Lesson 4: Momentum and Collisions

Momentum gives us a second great conservation law, and it solves problems — collisions, explosions, recoils — where energy methods alone fail because kinetic energy is not conserved. This lesson completes Theme A.2 and underpins the impulse ideas behind safety engineering and rocket propulsion.

Momentum and impulse

The momentum of an object is p = mv, a vector in kg m s⁻¹. Newton's second law in its general form says the resultant force equals the rate of change of momentum, F = Δp/Δt. Rearranged, the impulse of a force equals the change of momentum: FΔt = Δp. On a force–time graph, the impulse is the area under the curve — an essential technique when the force is not constant, as in bat–ball impacts.

Conservation of momentum: if no external impulse acts on a system (or external forces are negligible during a brief collision), the total momentum before equals the total momentum after. This holds in every collision. Kinetic energy, by contrast, is conserved only in elastic collisions; in inelastic collisions some kinetic energy becomes internal energy, sound and deformation, and in a perfectly inelastic collision the objects stick together. Explosions are the reverse: stored energy becomes kinetic energy while momentum is still conserved.

Worked problem 1 — perfectly inelastic collision

A car of mass 1200 kg travelling at 15 m s⁻¹ collides with a stationary car of mass 900 kg; the two lock together. Find the common speed and the kinetic energy lost.

Momentum before: 1200 × 15 = 18 000 kg m s⁻¹
Common speed: v = 18 000 / (1200 + 900) = 8.6 m s⁻¹
Ek before = ½ × 1200 × 15² = 135 000 J = 135 kJ
Ek after = ½ × 2100 × 8.57² = 77 100 J ≈ 77 kJ
Energy transferred to other forms ≈ 58 kJ (43% of the original)

The "lost" energy is not destroyed — it does the damage: crumpling metal, heating tyres and brakes, sound.

Worked problem 2 — impulse from a bounce

A 0.16 kg cricket ball arrives at 28 m s⁻¹, is struck, and leaves at 22 m s⁻¹ in the opposite direction. Contact lasts 0.012 s. Find the average force on the ball.

Take the initial direction as positive: u = +28, v = −22 m s⁻¹
Δp = m(v − u) = 0.16 × (−22 − 28) = 0.16 × (−50) = −8.0 kg m s⁻¹
F = Δp/Δt = −8.0 / 0.012 = −670 N

The average force is about 670 N — more than 400 times the ball's weight (1.6 N), which is why we may neglect weight during the impact, but the force from the bat dominates only because the contact is so brief. Note the sign handling: a bounce roughly doubles the momentum change compared with simply stopping the ball.

Practical worked example — testing conservation on an air track

Two gliders on a level air track collide and stick. Light gates give: glider 1 (0.420 kg) at 1.02 ± 0.02 m s⁻¹ before; glider 2 (0.380 kg) stationary; the pair move at 0.535 ± 0.012 m s⁻¹ after.

Momentum before = 0.420 × 1.02 = 0.428 kg m s⁻¹ (±2%)
Momentum after = 0.800 × 0.535 = 0.428 kg m s⁻¹ (±2.2%)

The values agree within uncertainty — conservation is verified, and the quality of the conclusion depends on the uncertainty analysis, not on the closeness of the two numbers alone. Levelling the track matters: a slight slope adds an external impulse, the classic systematic error in this experiment.

Nature of Science

Conservation of momentum survived the transition from Newtonian mechanics to relativity (with p = γmv) and to quantum physics, making it one of the most trusted principles in science. Its deep origin — the symmetry of space under translation (Noether's theorem) — is a beautiful example of mathematics revealing structure in nature.

Real-world application

Rocket propulsion is momentum conservation with an exhaust: the rocket gains momentum equal and opposite to the momentum carried away by the exhaust gas. Ion thrusters expel tiny masses at ~30 km s⁻¹, producing millinewton thrusts that, applied for months, change a probe's velocity by kilometres per second.

Common mistakes (examiner style)

• Treating momentum as a scalar: in 1-D problems, failing to give one direction a negative sign, especially in bounces and explosions.
• Using the pre-collision mass with the post-collision velocity in energy calculations.
• Claiming kinetic energy is conserved because momentum is.
• In impulse problems, using the contact force as the resultant (forgetting that during long contacts, weight also provides impulse).
• Confusing impulse (N s) with work (N m) because both involve force.

IB Exam Experience

Paper 1-style questions. 1. Two objects collide and stick together. Which is always conserved in the collision? A. kinetic energy B. momentum C. speed D. mechanical energy. Key: B. 2. A 0.20 kg ball moving at 6.0 m s⁻¹ hits a wall and rebounds at 6.0 m s⁻¹. The magnitude of the momentum change is: A. 0 B. 1.2 kg m s⁻¹ C. 2.4 kg m s⁻¹ D. 12 kg m s⁻¹. Key: C — Δp = 0.20 × (6 − (−6)) = 2.4. 3. A force–time graph for a collision is a triangle of base 0.040 s and peak 500 N. The impulse is: A. 5 N s B. 10 N s C. 20 N s D. 20 000 N s. Key: B — area = ½ × 0.040 × 500 = 10 N s.

Paper 2-style structured question (7 marks). A hockey puck of mass 0.17 kg slides at 12 m s⁻¹ and is struck by a stick exerting an average horizontal force of 85 N for 0.018 s in the opposite direction to the motion. (a) Calculate the impulse of the force. [2] (b) Calculate the final speed and direction of the puck. [3] (c) The contact time is doubled by a softer stick tape while the impulse stays the same. State the effect on the average force and on the puck's final speed. [2] *Marking notes:* (a) J = FΔt = 85 × 0.018 = 1.5 N s. (b) Initial p = 0.17 × 12 = 2.04; Δp = −1.53 → final p = 0.51 → v = 3.0 m s⁻¹ in the original direction (it does not rebound — a deliberate trap; accept 3.0 m s⁻¹ with direction stated). (c) Average force halves to ≈ 43 N; final speed unchanged because impulse is unchanged.

IA idea: Use a force sensor and motion sensor to investigate how the peak force in a collision depends on contact time (different bumpers on a trolley) at constant impulse.

The ballistic pendulum — momentum meets energy

A classic two-stage problem: a 12 g bullet embeds itself in a 1.5 kg wooden block hanging from wires; the block swings up 0.065 m. Stage 1 (collision) uses momentum because the impact is brief and external impulses are negligible: m_bullet × v = (m + M) × V. Stage 2 (swing) uses energy because tension does no work and gravity is conservative: ½(m + M)V² = (m + M)gh, so V = √(2 × 9.81 × 0.065) = 1.13 m s⁻¹ and v = (1.512/0.012) × 1.13 = 142 m s⁻¹. Using energy in stage 1 or momentum in stage 2 both fail — naming the stage boundary and the correct law for each is the entire skill, and it generalises (collisions then projectiles, explosions then orbits).

Explosions and the centre of mass

In an explosion at rest, the fragments' momenta sum to zero, so the centre of mass continues exactly as before — even while individual fragments fly apart. A 6.0 kg shell at rest bursts into 2.0 kg and 4.0 kg pieces; if the 2.0 kg piece leaves at 24 m s⁻¹, the 4.0 kg piece leaves at 12 m s⁻¹ oppositely, and the kinetic energy created (½ × 2 × 24² + ½ × 4 × 12² = 576 + 288 = 864 J) came from chemical energy. Note the lighter fragment takes the larger share of the energy — the same partition rule you met with the rifle and bullet, because at equal |p|, Ek = p²/2m favours the light mass.

Momentum in two dimensions

For collisions not in a straight line, momentum conservation applies component by component: Σpx and Σpy are each conserved. In snooker-style elastic collisions between equal masses, the two final velocity vectors are perpendicular — a result obtainable only by using momentum (vector) and energy (scalar) together. In the exam, set up x and y equations explicitly, define your axes on a diagram, and treat each equation as an independent accounting statement.

Impulse on graphs: the average is not the peak

Real impact forces spike: a force–time trace for a golf drive rises to a peak of about 18 000 N for 0.45 ms, roughly triangular. The impulse is the area: ½ × 18 000 × 0.00045 = 4.05 N s, accelerating the 46 g ball to 4.05/0.046 = 88 m s⁻¹. The *average* force (impulse/time) is 9000 N — half the peak for a triangle. Candidates who equate the peak force with F in F = Δp/Δt overstate the force by the shape factor; examiners deliberately use triangular and trapezoidal traces to test whether you integrate (count the area) rather than grab the largest number on the axis.