Check the barrier before accepting an endpoint
You will be able to: Find a one-dimensional turning point and distinguish a local energy check from path accessibility.
Can an endpoint have nonnegative kinetic energy but still be unreachable?
A ball may have enough energy to sit on the far side of a hill but not enough to get over the top. A positive charge in a potential barrier faces the same energy accounting: it must pass every intermediate point.
A useful starting point: A potential drop is not an energy drop for every charge →
Words and symbols before equations
- Total energy H
- The constant K + U for this ideal conservative one-dimensional motion.
- Turning point
- A position reached with K = 0 where the force reverses the particle’s motion.
- Barrier height
- The maximum U along the route relative to its starting value.
- Separatrix threshold
- At exactly the smooth barrier maximum, an ideal particle can approach the top without crossing it in finite time.
What this picture assumes
Ideal one-dimensional positive charge initially at x = 0, V(x) = 4x − x² V for 0 ≤ x ≤ 4 m. No other work. The candidate marker is not a claimed trajectory position. At exact peak energy the particle approaches the top asymptotically, without finite-time crossing.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Barrier 4 μJ; initial energy 3 μJ; local K(4 m)=3 μJ. Candidate is beyond the first turning point at 1 m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Take a positive charge moving initially rightward from x = 0 along V(x) = 4x − x² volts, x measured in meters from 0 to 4. U = qV rises to a maximum at x = 2 m and then falls. With q in μC, U in μJ equals q(4x − x²).
The local energy expression is K(x) = K_i − q(4x − x²). If K_i < 4q μJ, the first root is x_turn = 2 − √(4 − K_i/q) m. The particle cannot reach points beyond that first barrier even if the algebraic K becomes positive again farther right.
For K_i > 4q μJ it can cross this barrier. At exact equality, the smooth maximum is approached asymptotically in the ideal model, so do not label it a finite-time crossing. The diagram is an energy-versus-position graph, not a physical hill or a timed animation.
A worked example, step by step
A +1 μC particle starts at x = 0 with K_i = 3 μJ in the specified potential. Can it reach x = 4 m?
- At the barrier peak, U(2) = 4 μJ, greater than the initial total energy 3 μJ.
- Solve 3 = 4x − x², obtaining roots x = 1 and 3 m.
- The first root, x = 1 m, is the turning point reached from the left.
- Although the endpoint x = 4 has algebraic K = 3 μJ, the particle cannot reach it without extra energy because the intervening barrier blocks the path.
A nonnegative endpoint K is necessary but not sufficient. Check every barrier along the proposed route.
Why is the second root not the first turning point for a particle arriving from the left?
Compare with an explanation
It lies beyond the forbidden interval; the particle turns at the first root before reaching it.
Predict. Change one thing. Explain.
Move the candidate endpoint beyond the peak with K_i below, equal to and above the barrier. Compare local algebraic K with the actual accessibility label.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Barrier 4 μJ; initial energy 3 μJ; local K(4 m)=3 μJ. Candidate is beyond the first turning point at 1 m.
Ideal one-dimensional positive charge initially at x = 0, V(x) = 4x − x² V for 0 ≤ x ≤ 4 m. No other work. The candidate marker is not a claimed trajectory position. At exact peak energy the particle approaches the top asymptotically, without finite-time crossing.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant potential, energy, work, field-gradient or line-integral relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor q = +2 μC, compare K_i = 6 μJ and 10 μJ. Find the lower-energy turning point and say whether each case can cross the barrier.
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Compare with the answer and four-point rubric
- 1 point: The barrier height is 4q = 8 μJ.
- 1 point: For 6 μJ, x_turn = 2 − √(4 − 6/2) = 1 m.
- 1 point: The 6 μJ case turns back before the peak.
- 1 point: The 10 μJ case exceeds 8 μJ and can cross; its K at the peak is 2 μJ.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What do energy curves plot?
Potential and total energy against position, not a literal track.
RECALL 2Which root matters for the incoming particle?
The first accessible zero of K along its direction of motion.
RECALL 3What happens exactly at a smooth barrier threshold?
Approach to the top is asymptotic in this ideal model.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Check the barrier before accepting an endpoint
- Allowed local region: H − U(x) ≥ 0.
- For this model: U_max = 4q μJ at x = 2 m, q in μC.
- Below the barrier: x_turn = 2 − √(4 − K_i/q) m.
Remember: A nonnegative endpoint K is necessary but not sufficient. Check every barrier along the proposed route.
Conditions: Ideal one-dimensional positive charge initially at x = 0, V(x) = 4x − x² V for 0 ≤ x ≤ 4 m. No other work. The candidate marker is not a claimed trajectory position. At exact peak energy the particle approaches the top asymptotically, without finite-time crossing.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 2 (official Unit 9) · Objectives 9.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.3, objectives 9.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 2: Electric Potential, numbered Unit 9 in the official combined Physics C sequence. Topics 9.1–9.3 retain their official identifiers. Models assume electrostatic fields and state whether source charges are fixed or free. Potential integrals use the specified rods, ring, arc and infinite line or cylinder. Infinite-line examples use a finite reference radius, not zero potential at infinity. The optional 3D equipotential surface uses height to represent volts, not a particle trajectory. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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