The formula
How to calculate momentum
Momentum is mass times velocity — a measure of how hard something is to stop. In any collision the total momentum of the system is conserved, which is what makes it one of the most useful quantities in physics.
The force output applies the impulse-momentum theorem: the force required equals the change in momentum divided by the time over which it happens. Stopping in a shorter time requires proportionally more force.
Fill in the following:
- Mass (kg)
- Velocity (m/s)
- Time to stop (s)
No submit button: type and the answer moves. Your inputs end up in the link, so the page can be shared already filled in.
The calculation runs on exactly the numbers currently in the fields above, recomputed in full each time — there is no dependency on the order values are entered in, so adjusting one input to test a scenario and then changing it back leaves the result exactly where it started.
Why momentum matters
A momentum calculation gets used both to check work already done by hand and to explore how a formula behaves without redoing the algebra every time an input changes — this page exists for both, since the underlying arithmetic is the same either way.
Beyond a single check, the same calculation is worth rerunning whenever a measured input changes — a new reading, a corrected value, an updated assumption — since the result here always reflects exactly what is currently in the fields above rather than a value calculated once and then left stale.
It is worth remembering that a formula is only ever as good as the assumptions built into it, and most of the standard equations used across science and statistics carry at least one simplifying assumption — a linear approximation, an idealised gas, a normally distributed error term — that holds well in most ordinary cases and breaks down at the extremes. The result here reflects the standard formula exactly; whether that formula's assumptions are appropriate for your particular situation is a separate judgement worth making deliberately rather than assuming automatically.
It is worth keeping a note of which inputs were used to produce a given result, particularly where the figure is going into a report or a further calculation — reproducing a result later, or explaining how it was reached, is far easier with the original inputs to hand than by trying to reverse-engineer them from the output alone.
Worked example
Here is the calculation with the starting values:
- Mass: 1,400 kg
- Velocity: 13.9 m/s
- Time to stop: 0.15 s
That gives:
- Momentum: 19,460 kg·m/s
- Average force to stop it in that time: 129,733.3333 N
- Kinetic energy: 135,247 J
The figures above are the calculator's own default values, shown purely so the working is visible rather than hidden — the same steps apply exactly to your own numbers, entered in the fields at the top of this page.
Reading the result
That relationship is the entire basis of crash safety. A 1,400 kg car at 50 km/h carries about 19,500 kg·m/s; crumple zones and airbags extend the stopping time from milliseconds to tenths of a second, cutting the peak force by an order of magnitude.
Where this goes wrong. Momentum is a vector, so direction matters. In a head-on collision the momenta have opposite signs and partly cancel, which is why the arithmetic in collision problems needs a consistent sign convention.
A result that is wrong by an exact factor of ten, a hundred or a similar round number is almost always a units error rather than a mistake in the formula itself — checking each input against the unit stated beside it is the fastest way to track it down.
Momentum is proportional to velocity, kinetic energy to velocity squared. Doubling speed doubles momentum but quadruples energy, which is why stopping distances grow so much faster than speed.
In a closed system with no external forces, yes — in every collision, elastic or not. Kinetic energy is only conserved in elastic collisions; in real ones it goes into deformation, heat and sound.
It returns momentum. With 1,400 kg mass, 13.9 m/s velocity and 0.15 s time to stop, that comes to 19,460 kg·m/s. Change any field and the figure moves with it.
Generally, no more than the least precise input justifies — a result reported to six decimal places from inputs measured to two significant figures is implying a precision the calculation does not actually have. The calculator shows full precision so you can round appropriately for your own use.
Yes — the equation shown in the formula section above is the standard form used in textbooks and reference material for this calculation, not a simplified or approximate version.
Yes, in the sense that it applies the correct standard formula and returns an accurate result for the inputs given — but check your own course or publication's requirements for how results should be rounded, presented and referenced, since those conventions vary and are not something a calculator can know on your behalf.