PHYS2250 Chap.5 Impulse, Momentum and Collisions
Impulse, Momentum and Collisions
Let the system boundary decide what is conserved
Momentum methods depend on a declared system and the external impulse crossing its boundary. The chapter moves from force-time area to momentum change, then separates momentum conservation from kinetic-energy conservation in collisions.
Restitution supplies a relation along the impact line, while variable-mass problems require explicit momentum flux rather than an unqualified force-equals-mass-times-acceleration equation.
Reasoning route and exam method
Linear momentum p = mv is valuable because internal forces cancel when a complete system is chosen.
The general statement is that the rate of change of total momentum equals the net external force; this relationship belongs in the diagnostic checklist for the chapter. Integrating over a time interval gives impulse equal to the change in momentum; this is a useful checkpoint before any numerical substitution.
Conservation is therefore conditional rather than automatic: it applies when the external impulse on the chosen system is zero or negligible during the interaction; the same result should survive the relevant limiting-case test.
A short collision can justify that approximation even when external forces such as weight are present, because their impulse over the brief contact time is small; that distinction determines which quantities may be combined.
Momentum is conserved in an isolated collision whether the collision is elastic or inelastic; this observation fixes the physical interpretation of the algebra.
Kinetic energy is conserved only for an elastic collision; the statement supplies a direct check on the final result. The coefficient of restitution compares relative separation speed with relative approach speed along the line of impact; it adds a collision-specific relation but does not replace momentum conservation; this condition should remain visible beside the governing equation.
In two dimensions, resolve momentum by direction and apply restitution only along the impact normal for smooth bodies.
A 2.0 kg cart moving at 6.0 m s⁻¹ strikes a stationary 1.0 kg cart and they lock. Momentum gives 2(6) + 1(0) = 3v, hence v = 4.0 m s⁻¹. Initial kinetic energy is ½(2)(6²) = 36 J; final kinetic energy is ½(3)(4²) = 24 J. The 12 J difference becomes deformation, sound and internal energy.
Momentum is unchanged at 12 kg m s⁻¹, so the energy loss does not signal a momentum error.
Answer. J = m(v − u) = 0.20(−10 − 15) = −5.0 N s. The negative sign means opposite the initial positive direction.
SYSTEM BOUNDARY Variable mass. Define the system boundary before differentiating momentum.
Momentum carried across that boundary produces a flux term; writing F = ma for a changing-mass object usually omits it.
The shaded area beneath the force-time curve is the impulse delivered during contact. Its sign and area, rather than the peak force alone, determine the momentum change.
Conserve momentum for the system; conserve kinetic energy only when the collision condition earns it.
What this chapter covers
- 01
linear momentum
- 02
force-time impulse
- 03
collision conservation
- 04
coefficient of restitution
- 05
variable mass flux
Worked example · free
Compute an inelastic collision loss
- 3Conserve system momentum.
- 2Calculate initial and final kinetic energy.
- 2Interpret the difference.
Key terms
- Impulse
- The integral of force over time, equal to momentum change.
- Restitution
- The ratio of relative separation speed to relative approach speed along impact.
- External Impulse
- The time-integrated force crossing a system boundary and changing the system total momentum.
Impulse, Momentum and Collisions FAQ
When is momentum conserved?
When the net external impulse on the chosen system is zero or negligible over the interval. Internal collision forces then cancel in the total momentum balance for that system.
Is kinetic energy conserved in every collision?
No. It is conserved in an elastic collision and decreases in an inelastic one, while total energy is redistributed rather than destroyed through deformation, sound and heating.
Why are changing-mass systems special?
Mass crossing the boundary carries momentum. A correct balance includes that flux instead of differentiating mv as though the same particles remained inside throughout the process.
Exam move
State the system and interval above every collision solution. Use a momentum diagram, apply restitution only along impact, and calculate energy separately as a classification check. Work one sticking collision, one elastic head-on collision and one oblique smooth-sphere impact. Before writing equations, list the external impulses that might matter during the chosen interval and justify any neglected term by its short duration.
Draw approach and separation velocities with the same positive direction. After momentum is solved, calculate kinetic energy independently and classify the interaction from the result rather than assuming the label. For variable-mass practice, mark the boundary and the velocity of matter crossing it.
Check that the final momentum units and signs match the drawn directions, and explain where any kinetic-energy decrease appears physically. Repeat one calculation with the boundary narrowed to a single object and identify the external contact impulse replacing the internal cancellation available to the larger system.
Working through Impulse, Momentum and Collisions in PHYS2250? Sia is AskSia’s AI Physics tutor — ask any PHYS2250 Impulse, Momentum and Collisions question and get a clear, step-by-step explanation grounded in how PHYS2250 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
