= Equal interaction forces and unequal accelerations
{title2=$m_Aa_A=-m_Ba_B$}
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> gives
$$
a_A=\frac{-4\,\mathrm N}{1\,\mathrm{kg}}=-4\,\mathrm{m/s^2},\qquad a_B=\frac{+4\,\mathrm N}{2\,\mathrm{kg}}=+2\,\mathrm{m/s^2}.
$$
The 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 $d(p_A+p_B)/dt=m_Aa_A+m_Ba_B=-4+4=0$ 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 $T=1\,\mathrm s$:
$$
f(t)=\begin{cases}
8\,\mathrm{N/s}\;t,&0\leq t\leq T/2,\\
8\,\mathrm{N/s}\;(T-t),&T/2\leq t\leq T,\\
0,&\text{otherwise},
\end{cases}\qquad
F_{B\mathrm{\ on\ }A}(t)=-f(t),\quad F_{A\mathrm{\ on\ }B}(t)=f(t).
$$
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 $\tfrac12(1\,\mathrm s)(4\,\mathrm N)=2\,\mathrm{N\,s}$. The <impulse-momentum theorem> therefore gives
$$
J_A=-2\,\mathrm{N\,s},\quad J_B=+2\,\mathrm{N\,s},\qquad \Delta v_A=-2\,\mathrm{m/s},\quad \Delta v_B=+1\,\mathrm{m/s}.
$$
Opposite <impulses> give opposite <momentum> changes, while the <velocity> changes differ because of the unequal <masses>. The cumulative positive <impulse> is $I(t)=4\,\mathrm{N/s}\;t^2$ during the first half and $I(t)=2\,\mathrm{N\,s}-4\,\mathrm{N/s}\;(T-t)^2$ 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.
\Image[/newtons-third-law-opposite-pulse-and-impulse.png]
{title=Constructed opposite force pulses and impulses give unequal accelerations for unequal masses}
{description=Original prescribed triangular pulse, not measured data. Separate panels show signed <force>, cumulative <impulse>, and <acceleration> for $m_A=1\,\mathrm{kg}$ and $m_B=2\,\mathrm{kg}$ with no other horizontal <forces>. Total <momentum> change is zero throughout.}
Back to article page
