A spring launches a cart
You will be able to: Transfer elastic energy into kinetic energy and state the model’s limits.
How does stored spring energy set the launch speed?
A small cart touches a compressed spring on a level, smooth track. Release it: the spring pushes the cart while relaxing. At the relaxed length, the initially stored elastic energy has become the cart’s kinetic energy.
A useful starting point: Elastic potential energy →
Words and symbols before equations
- Compression x₀
- Initial deformation magnitude from relaxed length, in m.
- Launch point
- Here the relaxed spring position where an unattached cart loses contact.
- Ideal spring–cart system
- Negligible spring mass, no friction, no rotation, no energy lost to sound.
What this picture assumes
0.5 kg nonrotating cart, massless ideal spring k=200 N/m, initial compression 0.10 m. Horizontal frictionless launch; no other energy transfers. Slider selects configuration, not time.
Connect the picture to the physics
Choose cart plus spring. Initially K=0 and U_s=½kx₀². At relaxed length U_s=0, so ½mv²=½kx₀² and v=x₀√(k/m).
At an intermediate deformation x, ½kx₀²=½mv²+½kx². Some energy is still stored. The cart’s maximum speed in this launch occurs at relaxed length.
If the cart is attached, it can continue into stretching and later reverse; if merely touching the spring, it leaves at relaxed length. Neither case permits negative kinetic energy at a deformation beyond the initial turning magnitude without extra energy.
A worked example, step by step
A 0.5 kg cart is launched by a 200 N/m spring compressed 0.10 m. Find its speed at relaxed length and when 0.06 m of compression remains.
- Initial elastic energy =½(200)(0.10²)=1 J.
- At relaxed length, ½(0.5)v²=1, so v=2 m/s.
- At remaining compression 0.06 m, U_s=½(200)(0.06²)=0.36 J. Therefore K=0.64 J.
- Speed there is √(2×0.64/0.5)=1.6 m/s. It is lower because some energy remains elastic.
The spring force is largest at maximum compression, but the cart’s speed is initially zero.
At relaxed length the spring force is zero. Must the cart’s speed be zero?
Compare with an explanation
No. In this ideal launch the cart has maximum speed there; force controls acceleration, not the current speed.
Predict. Change one thing. Explain.
Keep initial compression 0.10 m, k=200 N/m and mass 0.5 kg. Reduce the remaining compression toward zero. Compare the force trend with the speed trend and account for both energy bars.
Elastic energy=0.36 J; kinetic energy=0.64 J; speed=1.6 m/s; spring force magnitude=12 N. Total remains 1 J.
0.5 kg nonrotating cart, massless ideal spring k=200 N/m, initial compression 0.10 m. Horizontal frictionless launch; no other energy transfers. Slider selects configuration, 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 2 kg cart is pushed by a 100 N/m spring compressed 0.20 m on a smooth horizontal track. (a) Find stored energy. (b) Find launch speed. (c) Predict launch speed for compression 0.40 m. (d) Explain how friction would change the result.
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Compare with the answer and four-point rubric
- 1 point: 2 J.
- 1 point: √2≈1.41 m/s.
- 1 point: 2√2≈2.83 m/s.
- 1 point: Some initial elastic energy would become thermal energy, leaving less kinetic energy and a smaller launch speed.
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 1Where is speed greatest in this launch?
At relaxed spring length.
RECALL 2Does maximum force mean maximum speed?
No; initially force is maximum and speed is zero.
RECALL 3Which boundary includes elastic U?
One containing the cart and spring.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A spring launches a cart
- ½kx₀²=½mv²+½kx² for the stated isolated ideal model.
- At relaxed length: v=x₀√(k/m).
- Real spring mass, friction or rotation require additional energy terms.
Remember: The spring force is largest at maximum compression, but the cart’s speed is initially zero.
Conditions: 0.5 kg nonrotating cart, massless ideal spring k=200 N/m, initial compression 0.10 m. Horizontal frictionless launch; no other energy transfers. Slider selects configuration, 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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