What this covers
Momentum is part of the Force and Motion I chapter. We teach it the way it is tested: the concept in plain English, then a worked example, then a question the student tries while the teacher checks the method.
Formulas you may need
How it is examined
It can appear in Paper 1 (objective) and Paper 2 (structured), and where an experiment applies, in Paper 3. We do not predict which questions appear; we prepare the technique for all of them.
A common mistake
How to study it
Learn the definition precisely, practise one or two SPM-style questions with full working, and link it to the rest of Force and Motion I. If it keeps costing marks, a one-to-one lesson fixes exactly that.
Force and Motion I · Formulas · Exam Papers
What you need to know
Momentum is the product of an object's mass and its velocity, expressed as p = mv, where p is momentum in kilogram metres per second (kg m s⁻¹), m is mass in kilograms, and v is velocity in m s⁻¹. Momentum is a vector quantity, so it has the same direction as the velocity.
For example, a 2 kg trolley moving at 3 m s⁻¹ has momentum p = mv = 2 kg × 3 m s⁻¹ = 6 kg m s⁻¹.
The principle of conservation of momentum states that in a closed system with no external force acting, the total momentum before an event equals the total momentum after the event. This applies to collisions and explosions alike.
In an elastic collision, both momentum and kinetic energy are conserved, and the objects separate after impact. In an inelastic collision, momentum is conserved but kinetic energy is not, and the objects may stick together and move with a common velocity after collision.
In an explosion, two objects initially at rest fly apart with equal and opposite momentum, since the total momentum must remain zero.
Worked example
Question: A 4 kg trolley moving at 3 m s⁻¹ collides with a stationary 2 kg trolley, and after the collision they stick together and move with a common velocity. Calculate this common velocity using conservation of momentum.
Given: m₁ = 4 kg, u₁ = 3 m s⁻¹, m₂ = 2 kg, u₂ = 0 m s⁻¹ (stationary). Since the trolleys stick together, this is an inelastic collision with a common final velocity v.
Applying conservation of momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v. Substituting: (4 kg × 3 m s⁻¹) + (2 kg × 0 m s⁻¹) = (4 kg + 2 kg) × v.
This gives 12 kg m s⁻¹ + 0 kg m s⁻¹ = 6 kg × v, so v = 12 kg m s⁻¹ ÷ 6 kg = 2 m s⁻¹. The two trolleys move together at 2 m s⁻¹ after the collision.
How it is examined
Paper 1 objective items commonly ask candidates to calculate momentum from mass and velocity, or to identify whether a described collision is elastic or inelastic. Paper 2 structured and essay questions frequently require candidates to apply the principle of conservation of momentum to a collision or an explosion, showing the full equation m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ with correct substitution and units, or to explain the difference between elastic and inelastic collisions in terms of kinetic energy.
Paper 3 practical work may involve trolleys colliding on a track, with candidates measuring velocities before and after collision to verify that momentum is conserved.
Common mistakes include forgetting that momentum is a vector, so a velocity in the opposite direction must be given a negative sign in the equation; treating a stationary object as though it has no mass term in the equation, which is incorrect since its initial momentum is simply zero; and omitting units on the final momentum or velocity value. Candidates must also remember that only momentum, not kinetic energy, is certain to be conserved in every type of collision.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)