Is momentum conserved in an inelastic collision?

Yes, the total momentum of an isolated system is conserved in an inelastic collision. The qualification about the system matters because an external force acting over time produces an external impulse and changes the system's momentum. Forces between the colliding objects are internal and produce equal and opposite momentum changes. Kinetic energy is not conserved because some of it is transferred into internal energy, heat, sound, vibration, or deformation. Therefore, momentum conservation and kinetic-energy conservation must be tested separately.

Can objects bounce apart in an inelastic collision?

Yes, objects can separate after impact and still undergo an inelastic collision. Bouncing only shows that their final relative speed is not zero; it does not prove that total kinetic energy was conserved. A ball that rebounds to a lower height than its release height is a familiar example because some of its mechanical energy has been transferred to other forms. To classify the collision reliably, calculate the total kinetic energy immediately before and after impact. If the final total is lower, the collision is inelastic even though the objects bounced apart.

What is the difference between inelastic and perfectly inelastic collisions?

An inelastic collision is any passive collision in which total kinetic energy decreases. A perfectly inelastic collision is the limiting case in which the colliding objects stick together and move with a common final velocity. This common-velocity outcome gives the maximum kinetic-energy decrease possible for the stated masses and initial velocities while conserving momentum. Ordinary inelastic collisions may involve objects that deform and then separate. Both types conserve total momentum when the system is isolated, but neither conserves kinetic energy.

Is total energy conserved in an inelastic collision?

Yes, total energy remains conserved in an inelastic collision. What is not conserved is the system's total kinetic energy between the selected before-and-after states. Some of the initial kinetic energy becomes internal energy, permanent deformation, thermal energy, sound, or vibration. Saying that energy is “lost” should therefore be understood as shorthand for kinetic energy being transferred into less mechanically useful forms. In an exam explanation, explicitly distinguishing total energy from kinetic energy demonstrates accurate conservation reasoning.

How can I prove that a collision is elastic from velocity data?

First calculate the total momentum before and after the collision using signed velocities. If the system is isolated, those values should agree within the precision of the data. Next calculate the total kinetic energy before and after using $Ek=\frac{1}{2}mv^2$ for every object. The collision is elastic if the two kinetic-energy totals are equal within reasonable rounding or experimental uncertainty. Momentum equality alone is insufficient because isolated inelastic collisions also conserve momentum.

Why is kinetic energy not conserved when objects stick together?

When objects stick together, their relative motion after impact is zero because they share one final velocity. During the impact, deformation and internal forces transfer some of the organized kinetic energy into internal energy, heat, sound, or permanent structural change. Momentum still determines the velocity of the combined mass when external impulse is negligible. The combined object may retain kinetic energy if it continues moving, so the final kinetic energy does not generally equal zero. For the given initial momentum, the common-velocity state has the smallest possible final translational kinetic energy and therefore the greatest kinetic-energy decrease.

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What is the difference between elastic and inelastic collisions?

• 29 Jul 2026•15

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What is the difference between elastic and inelastic collisions?

• 29 Jul 2026•15

The central difference between elastic and inelastic collisions is what happens to kinetic energy. In an elastic collision, the total kinetic energy of the colliding system is conserved. In an inelastic collision, some kinetic energy is transferred into internal energy, permanent deformation, sound, thermal energy, or other forms.

For an isolated system, momentum is conserved in both types of collision. This is the point IB Physics students most often confuse: momentum conservation does not by itself prove that a collision is elastic. You must compare the total kinetic energy before and after the collision.

This article focuses specifically on that distinction, the equations used to analyse it, and the reasoning expected in exam questions. For wider coverage of Newton's laws, impulse, and momentum, use the IB Physics forces and momentum explained exam-focused resources.

Elastic vs inelastic collisions at a glance

Feature	Elastic collision	Inelastic collision	Perfectly inelastic collision
Total momentum of an isolated system	Conserved	Conserved	Conserved
Total kinetic energy	Conserved	Decreases	Decreases by the maximum possible amount for the stated initial conditions
Total energy	Conserved	Conserved	Conserved
Objects after impact	Usually separate	May separate or remain together	Stick together and share one velocity
Typical idealized example	Collisions between gas particles	A bouncing ball that does not return to its original height	Two pieces of clay joining together
Main exam test	Check both momentum and kinetic energy	Check momentum, then calculate kinetic-energy loss	Use a common final velocity

The table contains an important qualification. Momentum is conserved only when the resultant external impulse on the chosen system is zero or negligible during the collision. In a short collision between two carts, friction may act, but its impulse during the brief impact can often be neglected.

Why momentum is conserved in both types of collision

Linear momentum is defined by

$p = mv$

where $m$ is mass and $v$ is velocity. Momentum is a vector, so its sign or direction must be included.

For a two-object system in one dimension, conservation of momentum gives

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

Here, $u_1$ and $u_2$ are the velocities before the collision, while $v_1$ and are the velocities afterward. A consistent positive direction must be selected before values are substituted.

During the collision, object 1 exerts a force on object 2, and object 2 exerts an equal and opposite force on object 1. These internal forces produce equal and opposite impulses. One object's momentum gain is therefore matched by the other's momentum loss, leaving the total momentum unchanged if external impulse is negligible.

This reasoning applies regardless of whether the objects rebound, deform, produce sound, or stick together. The type of collision is determined by the kinetic-energy outcome, not by whether momentum is conserved.

For a fuller review of this reasoning, see the linear momentum and impulse notes and the explanation of why momentum conservation is fundamental.

What makes a collision elastic?

A collision is elastic when both total momentum and total kinetic energy are unchanged between the states immediately before and immediately after impact. For two objects, the kinetic-energy equation is

$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

The velocities are squared, so kinetic energy is a scalar and cannot be negative. A velocity of $-4.0\text{ m s}^{-1}$ contributes the same kinetic energy as a velocity of $+4.0\text{ m s}^{-1}$ for an object of the same mass.

In a real impact, each object may temporarily compress and store elastic potential energy. In an ideal elastic collision, that stored energy is returned completely to kinetic energy as the objects regain their shapes. No net kinetic energy remains as permanent deformation, thermal energy, or sound after the interaction.

Perfectly elastic collisions are idealizations for most macroscopic objects. Collisions between atoms or molecules can often be modelled as elastic, while carefully designed carts or hard spheres may approximate elastic behaviour. Billiard balls are commonly used as an illustration, but an actual billiard-ball collision still transfers a small amount of energy into sound, heat, and deformation.

Worked elastic-collision example

A $0.20\text{ kg}$ cart moving at $3.0\text{ m s}^{-1}$ collides elastically with an identical stationary cart. For a one-dimensional elastic collision between equal masses where one object is initially stationary, the moving object transfers its velocity to the stationary object.

The final velocities are therefore

first cart: $v_1 = 0\text{ m s}^{-1}$
second cart: $v_2 = 3.0\text{ m s}^{-1}$

Check the momentum:

$p_i = (0.20)(3.0) = 0.60\text{ kg m s}^{-1}$

$p_f = (0.20)(0) + (0.20)(3.0) = 0.60\text{ kg m s}^{-1}$

Check the kinetic energy:

$E_{k,i} = \frac{1}{2}(0.20)(3.0)^2 = 0.90\text{ J}$

$E_{k,f} = \frac{1}{2}(0.20)(0)^2 + \frac{1}{2}(0.20)(3.0)^2 = 0.90\text{ J}$

Both quantities are conserved, so the collision is elastic. The apparent exchange of velocities is a special result for equal masses and should not be assumed when the masses differ.

What makes a collision inelastic?

An inelastic collision is one in which total kinetic energy is not conserved. For an ordinary passive collision, the final kinetic energy is less than the initial kinetic energy:

$E_{k,f} < E_{k,i}$

The missing kinetic energy has not disappeared. Total energy remains conserved, but some of the initially organized energy of motion becomes internal energy, permanent deformation, thermal energy, sound, or vibration.

Objects do not have to stick together for a collision to be inelastic. A tennis ball can rebound from the floor and still undergo an inelastic collision if its kinetic energy immediately after impact is lower than immediately before impact. Bouncing is therefore not sufficient evidence that a collision is elastic.

The energy transferred from kinetic energy can be calculated using

$E_{\text{transferred}} = E_{k,i} - E_{k,f}$

In an exam response, describe this quantity as energy transferred or converted, rather than energy that has vanished. The phrase “kinetic energy is lost” is acceptable shorthand when it is clear that total energy remains conserved.

What is a perfectly inelastic collision?

A perfectly inelastic collision is the limiting inelastic case in which the colliding objects stick together and move with a common final velocity. Momentum conservation becomes

$m_1u_1 + m_2u_2 = (m_1+m_2)v$

$v = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

For the specified masses and initial velocities, this outcome produces the maximum possible decrease in kinetic energy consistent with momentum conservation. It does not mean that all kinetic energy necessarily becomes zero. The combined object may still move after the collision and therefore retain translational kinetic energy.

Worked perfectly inelastic example

A $2.0\text{ kg}$ cart moving at $4.0\text{ m s}^{-1}$ collides with a stationary $1.0\text{ kg}$ cart. The carts lock together.

Apply momentum conservation:

$(2.0)(4.0) + (1.0)(0) = (2.0+1.0)v$

$v = \frac{8.0}{3.0} = 2.67\text{ m s}^{-1}$

The initial kinetic energy is

$E_{k,i} = \frac{1}{2}(2.0)(4.0)^2 = 16\text{ J}$

The final kinetic energy is

$E_{k,f} = \frac{1}{2}(3.0)(2.67)^2 \approx 10.7\text{ J}$

Therefore, approximately

$16 - 10.7 = 5.3\text{ J}$

has been transferred from kinetic energy into other forms. Momentum is conserved, but kinetic energy is not, so the collision is perfectly inelastic.

How to determine the collision type from data

When an IB question supplies masses and velocities, use a systematic test rather than relying on the appearance of the collision.

Define the system. Usually it consists of all colliding objects.
Choose a positive direction. Assign negative signs to velocities in the opposite direction.
Calculate total momentum before and after. This checks the collision data and your signs.
Calculate total kinetic energy before and after. Square each object's speed separately.
Classify the collision. Equal kinetic energies indicate an elastic collision; a decrease indicates an inelastic collision.
Check whether the objects share one final velocity. If they do, the collision is perfectly inelastic.

For example, suppose total kinetic energy changes from $12.0\text{ J}$ to $8.5\text{ J}$ . The collision is inelastic, and $3.5\text{ J}$ has been transferred into non-kinetic forms. It does not matter that the objects separate after impact.

If the question states that a collision is elastic, you may use both momentum and kinetic-energy conservation as simultaneous equations. If it states only that two bodies collide, do not assume kinetic-energy conservation. The A.2.3 collisions and explosions notes provide focused practice with these classifications.

Relative speed in a one-dimensional elastic collision

For a one-dimensional elastic collision, conservation of momentum and kinetic energy lead to a useful result:

relative speed of approach = relative speed of separation

If object 1 approaches object 2, this can be written with appropriate signs as

$u_1-u_2 = v_2-v_1$

This relation is not an additional conservation law. It follows from applying both momentum conservation and kinetic-energy conservation to a one-dimensional two-body collision.

A related quantity is the coefficient of restitution, defined as

$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$

For an ideal elastic collision, $e=1$ . For a perfectly inelastic collision, $e=0$ , while an ordinary inelastic collision usually has a value between zero and one. Treat this as a useful extension unless a question or your teacher specifically requires it; the safest general IB method is still to compare total kinetic energy before and after.

Collisions in two dimensions

Momentum conservation is vectorial, so a two-dimensional collision must be analysed separately in perpendicular directions:

$\sum p_{x,i} = \sum p_{x,f}$

$\sum p_{y,i} = \sum p_{y,f}$

If the collision is elastic, the scalar kinetic-energy equation provides another condition. Angles must be handled by resolving each velocity into components, such as $v\cos\theta$ and $v\sin\theta$ .

Kinetic energy itself is not separated into signed $x$ and $y$ quantities. Calculate it from the object's total speed, or equivalently from $v^2=v_x^2+v_y^2$ . A clear vector diagram usually prevents sign and trigonometry errors.

Common IB Physics exam mistakes

Saying momentum is always conserved

A more accurate statement is: total momentum is conserved in an isolated system, or when the net external impulse is negligible. If a vehicle collides with a rigid wall and the system contains only the vehicle, the wall exerts an external impulse and the vehicle's momentum is not conserved. Momentum can be conserved by expanding the system to include the vehicle, wall, and Earth.

Treating all rebounds as elastic

An object can bounce and still lose kinetic energy. Calculate the total kinetic energy on both sides of the collision rather than judging from whether the objects separate.

Using speed instead of signed velocity in momentum equations

Momentum contains velocity, so opposite directions require opposite signs. Kinetic energy contains speed squared, so each kinetic-energy contribution is non-negative.

Assuming sticking means kinetic energy becomes zero

Sticking identifies a perfectly inelastic collision, but the combined mass can continue moving. Only in a frame where the combined object is stationary would its final translational kinetic energy be zero.

Conserving kinetic energy without justification

Start with momentum conservation when external impulse is negligible. Add kinetic-energy conservation only if the collision is stated to be elastic or your calculation establishes that it is elastic.

The common mistakes in IB Physics forces and momentum article develops these exam traps further.

A practical exam-solving method

Before calculating, draw separate before and after diagrams. Label every mass and velocity, mark the positive direction, and write the momentum equation symbolically before inserting numbers.

Then ask what information identifies the collision:

Elastic stated: conserve momentum and kinetic energy.
Objects stick together: use one common final velocity.
Final velocities given: compare initial and final kinetic energy.
Energy transfer requested: calculate $E_{k,i}-E_{k,f}$ .
Two-dimensional motion: conserve momentum independently along each axis.

Retain unrounded values during intermediate calculations because kinetic energy depends on velocity squared. At the end, check units, significant figures, direction, and whether the result is physically plausible.

Targeted practice matters because these questions test both algebra and interpretation. The IB Physics A.2 questionbank can be used to practise collision setups, while Jojo AI can help identify where an incorrect sign or conservation assumption entered a solution.

Conclusion

The difference between elastic and inelastic collisions is the conservation of kinetic energy. In an isolated system, momentum is conserved in both: an elastic collision also conserves total kinetic energy, while an inelastic collision transfers some kinetic energy into other forms. A perfectly inelastic collision is the special case in which the objects stick together and move with one final velocity.

For IB exams, define the system, choose signs carefully, conserve vector momentum, and only conserve kinetic energy when elasticity is established. RevisionDojo's A.2 Study Notes and Questionbank are useful for reviewing the principle and applying it under exam conditions, with Jojo AI available for checking the reasoning in worked solutions.

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