Unit 8 is about momentum, a quantity based on an object's mass and velocity that can be transferred between objects during a collision. After this unit you should be able to calculate impulse and average force, apply conservation of momentum to collisions and explosions, and decide whether a collision is elastic or inelastic.
Read the overview first, then work through each topic note and answer its checkpoint question before moving on.
Momentum gives you a second way to analyze motion besides forces and energy. It works best when objects interact for a short, messy time, like a bat hitting a ball or two carts crashing. You don't need to know the force at every instant, only what the momentum was before and after.
Honors Physics Unit 8 covers linear momentum (), impulse (), and the impulse-momentum theorem, which says impulse equals the change in momentum. It then covers conservation of momentum for isolated systems and uses it to analyze collisions, explosions, and recoil. The unit finishes by comparing elastic collisions (momentum and kinetic energy both conserved) with inelastic and perfectly inelastic collisions (momentum conserved, kinetic energy lost), including problems in one and two dimensions.
A 2 kg ball moving east at 5 m/s has 10 kg·m/s of momentum pointing east. Because it is a vector, direction matters, and you show direction with positive and negative signs in one-dimensional problems. A slow freight train can have far more momentum than a fast tennis ball because its mass is so large.
The impulse-momentum theorem says . The same change in momentum can come from a large average net force over a short time or a small one over a long time. Airbags, padded gym floors, and bending your knees on a landing increase the stopping time, which reduces the average force.
Inside a system, objects push on each other with equal and opposite forces for the same amount of time, so one object's momentum gain matches the other's loss. That lets you set total momentum before equal to total momentum after. Kinetic energy is a separate question, and it is only conserved in elastic collisions.
Every topic in this unit comes back to Newton's laws. The second law in momentum form gives , and the third law explains why internal forces do not change a system's total momentum. When external impulse is negligible, conservation of momentum relates the initial and final velocities without requiring the forces during the collision. Some problems also need information about kinetic energy or how the objects move after impact.
Momentum is mass times velocity, p = mv, measured in kg·m/s.
The total momentum of a system stays constant when the net external force on the system is zero.
Every collision in an isolated system conserves momentum.
Momentum is mass times velocity, , measured in kg·m/s. It points the same direction as the velocity.
Net impulse is average net force multiplied by the time interval, , measured in N·s (the same unit as kg·m/s). The impulse-momentum theorem says net impulse equals the change in momentum. For constant mass, . On a net force vs. time graph, impulse is the signed area between the curve and the time axis.
Newton's second law gives over a time interval. For constant mass, its instantaneous form is . A rocket loses mass as it expels fuel, so analyzing it also requires accounting for the momentum carried by the exhaust.
Worked example. A 0.15 kg ball moves toward a wall at 30 m/s and bounces straight back at 20 m/s. Take toward the wall as positive, so m/s and m/s. Then kg·m/s. If the ball touches the wall for 0.005 s, the average force is N, pointing away from the wall. Notice the bounce makes equal to 50 m/s, not 10 m/s. A ball that bounces gets a bigger impulse than one that just stops.
Check yourself: A 1,500 kg car traveling at 20 m/s brakes to a stop in 5 s. What average net force acts on it? (Answer: kg·m/s, so N, which is 6,000 N opposite the car's motion.)
The total momentum of a system stays constant when the net external force on the system is zero. In equation form, .
This comes from Newton's third law. During a collision, object 1 pushes on object 2 exactly as hard as object 2 pushes back, for the same amount of time. Their impulses are equal and opposite, so whatever momentum one gains, the other loses.
The first step is always choosing your system. If you include both colliding carts, the forces between them are internal and cancel. Friction or a push from outside is external and can change the total. In most problems, collisions are so brief that external forces like friction barely matter during the impact, so you treat the system as isolated.
Worked example. Two skaters start at rest and push off each other. Skater A is 60 kg and skater B is 40 kg. B moves away at 3 m/s. Total momentum starts at zero, so , which gives m/s. A moves at 2 m/s in the opposite direction. The lighter skater goes faster, but the momenta (120 kg·m/s each) are equal in size and opposite in direction. Explosions and gun recoil work the same way.
Check yourself: A rifle and bullet start at rest. A 0.01 kg bullet leaves the 4 kg rifle at 400 m/s relative to the ground. If external impulse is negligible, what is the rifle's recoil velocity? (Answer: , so m/s, meaning 1 m/s backward relative to the ground.)
Every collision in an isolated system conserves momentum. Collision types differ only in what happens to kinetic energy, .
In an elastic collision, total kinetic energy after equals total kinetic energy before. Billiard balls and air-track gliders with magnetic bumpers come close. In an inelastic collision, some kinetic energy turns into heat, sound, or permanent deformation. A perfectly inelastic collision is the case where the objects stick together and share one final velocity, . It loses the most kinetic energy possible while still conserving momentum.
Worked example. A 2 kg cart at 3 m/s hits a stationary 4 kg cart and they stick. Then m/s. Kinetic energy before is J. After, it is J. So 6 J went into deformation, heat, and sound, while momentum stayed at 6 kg·m/s.
For an isolated system in a collision at an angle, momentum is conserved separately in x and y. Write one equation for each direction. If those equations and the other information in the problem determine the velocity components, use the Pythagorean theorem and inverse tangent to find speed and direction.
Check yourself: A 1,000 kg car moving at 10 m/s rear-ends a 1,500 kg car at rest, and they lock bumpers. How fast do they move together? (Answer: m/s.)
| Collision type | Momentum conserved? | Kinetic energy conserved? | What the objects do | Example |
|---|---|---|---|---|
| Elastic | Yes | Yes | Separate with total kinetic energy unchanged | Billiard balls, gas molecules |
| Inelastic | Yes | No, some is lost | Separate or stick together, with some kinetic energy converted to other forms | Fender-bender where cars separate |
| Perfectly inelastic | Yes | No, maximum loss | Stick together and share one velocity | Clay balls merging, a football tackle |
| Term | Definition |
|---|---|
| Momentum | Mass times velocity, $p = mv$, a vector quantity measured in kg·m/s. |
| Impulse | Average net force times the time interval, $J = F_{net,\mathrm{avg}}\Delta t$, equal to the change in momentum and measured in N·s. |
| Impulse-momentum theorem | The relationship $F_{net,\mathrm{avg}}\Delta t = \Delta p$, which relates net impulse to the change in momentum. |
| Conservation of momentum | The principle that total momentum of a system stays constant when no net external force acts on it. |
| Isolated system | A system with zero net external force, so its total momentum cannot change. |
| Internal force | A force between objects inside the chosen system, which cancels in pairs and cannot change the system's total momentum. |
| Elastic collision | A collision that conserves both momentum and total kinetic energy. |
| Inelastic collision | A collision that conserves momentum but converts some kinetic energy into heat, sound, or deformation. |
| Perfectly inelastic collision | A collision where the objects stick together, giving the largest possible kinetic energy loss. |
It's tempting to subtract speeds, so a ball going 30 m/s in and 20 m/s out looks like a change of 10 m/s. Momentum is a vector, so pick the incoming direction as positive and give the rebound a negative velocity. The change in velocity is -50 m/s, with a magnitude of 50 m/s. That gives a larger impulse than stopping the ball, but the average force also depends on the contact time.
Inelastic collisions lose kinetic energy, not momentum. As long as the system is isolated, momentum is conserved in every collision type. Use the momentum equation first, then check kinetic energy only if you need to classify the collision or find energy lost.
Most real collisions are inelastic, so you can't write a kinetic energy equation by default. Only use conservation of kinetic energy when the problem says the collision is elastic. If the objects stick together, the collision is definitely not elastic.
In angled collisions, adding the sizes of the momenta gives the wrong total. Break each velocity into x and y components and write a separate conservation equation for each direction. Recombine the components only at the end.