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Choose right as positive and let two carts have total moving masses $m_A=1\mathrm{kg}$ and $m_B=2\mathrm{kg}$, including any attached apparatus. At an instant when their repulsive interaction has magnitude $4\mathrm{N}$, assume no other horizontal forces. Then Newton's third law gives $F_{B\mathrm{on}A}=-4\mathrm{N}$ and $F_{A\mathrm{on}B}=+4\mathrm{N}$, while Newton's second law givesThe smaller mass has the larger acceleration magnitude. For a nonzero interaction and zero other horizontal forces, $a_A/a_B=-m_B/m_A$. Meanwhile $d(p_A+p_B)/dt=m_A{a}_A+m_B{a}_B=-4+4=0$ in force units. The two carts' velocities change while total momentum stays constant.

To distinguish an instantaneous force from its accumulated impulse, prescribe an original triangular interaction pulse lasting $T=1\mathrm{s}$:This chosen pulse reaches $4\mathrm{N}$ at $t=0.5\mathrm{s}$; it is a constructed model, not an experimental trace or a claim about a particular bumper's material response. Its positive area is $\frac{1}{2}(1\mathrm{s})(4\mathrm{N})=2\mathrm{N}\mathrm{s}$. The impulse-momentum theorem therefore givesOpposite impulses give opposite momentum changes, while the velocity changes differ because of the unequal masses. The cumulative positive impulse is $I(t)=4N/s{t}^2$ during the first half and $I(t)=2\mathrm{N}\mathrm{s}-4N/s(T-t{)}^2$ during the second half. Thus $J_A(t)=-I(t)$ and $J_B(t)=I(t)$ cancel throughout the pulse, not merely at its endpoint.

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.