Conservation of Momentum: Elastic vs Inelastic Collisions

Total momentum is conserved in every collision when the net external force on the system is zero. Kinetic energy is conserved only in elastic collisions. When objects stick together (perfectly inelastic), the final velocity is vf = (m1v1 + m2v2)/(m1 + m2).

AP Physics: Unit 4 (topics 4.3 Conservation of Linear Momentum, 4.4 Elastic and Inelastic Collisions). Covers Topics 4.3 and 4.4 of AP Physics 1 Unit 4 (Linear Momentum), worth 10-15% of the multiple-choice section. AP Physics C: Mechanics tests the same topics in its own Unit 4.

What Conservation of Momentum Says

Conservation of momentum says that when the net external force on a system is zero, the system's total momentum does not change. For a two-object collision, total momentum before impact equals total momentum after:

m1v1i+m2v2i=m1v1f+m2v2fm_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

Momentum is p=mvp = mv, and it is a vector. On the AP exam that mostly means keeping track of signs: pick a positive direction, give every velocity a sign, and never drop a negative. A 2.0 kg ball moving left at 3.0 m/s has momentum −6.0 kg⋅m/s-6.0\ \text{kg}\cdot\text{m/s}, not +6.0+6.0.

This law holds for every collision type: elastic, inelastic, and perfectly inelastic. What separates those categories is kinetic energy, not momentum.

Why It Works: Newton's Third Law and Impulse

During a collision, each object pushes on the other with forces that are equal in magnitude and opposite in direction (Newton's third law). Both forces act for the same contact time, so the impulses J=FavgΔtJ = F_{avg}\Delta t are also equal and opposite. Since impulse equals change in momentum (the impulse-momentum theorem), whatever momentum one object gains, the other loses. The total never changes.

External forces like friction technically act during the collision, but contact times are so short (often milliseconds) that their impulse is tiny compared with the impulse from the huge collision forces. That is why you can treat momentum as conserved even for cars skidding on pavement, as long as you compare the instants just before and just after impact.

Elastic vs Inelastic vs Perfectly Inelastic

All three collision types conserve momentum. The difference is what happens to kinetic energy.

TypeMomentumKinetic energyWhat you see
ElasticConservedConservedObjects bounce apart with no energy lost to heat or deformation
InelasticConservedDecreasesObjects bounce apart, but some KE becomes thermal energy or deformation
Perfectly inelasticConservedMaximum lossObjects stick together and move with one shared velocity

Truly elastic collisions are rare in everyday life. Billiard balls and air-track gliders with magnetic bumpers come close, and collisions between gas molecules are effectively elastic. Most real collisions are inelastic. Note that perfectly inelastic does not mean all kinetic energy disappears: the combined object usually keeps moving, so it keeps some KE. It means the collision loses the maximum amount of KE that momentum conservation allows.

How to Find Final Velocity After a Collision

If the objects stick together, there is only one final velocity, so momentum conservation alone solves the problem:

vf=m1v1i+m2v2im1+m2v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}

Substitute each velocity with its sign. If the numerator comes out negative, the combined object moves in the negative direction.

If the objects bounce apart, you have two unknown final velocities but only one momentum equation, so you need one more piece of information. AP problems supply it in one of three ways: they give you one of the final velocities, they tell you the collision is elastic (so KE conservation gives a second equation), or they tell you how much energy was lost. Either way, set up pi=pfp_i = p_f first, plug in what you know, and solve for the remaining unknown. You can check any answer with the momentum collision calculator.

Checking Kinetic Energy

To classify a collision, compute total kinetic energy before and after using K=12mv2K = \frac{1}{2}mv^2. KE is a scalar, so signs on velocity do not matter here: a leftward-moving object still contributes positive KE. If Kf=KiK_f = K_i, the collision is elastic. If Kf<KiK_f < K_i, it is inelastic, and the missing energy became thermal energy, sound, and permanent deformation. Total energy is still conserved (see conservation of energy); it just left the kinetic category.

A common trap: never assume KE is conserved just because a problem looks clean. Unless the problem says elastic, or the numbers prove it, momentum is the only quantity you can bank on during a collision.

How AP Physics 1 Tests This

Conservation of momentum is Topics 4.3 and 4.4 in Unit 4 (Linear Momentum), which makes up 10 to 15% of the AP Physics 1 multiple-choice section. Expect qualitative questions (which quantities are conserved in this collision?), calculations like the worked examples below, and lab-style questions about measuring cart velocities before and after impact.

Free-response questions often chain momentum with energy: a ball of clay embeds in a block (perfectly inelastic, momentum conserved), then the block swings upward or compresses a spring (energy conserved). Treat the two phases separately and never carry kinetic energy across the collision itself. A calculator is allowed on both sections of the exam. For the full unit picture, see the Unit 4 overview.

Perfectly Inelastic: Two Cars Lock Bumpers

A 1200 kg car traveling at 20.0 m/s rear-ends a stationary 800 kg car. The bumpers lock and the two cars slide together. Find their common final velocity and the kinetic energy lost in the collision.

  1. Take the moving car's direction as positive. Initial momentum: pi=(1200 kg)(20.0 m/s)+(800 kg)(0)=24,000 kg⋅m/sp_i = (1200\ \text{kg})(20.0\ \text{m/s}) + (800\ \text{kg})(0) = 24{,}000\ \text{kg}\cdot\text{m/s}

  2. The cars stick together, so one final velocity: vf=pim1+m2=24,000 kg⋅m/s2000 kg=12.0 m/sv_f = \frac{p_i}{m_1 + m_2} = \frac{24{,}000\ \text{kg}\cdot\text{m/s}}{2000\ \text{kg}} = 12.0\ \text{m/s}

  3. Initial kinetic energy (only the first car moves): Ki=12(1200 kg)(20.0 m/s)2=240,000 JK_i = \frac{1}{2}(1200\ \text{kg})(20.0\ \text{m/s})^2 = 240{,}000\ \text{J}

  4. Final kinetic energy: Kf=12(2000 kg)(12.0 m/s)2=144,000 JK_f = \frac{1}{2}(2000\ \text{kg})(12.0\ \text{m/s})^2 = 144{,}000\ \text{J}

  5. Energy lost: ΔK=240,000 J−144,000 J=96,000 J\Delta K = 240{,}000\ \text{J} - 144{,}000\ \text{J} = 96{,}000\ \text{J}, which is 40% of the original KE.

The wreckage moves at 12.0 m/s in the first car's original direction. 96,000 J (40% of the initial kinetic energy) is converted to thermal energy and deformation, so the collision is perfectly inelastic.

Head-On Collision with Signs

A 3.0 kg cart moves right at 4.0 m/s while a 2.0 kg cart moves left at 3.0 m/s. They collide head-on and stick together. Find the velocity of the pair after the collision.

  1. Define rightward as positive: v1i=+4.0 m/sv_{1i} = +4.0\ \text{m/s} and v2i=−3.0 m/sv_{2i} = -3.0\ \text{m/s}.

  2. Total initial momentum: pi=(3.0 kg)(+4.0 m/s)+(2.0 kg)(−3.0 m/s)=12.0−6.0=6.0 kg⋅m/sp_i = (3.0\ \text{kg})(+4.0\ \text{m/s}) + (2.0\ \text{kg})(-3.0\ \text{m/s}) = 12.0 - 6.0 = 6.0\ \text{kg}\cdot\text{m/s}

  3. Sticking means one final velocity: vf=6.0 kg⋅m/s3.0 kg+2.0 kg=1.2 m/sv_f = \frac{6.0\ \text{kg}\cdot\text{m/s}}{3.0\ \text{kg} + 2.0\ \text{kg}} = 1.2\ \text{m/s}. Positive, so the pair moves right.

  4. KE check: Ki=12(3.0)(4.0)2+12(2.0)(3.0)2=24.0+9.0=33.0 JK_i = \frac{1}{2}(3.0)(4.0)^2 + \frac{1}{2}(2.0)(3.0)^2 = 24.0 + 9.0 = 33.0\ \text{J}, while Kf=12(5.0)(1.2)2=3.6 JK_f = \frac{1}{2}(5.0)(1.2)^2 = 3.6\ \text{J}.

1.2 m/s to the right. The collision converts 29.4 J of the original 33.0 J of kinetic energy (about 89%) into other forms, typical of a perfectly inelastic head-on crash.

Is This Collision Elastic?

On a frictionless air track, a 0.30 kg glider moving at 1.2 m/s hits a stationary 0.60 kg glider. Afterward, the 0.30 kg glider rebounds backward at 0.40 m/s. Find the 0.60 kg glider's final velocity and decide whether the collision is elastic.

  1. Take the incoming glider's direction as positive. Initial momentum: pi=(0.30 kg)(1.2 m/s)+0=0.36 kg⋅m/sp_i = (0.30\ \text{kg})(1.2\ \text{m/s}) + 0 = 0.36\ \text{kg}\cdot\text{m/s}

  2. Final momentum with the rebound as negative: pf=(0.30 kg)(−0.40 m/s)+(0.60 kg) v2f=−0.12+0.60 v2fp_f = (0.30\ \text{kg})(-0.40\ \text{m/s}) + (0.60\ \text{kg})\,v_{2f} = -0.12 + 0.60\,v_{2f}

  3. Set pf=pip_f = p_i: −0.12+0.60 v2f=0.36-0.12 + 0.60\,v_{2f} = 0.36, so v2f=0.480.60=0.80 m/sv_{2f} = \frac{0.48}{0.60} = 0.80\ \text{m/s} forward.

  4. Initial KE: Ki=12(0.30 kg)(1.2 m/s)2=0.216 JK_i = \frac{1}{2}(0.30\ \text{kg})(1.2\ \text{m/s})^2 = 0.216\ \text{J}

  5. Final KE: Kf=12(0.30)(0.40)2+12(0.60)(0.80)2=0.024+0.192=0.216 JK_f = \frac{1}{2}(0.30)(0.40)^2 + \frac{1}{2}(0.60)(0.80)^2 = 0.024 + 0.192 = 0.216\ \text{J}

The 0.60 kg glider moves forward at 0.80 m/s. Kinetic energy is the same before and after (0.216 J), so the collision is elastic.

Frequently asked questions

Is momentum conserved in an inelastic collision?

Yes. Momentum is conserved in every type of collision (elastic, inelastic, and perfectly inelastic) as long as the net external force on the system is zero. Kinetic energy is the quantity that decreases in an inelastic collision, not momentum.

How do I know if a collision is elastic or inelastic?

Compute the total kinetic energy before and after the collision. If the two totals are equal, the collision is elastic. If kinetic energy decreased, it is inelastic. If the objects stick together and move as one, it is perfectly inelastic, which loses the largest amount of kinetic energy that momentum conservation allows.

What is the formula for final velocity when two objects stick together?

vf = (m1v1 + m2v2)/(m1 + m2). Substitute each initial velocity with its sign (negative if that object moves in the negative direction). The result is the shared velocity of the combined object after the collision.

Where does the kinetic energy go in an inelastic collision?

It converts to other forms: thermal energy, sound, and the work of permanently deforming the objects. Total energy is still conserved; only the kinetic portion shrinks. That is why crumpled bumpers and skid marks are evidence of an inelastic crash.

Do I need both momentum and energy equations for every collision problem?

No. If the objects stick together, momentum conservation alone gives the final velocity. You only bring in the kinetic energy equation when the problem says the collision is elastic, or when it asks how much energy was lost.