Skip to main content
Next exam on November 14 – 15, 2026:
All concepts

Conservation of Mechanical Energy

机械能守恒定律

Gravitational potential energy and its relation to the work done by gravity, testing whether mechanical energy is conserved, and applying the conservation law.

7 real-exam questions · 4 sittings12 in mock papers16 practice

Formulas

  • Gravitational potential energy

    Definition
    Ep=mghE_p=mgh

    Applies when:h is measured from the chosen zero level; the sign of Eₚ depends on that choice

  • Conservation of mechanical energy

    Law
    Ek+Ep=constE_k+E_p=\text{const}

    Applies when:Only gravity (or elastic force) does work; all other forces do zero net work

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Physics December 2025 Past Paper
    Question: A basketball with mass m=0.5 kg\mathrm { m } = 0.5 \mathrm {~kg} bounces from the ground to a height h=1 m\mathrm { h } = 1 \mathrm {~m} before falling back to the ground. The change in gravitational potential energy during this entire process is ( ). (g=10 m/s2)\left( \mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)
    • A.10 J
    • B.0.5 J
    • C.5 J
    • D.0 J
    Show answer & solution

    Answer: D

    1 Initial and final positions are both on the ground, so gravitational potential energy change over the whole bounce cycle is ΔU=0\Delta \mathrm { U } = 0. 2 Therefore the net change in gravitational potential energy is 0 J .
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    Mass mm is thrown upward at v from ground level, no air resistance. Mechanical energy at the highest point is (ground == zero PE)
    • A.12mv2\frac { 1 } { 2 } m v ^ { 2 }
    • B.mgh
    • C.1/2mv2+mgh1 / 2 m v ^ { 2 } + m g h
    • D.0
    Show answer & solution

    Answer: A

    No air resistance → mechanical energy conserved. At launch: E=1/2mv2+0=1/2mv2\mathrm { E } = 1 / 2 \mathrm { mv } ^ { 2 } + 0 = 1 / 2 \mathrm { mv } ^ { 2 }. At top: KE=0\mathrm { KE } = 0, PE=mgh\mathrm { PE } = \mathrm { mgh }, but total E still =1/2mv2= 1 / 2 \mathrm { mv } ^ { 2 }. B and C are not correct expressions; D is clearly wrong. Answer: A.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    A 0.2 kg ball falls freely from a height of 2 m to the ground. The change in its gravitational potential energy is (g=10 m/s2)\left( g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)
    • A.0 J
    • B.- 4 J
    • C.4 J
    • D.2 J
    Show answer & solution

    Answer: B

    As the ball descends, its gravitational PE decreases. ΔEp=−mgh=−0.2×10×2=−4 J\Delta E _ { p } = - m g h = - 0.2 \times 10 \times 2 = - 4 \mathrm {~J}.
    Take the full CSCA Physics April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

7 questions across 4 sittings1 of 4 sittings available to verify free online
  • CSCA Physics April 2026 Past Paper2 questionsreading page not yet public
  • CSCA Physics March 2026 Past Paper2 questionsreading page not yet public
  • CSCA Physics January 2026 Past Paper2 questionsreading page not yet public
  • CSCA Physics December 2025 Past Paper1 questionView question 36 and answer free (full paper)

12 questions across 5 mock papers — our own papers, so a coverage signal rather than real-exam evidence

16 related practice questions availablePractise free: Work and Energy

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Related concepts