d = 0.167$ m and $2.94$ J is destroyed in the impact.
Show solution
The unrounded 3.9598 m/s is carried into the collision step; using 3.96 there would move the sliding distance in the third digit, and the question asks for three.
three quarters of the energy destroyed, and the surviving quarter is exactly $m_1/(m_1+m_2)$
Answer $$\boxed{\;v = 3.96\ \mathrm{m/s},\ v' = 0.990\ \mathrm{m/s},\ d = 0.167\ \mathrm{m},\ \Delta KE = -2.94\ \mathrm{J}\;}$$
Check
Check on (d) with the loss formula: $\tfrac12\frac{(0.500)(1.50)}{2.00}(3.9598)^{2} = \tfrac12(0.375)(15.68) = 2.94$ J, agreeing with the difference of the two kinetic energies. Size check on (c): 17 cm is a short skid, which is right for a pair moving at less than a metre per second.
D · interleaved 4 questions
1§11 mixed — a swinging bob and a block●●●●○
A bob hangs on a light string. It is pulled aside, released from rest, and at the lowest point of its swing it strikes a block resting on a smooth table at exactly that height. The bumpers are springy.
Given
bob $0.300\ \mathrm{kg}$, released from rest $0.450\ \mathrm{m}$ above its lowest point
block $0.200\ \mathrm{kg}$, at rest
the collision is elastic and head-on
no friction anywhere; $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of the bob just before it strikes the block.
(b) Find the velocity of each body just after the collision.
(c) Find how high the bob swings back.
Hint 1/4
The problem has three parts and each has its own conserved quantity. Identify the two instants that separate them before choosing any equation.
Hint 2/4
Swing: $v = \sqrt{2gh}$. Collision: elastic with a target at rest. Swing back: $h = v^{2}/2g$ with the new speed.
Hint 3/4
With $h = 0.450$ m, $m_1 = 0.300$ kg and $m_2 = 0.200$ kg, the fractions in the elastic pair are $0.100/0.500$ and $0.600/0.500$.
Hint 4/4
$v = 2.97$ m/s, then $v_1' = 0.594$ m/s and $v_2' = 3.56$ m/s, and the bob rises only $0.0180$ m.
Show solution
The elastic formulas are legal here because the block really is at rest; the return height is done with energy because the string is smooth.
Both conservation laws on the collision: momentum $(0.300)(0.594)+(0.200)(3.5638) = 0.891$ kg m/s against $(0.300)(2.9698) = 0.891$; kinetic energy $0.0529+1.270 = 1.32$ J against $\tfrac12(0.300)(8.82) = 1.32$ J. Both hold, so the collision step is right, and the height then follows from one division.
2§11 mixed — sticking, then a vertical loop●●●●●
A small ball slides along a smooth horizontal track, sticks to a second ball at rest, and the pair must then complete a vertical circular loop in the track without leaving it at the top.
Given
ball 1: $0.100\ \mathrm{kg}$, ball 2: $0.200\ \mathrm{kg}$ at rest
the two stick together on contact
the loop has radius $0.400\ \mathrm{m}$ and the whole track is smooth
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the smallest speed the pair may have at the top of the loop.
(b) Find the speed the pair needs at the bottom.
(c) Find the smallest speed ball 1 must have had before the collision.
Hint 1/4
Work backwards. The hardest condition is at the top of the loop, so start there and travel back to the moment before the collision.
Hint 2/4
At the top with the track only just in contact, gravity supplies the whole centripetal force: $mg = mv_{\rm top}^{2}/R$. Then mechanical energy from bottom to top, then momentum through the collision.
Hint 3/4
With $R = 0.400$ m the top condition gives $v_{\rm top}^{2} = gR = 3.92$; the rise to the top is $2R = 0.800$ m; the masses are $0.100$ kg and $0.200$ kg.
Hint 4/4
$v_{\rm top} = 1.98$ m/s, $v' = 4.43$ m/s at the bottom, and ball 1 must arrive at $13.3$ m/s.
Show solution
Working backwards is chosen because only the loop imposes a condition; running forwards would mean guessing the approach speed and testing it.
Energy audit as a check on the size of (c): ball 1 arrives with $\tfrac12(0.100)(13.28)^{2} = 8.82$ J and the pair leaves the collision with $\tfrac12(0.300)(4.4272)^{2} = 2.94$ J, a surviving third, which is $m_1/(m_1+m_2) = 0.100/0.300$ exactly.
3§11 mixed — a firework that bursts at the top●●●●●
A firework shell is fired and bursts at the very top of its flight, where its velocity is horizontal. Two pieces fly apart horizontally and then fall freely to the ground.
Given
shell $3.00\ \mathrm{kg}$, moving east at $6.00\ \mathrm{m/s}$ at the burst
the burst happens $25.0\ \mathrm{m}$ above level ground
piece 1: $1.00\ \mathrm{kg}$, thrown west at $4.00\ \mathrm{m/s}$ just after the burst
piece 2: $2.00\ \mathrm{kg}$; both pieces leave horizontally; air resistance ignored
Find
(a) Find the velocity of the heavier piece just after the burst.
(b) Find how long the pieces take to reach the ground.
(c) Find how far east of the burst point the heavier piece lands.
Hint 1/4
The burst is one problem and the fall is another. The burst is over in an instant, and afterwards each piece is an ordinary projectile.
Hint 2/4
Momentum through the burst, since the explosive force is internal; then $y = \tfrac12 gt^{2}$ for the fall and $x = v_xt$ for the horizontal travel.
Hint 3/4
The shell carries $(3.00)(6.00) = 18.0$ kg m/s east, piece 1 carries $-4.00$ kg m/s, and the drop is $25.0$ m.
Hint 4/4
$v_2 = 11.0$ m/s east, the fall takes $2.26$ s, and the heavy piece lands $24.8$ m east of the burst point.
Show solution
Momentum is used for the burst because the explosive force is unknown, and kinematics for the fall because the burst is over long before the pieces move any distance.
(a) The burst
$$(3.00)(6.00) = (1.00)(-4.00)+(2.00)v_2$$
the explosive is internal, so the total momentum is the same immediately after as immediately before
the heavy piece has to make up for the momentum the light one took the other way
(b) The fall
$$25.0 = \tfrac12(9.80)t^{2} \Rightarrow t = 2.26\ \mathrm{s}$$
both pieces start with no vertical velocity, so the drop is a free fall from rest and the same for both
(c) The horizontal travel
$$x = v_2 t = (11.0)(2.2588) = 24.8\ \mathrm{m}$$
nothing acts horizontally after the burst, so the horizontal velocity is unchanged all the way down
Answer $$\boxed{\;v_2 = 11.0\ \mathrm{m/s\ east},\quad t = 2.26\ \mathrm{s},\quad x = 24.8\ \mathrm{m\ east}\;}$$
Check
Centre of mass check: the light piece lands $(4.00)(2.2588) = 9.04$ m west, and the mass weighted average of the two landing points is $[(1.00)(-9.04)+(2.00)(24.8)]/3.00 = 13.5$ m east, which is exactly where the unexploded shell would have landed, $(6.00)(2.2588) = 13.6$ m east, to within rounding. The burst did not move the centre of mass.
4§11 mixed — the same push, measured two ways●●●○○
A 1200 kg car is pushed from rest along a level road by a constant net force. Two students answer different questions about the same push, one asking how fast and the other asking how much energy.
Given
$m = 1200\ \mathrm{kg}$, starting from rest
constant net force $3000\ \mathrm{N}$ along the motion
the force acts for $5.00\ \mathrm{s}$
level road, and every other force is already included in the net figure
Find
(a) Find the speed of the car after 5.00 s, using the impulse.
(b) Find the distance covered in that time, and the work done by the net force.
(c) Check that the work done equals the kinetic energy gained.
Hint 1/4
One of the two questions is answered by a force acting for a time and the other by the same force acting over a distance. Decide which is which before computing anything.
Hint 2/4
$\bar F\Delta t = \Delta p$ for the speed; $W = Fd$ and $W_{\rm net} = \Delta KE$ for the energy.
Hint 3/4
$F = 3000$ N, $\Delta t = 5.00$ s, $m = 1200$ kg from rest, so $a = 2.50$ m/s$^{2}$.
Hint 4/4
$v = 12.5$ m/s, $d = 31.25$ m, $W = 9.38\times10^{4}$ J, and the kinetic energy gained matches it exactly.
Show solution
Both routes are carried out on purpose, because the point of the question is that a time question and a distance question about the same push have different tools and the same underlying force.
(a) Time route
$$\Delta p = F\Delta t = (3000)(5.00) = 1.50\times10^{4}\ \mathrm{kg\,m/s}$$
the impulse relation needs nothing except the force and the interval
force and displacement are parallel, so no cosine is needed
(c) The two routes meeting
$$\Delta KE = \tfrac12(1200)(12.5)^{2} = 9.38\times10^{4}\ \mathrm{J}$$
equal to the work, as the work-energy principle requires
Answer $$\boxed{\;v = 12.5\ \mathrm{m/s},\quad d = 31.3\ \mathrm{m},\quad W = \Delta KE = 9.38\times10^{4}\ \mathrm{J}\;}$$
Check
Independent check on (a) through Newton's second law: $a = 2.50$ m/s$^{2}$ for 5.00 s from rest gives $v = 12.5$ m/s, agreeing with the impulse route. The two routes are genuinely different: one never mentions distance and the other never mentions time.
Mistake ledger (16 entries)
⚠ Treating momentum as a size with no direction
Kinetic energy has no direction and was the quantity in use for two sections, so the habit of adding sizes carries over to a vector.
Contact times are quoted in milliseconds because that is how they are measured, and the conversion is the least interesting step in the calculation, so it is the one that is skipped.
wrong$$\bar F = \frac{12.3}{1.5} = 8.2\ \mathrm{N}$$
right$$\bar F = \frac{12.3}{1.5\times10^{-3}} = 8.2\times10^{3}\ \mathrm{N}$$
⚠ Writing the conservation line without naming the system
The equation looks identical in every problem, so it gets written before the question of what is inside the system is asked.
wrong$$\text{ball on a wall: } mv = mv'\;\Rightarrow\; v' = v$$
right$$\text{ball alone is not isolated: } \Delta p_{\rm ball} = J_{\rm wall\ on\ ball} \ne 0$$
⚠ Adding momenta without signs when the bodies approach each other
The word total suggests adding, and the question quotes both speeds as positive numbers with the direction hidden in a word.
right$$p_{\rm tot} = (3)(+4)+(2)(-5) = +2\ \mathrm{kg\,m/s}\ \text{when they approach}$$
⚠ Conserving kinetic energy through a collision that was not elastic
The previous section conserved energy in every single problem and the habit is only two weeks old; here it can destroy 99% of the answer.
wrong$$\tfrac12 mv^{2} = \tfrac12 (M+m)v'^{2}\ \text{for a bullet embedding}$$
right$$mv = (M+m)v'\ \text{, and the energy falls by the factor } \frac{m}{M+m}$$
⚠ Leaving the embedded body out of the mass afterwards
The bullet is small, so it feels safe to ignore, and it is written on the other side of the equation from the mass it joined.
wrong$$v' = \frac{mv}{M}$$
right$$v' = \frac{mv}{M+m}$$
⚠ Dividing by one mass instead of the total after the bodies lock
The last mass written down is the one still in mind, and the phrase they lock together reads as a description rather than as an instruction about the denominator.
wrong$$v' = \frac{p_{\rm tot}}{m_2}$$
right$$v' = \frac{p_{\rm tot}}{m_1+m_2}$$
⚠ Using the kinetic energy equation and picking the wrong root
The energy equation is the one quoted in the definition of an elastic collision, so it feels like the one to solve, and it is quadratic.
wrong$$v_1' = v_1,\ v_2' = v_2\ \text{(a root of the quadratic)}$$
right$$v_2'-v_1' = v_1-v_2\ \text{, the linear rule, which discards the no-collision root}$$
⚠ Using the target-at-rest formulas when the target is moving
The boxed pair is short and memorable, and the condition attached to it is one line of small print beside it.
wrong$$v_1' = \frac{m_1-m_2}{m_1+m_2}v_1\ \text{ with } v_2 \ne 0$$
axes named before use; each axis is a separate equation with its own licence
Centre of mass of particles
$$x_{\rm cm} = \frac{\sum m_ix_i}{M}$$
positions from a chosen origin; the same formula with $y$ for the other coordinate
The law the centre of mass obeys
$$\sum \vec F_{\rm ext} = M\vec a_{\rm cm}$$
any system at all; internal forces cancel in third-law pairs
Check yourself
Close the page and write, from memory: the two forms of the momentum law, the condition under which a total is conserved and what has to be named before you may say so, the one line that distinguishes an elastic collision from a sticking one, and what the centre of mass of an exploding shell does. Then reopen and check which of the four came out incomplete.
Compute the momentum of a body and the change in it when the velocity reverses, keeping the sign the reversal puts in?
c-momentum
Turn a contact time and a momentum change into an average force, and say why an airbag lowers the force without touching the impulse?
c-impulse
Name a system, list its external forces, and decide whether its total momentum is conserved over the interval in question?
c-conservation
Solve a collision in which the bodies stick, find the energy destroyed, and attach an energy calculation on either side of the impact?
c-inelastic
Use the reversal of relative velocity, and say what happens for equal, heavy and light masses without solving anything?
c-elastic-1d
Split a collision into components, solve the two equations, and rebuild a speed and a direction from them?
c-2d-collisions
Locate the centre of mass of a set of particles and use its refusal to notice internal forces to solve a problem?
c-center-of-mass
Glossary (15 terms)
linear momentumdoğrusal momentum
The product of the mass of a body and its velocity, a vector pointing along the motion, measured in kilogram metres per second. It is the quantity a net force changes at a rate equal to that force.
impulseitme
The effect of a force acting over an interval of time, equal to the area under the graph of force against time and equal to the change of momentum it produces. Measured in newton seconds.
average forceortalama kuvvet
The constant force that would deliver the same impulse over the same interval as the real, varying force. It is smaller than the peak force, typically by a factor between one and a half and two in a collision.
newton secondnewton saniye
The unit of impulse, identical to the kilogram metre per second used for momentum, since one newton is one kilogram metre per second squared.
systemsistem
The set of bodies chosen for study, listed explicitly before any conservation statement is written. Whether a force counts as internal or external depends entirely on this choice.
internal forceiç kuvvet
A force exerted by one member of the system on another. Internal forces occur in equal and opposite pairs inside the system, so they cancel from the total and can never change it.
external forcedış kuvvet
A force exerted on a member of the system by something outside it. Only these can change the total momentum of the system or accelerate its centre of mass.
isolated systemyalıtılmış sistem
A system on which the external forces sum to zero, so that its total momentum is constant. A system can be isolated in one direction and not in another.
elastic collisionesnek çarpışma
A collision in which the total kinetic energy after equals the total before. The bodies separate at the same relative speed at which they approached.
esnek olmayan çarpışma
A collision in which some kinetic energy is converted into heat, sound or permanent deformation. Momentum is still conserved; only the energy has fallen.
perfectly inelastic collisiontam esnek olmayan çarpışma
The extreme case in which the bodies move off together with a single common velocity. It destroys the largest amount of kinetic energy that conservation of momentum permits.
relative velocity of approachyaklaşma bağıl hızı
The rate at which the gap between two bodies closes before a collision, equal to the difference of their velocities. In an elastic collision the separation velocity afterwards has the same size.
recoilgeri tepme
The backward motion acquired by a body that throws part of itself, or another body, forwards. It follows from a total momentum that was zero before and must stay zero.
centre of masskütle merkezi
The average position of the material of a system, each position weighted by the mass sitting there. It moves as though the whole mass were concentrated at it and only the external forces acted.
ballistic pendulumbalistik sarkaç
A hanging block used to measure the speed of a projectile that embeds itself in it. Momentum gives the speed just after the impact and energy converts the swing height into that speed.
What comes next
§12 · Rotational Motion
Every body on this page was treated as a point, and every collision was arranged so that nothing was left spinning. The next section drops that restriction and asks what happens when a force arrives off centre.
Sources
Physics for Scientists and Engineers with Modern Physics, D. Giancoli, 5th edition The set textbook for the course. The week line for this section carries no chapter numbers, so none are quoted here; the scope taken is the standard content of the linear momentum material in that book.
SI units and the value of g used throughout $g = 9.80\ \mathrm{m/s^{2}}$ near the ground, the same value as in the earlier sections. Momentum in kg m/s, impulse in N s, and the two units are identical.
Course catalogue description and assessment weights The catalogue lists the dynamics of a system of particles and collisions, which is what places the centre of mass in this section. The assessment weights quoted in the exam note come from the same source and nothing beyond them is claimed.