Momentum, impulse and collisions: elastic, inelastic, and two dimensions

Momentum is mass times velocity, p = mv, and it is a vector: a sign or a direction rides with every number. Impulse is the change in momentum, J = Δp = FΔt, which is why a longer stop needs a smaller force. In a collision with no outside force, the total momentum before equals the total after, always; kinetic energy is conserved only in an elastic collision. That is the whole topic. Below, one crash worked line by line, the energy accounting, the two equations of an elastic collision, and how two dimensions change nothing but the bookkeeping.

Momentum is a vector, and the sign is the physics

The momentum formula is p = mv, in kg·m/s. A 2.0 kg object moving right at 3.0 m/s has p = +6.0 kg·m/s if right is positive. The same object moving left at 3.0 m/s has p = −6.0 kg·m/s. Those two are not the same momentum, and a collision problem that drops the sign is already wrong before the first line.

So the first thing written on the page, before any number, is the positive direction. Everything after it carries a sign against that choice.

Why momentum, and not speed, is the thing that is conserved comes straight from Newton's third law. When two objects collide, each pushes on the other with the same size of force for the same length of time, in opposite directions. So each gets the same size of Δp as the other, with opposite signs, and the two changes cancel. The total momentum of the pair does not move. Speed has no sign and cannot do that cancelling; momentum can, which is why it is the conserved quantity.

Impulse: the same Δp, a long time or a big force

The impulse–momentum theorem says J = FΔt = Δp = m(v − v₀). The units agree: a newton-second is a kg·m/s² × s, which is a kg·m/s. Impulse is what a force does to momentum over the time it acts.

The useful reading of the theorem is backwards. If the change in momentum is fixed, because an object has to stop, then F × Δt is fixed, and a longer Δt means a smaller F. Airbags, crumple zones, bent knees on landing and a catcher's mitt moving backward all work this one way: they stretch the stopping time, so the stopping force falls.

A number to carry into the worked example. A 1,800 kg truck is taken from rest to 8.00 m/s by a collision.

• Δp = m(v − v₀) = 1,800 kg × (8.00 m/s − 0) = 14,400 kg·m/s (the change in momentum)
• If the contact lasts 0.150 s: F = Δp ÷ Δt = 14,400 kg·m/s ÷ 0.150 s = 96,000 N (the average force; kg·m/s ÷ s = kg·m/s² = N)

If the same change took 1.50 s instead, the force would be 9,600 N. Same impulse, a tenth of the force.

Conservation of momentum: the one equation

For two objects with masses m₁ and m₂, speeds u₁ and u₂ before and v₁ and v₂ after:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

written with every velocity carrying its sign. It holds for any collision, and for any explosion or separation, as long as no outside force acts on the pair during the brief contact.

"No outside force" is worth saying in practice. Friction and gravity do act on two colliding cars. But the collision lasts a fraction of a second, and the forces between the cars during that fraction are enormous compared with friction, so the momentum friction removes during the crash itself is negligible. Momentum just before the crash equals momentum just after. Afterwards, over the next several seconds, friction brings the wreck to a stop and the momentum is gone, transferred to the road and Earth.

Worked example: a 1,200 kg car at 20.0 m/s hits a 1,800 kg truck at rest; they lock together

A 1,200 kg car travelling east at 20.0 m/s hits a stationary 1,800 kg truck. The two lock together. Find their common speed, the kinetic energy lost, and the average force on each if the contact lasts 0.150 s.

Givens: m₁ = 1,200 kg, u₁ = +20.0 m/s, m₂ = 1,800 kg, u₂ = 0. East is positive. Three significant figures.

Line 1, momentum before.

• p_before = m₁u₁ + m₂u₂ = 1,200 kg × 20.0 m/s + 1,800 kg × 0 = 24,000 kg·m/s, east (adding the two momenta with their signs)

Line 2, the shared speed after. The two move as one object of mass 3,000 kg at a speed v.

• 24,000 kg·m/s = 3,000 kg × v (conservation: before equals after)
• v = 24,000 ÷ 3,000 = 8.00 m/s, east (dividing; the positive sign says east)

Line 3, kinetic energy before. Only the car was moving.

• KE_before = ½m₁u₁² = ½ × 1,200 kg × (20.0 m/s)² = ½ × 1,200 × 400 = 240,000 J (kg × m²/s² = J)

Line 4, kinetic energy after.

• KE_after = ½ × 3,000 kg × (8.00 m/s)² = ½ × 3,000 × 64.0 = 96,000 J

Line 5, the energy lost.

• KE_lost = 240,000 J − 96,000 J = 144,000 J (subtracting)
• 144,000 ÷ 240,000 = 0.600, so 60.0 % of the kinetic energy went into bending metal, heat and sound (the fraction)

Momentum was conserved exactly and kinetic energy was not. Both are true at once, and that is the normal case, not a contradiction. Work and kinetic energy is where the 144,000 J is accounted for: it was transferred, not destroyed.

Line 6, the impulse on each vehicle. Newton's third law in numbers.

• J on the truck = m₂(v − u₂) = 1,800 kg × (8.00 − 0) m/s = +14,400 N·s (the truck gained momentum eastward)
• J on the car = m₁(v − u₁) = 1,200 kg × (8.00 − 20.0) m/s = 1,200 × (−12.0) = −14,400 N·s (the car lost the same amount)

Equal in size, opposite in sign. The car pushed the truck east and the truck pushed the car west, for the same 0.150 s, with the same force.

Line 7, the average force.

• F = J ÷ Δt = 14,400 N·s ÷ 0.150 s = 96,000 N on each, in opposite directions (the impulse–momentum theorem)

For scale, the truck's weight is 1,800 kg × 9.8 m/s² = 17,640 N, and 96,000 ÷ 17,640 = 5.4, so the collision force on it was over five times its own weight.

Elastic collisions: two equations, or the one shortcut

An elastic collision is one in which kinetic energy is conserved as well as momentum. Hard spheres, air-track carts with spring bumpers and gas molecules come close. Cars do not.

With both conserved you have two equations and, for two unknown final speeds, enough to solve. For a target at rest (u₂ = 0) the algebra gives:

• v₁ = (m₁ − m₂)/(m₁ + m₂) × u₁
• v₂ = 2m₁/(m₁ + m₂) × u₁

A check with numbers. A 2.0 kg cart at 3.0 m/s hits a 1.0 kg cart at rest, elastically.

• v₁ = (2.0 − 1.0)/(2.0 + 1.0) × 3.0 = (1.0/3.0) × 3.0 = 1.0 m/s (the heavier cart keeps going, slower)
• v₂ = 2 × 2.0/(3.0) × 3.0 = (4.0/3.0) × 3.0 = 4.0 m/s (the lighter cart is sent off faster than the first was going)
• Momentum: 2.0 × 3.0 + 0 = 6.0 before; 2.0 × 1.0 + 1.0 × 4.0 = 6.0 after, in kg·m/s
• Kinetic energy: ½ × 2.0 × 3.0² = 9.0 J before; ½ × 2.0 × 1.0² + ½ × 1.0 × 4.0² = 1.0 + 8.0 = 9.0 J after

Both conserved, so the collision was elastic and the formulas hold. Once you have derived the pair once, there is a shortcut your teacher may let you use: in a one-dimensional elastic collision the relative speed reverses, u₁ − u₂ = −(v₁ − v₂). Here 3.0 − 0 = 3.0 before and 1.0 − 4.0 = −3.0 after.

"Elastic" is a test you apply, not a label you assume. If a question does not say the collision is elastic, do not conserve kinetic energy. If it does, check that KE comes out equal at the end; if it does not, a line is wrong.

The three wrong lines a marker sees

1. A dropped sign in a head-on case. Two objects moving toward each other have momenta of opposite sign, and the total is the difference, not the sum. The symptom is a combined speed that is too large, sometimes larger than either original speed, which a locked pair can never have.
2. Kinetic energy conserved in a collision that locks. Every perfectly inelastic collision loses kinetic energy, and the test for "perfectly inelastic" is whether the objects stick. The symptom in the worked example would be a speed of about 13 m/s instead of 8.00 m/s, from setting ½ × 3,000 × v² = 240,000 J and taking √160.
3. Momentum treated as a scalar in two dimensions. Adding magnitudes of momenta that point different ways. The symptom is a total momentum larger than the vector sum can be. The fix is the next section.

A fourth, briefer: J = F/Δt. Impulse is force times time, and the units tell you so: N × s, not N/s.

Two dimensions: conserve x and y separately

A glancing collision changes nothing about the physics and one thing about the bookkeeping. Momentum is conserved along x and along y, separately, so the one equation becomes two:

• m₁u₁ₓ + m₂u₂ₓ = m₁v₁ₓ + m₂v₂ₓ
• m₁u₁ᵧ + m₂u₂ᵧ = m₁v₁ᵧ + m₂v₂ᵧ

Each velocity is split into components with sin and cos, exactly as a projectile's launch velocity is split in the component habit from projectile motion. Two equations give two unknowns, usually one object's final speed and direction. Then the components are recombined with Pythagoras and tan⁻¹ for the answer. This page keeps one long example; a two-dimensional one is the same seven lines done twice.

A sibling to try cold

Cover the example. A 900 kg car at 15.0 m/s hits a 2,100 kg van at rest, and they lock together. Find the common speed, the kinetic energy before and after, and the fraction lost. Write the sign convention before the first number. Mix the collision types when you practise: Rohrer and Taylor (2007) found that students who practised problem types shuffled together, rather than in blocks of one type, did better on a later test, because the test asks you to decide which kind of collision you are looking at, and blocked practice never asks that.

Check

• p_before = 900 × 15.0 = 13,500 kg·m/s
• v = 13,500 ÷ 3,000 = 4.50 m/s
• KE_before = ½ × 900 × 15.0² = 101,250 J
• KE_after = ½ × 3,000 × 4.50² = 30,375 J
• Lost = 70,875 J, which is 70.0 % of the original

More of the energy was lost than in the worked example, because the van is a bigger fraction of the total mass. In a perfectly inelastic collision with a target at rest, the fraction lost is m₂/(m₁ + m₂), which is 2,100/3,000 = 0.700 here and 1,800/3,000 = 0.600 above. That is a check you can run on any such problem in one line.

Where this sits in the course

This page covers ideas 9 to 12 of 25 in SPH4U, the second half of strand C: Impulse and momentum, Conservation of momentum in one dimension; inelastic collisions, Elastic collisions and Momentum in two dimensions. Before them is idea 8, Power and efficiency, which closes the energy half of the strand; this is the second thing the course calls conserved, and this time the conserved thing is a vector. After them the fields strand opens with idea 13, Newton's law of universal gravitation, on the gravitation and orbits page. The SPH4U course map shows all 25 ideas in teaching order, and every idea of the three science courses has a page like this one.

Momentum on one screen

• p = mv, with a sign. Choose the positive direction first; every momentum carries a sign against it.
• J = FΔt = Δp. The same change in momentum over a longer time is a smaller force.
• Conservation, always. m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ for any collision with no outside force during the contact. Kinetic energy is conserved only if the collision is elastic, and that is a test, not an assumption.
• The worked crash. 8.00 m/s shared; 60 % of the kinetic energy transferred to deformation, heat and sound; impulses of ±14,400 N·s; 96,000 N on each.
• Two dimensions. The same equation on x and on y, components in, Pythagoras out.
• The three wrong lines. A dropped sign head-on; KE conserved when the objects lock; magnitudes added in two dimensions.
• The habit. Derive the elastic-collision formulas once from the two conservation laws rather than memorising them; how to study physics is built on that.