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Multi-stage problems split at the joints: energy, momentum, angular momentum and the second law, each where its condition holds, with results passed from stage to stage.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to split a mechanics problem into stages, justify the principle for each, and chain the results.
This lesson closes the mechanics half of the course. You have Newton's second law as a differential equation, work and potential energy, impulse and momentum, collisions, rotation and angular momentum, gravitation, oscillation and drag. Each has been practiced on its own. Here they are used together, which is how real problems, and the free-response questions of the AP exam, present them.
| Term | What it means |
|---|---|
| Stage | A part of a process during which one set of rules applies. |
| Conserved quantity | One that stays fixed in a stage: energy, momentum or angular momentum. |
| Isolated system | One with no net external force, so its momentum is conserved. |
| Conservative force | One whose work depends only on the endpoints, such as gravity or a spring. |
| Dissipative process | One that turns mechanical energy into heat: friction, drag, sticking collisions. |
| Ballistic pendulum | A block that catches a projectile and swings up, used to measure its speed. |
| Working backward | Starting from what is measured at the end and reasoning to the start. |
A multi-stage problem is several simple problems joined end to end. The skill is to find the joints: the instants where the forces change character. A collision, a release, a string going slack, the moment of leaving a ramp. Between joints, one principle governs:
The output of one stage, a speed or an angular speed, is the input to the next. Dissipation, from friction or sticking, is allowed within a stage only if you account for it: $K_1 + U_1 = K_2 + U_2 + E_{\text{lost}}$.
Another way: picture
Picture the problem as a comic strip. Each panel is a stage: the block sliding, the impact, the swing. Under each panel write which quantity is conserved and why. The arrows between panels carry a single number, usually a speed, from one to the next. Most errors come from a conservation law that leaks across a panel border where it does not hold.
Another way: steps
Ask two questions. First, are there external forces or torques during this stage? If not, or if the stage is so brief that their impulse is negligible, momentum or angular momentum is conserved. Second, does any non-conservative force do work? If not, mechanical energy is conserved.
A collision is brief and internal, so momentum holds, but unless it is elastic, energy does not. A frictionless slide has an external force, gravity, so momentum changes, but gravity is conservative, so energy holds. A skater pulling in her arms has no external torque about the vertical, so angular momentum holds, but her muscles do work, so kinetic energy grows. Each stage gets exactly the principles that are true for it.
The classic chain is a bullet fired into a hanging block. Stage one is the impact: it lasts a fraction of a millisecond, the strings stay vertical, and momentum is conserved horizontally, $mv = (m + M)v'$. Stage two is the swing: the strings do no work, gravity is conservative, so $\tfrac{1}{2}(m + M)v'^2 = (m + M)gh$.
Putting them together, $v = \dfrac{m + M}{m}\sqrt{2gh}$. Using energy across the impact instead would give $v = \sqrt{2gh(m + M)/m}$, far too small a speed. The fraction of kinetic energy that survives the impact is $m/(m + M)$, typically under one percent; the rest heats the block. The device was used to measure bullet speeds long before electronic timers.
Problems with vertical circles, loops, swings, riders on Ferris wheels, pair energy with the second law. Energy gives the speed at a point; the second law along the radius gives the force: $\sum F_{\text{toward center}} = mv^2/R$.
At the top of a loop, the track and gravity both push toward the center, $N + mg = mv^2/R$. The cart stays on only if $N \ge 0$, which needs $v^2 \ge gR$ at the top. Energy then says the starting height must be at least $2.5R$ for a sliding block. A rolling ball needs more, $2.7R$, because some of its energy goes into spinning.
When a body rolls without slipping, its kinetic energy has two parts, $\tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$, linked by $v = R\omega$. Static friction does no work on a rolling body, since the contact point is momentarily at rest, so energy is still conserved.
When a spinning body collides with something or something lands on it, angular momentum about the pivot carries through the impact, just as linear momentum does. A lump of clay thrown at the end of a rod on a pivot: angular momentum about the pivot is conserved in the impact, $mvr = (I + mr^2)\omega$, and energy then governs the swing that follows. The same splitting into stages works for rotation as for straight-line motion.
Accident investigators, forensic scientists and engineers analyzing failures often know the end of a story and want the beginning. The stages are the same; they are just traversed in reverse. From skid marks, energy and friction give the speed just after a crash; momentum then gives the speeds just before.
Working backward requires care at each joint to use the principle for that stage, not the one that happens to be convenient. It also means uncertainties compound: a 10 percent uncertainty in the friction coefficient becomes 5 percent in the speed after the crash, through the square root, and passes unchanged through momentum.
Checking an answer. Kinetic energy must not increase across a collision unless something, such as an explosion or a spring, supplies it. Limits must behave: a very heavy target should stop the projectile; a very light one should change nothing. Symbols should be kept until the end, when masses and $g$ often cancel. And each stage's result should be sensible on its own.
Some stages have no conserved quantity: a boat slowing under drag, a rocket burning fuel, a mass on a spring with friction. Then the second law must be solved as a differential equation for that stage, as in the drag and oscillation lessons. The stage's output, the speed after a time or the distance to stop, still passes to the next stage in the usual way.
Recognizing that a stage needs this is part of the skill. If a force depends on speed or time, energy and momentum give no shortcut, and the equation of motion must be integrated. Most exam problems keep such stages short and pair them with conservation laws for the rest.
Free-response questions on the AP Physics C: Mechanics exam typically set one physical situation and ask a sequence of parts that walk through its stages: a force diagram, a speed from energy, a speed after a collision from momentum, a distance from kinematics or a differential equation, and a question asking you to justify which principle applies.
The justification is scored. Saying "momentum is conserved because the collision is brief and the external forces are finite, so their impulse is negligible" earns the point; "momentum is always conserved" does not. Naming the stage, the principle and the reason, as this lesson practices, is exactly what is asked.
Some joints are obvious, like a collision. Others are easy to miss. A string that goes slack turns a circular motion into a projectile. A block that reaches the end of a rough patch of floor switches from a stage with friction to one without. A ball that starts rolling after skidding switches from kinetic friction, which does work, to rolling, where static friction does none. A rocket that burns out switches from a stage with thrust to free flight.
A good habit is to ask, at every point in the story, whether any force has just appeared, disappeared or changed its law. If so, that point is a joint, and the quantities to carry across it are the ones that cannot jump: position always, and velocity unless an impulse acts there. Writing the joint quantities explicitly, as a list between stages, catches most errors before any equation is solved.
When a crash is serious, state police reconstruction units, such as the Ohio State Highway Patrol's, measure skid marks, final positions and crush damage, and work backward. Friction and the skid length give the speeds just after impact; momentum, applied in two dimensions with the angles of departure, gives the speeds just before.
These analyses are used in court, so every stage must be justified. Modern cars add an event data recorder that logs speed in the seconds before a crash, and investigators check it against the physics. When both agree, the reconstruction is strong evidence; when they disagree, the discrepancy itself is informative.
Designers of roller coasters, such as those at Cedar Point in Sandusky, Ohio, chain the same stages: a lift or launch supplies energy, then energy conservation, with an allowance for friction and drag, gives the speed at every point. The second law along the radius at each curve gives the force on riders, which must stay within limits of a few $g$.
Modern coaster loops are not circles but clothoids, whose curvature grows gradually. A circular loop entered at the speed needed to clear the top would press riders with six times their weight at the bottom; a clothoid, tighter at the top, lets the cars go slower there and keeps the forces tolerable everywhere.
Energy is conserved in the universe, so it is tempting to write $\tfrac{1}{2}mv^2 = mgh$ straight across a ballistic pendulum. But mechanical energy is not conserved when bodies stick: most of it turns into heat, sound and deformation. Momentum carries through the collision; energy is used only on either side of it.
The reverse error also happens: using momentum during a slide down a ramp. Gravity and the ramp are external forces acting for a long time, so momentum changes. Each principle belongs to the stages where its condition holds.
A $10$ g bullet embeds in a $2.0$ kg block that swings up $0.20$ m. Find the block's speed after the impact.
$v' = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.20} = 1.98\ \text{m/s}$
Energy for the swing.
Conserve momentum through the impact.
$0.010v = 2.010 \times 1.98$
The impact is brief.
Solve for the bullet's speed.
$v = \dfrac{2.010 \times 1.98}{0.010} = 398\ \text{m/s}$
About 900 mph.
Find the energy lost in the impact.
$\tfrac{1}{2}(0.010)(398)^2 - \tfrac{1}{2}(2.010)(1.98)^2 = 792 - 3.9 = 788\ \text{J}$
Over 99 percent becomes heat.
Check what energy alone would have given.
$v = \sqrt{\dfrac{2.010 \times 2 \times 9.8 \times 0.20}{0.010}} = 28\ \text{m/s}$
Wildly wrong.
A $2.0$ kg block slides from rest down a frictionless ramp $0.80$ m high. Find its speed at the bottom.
$v = \sqrt{2 \times 9.8 \times 0.80} = 3.96\ \text{m/s}$
Energy for the slide.
It hits and sticks to a $2.0$ kg block at rest. Find their speed.
$v' = \dfrac{2.0 \times 3.96}{4.0} = 1.98\ \text{m/s}$
Momentum for the collision.
Find the energy the pair carries.
$K = \tfrac{1}{2} \times 4.0 \times 1.98^2 = 7.84\ \text{J}$
Half of the $15.7$ J the block had.
They slide on and compress a $400$ N/m spring. Set up energy.
$\tfrac{1}{2} \times 400 \times x^2 = 7.84$
Energy for the compression.
Solve for the compression.
$x = \sqrt{\dfrac{2 \times 7.84}{400}} = 0.198\ \text{m}$
About $20$ cm.
Find the spring's greatest force.
$F = kx = 400 \times 0.198 = 79\ \text{N}$
At full compression.
A $1.2$ m, $3.0$ kg rod hangs from a pivot at its top. A $0.50$ kg lump of clay moving at $6.0$ m/s hits its bottom end and sticks. Find the rod's inertia.
$I_{\text{rod}} = \tfrac{1}{3} \times 3.0 \times 1.2^2 = 1.44\ \text{kg·m}^2$
A rod about one end.
Find the clay's angular momentum about the pivot.
$L = mvr = 0.50 \times 6.0 \times 1.2 = 3.6\ \text{kg·m}^2\text{/s}$
Its line of motion is $1.2$ m from the pivot.
Find the total inertia after the impact.
$I = 1.44 + 0.50 \times 1.2^2 = 2.16\ \text{kg·m}^2$
Clay at the end.
Conserve angular momentum about the pivot.
$\omega = \dfrac{3.6}{2.16} = 1.67\ \text{rad/s}$
The pivot's force has no torque about the pivot.
Find the kinetic energy after the impact.
$K = \tfrac{1}{2} \times 2.16 \times 1.67^2 = 3.0\ \text{J}$
The clay brought $9.0$ J.
Find how high the center of mass of rod and clay rises.
$\Delta y_{\text{cm}} = \dfrac{3.0}{3.5 \times 9.8} = 0.087\ \text{m}$
Energy for the swing.
Find the angle of swing.
$y_{\text{cm}} = \dfrac{3.0 \times 0.60 + 0.50 \times 1.2}{3.5} = 0.686, \quad \cos\theta = 1 - \dfrac{0.087}{0.686} \Rightarrow \theta = 29°$
The center of mass rises $y_{\text{cm}}(1 - \cos\theta)$.
Conserve momentum through the impact.
$v' = \dfrac{0.20 \times 5.0}{0.50} = 2.0\ \text{m/s}$
They stick.
Use energy for the climb.
$h = \dfrac{v'^2}{2g} = \dfrac{4.0}{19.6}$
Only gravity does work.
Evaluate the height.
A ball of clay slides from rest down a frictionless ramp $0.3$ m high, hits a cart and sticks to it, and the pair rolls up a second ramp. Which principle connects the clay's speed just before the impact to the pair's speed just after?
Complete the worked solution: a skater spinning at $4$ rad/s with rotational inertia $6$ kg·m² pulls in her arms, reducing it to $2$ kg·m². Find her new angular speed in rad/s, and her rotational kinetic energy before and after, in J.
Conserve angular momentum about the spin axis.
$\omega_2 = \dfrac{I_1\omega_1}{I_2} =$ o
The ice exerts no torque about the vertical axis.
Evaluate the kinetic energy before.
$K_1 = \tfrac{1}{2}I_1\omega_1^2 =$ k
Rotational kinetic energy.
Evaluate the kinetic energy after.
$K_2 = \tfrac{1}{2}I_2\omega_2^2 =$ e
Larger, by the ratio of the inertias.
Account for the extra energy.
$\text{the work her arms do pulling inward}$
Energy is not conserved here; angular momentum is.
Match each stage of a problem to the principle that governs it.
| conservation of momentum | conservation of mechanical energy | conservation of angular momentum | Newton's second law | |
|---|---|---|---|---|
| carts couple | ||||
| sled on ice | ||||
| diver tucks | ||||
| force at top of loop |
A $10$ g bullet moving at $300$ m/s embeds itself in a $0.99$ kg block hanging on strings. With $g = 10$ m/s², fill in the block's speed just after the impact in m/s, the height it swings up in m, and the fraction of the bullet's kinetic energy that survives the impact.
| value | |
|---|---|
| speed after impact (m/s) | |
| height of swing (m) | |
| fraction of energy kept |
A $3$ kg block slides from rest down a frictionless track from height $h$, collides with a $2$ kg block at rest at the bottom, and sticks to it. The pair slides up the other side. Write the height they reach as a formula in $h$.
Answer:
A $2$ kg cart starts from rest at a height of $3.5R$ on a frictionless track and runs through a vertical loop of radius $R$. With $g = 9.8$ m/s², how hard does the track push on the cart at the top of the loop, in newtons?
Answer: N from the track at the top
An Ohio State Highway Patrol reconstructionist finds that a $2000$ kg car struck a parked $1500$ kg car, the two locked together, and they skidded $12$ m to a stop. With a friction coefficient of $0.70$ and $g = 9.8$ m/s², how fast was the moving car going just before the crash, in m/s?
Answer: m/s before impact
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $3$ kg block slides from rest down a frictionless track from height $h$, collides with a $2$ kg block at rest at the bottom, and sticks to it. The pair slides up the other side. Write the height they reach as a formula in $h$.
Answer:
You can solve multi-stage problems. Explain to someone why a ballistic pendulum needs momentum for the impact and energy for the swing.
20. Your turn: a $0.20$ kg ball at $5.0$ m/s hits and sticks to a $0.30$ kg cart at rest, which then rolls up a ramp. How high does it go?, step 3
$h = 0.20\ \text{m}$
About $20$ cm.