Free Pearson Edexcel International A Level Physics lesson

Interactions and momentum conservation

Choose the system before adding its motion.

What you’ll learn

  • Identify Newton’s third-law pairs on different bodies.
  • Use signed momentum and conservation for one-dimensional interactions.
  • Design and judge a light-gate collision test without confusing momentum and kinetic energy.

Equal interaction forces act on different bodies

Newton’s third law describes an interaction: when body A exerts a force on body B, B simultaneously exerts an equal-magnitude opposite-direction force on A. The pair belongs to the same interaction, has the same type of force and acts on different bodies. The two arrows therefore do not cancel on either body’s individual free-body diagram. A surface force on a resting book and the book’s weight may balance, but they are not a third-law pair because both act on the book.

Equal forces do not imply equal accelerations. For each constant-mass body, use its own resultant and mass in ΣF = ma. When two initially stationary skaters push apart with no significant external horizontal force, their interaction forces have equal size at every instant. A lighter skater can have the greater acceleration. Momentum p = mv combines mass with signed velocity, so its direction is the velocity direction and its unit is kg m s⁻¹.

For a constant mass, change in momentum over a short interval is mΔv. Newton’s second law relates this change to the force and duration. The equal/opposite interaction forces over the same interval produce equal/opposite changes in momentum, although the resulting velocities need not have equal magnitudes. Choose a system boundary carefully: internal forces can cancel in the total system sum even while changing the motions of its individual members. External forces must still be considered.

p = mv FA on B = −FB on A Equal interaction forces need not give equal accelerations

Symbols and units

  • p is momentum in kg m s⁻¹; m is mass in kg; v is signed velocity in m s⁻¹.
  • FA on B and FB on A are equal and opposite interaction forces in N acting on different bodies.

How the equation explains the idea

  1. Multiplying mass by signed velocity gives momentum in that direction. Equal and opposite interaction forces cause equal and opposite momentum changes over the same time.
  2. Neglecting other forces along the chosen axis, their accelerations need not match: aA = FB on A/mA and aB = FA on B/mB. Larger mass means smaller acceleration magnitude for this equal force magnitude.

Worked example

Two carts push apart

  1. Carts of 0.50 kg and 1.00 kg exert equal horizontal interaction forces of magnitude 2.0 N during a push. Neglect other horizontal forces.
  2. The 0.50 kg cart has acceleration magnitude 2.0/0.50 = 4.0 m s⁻²; the 1.00 kg cart has magnitude 2.0/1.00 = 2.0 m s⁻².
  3. Directions are opposite. Over the same 0.20 s interval their momentum changes are equal and opposite, magnitude 2.0 × 0.20 = 0.40 kg m s⁻¹.

Equal forces cause different acceleration magnitudes because the masses differ.

Keep this idea. Name both bodies in a third-law pair; cancel internal forces only when adding the whole chosen system.

Conserve a signed total for an isolated system

The total linear momentum of a system stays constant when its resultant external force is zero, or changes negligibly during the interaction considered. Internal forces occur in equal/opposite pairs, so their effects on the system’s total momentum cancel. In one dimension, choose a positive direction and add mv with signs for every included object before and after. Saying only that there is no force is misleading: internal collision forces can be very large.

If two objects stick together, their final velocities are the same and the final momentum is (m₁ + m₂)v. If they separate or bounce, retain separate final velocities. One conservation equation cannot determine two unspecified final velocities without further information. An explosion or release from a spring can increase kinetic energy while conserving total momentum; sticking can reduce kinetic energy while also conserving momentum. Track energy transfer separately rather than assuming momentum conservation means kinetic energy conservation.

A useful system contains all interacting objects whose internal forces you want to cancel. For a cart collision, friction with the track is external to the two-cart system; over a short interaction it may be negligible, but over a long journey it may not. A system initially at rest has zero total momentum, so two parts can move apart with opposite momenta. Equal and opposite momentum does not imply equal speed unless the masses are equal.

For negligible external effect: Σmᵢuᵢ = Σmᵢvᵢ Sticking: m₁u₁ + m₂u₂ = (m₁ + m₂)v

Symbols and units

  • m₁ and m₂ are masses in kg; u₁ and u₂ are initial signed velocities in m s⁻¹.
  • v is their shared final velocity in m s⁻¹ if they stick together; Σ denotes the system total.

How the equation explains the idea

  1. Negligible external impulse makes total initial and final momentum equal. Sticking gives m₁u₁ + m₂u₂ = m₁v + m₂v = (m₁ + m₂)v.
  2. Divide by total mass: v = (m₁u₁ + m₂u₂)/(m₁ + m₂). Opposite velocities must keep their signs.
  3. This conserves momentum but not necessarily kinetic energy; sticking normally transfers kinetic energy into internal forms.
Adding a stationary mass lowers the shared collision speed

Conditions. A 2 kg object travelling at +6 m s⁻¹ sticks to an initially stationary mass m₂. External impulse is negligible. For the m₂ = 0 endpoint, use the no-added-mass limiting case.

Initially stationary mass / kg; Shared final velocity / m s⁻¹.

Momentum conservation gives v = 12/(2 + m₂). Adding 2 kg gives 3 m s⁻¹, half the initial speed; adding 4 kg gives 2 m s⁻¹. Total momentum remains 12 kg m s⁻¹, but kinetic energy is reduced when a positive stationary mass sticks.

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

Worked example

A moving cart collects another

  1. A 0.60 kg cart at +2.0 m s⁻¹ sticks to a stationary 0.40 kg cart. Neglect external horizontal effects during impact.
  2. Initial momentum = 0.60 × 2.0 + 0.40 × 0 = 1.20 kg m s⁻¹.
  3. Combined final mass is 1.00 kg, so v = 1.20/1.00 = +1.20 m s⁻¹.
  4. Initial kinetic energy is 1.20 J; final kinetic energy is 0.72 J. The 0.48 J difference transfers to deformation, thermal energy and sound.

Momentum is conserved even though translational kinetic energy decreases.

Keep this idea. Choose the system and direction first, then conserve its signed momentum only when external effects are negligible.

Test a one-dimensional collision with light gates

Use two carts on a level low-friction track or air track, fitted with measured short interrupt flags. Measure the total mass of each moving assembly, including attached flags and fittings. Gates near the collision can estimate speeds immediately before and after from flag length divided by blocking time. A speed measurement needs a separate direction record; many timers display positive speed only. Use video or the known motion to assign signs. Prevent carts running off the track and keep hands clear of the collision region.

Record the individual velocities and masses before calculating total momenta. State why external effects are small: a level track minimises an unbalanced component of weight along the motion, and nearby measurements reduce frictional losses between measurement and collision. Check the level with a stationary-cart test or motion in both directions; an air track is not automatically force-free. Repeat collisions and vary a meaningful condition, such as one cart’s initial speed or mass, while keeping the interpretation comparable.

Judge a momentum difference against measurement uncertainty rather than demanding exact numerical equality. Propagate each mass/velocity uncertainty to its momentum, and combine cautious absolute uncertainty bounds for the before/after totals. A percentage difference becomes unhelpful when the initial total is close to zero; use an absolute comparison instead. If a difference exceeds the bounds, check external forces, timing/flag geometry, missed moving mass and direction signs. Evidence may support the conservation model without proving the apparatus has no limitations.

Speed estimate = flag length/blocking time Momentum difference = pafter − pbefore

Symbols and units

  • L is flag length in m and Δt is gate-blocking time in s; L/Δt is the interval mean speed in m s⁻¹.
  • p is signed system momentum in kg m s⁻¹. Before and after totals use all chosen system objects.

How the equation explains the idea

  1. The flag travels its length during blocking, so speed = L/Δt. Assign direction separately before multiplying by mass.
  2. Compute pafter − pbefore and compare its magnitude with propagated uncertainty. A nonzero difference alone is not proof that momentum conservation fails; external impulse and measurement limitations must be examined.

Worked example

A collision within measurement bounds

  1. A pair of cart measurements gives pbefore = +0.800 ± 0.030 kg m s⁻¹ and pafter = +0.770 ± 0.030 kg m s⁻¹.
  2. Measured difference = 0.770 − 0.800 = −0.030 kg m s⁻¹.
  3. A conservative uncertainty in the difference is 0.030 + 0.030 = 0.060 kg m s⁻¹.

The difference is compatible with zero within these stated bounds; it does not establish perfect equality.

Keep this idea. Measure velocities close to the interaction, record their signs and assess an absolute momentum difference against uncertainty.

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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