From height to speed
You will be able to: Use mechanical-energy conservation between two positions on a smooth track.
How can you predict speed without finding the time?
A small nonrotating cart rolls down a smooth track from a point 2 m above the bottom. As it descends, gravitational energy decreases while kinetic energy increases. The track’s shape can change the trip time without changing the final speed at the same height.
A useful starting point: Energy system boundaries →
Words and symbols before equations
- Conserved
- The total remains constant over the chosen process.
- Energy budget
- An account of each energy form at the initial and final states.
- Smooth fixed track
- Here: no friction, a stationary track, and no rotational-energy contribution.
What this picture assumes
2 kg nonrotating particle/cart released from rest at 3 m on a fixed smooth track; g=10. U=0 at y=0. No friction, air resistance or external work. Slider selects position, not time.
Connect the picture to the physics
Choose cart plus Earth and an energy zero. If external work is zero and no mechanical energy is converted into thermal energy or sound, K_i+U_i=K_f+U_f.
For a cart released from rest through height drop h, mgh=½mv² gives v=√(2gh). Mass cancels because both gravitational energy and inertia scale with mass. Do not cancel it before writing a consistent equation.
Energy determines which speeds are compatible with the heights. It does not by itself determine velocity direction, travel time or whether a one-sided track can maintain contact. A vertical-loop problem needs both an energy equation and a radial-force contact condition.
A worked example, step by step
A 2 kg nonrotating cart starts from rest at y=3 m on a frictionless fixed track. Find its speed at y=1 m using g=10.
- System is cart+Earth. Set U=0 at y=0. Initially K=0 and U=2×10×3=60 J.
- At y=1 m, U=20 J, so K=60−20=40 J.
- ½(2)v²=40 gives v=√40≈6.32 m/s.
- The 2 m height drop gives the same result directly: √(2×10×2). No time calculation is needed.
A smooth track conserves this mechanical energy only under the stated no-loss, no-rotation and no-external-work assumptions.
If you double the cart’s mass, does its speed at the same height change?
Compare with an explanation
Not in this ideal model. Both energy terms double, leaving the same speed.
Predict. Change one thing. Explain.
A 2 kg cart is released from rest at 3 m. Scrub its current height from 3 m to 0 m. Explain why the energy bars exchange equal amounts while their sum stays 60 J. The control selects a position, not elapsed time.
K=40 J; U=20 J; total=60 J. Speed=6.32 m/s. Track drawing is schematic; labeled height defines the energy calculation.
2 kg nonrotating particle/cart released from rest at 3 m on a fixed smooth track; g=10. U=0 at y=0. No friction, air resistance or external work. Slider selects position, not time.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a work, energy or power 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 questionA 1 kg cart starts at speed 2 m/s and height 4 m. It reaches height 1 m on a smooth fixed track. (a) Find initial mechanical energy. (b) Find final U. (c) Calculate final speed. (d) State two assumptions.
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Compare with the answer and four-point rubric
- 1 point: K_i+U_i=2+40=42 J.
- 1 point: 10 J.
- 1 point: K_f=32 J, so v_f=8 m/s.
- 1 point: No friction or air resistance; no rotation; stationary track/no external energy input. Any two relevant stated assumptions earn this point.
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 stays constant in this ideal model?
K+U, not K or U separately.
RECALL 2Why can mass cancel?
Both K and gravitational U are proportional to the same mass.
RECALL 3What does energy not establish by itself?
Timing, velocity direction, or track contact.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
From height to speed
- K_i+U_i=K_f+U_f when mechanical energy is conserved.
- Near Earth: ½mv_i²+mgy_i=½mv_f²+mgy_f.
- From rest: v=√(2gΔh), if Δh is the downward height drop.
Remember: A smooth track conserves this mechanical energy only under the stated no-loss, no-rotation and no-external-work assumptions.
Conditions: 2 kg nonrotating particle/cart released from rest at 3 m on a fixed smooth track; g=10. U=0 at y=0. No friction, air resistance or external work. Slider selects position, not time.
Refresh Kid · Unit 3 · Objectives 3.4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.4, objectives 3.4.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. 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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