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Next exam on November 14 – 15, 2026:
All concepts

Impulse and the Impulse–Momentum Theorem

冲量与动量定理

The definitions and vector nature of impulse and momentum, the impulse–momentum theorem, and its use in cushioning and bungee-type problems.

8 real-exam questions · 4 sittings19 in mock papers22 practice

Formulas

  • Definition of impulse

    Definition
    I=FΔtI=F\Delta t

    Applies when：For a constant force; a vector along the force

  • Definition of momentum

    Definition
    p=mvp=mv

    Applies when：A vector, directed along the velocity

  • Impulse–momentum theorem

    Law
    I=Δp=mv−mv0I=\Delta p=mv-mv_0

    Applies when：A vector equation; in one dimension fix a positive direction first

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
    An object with mass m=0.2 kg\mathrm { m } = 0.2 \mathrm {~kg} moves with constant velocity v=20 m/s\mathrm { v } = 20 \mathrm {~m} / \mathrm { s } on a smooth horizontal surface. The magnitude of its momentum is ().
    • A．0.8 kg⋅ m/s0.8 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    • B．4 kg⋅ m/s4 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    • C．0.4 kg⋅ m/s0.4 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    • D．8 kg⋅ m/s8 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    Show answer & solution

    Answer: B

    1 Momentum p=mv=0.2 kg×20 m/s=4 kg⋅ m/s\mathrm { p } = \mathrm { m } \mathrm { v } = 0.2 \mathrm {~kg} \times 20 \mathrm {~m} / \mathrm { s } = 4 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    A 3 kg object moves at 4 m/s4 \mathrm {~m} / \mathrm { s }. Its momentum is
    • A．7 kg⋅ m/s7 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    • B．12 kg⋅ m/s12 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    • C．24 kg⋅ m/s24 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    • D．48 kg⋅ m/s48 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }
    Show answer & solution

    Answer: B

    p=mv=3×4=12 kg⋅ m/sp = m v = 3 \times 4 = 12 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }. Don't confuse with kinetic energy E=1/2mv2=24 J\mathrm { E } = 1 / 2 \mathrm { mv } ^ { 2 } = 24 \mathrm {~J} - option C is a trap. Answer: B.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    A 4 kg object slides from rest down an incline at 30∘30 ^ { \circ } from the horizontal. The magnitude of the impulse of gravity over 3 s is (g=10 m/s2)\left( g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)
    • A．60 N⋅ s60 \mathrm {~N} \cdot \mathrm {~s}
    • B．40 N⋅ s40 \mathrm {~N} \cdot \mathrm {~s}
    • C．80 N⋅ s80 \mathrm {~N} \cdot \mathrm {~s}
    • D．120 N⋅ s120 \mathrm {~N} \cdot \mathrm {~s}
    Show answer & solution

    Answer: D

    Gravity is constant in magnitude ( mgm g ) and direction (downward). Impulse I=mg⋅t=4×10×3=120 N⋅ sI = m g \cdot t = 4 \times 10 \times 3 = 120 \mathrm {~N} \cdot \mathrm {~s}. (The incline is irrelevant: gravity itself does not change.)
    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.

8 questions across 4 sittings1 of 4 sittings available to verify free online

19 questions across 6 mock papers — our own papers, so a coverage signal rather than real-exam evidence

22 related practice questions availablePractise free: Momentum and Impulse

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.

Glossary

velocity

Related concepts

Conservation of Momentum