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CAMBRIDGE INTERNATIONAL IGCSE™ PHYSICS 0625 · EXAMINATIONS 2026–2028 · SUPPLEMENT / EXTENDED

Momentum and impulse

Follow the momentum before and after a collision. Learn why direction matters, why carts can push apart, and why giving an object longer to stop reduces its average stopping force.

This whole lesson is Supplement / Extended, not required for Core-only candidates. It belongs to Cambridge IGCSE Physics 0625, not Egyptian Baccalaureate / Thanaweya Amma or Year 3.

Objectives 1.6.1–1.6.4 · Source PDF pp. 90–95 (printed pp. 82–87).

LEARN · STEP 1 OF 6 · SUPPLEMENT

Momentum combines mass and velocity

A moving shopping trolley is harder to stop when it is loaded. The quantity momentum combines mass and velocity: p = mv. Mass is in kg, velocity in m/s and momentum in kg m/s. Momentum is not a force, and “hard to stop” is only an introduction, not its definition.

Worked example

A 0.80 kg cart moving at 3.0 m/s has momentum 0.80 × 3.0 = 2.4 kg m/s in its direction of travel. Double the mass at the same velocity and momentum doubles.

Think first: Two carts have equal momentum. Must they have the same velocity?

No. A 1 kg cart at 4 m/s and a 2 kg cart at 2 m/s both have 4 kg m/s, if moving in the same direction.

A reliable method for a collision question

  1. Choose the system (both carts) and state the positive direction.
  2. List masses in kg and signed velocities in m/s; draw a before-and-after sketch.
  3. Check the condition: negligible net external impulse along the line of motion.
  4. Write mA uA + mB uB = mA vA + mB vB. If they stick, replace the right side with (mA + mB)v.
  5. Solve and interpret the sign. For an individual impulse, use that object’s own mass and final-minus-initial velocity.
Worked comparison: same stop, different time

A 0.25 kg ball at +12 m/s stops without rebounding. Its momentum changes by 0.25(0 − 12) = −3 kg m/s. Over 0.030 s its average resultant force is −100 N; over 0.150 s it is −20 N. Five times the stopping time gives one-fifth the average force magnitude, for the same change. These values do not tell us the peak force.

Read a momentum-time graph

The slope is Δp/Δt, so it gives the average resultant force over that interval. A straight line means constant slope; a horizontal line means zero resultant force, not necessarily zero momentum or zero velocity. Practise this in Question 14.

Practical caution: use supervised, low-speed laboratory carts and a secure track. Do not recreate vehicle crashes or use people as collision objects. A level track still has weight and support forces; these balance vertically. The relevant approximation is negligible net external horizontal impulse during the short collision.

INTERACTIVE INVESTIGATION · SUPPLEMENT

What happens when the carts stick?

Predict the common velocity

Two carts approach on a straight, level, low-friction track and stick. Choose right as positive. Neglect external horizontal impulse during impact. This idealised model does not simulate collision time or peak force. Arrow direction and length represent velocity; cart positions are a before/after sketch, not an animation.

Before · right is positive →A: 2 kgB: 1 kgat restAfter diagram hidden until you check or study.
Use the signed momenta, not the arrow lengths alone, to predict the shared velocity.

Investigate: keep both velocities fixed and increase only B’s mass. Predict before checking. Then try to make the final velocity zero by balancing the two opposing momenta. Changing any control hides the previous answer.

18 ORIGINAL QUESTIONS · SUPPLEMENT / EXTENDED · MIXED FORMATS

Practise, then explain

Six multiple-choice questions, six numerical responses and six written tasks. Numerical answers are checked automatically; written answers use a self-assessment guide. The suggested marks are for this original practice, not an official Cambridge mark scheme.

Practice stays only while this page is open. Refreshing or leaving clears these responses. Copy useful written work into your notes before leaving.

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Question 1 · Meaning and direction · Supplement · 1 mark · Multiple choice

Which statement about momentum is correct?

Optional hint

Velocity contains direction; a sign depends on your chosen positive direction.

Source alignment: supplied Physics-0625 Volume A, PDF pp. 90–95 (including Figure 13.1), and official Cambridge 2026–2028 syllabus, section 1.6. Original teaching, diagrams and practice. Clarifications to the source: conservation needs zero net external impulse, not absence of every force; FΔt uses a constant force or its time average; average force is not peak force. Signed change and change magnitude are distinguished throughout. The source PDF is not modified.

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