Momentum and collisions: which principle

Late September 2026. In Year 4 at the classical high school (liceo classico), with two hours of physics a week, the chapter on work and energy is closing, and in November comes the next block: momentum, collisions, and then gravitation. The test falls between seven and ten weeks from now, so between mid-November and mid-December.

The chapter that closes at the end of October and the one that begins in November share a question that comes up again in every mixed problem: is energy conserved or momentum? It is not a detail — it is the decision that, if wrong, makes you set up the whole exercise the wrong way. I talk about it in the previous piece, physics in Year 4 at the classical high school: energy and circular motion; here I pick up exactly where that chapter leaves the problem open.

Which principle to use: the deciding criterion

The total momentum of a system is conserved whenever the forces external to the system are negligible during the collision — and this holds always in collisions, even inelastic ones, even perfectly inelastic ones. It is the most general law of the chapter, and it has no exceptions tied to the type of collision.

Total kinetic energy, on the other hand, is conserved only in elastic collisions. It is not a special case to check each time with a calculation: it is the very definition of an elastic collision. In an inelastic collision part of the kinetic energy is turned into other forms (heat, deformation, sound) and you no longer find it as energy of motion — but momentum never disappears.

Type of collision	Momentum	Kinetic energy
Elastic	conserved	conserved
Inelastic (general)	conserved	not conserved
Perfectly inelastic (the bodies stay together)	conserved	not conserved, the minimum possible

The practical criterion for a mixed problem — when the same exercise contains both a stretch with no collisions and a collision — is this: use energy to describe the motion before and after the collision (a fall, a descent along a plane, a climb), and use momentum at exactly the instant of the collision, and at no other moment. The third exercise below shows this step in full.

Momentum and impulse

The momentum of a body is the product of its mass and its velocity:

p = m v

It is a vector quantity: it has the same direction and sense as the velocity. In a one-dimensional problem this translates into a practical but easily forgotten rule: velocities have a sign, and two bodies moving in opposite directions have momenta of opposite sign, even if they are identical in magnitude.

The impulse of a force is the product of the force and the time interval over which it acts, and the impulse theorem links it to the change in momentum:

J = F Δ t = Δ p = m v_{f} - m v_{i}

Don’t confuse impulse and force: force is what acts instant by instant, impulse is its effect accumulated over the time it acts. A collision, even a very brief one, can generate an enormous impulse if the force during the contact is very large — and that is exactly what happens physically in a collision.

Conservation of momentum in an isolated system

If no appreciable external forces act on a system of two bodies (or they cancel, like weight and the normal force in a collision on a horizontal plane), the total momentum before the collision equals that after:

m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}

This equation, on its own, has two unknowns if you don’t know either of the final velocities: to solve it you need a second equation, which depends on the type of collision.

Collisions: elastic, inelastic, perfectly inelastic

A perfectly inelastic collision is one in which the two bodies stay together after the collision and move with the same final velocity $v_{f}$ . The second equation, in this case, is simply $v_{1 f} = v_{2 f} = v_{f}$ , and conservation of momentum alone is enough to solve the problem.

An elastic collision is one in which, besides momentum, the total kinetic energy is also conserved:

\frac{1}{2} m_{1} v_{1 i}^{2} + \frac{1}{2} m_{2} v_{2 i}^{2} = \frac{1}{2} m_{1} v_{1 f}^{2} + \frac{1}{2} m_{2} v_{2 f}^{2}

Here the unknowns are two ( $v_{1 f}$ and $v_{2 f}$ ) and the equations are two (momentum and kinetic energy): the system can always be solved, even if it takes more calculation — you see them worked out step by step in the second exercise.

A general inelastic collision (not perfectly inelastic: the bodies don’t stay together but some energy is lost anyway) requires an additional piece of data from the problem — for example one of the two final velocities, or the fraction of energy lost — because a single conservation-of-momentum equation isn’t enough to find two unknowns.

A note on the centre of mass

The centre of mass of a system of bodies moves, during a collision, as if the whole mass of the system were concentrated there and no external force acted — it is another way of seeing why the total momentum doesn’t change: the velocity of the centre of mass stays constant during the collision, exactly because $p_{t o t} = (m_{1} + m_{2}) v_{c m}$ is the quantity that is conserved.

Gravitation: Newton’s law and Kepler’s third law

Gravitation closes the term. Newton’s law of universal gravitation says that two masses attract each other with a force proportional to the product of the masses and inversely proportional to the square of the distance:

F = G \frac{m _{1} m _{2}}{r ^{2}}

with $G \approx 6.674 \times 1 0^{- 11} N m^{2} /k g^{2}$ . It is a law in which the most frequent mistakes are not conceptual but about powers of ten: $G$ is an extremely small number, the masses of planets are extremely large numbers, and a sign error in the exponent changes the result by several orders of magnitude without the calculation “looking” wrong at first glance.

Kepler’s third law links the orbital period $T$ to the radius of the orbit $r$ :

\frac{T ^{2}}{r ^{3}} = constant (for bodies orbiting the same central body)

The quickest way to use it without calculating the constant is to compare two orbits, as in the fourth exercise below.

Mistakes that cost marks

Applying conservation of kinetic energy to an inelastic collision. Kinetic energy is conserved only in elastic collisions: using it in an inelastic collision (including the perfectly inelastic one) leads to a wrong result, because that equation, quite simply, is not true in that case.
Forgetting the signs of the velocities. In a one-dimensional problem momentum remains a vector: two bodies moving towards each other have velocities of opposite sign, and they must be added algebraically with that sign, not with the magnitudes.
Confusing impulse and force. Impulse is $F Δ t$ , not $F$ : in a collision lasting a fraction of a second, a huge force can still give a finite and reasonable impulse.
Getting the exponents wrong in gravitation. $G$ and the masses involved are extreme numbers: it is always worth writing everything in scientific notation before multiplying, never in your head.

Four worked exercises

1. Perfectly inelastic collision between two trolleys

A trolley of mass $m_{1} = 3 kg$ travels at $v_{1 i} = 4 m/s$ and hits a second trolley of mass $m_{2} = 2 kg$ moving in the opposite direction at $v_{2 i} = - 1 m/s$ . After the collision they stay together. Find the common final velocity and the percentage of kinetic energy lost in the collision.

Conservation of momentum, with $v_{1 f} = v_{2 f} = v_{f}$ :

m_{1} v_{1 i} + m_{2} v_{2 i} = (m_{1} + m_{2}) v_{f}

3 \cdot 4 + 2 \cdot (- 1) = 12 - 2 = 10 kg m/s

v_{f} = \frac{10}{3 + 2} = \frac{10}{5} = 2 m/s

The positive sign says that the system, after the collision, moves in the direction the heavier trolley was travelling. Kinetic energy before and after:

E_{k, i} = \frac{1}{2} \cdot 3 \cdot 4^{2} + \frac{1}{2} \cdot 2 \cdot (- 1)^{2} = 24 + 1 = 25 J

E_{k, f} = \frac{1}{2} \cdot 5 \cdot 2^{2} = \frac{1}{2} \cdot 5 \cdot 4 = 10 J

The energy lost is $25 - 10 = 15 J$ , that is $15/25 = 60%$ of the initial energy — lost in deformation and heat in the collision, not in momentum, which stays exactly $10 kg m/s$ before and after.

If you want to check for yourself whether you have understood the method, in the Year 3 physics test — work and energy you’ll also find questions on collisions and conservation of momentum: the correct answer is free, and the guided step-by-step solution unlocks with your first lesson. It is catalogued under Year 3 because at the science high school this block (work, energy and collisions) is done in a single year, a year earlier than at the classical high school.

2. Head-on elastic collision between two different masses

A body of mass $m_{1} = 2 kg$ travels at $v_{1 i} = 6 m/s$ and collides elastically with a stationary body of mass $m_{2} = 4 kg$ ( $v_{2 i} = 0$ ). Find the final velocities of both.

I write the two conservation equations. Momentum:

m_{1} v_{1 i} = m_{1} v_{1 f} + m_{2} v_{2 f} ⟹ 2 \cdot 6 = 2 v_{1 f} + 4 v_{2 f} ⟹ 6 = v_{1 f} + 2 v_{2 f} (1)

Kinetic energy:

\frac{1}{2} m_{1} v_{1 i}^{2} = \frac{1}{2} m_{1} v_{1 f}^{2} + \frac{1}{2} m_{2} v_{2 f}^{2} ⟹ 36 = v_{1 f}^{2} + 2 v_{2 f}^{2} (2)

From (1): $v_{1 f} = 6 - 2 v_{2 f}$ . I substitute into (2):

36 = (6 - 2 v_{2 f})^{2} + 2 v_{2 f}^{2} = 36 - 24 v_{2 f} + 4 v_{2 f}^{2} + 2 v_{2 f}^{2} = 36 - 24 v_{2 f} + 6 v_{2 f}^{2}

0 = 6 v_{2 f}^{2} - 24 v_{2 f} = 6 v_{2 f} (v_{2 f} - 4) ⟹ v_{2 f} = 0 or v_{2 f} = 4 m/s

The solution $v_{2 f} = 0$ is the “trivial” one (no collision took place): I discard it and keep $v_{2 f} = 4 m/s$ . From which $v_{1 f} = 6 - 2 \cdot 4 = - 2 m/s$ .

The lighter body bounces back at $2 m/s$ (negative sign, opposite to the starting direction) and the heavier body moves off forwards at $4 m/s$ . Check: final momentum $= 2 \cdot (- 2) + 4 \cdot 4 = - 4 + 16 = 12 kg m/s$ , equal to the initial $2 \cdot 6 = 12$ ; final kinetic energy $= \frac{1}{2} \cdot 2 \cdot 4 + \frac{1}{2} \cdot 4 \cdot 16 = 4 + 32 = 36 J$ , equal to the initial.

3. First energy, then momentum

A block of mass $m_{1} = 1 kg$ starts from rest at the top of a smooth (frictionless) ramp of height $h = 1.25 m$ and slides to the bottom. There it hits a second, stationary block of mass $m_{2} = 4 kg$ and the two stay together. Find the common velocity immediately after the collision.

The problem has two distinct phases, and they must be treated with two different principles.

Phase 1 — the descent, no collision: I use energy. The plane is smooth, so mechanical energy is conserved:

m_{1} g h = \frac{1}{2} m_{1} v_{1}^{2} ⟹ v_{1} = 2 g h = 2 \cdot 9.8 \cdot 1.25 = 24.5 \approx 4.95 m/s

Phase 2 — the collision: I use momentum. During the collision momentum is conserved (the collision lasts an instant, weight and the normal force have no time to change it):

m_{1} v_{1} = (m_{1} + m_{2}) v_{f} ⟹ v_{f} = \frac{1 \cdot 4.95}{1 + 4} = \frac{4.95}{5} \approx 0.99 m/s

Note what you cannot do: use conservation of energy for the collision as well. Before the collision the system’s kinetic energy is $\frac{1}{2} \cdot 1 \cdot 4.9 5^{2} \approx 12.3 J$ ; after, it is $\frac{1}{2} \cdot 5 \cdot 0.9 9^{2} \approx 2.45 J$ — almost 80% of the energy was lost in the collision (it is perfectly inelastic), exactly as in the first exercise. Momentum, on the other hand, is the same before and after the collision: $1 \cdot 4.95 \approx 4.95 kg m/s$ against $5 \cdot 0.99 \approx 4.95 kg m/s$ .

4. Kepler’s third law: comparing two satellites

A satellite $A$ orbits a planet with period $T_{A} = 90 min$ at a distance $r_{A}$ from the planet’s centre. A satellite $B$ orbits the same planet at a distance $r_{B} = 4 r_{A}$ . Find the period $T_{B}$ .

By Kepler’s third law, $T^{2} / r^{3}$ is the same constant for both satellites (same central body):

\frac{T _{A}^{2}}{r _{A}^{3}} = \frac{T _{B}^{2}}{r _{B}^{3}} ⟹ (\frac{T _{B}}{T _{A}})^{2} = (\frac{r _{B}}{r _{A}})^{3} = 4^{3} = 64

\frac{T _{B}}{T _{A}} = 64 = 8 ⟹ T_{B} = 8 \cdot 90 = 720 min = 12 h

A satellite four times further away takes eight times as long to complete an orbit — not four: it is the cube of the ratio of the distances that counts, not the ratio itself, and it is the point where a factor is most often lost in the test.

How we work on this chapter

The work in lessons, on this chapter, always starts from the same question before any calculation: what is happening physically, a collision or a stretch of free motion? Only after answering do you choose the right principle — energy on one side, momentum on the other — and write the equation. The full picture of how I work on physics online is on the physics page; the chapter immediately before, on circular motion and work-energy, is the piece before this one: physics in Year 4 at the classical high school: energy and circular motion.

To practise on other questions of this type, the Year 3 physics practice test — work and energy also includes collisions and momentum — same label on the practice test, content valid for your Year 4 syllabus at the classical high school.

The same criterion, with the whole December test, is in work, energy and collisions in Year 3; energy conservation in several phases, in another track, is in work, power and energy.

If you want a hand before the November-December test, write to me from the contact page.

Momentum and collisions in Year 4: which principle to use