Inelastic Collisions — Momentum Conserved, KE Lost

Spec mapping: OCR H556 Module 3.3 — Distinction between elastic and inelastic collisions; perfectly inelastic ("sticky") collisions as the limit case; calculations of kinetic energy converted in inelastic interactions. (Refer to the official OCR H556 specification document for exact wording.)

In the real world, collisions are almost always inelastic: they conserve linear momentum (no resultant external force acts during the brief interaction), but the total kinetic energy decreases. The "lost" KE is converted into heat, sound, permanent deformation and elastic vibrations within the colliding bodies. A head-on car crash is inelastic; a lump of putty hitting a wall is perfectly inelastic; even a high-quality bouncing ball is mildly inelastic because it never quite returns to the height from which it was dropped. The OCR mark scheme keenly tests whether candidates can identify which energy law to use when.

This lesson completes the classification of collision types for OCR Module 3.3. We work through standard inelastic and perfectly inelastic examples, derive the fraction of KE lost in the stuck-together case, set up the ballistic-pendulum problem (the foundational "split the analysis into two phases" technique), and tabulate the contrast with Lesson 7's elastic case so that the right tools are clearly attached to the right problem type.

Key Definition: A collision is inelastic if the total kinetic energy decreases during the interaction — some macroscopic KE is converted into internal energy (heat, sound, deformation). It is perfectly inelastic if the two bodies move together as a single composite mass afterwards (i.e. relative velocity of separation is zero); this is the limit case that loses the maximum KE consistent with momentum conservation.

Inelastic vs Perfectly Inelastic

Two distinct types of inelastic collision appear in A-Level questions:

Inelastic (general): the two bodies separate after the collision (their relative velocity is non-zero) but total kinetic energy has decreased. Examples: a bouncing ball, most everyday collisions, a low-velocity car-to-car prang where the vehicles dent but do not lock.
Perfectly inelastic: the two bodies stick together after the collision and move as a single object with a single common velocity. The relative velocity of separation is zero. This case loses the maximum possible KE consistent with momentum conservation (proof below). Examples: railway trucks coupling automatically on impact, a dart embedding in a dartboard, a bullet lodging in a wooden block, two clay balls colliding and merging.

Both types conserve linear momentum (closed system, no external resultant force). Neither conserves kinetic energy. The OCR spec uses "inelastic" loosely to mean "any collision in which KE is not conserved", and "perfectly inelastic" or "the bodies stick together" for the maximum-loss limit case.

Common Exam Mistake: Treating "inelastic" and "perfectly inelastic" as synonymous. They are NOT. Perfectly inelastic is the sticky limit; general inelastic can have any KE loss between zero (elastic limit) and the perfectly-inelastic maximum.

Diagram — Collision Type Decision Tree

flowchart TD
  A[Two bodies collide] --> B{Total KE after = Total KE before?}
  B -- Yes --> C[Elastic — Lesson 7]
  B -- No --> D{Do they stick together (relative v_sep = 0)?}
  D -- Yes --> E["Perfectly inelastic — MAX KE lost"]
  D -- No --> F["Inelastic but separating — partial KE lost"]
  C --> G["2 conservation laws: p AND KE"]
  E --> H["1 unknown (common v_f); use p only"]
  F --> I["2 unknowns; use p + extra data (e.g. one final v)"]

Summary Table — Three Collision Types

Feature	Elastic	Inelastic (separating)	Perfectly inelastic
Momentum conserved?	Yes	Yes	Yes
Kinetic energy conserved?	Yes	No (some lost)	No (maximum lost)
Do the bodies stick?	No	No	Yes
Relative velocity of separation	$= -(u_1 - u_2)$	$	v_{\text{sep}}
Equations needed	Both p and KE	p, plus extra data	p only (common $v_f$ )
Real examples	Atomic collisions; snooker balls (approx)	Most everyday collisions; bouncing balls	Train trucks coupling; dart in board; bullet in block

Perfectly Inelastic — The One-Equation Case

When two bodies stick together, the problem has a single unknown (the common final velocity $v_f$ ), so a single equation — momentum conservation — is enough to solve it. From momentum conservation:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2)\,v_f$

Solving for the common final velocity:

$v_f = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

The kinetic energy converted to other forms is:

$\Delta \text{KE} = \text{KE}_{\text{before}} - \text{KE}_{\text{after}}$

which is always positive for any genuine collision (the sum of squares is minimised by the equal-velocity case, as the proof below shows).

Worked Example 1 — Coupling Trucks

A $5000$ kg railway truck moves at $4.0$ m s $^{-1}$ and collides with a stationary $3000$ kg truck on the same straight track. They couple together automatically. Find the common velocity and the kinetic energy converted.

Momentum: $5000 \times 4.0 + 3000 \times 0 = 8000\,v_f \Rightarrow 20\,000 = 8000\,v_f \Rightarrow v_f = 2.5$ m s $^{-1}$ .

Kinetic energy: Before $= \frac{1}{2}(5000)(4.0)^2 = 40\,000$ J $= 40$ kJ. After $= \frac{1}{2}(8000)(2.5)^2 = 25\,000$ J $= 25$ kJ.

$\Delta$ KE = 15 kJ converted (about $37.5\%$ of the original kinetic energy). The lost energy is converted into sound (the audible "clank" of the coupler), heat in the coupling mechanism, and small permanent deformation of the linkage.

Note that this is the maximum KE loss consistent with momentum conservation. Any other pair of final velocities that satisfy $5000 v_1' + 3000 v_2' = 20\,000$ — say $v_1' = 3.0$ , $v_2' = 1.67$ — would have higher total final KE, so less lost. The sticky outcome is the energetic extreme.

Proof — Perfectly Inelastic Collisions Lose the Maximum KE

A useful A* result. For two masses with total momentum $p_{\text{total}} = m_1 u_1 + m_2 u_2$ , define the centre-of-mass velocity $v_{\text{cm}} = p_{\text{total}}/(m_1 + m_2)$ . The total KE can be decomposed as:

$\text{KE}_{\text{total}} = \tfrac{1}{2}(m_1 + m_2) v_{\text{cm}}^2 + \tfrac{1}{2} \mu (v_1 - v_2)^2$

where $\mu = m_1 m_2 / (m_1 + m_2)$ is the reduced mass and $(v_1 - v_2)$ is the relative velocity. The first term is the KE of the centre of mass (fixed by momentum conservation, so unchanged). The second term, $\frac{1}{2} \mu v_{\text{rel}}^2$ , is the "relative kinetic energy" of the two bodies as they move with respect to each other.

For any collision, $p_{\text{total}}$ is conserved, so $v_{\text{cm}}$ is unchanged → first term unchanged. The KE loss is therefore entirely from the second term:

$\Delta\text{KE} = \tfrac{1}{2}\mu \left[(u_1 - u_2)^2 - (v_1 - v_2)^2\right]$

This is maximised when $(v_1 - v_2) = 0$ — exactly the perfectly inelastic case. Any other final state has non-zero relative velocity and therefore some residual relative KE. So perfectly inelastic = maximum KE loss. ∎

(OCR does not require this proof, but the result is examinable: "Explain why a perfectly inelastic collision loses the maximum possible kinetic energy.")

Worked Example 2 — Bullet and Block

A $0.020$ kg bullet travels at $400$ m s $^{-1}$ horizontally and embeds in a $2.0$ kg wooden block resting on a smooth horizontal surface. Find the common velocity of the bullet-block system and the energy converted.

Momentum: $(0.020)(400) + (2.0)(0) = (2.02)\,v_f \Rightarrow 8.0 = 2.02\,v_f \Rightarrow v_f = 3.96$ m s $^{-1}$ .

KE before $= \frac{1}{2}(0.020)(400)^2 = 1600$ J. KE after $= \frac{1}{2}(2.02)(3.96)^2 = \frac{1}{2}(2.02)(15.7) = 15.8$ J.

$\Delta$ KE = 1584 J converted (about $99\%$ of the original KE). Almost all the bullet's KE has gone into heat (the bullet and the wood are noticeably hot), permanent deformation (the bullet is flattened, the wood splintered) and sound. Only $\sim 1\%$ of the bullet's KE has become macroscopic motion of the block.

This is an extreme but common case: when a light fast object hits a heavy slow (or stationary) object and sticks, the fraction of KE retained is approximately $m_1 / (m_1 + m_2) \approx m_1 / m_2$ , which is tiny when one mass is much larger. The exact fraction (derived below) is:

$\frac{\text{KE}_{\text{after}}}{\text{KE}_{\text{before}}} = \frac{m_1}{m_1 + m_2}$

In our example: $0.020 / 2.02 \approx 0.0099$ , or $0.99\%$ — agreeing with the direct calculation.

Worked Example 3 — Inelastic but Not Sticking

A $2.0$ kg trolley at $4.0$ m s $^{-1}$ collides with a $3.0$ kg trolley at rest. After the collision, the $2.0$ kg trolley moves at $0.5$ m s $^{-1}$ in the original direction. Find the velocity of the second trolley and the KE converted.

Momentum conservation: $(2.0)(4.0) = (2.0)(0.5) + (3.0)\,v_2 \Rightarrow 8.0 = 1.0 + 3.0\,v_2 \Rightarrow v_2 = 7.0/3.0 = 2.33$ m s $^{-1}$ .

KE before $= \frac{1}{2}(2.0)(16) = 16.0$ J. KE after $= \frac{1}{2}(2.0)(0.25) + \frac{1}{2}(3.0)(5.44) = 0.25 + 8.17 = 8.42$ J.

$\Delta$ KE = 7.58 J converted (about $47\%$ of initial KE).

Contrast with Lesson 7: if this same pair had collided elastically, the 2.0 kg trolley would have ended up at $v_1 = -0.80$ m s $^{-1}$ (bouncing back) and the 3.0 kg trolley at $v_2 = 3.20$ m s $^{-1}$ , with KE fully preserved. Different final speeds, different energy balance — but in both cases momentum conservation holds. This is the key conceptual contrast Lesson 7 vs Lesson 8: momentum is always conserved; KE is only sometimes conserved.

Note: this is not perfectly inelastic — the trolleys separate. We needed the extra information "the 2.0 kg trolley moves at 0.5 m s $^{-1}$ afterwards" because momentum conservation alone gives one equation in two unknowns ( $v_1, v_2$ ); the extra datum fixes the system.

Worked Example 4 — Ballistic Pendulum (The Classical Two-Phase Problem)

A $0.010$ kg bullet is fired at $500$ m s $^{-1}$ into a $2.0$ kg block suspended at rest from a long light string. The bullet embeds in the block, and the combined mass swings upwards. Find (a) the initial speed of the block+bullet just after impact, and (b) the maximum height to which the pendulum swings.

This is the canonical "split-into-phases" problem, and OCR examiners adore it. The crucial insight: the collision phase and the swing phase use different conservation laws.

Phase 1 — collision (use momentum only; KE is NOT conserved):

$0.010 \times 500 + 2.0 \times 0 = 2.010\,v_f$

$v_f = \frac{5.0}{2.010} = 2.49 \text{ m s}^{-1}$

Phase 2 — swing (use mechanical energy conservation; no further impact, only gravity):

$\tfrac{1}{2}(2.010)(2.49)^2 = (2.010)(9.81)\,h$

$\tfrac{1}{2}(2.49)^2 = 9.81\,h \quad (\text{mass cancels})$

$h = \frac{3.10}{9.81} = 0.316 \text{ m}$

So the pendulum rises by about 32 cm.

Crucial point — the most common error: do NOT equate the initial bullet KE ( $\frac{1}{2}(0.010)(500)^2 = 1250$ J) directly to $mgh$ . Most of those 1250 J were lost to heat and deformation during the embedding (Phase 1). Only the $\frac{1}{2}(2.010)(2.49)^2 = 6.20$ J of KE held by the block-bullet system just after impact is available to raise the pendulum. Ignoring the Phase 1 KE loss gives a completely wrong (and far too large) answer.

Energy accounting:

Original bullet KE: $1250$ J
KE of block+bullet just after impact: $6.20$ J
KE → heat/sound/deformation in Phase 1: $1250 - 6.20 = 1244$ J ( $99.5\%$ lost)
KE → gravitational PE at top of swing: $6.20$ J ( $100\%$ converted in Phase 2)

Worked Example 5 — Fraction of KE Lost (Derivation)

Derive the general formula for the fraction of kinetic energy lost when a mass $m_1$ at velocity $u_1$ sticks to a stationary mass $m_2$ .

Perfectly inelastic: $v_f = m_1 u_1 / (m_1 + m_2)$ .

$\text{KE}_{\text{before}} = \tfrac{1}{2} m_1 u_1^2$

$\text{KE}_{\text{after}} = \tfrac{1}{2}(m_1 + m_2)\,v_f^2 = \tfrac{1}{2}(m_1 + m_2) \cdot \frac{m_1^2 u_1^2}{(m_1 + m_2)^2} = \frac{m_1^2 u_1^2}{2(m_1 + m_2)}$

Fraction retained $= \text{KE}_{\text{after}} / \text{KE}_{\text{before}} = m_1 / (m_1 + m_2)$ .

Fraction lost $= 1 - m_1/(m_1+m_2) = m_2 / (m_1 + m_2)$ .

This confirms the earlier intuition: if $m_2 \gg m_1$ (a bullet into a heavy block), most of the KE is lost — the heavy block barely accelerates. If $m_2 \ll m_1$ (a lorry hitting a fly), almost none is lost — the lorry barely notices. The fraction lost is the target mass divided by the total mass.

Where Does the Lost Energy Go?

In a real inelastic collision, the missing KE becomes:

Thermal energy — the colliding bodies heat up. In a head-on car crash at 30 mph, the metal panels can warm by several °C. In the bullet-into-block example, both the bullet and the wood are noticeably hot after the embedding.
Sound — the bang of the collision carries off a small but real fraction of energy as pressure waves in the surrounding air. The audible "clank" of railway-truck coupling is energy.
Permanent deformation — crushed metal, splintered wood, flattened bullets. This requires real work against internal forces (breaking inter-atomic bonds, plastically rearranging the lattice) and often accounts for most of the loss in macroscopic collisions.
Elastic vibrations — waves travelling through the bodies after the collision, eventually damping out to heat. Steel-on-steel collisions ring audibly because of this.

Total energy is always conserved — the KE just stops being macroscopic kinetic energy and becomes internal energy. This is the First Law of Thermodynamics in disguise. The OCR specification expects A-Level candidates to state explicitly that "kinetic energy is not conserved in an inelastic collision, but total energy is, and the lost KE goes to heat, sound, deformation and vibration."

Real-World Applications

Car Crash Safety

Car designers actively want head-on collisions to be inelastic — specifically, they want the crumple zones to convert as much kinetic energy as possible into permanent deformation, so that the passenger cabin (a stiff "safety cage") does not have to absorb that energy with the passengers' bodies. A perfectly elastic car-on-car collision would be far more dangerous, because both cars (and passengers) would rebound at nearly the original speed, doubling the deceleration.

Inelastic Collisions — Momentum Conserved, KE Lost

Inelastic Collisions — Momentum Conserved, KE Lost

Inelastic vs Perfectly Inelastic

Diagram — Collision Type Decision Tree

Summary Table — Three Collision Types

Perfectly Inelastic — The One-Equation Case

Worked Example 1 — Coupling Trucks

Proof — Perfectly Inelastic Collisions Lose the Maximum KE

Worked Example 2 — Bullet and Block

Worked Example 3 — Inelastic but Not Sticking

Worked Example 4 — Ballistic Pendulum (The Classical Two-Phase Problem)

Worked Example 5 — Fraction of KE Lost (Derivation)

Where Does the Lost Energy Go?

Real-World Applications

Car Crash Safety

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