Impulse ($\mathbf{J}$)

The impulse of a force over a time interval is its vector integral:Its International System of Units unit is $\mathrm{N}\mathrm{s}=\mathrm{kg}m/s$. The impulse-momentum theorem equates the net impulse to the change in momentum. For one fixed-mass point mass, integrate its net force; for a closed collection obeying Newton's third law, integrate the sum of external forces. In one dimension, the signed area under the force versus time graph gives the impulse; negative force contributes negative area. ^[ref]

Three stipulated rightward net force pulses illustrate the same $J=6\mathrm{N}\mathrm{s}$: constant $12\mathrm{N}$ for $0.5\mathrm{s}$, a triangle rising from zero to $24\mathrm{N}$ at $0.25\mathrm{s}$ and falling to zero at $0.5\mathrm{s}$, and constant $3\mathrm{N}$ for $2\mathrm{s}$. Their areas areApplied separately to the same fixed mass $m=2\mathrm{kg}$, each gives $\Delta v=J/m=3m/s$. Their peak forces differ: $12$, $24$, and $3\mathrm{N}$. Thus impulse fixes the final momentum change, while the force history determines how abruptly it occurs.

The mean net force over the chosen interval is $\overline{\mathbf{F}}=\mathbf{J}/(t_f-t_i)$. Extending the interval for a fixed impulse reduces this mean force. A lower peak force additionally depends on the pulse shape: the two short pulses above have the same mean $12\mathrm{N}$ but different peaks. This explains why controlling the duration and shape of a stopping impulse matters in cushioning a collision.

An impulse directed along the position vector relative to a specified origin. Its angular impulse vanishes because $\Delta \mathbf{J}=\mathbf{r}\times \Delta \mathbf{p}=0$. It can nevertheless change kinetic energy; if an initially circular Kepler orbit receives radial velocity $w$, the added specific orbital energy is $w^2/2$.

A radial impulse that multiplies the initial circular kinetic energy by $\beta >1$ adds radial velocity $\pm \sqrt{GM/r_0}\sqrt{\beta -1}$ and leaves specific angular momentum $j^2=GM{r}_0$. The new specific orbital energy is $GM(\beta -2)/(2r_0)$, yielding orbital eccentricity $e=\sqrt{\beta -1}$ and semilatus rectum $p=r_0$. For $1<\beta <2$, the apsidal radii are $r_0/(1\pm e)$; for $\beta \ge 2$ there is no finite apocentre. A nonzero radial kick occurs away from an apsis, so the old circular radius is not automatically the new periapsis.

Integrating the closed-collection balance in momentum conservation and applying the fundamental theorem of calculus givesFor one fixed-mass point mass, this becomes ${\mathbf{J}}_{\mathrm{net}}=\Delta \mathbf{p}=m\Delta \mathbf{v}$. Zero external impulse implies equal initial and final total momenta; momentum conservation throughout the interval requires zero net external force throughout it.