Free Pearson Edexcel International A Level Physics lesson

Impulse and collisions in a plane

Track momentum as a vector and test whether kinetic energy survives.

What you’ll learn

  • Obtain impulse from a force record.
  • Conserve two perpendicular momentum components.
  • Evaluate collision evidence from calibrated video.

A short force can produce a lasting velocity change

Momentum changes when a resultant external force acts. For constant mass, F = Δp/Δt describes the average force over an interval; multiplying by the interval gives impulse. A rebound requires signed velocities: an object travelling towards a barrier and then away changes momentum by more than it would if it merely stopped. Choose a positive direction before reading either velocity.

During real contact the force rises and falls. The signed area under the force–time curve is the impulse; a peak force multiplied by the whole duration usually overestimates that area. Two cushions can deliver the same momentum change with different peak forces and durations. Extending stopping time lowers average force when the initial and final momenta are fixed.

For core practical 9, zero a force sensor, align its axis with a low-friction trolley and log the impact quickly enough to resolve the pulse. Obtain velocities just before and after contact using a calibrated light gate or video. Integrate after subtracting the baseline, compare with mΔv and repeat across several incident velocities. Sensor saturation or a poorly sampled peak is an instrument limitation, not evidence that momentum fails.

Impulse J = ∫F dt = Δp = m(v − u); average F = Δp/Δt

Symbols and units

  • J is impulse in N s = kg m s⁻¹
  • F is resultant force in N
  • t is time in s
  • m is constant mass in kg
  • u and v are signed velocities in m s⁻¹.

How the equation explains the idea

  1. Each short interval contributes FΔt to momentum change, so the signed force–time area gives Δp. For constant mass, Δp = m(v − u).
  2. Keeping Δp fixed while doubling contact duration halves the average force Δp/Δt. Peak force need not equal this average, so use the pulse shape if the peak is asked for.
Different peaks can deliver the same impulse

Conditions. Two illustrative triangular positive-force pulses: an 8 N peak over 1 s, and a 4 N peak over 2 s. Mass is constant; plotted force is the resultant along one axis.

Time / s; Resultant force / N.

Each triangle has area 4 N s: ½ × 1 × 8 = ½ × 2 × 4. They therefore produce the same momentum change despite different peak and average forces during contact.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Rebounding inspection cart

  1. A 0.40 kg cart has u = +1.5 m s⁻¹ and v = −0.50 m s⁻¹.
  2. Δp = 0.40(−0.50 − 1.5) = −0.80 kg m s⁻¹.
  3. Over 0.080 s, average force = −0.80/0.080 = −10 N.

The barrier gives an impulse of −0.80 N s.

Keep this idea. Compare signed force area with the signed momentum change of the same object.

Conserve a vector one component at a time

For two colliding objects, internal contact forces exchange momentum but cancel in the system total. If external impulse is negligible during the event, the total momentum vector before equals that afterwards. A two-dimensional collision therefore supplies two equations. Resolve every velocity along fixed perpendicular axes; conservation of momentum magnitudes is generally false.

A sphere initially moving east can emerge north-east while a second sphere leaves south-east. Their north–south momentum components must cancel if the initial component was zero. The east components must add to the original eastward momentum. A vector diagram gives a useful directional check before solving simultaneous equations, and helps expose a missing minus sign.

Core practical 10 uses ICT to analyse small-sphere collisions. Place the camera perpendicular to the motion plane, include a scale in that plane and check frame timing. Fit positions across several frames immediately before and after the collision to obtain component velocities. Record masses and propagate uncertainties; rolling, friction and perspective can affect the result. Compare both components independently instead of declaring agreement from the speed totals.

Σm uₓ = Σm vₓ; Σm uᵧ = Σm vᵧ when external impulse ≈ 0

Symbols and units

  • m is mass in kg
  • uₓ,uᵧ and vₓ,vᵧ are signed initial and final components in m s⁻¹
  • momentum components have units kg m s⁻¹.

How the equation explains the idea

  1. Negligible external impulse conserves the total momentum vector. Resolve every velocity using one common pair of perpendicular axes, then conserve each component independently.
  2. Adding momentum magnitudes can fail: momenta of 3 kg m s⁻¹ east and 4 kg m s⁻¹ north give magnitude 5 kg m s⁻¹, not 7. The direction also matters.

Worked example

A sideways split

  1. A 0.20 kg sphere starts with velocity (3.0, 0) m s⁻¹; another 0.20 kg sphere is stationary.
  2. Afterwards the first has velocity (1.0, 1.5) m s⁻¹.
  3. Equal masses give second velocity (3.0 − 1.0, 0 − 1.5) = (2.0, −1.5) m s⁻¹.

The second sphere travels with speed 2.5 m s⁻¹, below the original direction.

Keep this idea. Use one coordinate system for all objects and conserve each momentum component.

Momentum agreement does not prove elasticity

An isolated collision conserves total momentum even when deformation, sound and heating reduce the system’s kinetic energy. Elastic means total kinetic energy is also conserved. Inelastic does not necessarily mean the objects stick together; a rebound can still be inelastic. Compare the sums of ½mv² before and after, including every participating object.

For non-relativistic motion, substitute v = p/m into ½mv² to obtain Ek = p²/(2m). Here p is the magnitude of one particle’s momentum. Squaring each object’s momentum and summing its energy is different from squaring the total vector momentum. Opposite momenta may cancel to zero while the objects still carry positive kinetic energies.

A practical energy test needs uncertainty as well as a calculated ratio. Speeds enter squared, so a small percentage error in speed makes approximately twice that percentage error in kinetic energy. For rolling spheres, translational energy alone may omit rotational energy exchanged in contact. Report what energy model was measured and avoid calling a small apparent gain evidence of energy creation.

Ek = ½mv² = p²/(2m) for non-relativistic motion; elastic if ΣEk,before = ΣEk,after

Symbols and units

  • Ek is kinetic energy in J
  • m is mass in kg
  • speed v is in m s⁻¹
  • momentum magnitude p is in kg m s⁻¹.

How the equation explains the idea

  1. For nonrelativistic motion, p = mv, hence v = p/m. Substitute into Ek = mv²/2 to get Ek = p²/(2m).
  2. Conserved total momentum does not imply conserved kinetic energy: two equal masses with equal opposite speeds can stick at rest. Their net momentum was zero, but their initially positive kinetic energies become internal energy.

Worked example

Two carts latch together

  1. A 0.50 kg cart at 2.0 m s⁻¹ hits an identical stationary cart and both move at 1.0 m s⁻¹.
  2. Before: Ek = ½ × 0.50 × 2.0² = 1.0 J.
  3. After: Ek = ½ × 1.0 × 1.0² = 0.50 J; momentum remains 1.0 kg m s⁻¹.

The collision is inelastic; 0.50 J leaves the translational kinetic store.

Keep this idea. Test elasticity with the total kinetic energy of the stated model, separately from momentum.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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