Lessons · momentum

Collisions: why momentum always survives but kinetic energy usually doesn't

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 $m_1 = 2$ kg cart moving at $v_1 = 3$ m/s strikes a stationary $m_2 = 1$ kg cart. Slide the coefficient of restitution $e$ from $1$ (perfectly elastic — the carts bounce apart with no energy lost) down to $0$ (perfectly inelastic — they latch together). The momentum bars are the same total height before and after at every setting; the kinetic-energy bars match only when $e = 1$ , and the gap that opens as $e \to 0$ is energy lost to deformation. Momentum is conserved because the contact forces are equal and opposite (Newton's third law); kinetic energy is not, because the deformation work need not come back.

Elastic (

e=1

): body 1 final

v_1'

1 m/s

Elastic (

e=1

): body 2 final

v_2'

4 m/s

Inelastic (

e=0

): common velocity

v_{cm}

2 m/s

Inelastic (

e=0

): kinetic energy lost

3 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: “Momentum is only conserved when no kinetic energy is lost — so in a 'sticky' (inelastic) collision, momentum isn't conserved either.”

Momentum is conserved in every collision, elastic or not. It follows from Newton's third law: during contact the two bodies push on each other with equal and opposite forces at every instant, so the impulses $J = \int F\,dt$ they receive are equal and opposite and cancel — the total momentum change is zero regardless of how much energy is lost. Kinetic energy is the quantity that is only conditionally conserved (when $e = 1$ ). The two laws are independent: drag $e$ to $0$ and watch the momentum total bar stay pinned while the kinetic-energy total bar collapses by $\tfrac12\mu(1-e^2)(\Delta v)^2$ . People conflate the two because the elastic case happens to conserve both.

Modeling assumptions — author-asserted, disclosed not discharged

An isolated 1D collision: no external horizontal force during contact, so the total momentum is conserved (the carts' mutual forces are internal).
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

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):