Is energy conserved in an inelastic collision?

Total energy is always conserved. What is not conserved is kinetic energy, part of which becomes heat, sound or deformation.

How do I tell whether a collision is elastic?

Calculate total kinetic energy before and after. Equal means elastic; less afterwards means inelastic.

Why resolve momentum in two-dimensional collisions?

Momentum is a vector, conserved independently in each perpendicular direction. Resolving gives two separate equations that can be solved.

How did momentum conservation predict the neutrino?

In beta decay the detected particles’ momenta did not balance. Rather than abandon conservation, physicists proposed an undetected neutral particle carrying the missing momentum, later confirmed.

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99–102MechanicsStretch

Elastic and Inelastic Collisions

Momentum is conserved in every collision in a closed system. $\textbf{Kinetic energy}$ is not — some is usually transferred to heat, sound and deformation. Whether it survives is what distinguishes an $\textbf{elastic}$ collision from an $\textbf{inelastic}$ one.

40 min Topic 6 · Further Mechanics

What you'll be able to do

Distinguish elastic from inelastic collisions using kinetic energy
Calculate the kinetic energy lost in an inelastic collision
Use $E_{k} = p^{2}/2m$
Apply conservation of momentum in two dimensions by resolving

Elastic and inelastic

In an $\textbf{elastic}$ collision, total kinetic energy is conserved as well as momentum. Collisions between gas molecules are elastic, and so to a good approximation are collisions between snooker balls.

In an $\textbf{inelastic}$ collision, some kinetic energy is transferred to other forms — heat, sound, permanent deformation. Total energy is still conserved; it is only $\textit{kinetic}$ energy that is not. A $\textbf{perfectly inelastic}$ collision is one where the objects stick together, losing the maximum kinetic energy consistent with momentum conservation.

The test is always the same: compute total kinetic energy before and after and compare.

\Sigma E_{k,\text{before}} = \Sigma E_{k,\text{after}}

Elastic only.

E_{k} = \tfrac{1}{2}mv^{2}

Kinetic energy.

Example 1A 2.0 kg trolley at

3.0\ \mathrm{m\,s^{-1}}

sticks to a stationary 1.0 kg trolley, moving off at

2.0\ \mathrm{m\,s^{-1}}

. Is the collision elastic? Find the kinetic energy lost.

1Before:

\tfrac12 \times 2.0 \times 3.0^{2} = 9.0

2After:

\tfrac12 \times 3.0 \times 2.0^{2} = 6.0

3Loss:

9.0 - 6.0

AnswerInelastic; 3.0 J of kinetic energy is transferred to other forms.

Tip — Momentum being conserved says nothing about whether a collision is elastic. Only a kinetic energy comparison does.

Kinetic energy from momentum

Substituting $v = p/m$ into $E_{k} = \tfrac12 mv^{2}$ gives $E_{k} = \dfrac{p^{2}}{2m}$ .

This form is very useful for explosions and recoils, where momenta are equal and opposite. With equal momentum magnitudes, kinetic energy is inversely proportional to mass — so the lighter fragment carries most of the energy.

E_{k} = \dfrac{p^{2}}{2m}

Kinetic energy in terms of momentum.

Example 1A stationary nucleus of mass

3.9\times10^{-25}

kg emits an alpha particle of mass

6.6\times10^{-27}

kg. What fraction of the released kinetic energy does the alpha particle carry?

1Momenta are equal and opposite, so

E_{k} \propto 1/m

2Remaining nucleus mass:

3.9\times10^{-25} - 6.6\times10^{-27} = 3.834\times10^{-25}

kg.

3Fraction for alpha:

\dfrac{m_{\text{nucleus}}}{m_{\text{nucleus}} + m_{\alpha}} = \dfrac{3.834}{3.834 + 0.066}

AnswerAbout 98% — the light alpha particle takes almost all the energy.

Collisions in two dimensions

Momentum is a vector, so in a glancing collision it is conserved $\textbf{separately in each of two perpendicular directions}$ .

Resolve every velocity into components along chosen $x$ and $y$ axes, then write one conservation equation for each. Commonly the $x$ -axis is along the initial motion, so the initial $y$ -momentum is zero.

This gives two equations, enough to find two unknowns — typically a speed and an angle.

\Sigma p_{x,\text{before}} = \Sigma p_{x,\text{after}}

Along

x

\Sigma p_{y,\text{before}} = \Sigma p_{y,\text{after}}

Along

y

Example 1A 0.20 kg puck at

5.0\ \mathrm{m\,s^{-1}}

strikes a stationary 0.20 kg puck. The first moves off at

3.0\ \mathrm{m\,s^{-1}}

53^{\circ}

to its original path. Find the velocity of the second puck.

x

0.20 \times 5.0 = 0.20 \times 3.0\cos53^{\circ} + 0.20v_{x}

, so

5.0 = 1.805 + v_{x}

and

v_{x} = 3.195

y

0 = 0.20 \times 3.0\sin53^{\circ} + 0.20v_{y}

, so

v_{y} = -2.396

3Speed:

\sqrt{3.195^{2} + 2.396^{2}} = 3.99

; angle

\tan^{-1}(2.396/3.195) = 36.9^{\circ}

Answer

4.0\ \mathrm{m\,s^{-1}}

37^{\circ}

on the other side of the original path.

Using collisions as evidence

Particle physics relies on these conservation laws. If the momentum of detected particles after a decay does not balance, a particle must have carried the missing momentum undetected.

That reasoning led to the prediction of the neutrino: in beta decay the emitted electrons had a spread of energies and the momenta did not balance, which only made sense if a third, neutral particle was also emitted.

The same bookkeeping is applied to collisions in accelerators to identify the products and to infer particles that leave no track.

Tip — When a question says momentum "appears not to be conserved", the expected answer is that an undetected particle carries the difference — not that the law fails.

Equation recap

\Sigma p \text{ conserved}

All collisions in a closed system.

\Sigma E_{k} \text{ conserved}

Elastic collisions only.

E_{k} = p^{2}/2m

KE from momentum.

p_{x}, p_{y} \text{ conserved separately}

Two dimensions.

Common mistakes to avoid

Saying momentum is lost in an inelastic collision.

Momentum is conserved; kinetic energy is what is lost.

Claiming energy is destroyed in inelastic collisions.

Kinetic energy is transferred to heat, sound and deformation.

Adding speeds directly in a 2D collision.

Resolve into components and conserve momentum in each direction separately.

Assuming equal momenta mean equal kinetic energies.

E_{k} = p^{2}/2m

, so the lighter object has more kinetic energy.

Key takeaways

Momentum is conserved in all collisions within a closed system.
Elastic collisions also conserve kinetic energy; inelastic ones do not.
Objects that stick together undergo a perfectly inelastic collision.
$E_{k} = p^{2}/2m$ — equal momenta give the lighter object more energy.
In 2D, conserve momentum separately along perpendicular axes.

Test yourself

Answer a few questions, then review anything you missed.

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