Why this unit matters
Momentum conservation explains outcomes of short, strong interactions where force details are hard to track directly.
What you will learn
- Calculate linear momentum for single objects and multi-object systems.
- Relate impulse to momentum change using force-time reasoning.
- Apply momentum conservation in one-dimensional collision and explosion problems.
Understand the core ideas
Momentum is a vector quantity defined as p = mv. It is especially useful for short interactions where forces are large but act briefly, making full force-time modeling inconvenient. Impulse connects force and momentum change through J = delta p, and graphically impulse is the area under a force-time curve. Because momentum carries sign and direction, consistent axis choice is essential in every collision equation.
For an isolated system with negligible external impulse, total momentum is conserved during collisions and explosions. This conservation applies whether objects stick together or separate, and whether kinetic energy is conserved or not. Elastic collisions conserve both momentum and kinetic energy, while inelastic collisions conserve momentum but lose some kinetic energy to deformation, heat, or sound.
A strong problem setup identifies system boundary, positive direction, and interaction type first. Then write initial equals final momentum with explicit signs and units. That short planning sequence catches most errors before algebra begins.
Key terms
- linear momentum
- Vector quantity p = mv that combines inertia and motion direction.
- impulse
- Change in momentum, equal to force times time interval for constant force.
- isolated system
- System with negligible net external impulse during the interval of interest.
- inelastic collision
- Collision where momentum is conserved but kinetic energy is not fully conserved.
Perfectly inelastic cart collision
Cart A: 2.0 kg at +3.0 m/s. Cart B: 1.0 kg at 0 m/s. They collide on a track and stick together. Right is positive.
- Compute initial total momentum: p_i = (2.0 kg)(+3.0 m/s) + (1.0 kg)(0) = +6.0 kg*m/s.
- Use conservation of momentum for an isolated collision: p_f = p_i = +6.0 kg*m/s.
- Because carts stick, final mass is 3.0 kg, so v_f = p_f / m_total = 6.0 / 3.0 = +2.0 m/s.
- Interpret sign and magnitude: positive means rightward motion; speed is between 0 and 3.0 m/s as expected for sticking.
- Check units: kg*m/s divided by kg gives m/s, consistent for velocity.
A common misconception
Claim: Kinetic energy is always conserved in collisions.
Correction: Momentum is conserved in isolated collisions; kinetic energy is conserved only in elastic collisions.
Lessons in this unit
- Momentum and system choiceDefine system boundaries and momentum signs before solving.
- Impulse and force-time graphsUse graph area to compute impulse and predict momentum change.
- Conservation in collisionsSet up momentum equations for elastic and inelastic cases.
- Recoil and explosion modelsAnalyze separation events from total momentum constraints.
Study task
Unit checkpoint
A 2.0 kg cart moving at 3.0 m/s collides and sticks to a 1.0 kg cart at rest. What is their final speed?
Conserve momentum: (2.0)(3.0) + (1.0)(0) = (3.0)vf, so vf = 6.0/3.0 = 2.0 m/s.