Lessons · momentum

Head-on collision: momentum can be zero — and still conserved

Regime 2 — calculus does what algebra cannot. The acceleration isn't constant, so the algebra formulas don't apply; calculus is the only road in, and SymPy proves the closed form solves the equation of motion.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A $m_1 = 1$ kg block moving right at $v_1 = +4$ m/s meets a $m_2 = 2$ kg block moving left at $v_2 = -2$ m/s, head-on. Their momenta are equal and opposite, so the total momentum is exactly zero. Slide the coefficient of restitution $e$ from $1$ (perfectly elastic — the blocks bounce apart) down to $0$ (perfectly inelastic — they latch together): the momentum total stays pinned at zero the whole way (only the split between the two bodies flips, above and below the line), while the kinetic-energy total collapses — at $e = 0$ the pair comes to a dead stop and all $12$ J is lost. Zero total momentum is still a conserved momentum; it does not mean the collision does nothing.

Elastic (

e=1

): body 1 final

v_1'

-4 m/s

Elastic (

e=1

): body 2 final

v_2'

2 m/s

Inelastic (

e=0

): common velocity

v_{cm}

0 m/s

Inelastic (

e=0

): kinetic energy lost

12 J

Momentum conservation is the time-integral of Newton's third law

F_{12} = -F_{21} \ \Longrightarrow\ J_1 = \int F_{12}\,dt = -\int F_{21}\,dt = -J_2 \ \Longrightarrow\ \Delta p_1 = -\Delta p_2

During contact the two bodies push on each other with equal and opposite forces at every instant (Newton's third law). The impulse each receives is the time-integral of that force, $J=\int F\,dt$ — and since the forces are equal and opposite, so are the impulses. They cancel: $\Delta p_1+\Delta p_2=0$ . The total momentum is unchanged no matter how complicated the contact force is, elastic or not. This is the impulse–momentum theorem doing the work the algebra states as a rule.

check $m_1v_1'+m_2v_2'=m_1v_1+m_2v_2$ ; check $v_2'-v_1'=e(v_1-v_2)$ ; check $\Delta\text{KE}=\tfrac12\mu(1-e^2)(\Delta v)^2$ ; $e=1\Rightarrow\text{KE}$ conserved; $e=0\Rightarrow$ common velocity $v_{cm}$

✓ Total momentum is unchanged for every $e$ : $m_1v_1'+m_2v_2' = m_1v_1+m_2v_2$ . The contact forces are equal and opposite (Newton's third law), so the impulses $J=\int F\,dt$ cancel — true for any force profile, elastic or not. [simplify]
✓ The solved finals obey the restitution relation $v_2'-v_1' = e\,(v_1-v_2)$ — the relative speed of separation is $e$ times the relative speed of approach. [simplify]
✓ The kinetic energy lost is $\tfrac12\mu(1-e^2)(v_1-v_2)^2$ with reduced mass $\mu=m_1m_2/(m_1+m_2)$ — so KE is conserved only when $e=1$ , and the loss is greatest when $e=0$ . [numeric]
✓ At $e=1$ (perfectly elastic) the kinetic energy is unchanged: $\text{KE}_f = \text{KE}_i$ . [simplify]
✓ At $e=0$ (perfectly inelastic) the bodies move off together at the common velocity $v_{cm}=(m_1v_1+m_2v_2)/(m_1+m_2)$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “When two equal-and-opposite momenta cancel to a total of zero, there is 'no momentum' to conserve — so a head-on collision like this is a special case where conservation doesn't really say anything.”

Zero is a perfectly good value of the total momentum, and it is conserved: it stays zero before, during, and after, at every $e$ . Conservation is not silent here — it is what forces the two final momenta to remain equal and opposite. And it is independent of kinetic energy: drag $e$ to $0$ and the blocks stop dead (the common velocity is $v_{cm} = (m_1v_1+m_2v_2)/(m_1+m_2) = 0$ ), losing all $\tfrac12\mu(v_1-v_2)^2 = 12$ J of kinetic energy, while the momentum total never budges from zero. A zero total momentum is the cleanest possible proof that momentum conservation and kinetic-energy conservation are two different statements.

Modeling assumptions — author-asserted, disclosed not discharged

An isolated 1D collision: no external horizontal force during contact, so the total momentum is conserved (the blocks' mutual forces are internal). Up the track is positive, so $v_2 < 0$ is genuinely the leftward block.
The outcome is summarised by a single coefficient of restitution $e \in [0, 1]$ — the ratio of the relative speed of separation to the relative speed of approach — rather than modelling the contact force in detail.

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

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