A steady force acting for a time delivers an impulse, and that impulse is exactly the change in momentum it causes.

Example

A steady force acting for a time delivers an impulse, and that impulse is exactly the change in momentum it causes. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A force over time is an impulse

Push with a steady force for a while and you deliver an impulse: force times the time it acts. A 4 newton push lasting 2 seconds is an impulse of 8 newton seconds.

J=Ft=4 N2 s=8 NsJ = F\,t = 4\ \text{N} \,\cdot\, 2\ \text{s} = \hl{8}\ \text{N}\,\text{s}
A steady push acts for a whileA cart on a track with a force arrow pushing it to the right.mF

Same time, larger force, larger impulse

Hold contact time fixed. A larger force delivers more impulse.

FtJ2 N2 s4 kgm/s4 N2 s8 kgm/s6 N2 s12 kgm/s\begin{array}{c|c|c}F & t & J \\ \hline 2\ \text{N} & 2\ \text{s} & 4\ \text{kg}\,\text{m}/\text{s} \\ 4\ \text{N} & 2\ \text{s} & 8\ \text{kg}\,\text{m}/\text{s} \\ 6\ \text{N} & 2\ \text{s} & 12\ \text{kg}\,\text{m}/\text{s}\end{array}

Same force, longer time, larger impulse

Hold force fixed. Let it act longer and it delivers more impulse.

FtJ4 N1 s4 kgm/s4 N2 s8 kgm/s4 N3 s12 kgm/s\begin{array}{c|c|c}F & t & J \\ \hline 4\ \text{N} & 1\ \text{s} & 4\ \text{kg}\,\text{m}/\text{s} \\ 4\ \text{N} & 2\ \text{s} & 8\ \text{kg}\,\text{m}/\text{s} \\ 4\ \text{N} & 3\ \text{s} & 12\ \text{kg}\,\text{m}/\text{s}\end{array}

Impulse is the change in momentum

An impulse changes momentum by exactly that amount, because force is mass times acceleration and acceleration over time is the velocity change. Newton seconds and kilogram metres per second are the same unit.

J=ΔpJ = \Delta p

From rest, find the new speed

Starting from rest, the cart gains the whole impulse as momentum, 8 kilogram metres per second. Divide by the 2 kilogram mass to get the speed: 4 metres per second.

v=Jm=8 kgm/s2 kg=4 m/sv = \frac{J}{m} = \frac{8\ \text{kg}\,\text{m}/\text{s}}{2\ \text{kg}} = \hl{4}\ \text{m}/\text{s}
The cart speeds upThe cart further along the track with a velocity arrow showing its new speed.mv

Same mass, more impulse, more speed change

Starting from rest with the same cart, a larger impulse becomes a larger velocity change.

JmΔv4 kgm/s2 kg2 m/s8 kgm/s2 kg4 m/s12 kgm/s2 kg6 m/s\begin{array}{c|c|c}J & m & \Delta v \\ \hline 4\ \text{kg}\,\text{m}/\text{s} & 2\ \text{kg} & 2\ \text{m}/\text{s} \\ 8\ \text{kg}\,\text{m}/\text{s} & 2\ \text{kg} & 4\ \text{m}/\text{s} \\ 12\ \text{kg}\,\text{m}/\text{s} & 2\ \text{kg} & 6\ \text{m}/\text{s}\end{array}
mechanics A 4 N push for 2 s is an 8 N s impulse, which from rest gives a 2 kg cart a clean 4 m/s.