What linear momentum is
Linear momentum is the product of an object's mass and its velocity. It measures how hard the object is to stop, and unlike kinetic energy it is a vector: it points in the direction of motion, and it changes sign when the object reverses.
Its importance comes from a single fact: in any interaction between objects, the total momentum of the system is conserved, provided no external force acts. Two cars colliding, a bullet leaving a rifle, a rocket expelling exhaust, two billiard balls kissing — all of them redistribute momentum without creating or destroying any of it. That conservation law is more fundamental than F = ma, and it holds in situations where F = ma does not, such as when the mass of the system changes.
The distinction from kinetic energy is worth dwelling on because it explains a lot. A 145 g baseball at 40 m/s and an 8 g bullet at 850 m/s carry almost the same momentum, 5.8 and 6.8 kg·m/s. Their kinetic energies are 116 J and 2,890 J — a factor of 25 apart. Momentum, being linear in speed, treats them as comparable; energy, being quadratic, does not. Which one matters depends on the question: catching them is a momentum problem, and the damage they do is an energy problem.
The unit, kg·m/s, is identical to the newton-second, because a newton-second is exactly the momentum a one-newton force imparts in one second. That equivalence is not a coincidence — it is the impulse–momentum theorem written as a unit.
Momentum, impulse and where the impact force comes from
Newton's second law in its original form says that net force equals the rate of change of momentum: ΣF = dp/dt. Multiply both sides by a time interval and integrate, and you get the impulse–momentum theorem: the impulse delivered, force times time, equals the change in momentum.
Favg · Δt = Δp = m(v₂ − v₁)
Read that equation from right to left and it becomes an engineering tool. The velocity change in a collision is fixed by the physics of the situation — a car at 20 m/s hitting a wall will end up at zero, and the impulse required is therefore fixed at m × 20. The only free variable is the time over which it happens, and force is inversely proportional to that time. Stretch the stop from 15 ms to 150 ms and the average force falls by a factor of ten.
Every impact-mitigation device works this way and only this way. A crumple zone, an airbag, a crash barrier, a boxer rolling with a punch, a gymnast bending their knees on landing, packaging foam around a hard drive — none of them reduce the impulse, because the impulse is set by the velocity change. All of them lengthen Δt. The table this calculator generates makes the trade explicit: it is the same impulse spread over nine different contact times, and the force column falls in exact inverse proportion.
The word average in Favg deserves attention. Real impact forces rise and fall through the contact, often peaking at two or three times their average. The impulse–momentum theorem gives you the average exactly and tells you nothing about the peak, which depends on the stiffness of the structures involved.
Worked example: catching a cricket ball, hands soft and hands hard
A 160 g cricket ball arrives at 25 m/s and you catch it. Compare a stiff catch, where the ball stops in 20 ms, with a soft one where you draw your hands back and the stop takes 150 ms.
- Momentum before. p₁ = 0.160 × 25 = 4.0 kg·m/s.
- Momentum after. The ball is at rest, so p₂ = 0.
- Impulse required. Δp = 0 − 4.0 = −4.0 N·s. The minus sign says the impulse opposes the ball's original direction. This value is the same for both catches — you cannot change it.
- Stiff catch. Favg = 4.0 ÷ 0.020 = 200 N, about the weight of a 20 kg mass concentrated on your fingers.
- Soft catch. Favg = 4.0 ÷ 0.150 = 26.7 N, smaller by exactly the ratio of the two contact times, 150 ÷ 20 = 7.5, for exactly the same ball at exactly the same speed.
- Check against energy. The ball's kinetic energy is ½ × 0.160 × 625 = 50 J in both cases. The stiff catch absorbs it over a short distance and the soft one over a long distance — 50 J ÷ 200 N = 0.25 m against 50 J ÷ 26.7 N = 1.87 m of hand travel. Energy and momentum give consistent answers, as they must.
Now push the same arithmetic to a collision. A 1,500 kg car at 20 m/s has 30,000 kg·m/s of momentum. Stopping against a rigid wall in 0.15 s needs an average force of 200,000 N — about 20 tonnes-force. Doubling the crush distance doubles the stopping time and halves that force, which is why a car's front structure is designed to deform progressively rather than to resist.
Reading the sign, and knowing which quantity you actually need
Signs carry the physics here, so read them. A negative momentum means motion opposite to your chosen positive direction. A negative impulse means the net force acted opposite to that direction. When an object reverses — a ball bouncing off a bat, a wave hitting a wall — the impulse is larger in magnitude than either momentum alone, because the interaction must first stop the object and then send it the other way. Getting a bounce problem wrong by taking the difference of magnitudes instead of the difference of signed values is the classic error, and it undercounts the impulse.
Then choose the right quantity for your question. Ask whether you care about stopping or about damage. Stopping is momentum: the impulse required depends on mv and the force depends on how long you have. Damage is energy: the work absorbed depends on ½mv² and the force depends on how far you have. In practice both frames give consistent answers, but the shorter route is usually momentum when you know the time and energy when you know the distance.
The average force figure deserves suspicion in one specific way: it is only as good as your contact time. Contact times in real collisions are short and hard to estimate — milliseconds for a bat on a ball, tens of milliseconds for a boxing glove, roughly a tenth of a second for a car crushing its front structure. If you are unsure of it to a factor of two, your force is uncertain by the same factor. Use the calculator's table to see the whole range instead of committing to one number.
Momentum and kinetic energy of familiar moving objects
| Object | Mass | Speed (m/s) | Momentum (kg·m/s) | Energy (J) |
|---|---|---|---|---|
| Hammer head | 0.5 kg | 10 | 5.0 | 25 |
| Baseball, fast pitch | 0.145 kg | 40 | 5.8 | 116 |
| Rifle bullet | 0.008 kg | 850 | 6.8 | 2,890 |
| Cyclist and bike | 80 kg | 8.33 | 666 | 2,776 |
| Sprinter | 80 kg | 10 | 800 | 4,000 |
| Car on the motorway | 1,500 kg | 27.78 | 41,667 | 578,704 |
| Loaded articulated lorry | 40,000 kg | 25 | 1,000,000 | 12,500,000 |
Compare the bullet and the baseball: nearly identical momentum, energies a factor of 25 apart. Momentum is linear in speed and energy is quadratic, which is why they rank fast light objects so differently.
Mistakes that produce wrong momentum and impulse figures
- Dropping the sign on a reversal. A ball arriving at +30 m/s and leaving at −20 m/s has a velocity change of −50 m/s, not 10. This single error is the most common in the whole topic.
- Using speed instead of velocity when summing. Momentum of a system is the vector sum. Two equal masses moving towards each other have zero total momentum, not double.
- Confusing impulse with force. Impulse is force multiplied by time and has units of N·s. Quoting an impulse in newtons is a dimensional error.
- Guessing the contact time. The average force is inversely proportional to it, so an order-of-magnitude guess gives an order-of-magnitude answer. Where possible measure it or bound it.
- Assuming the average force is the peak force. Real force–time curves peak well above their mean. The impulse–momentum theorem gives the mean exactly and the peak not at all.
- Applying conservation of momentum with an external force present. Momentum is conserved for an isolated system. If friction, gravity or a wall acts from outside the system you defined, include it or redraw the system boundary.
Momentum, collisions and the rest of dynamics
Conservation of momentum is what lets you solve collisions without knowing anything about the forces involved. Write down the total momentum before and set it equal to the total afterwards, and you have one equation per dimension. For a perfectly inelastic collision, where the objects stick together, that is enough on its own. For an elastic collision you add conservation of kinetic energy as a second equation and solve the pair.
The link to forces runs through Newton's second law: F = ma is what the momentum form reduces to when mass is constant, and the momentum form is the one to use when it is not. Rockets, hoppers and lifted chains all need dp/dt. Conversely, once you have an average force you can hand it to the second law for an acceleration and then to the final velocity calculator for the distance travelled during the stop.
The rotational analogue is angular momentum, L = Iω, with moment of inertia in place of mass and angular velocity in place of velocity. It obeys its own conservation law and changes only when a net torque acts — which is why a spinning skater speeds up when they pull their arms in, and why a gyroscope resists being tilted.
Finally, note the limit. Above about a tenth of the speed of light, momentum is γmv rather than mv, and the discrepancy grows without bound as the speed approaches c. Conservation still holds exactly; it is the expression for p that changes.