Momentum and impulse

Momentum, and the general second law

The momentum of an object is

a vector pointing along the velocity, measured in kg m s⁻¹. Newton's second law, stated properly, is a statement about momentum. The resultant force equals the rate of change of momentum.

For constant mass this collapses to the familiar F = ma, so the shortcut is safe there. The momentum form is the more general statement, and it applies to a fixed collection of matter, which is where care is needed.

A rocket needs care. Applying F = Δ(mv)/Δt to the rocket alone, with m falling as fuel burns, gives a wrong answer, because the exhaust leaves carrying momentum with it. Take the rocket and its exhaust together instead. That system has no external force on it in deep space, so its total momentum is conserved, and the momentum the exhaust takes backwards is the momentum the rocket gains forwards. Conservation handles the whole problem, and the shrinking mass never has to go into an equation at all.

Conservation

In a closed system, one whose external forces contribute no net impulse over the interval, total momentum is conserved. The total before equals the total after, in every collision and every explosion. In practice the impact forces dwarf anything external, which is why the law holds across the instant of a crash even with friction and gravity still acting.

During a collision, the two bodies exert equal and opposite forces for the same time. Their impulses, and therefore their changes in momentum, are equal and opposite, so total momentum is conserved in a closed system.

Watch Conservation of momentum and collisions on the InkPhysics YouTube channel

Describe the drawing

Above: the 2.0 kg trolley at 1.5 m/s couples with the parked 1.0 kg and the pair moves off at 1.0 m/s, exactly what conservation demands; the amber ledger never flinches. Below, the film in reverse spirit: a resting pair springs apart, the light trolley leaving twice as fast as the heavy one, and the two momentum arrows grow equal and opposite so the sum stays pinned at zero. Kinetic energy is not conserved in the coupling; momentum is, both times.

A one-dimensional collision turns momentum into a signed number. Choose a positive direction in the first line and let the signs do the vector work for you. When a moving object “loses” its momentum by stopping, the momentum is transferred to whatever exerted the stopping force, often the Earth, whose large mass makes the resulting change in velocity negligible.

The same law applies to explosions. A stationary system has zero total momentum, so once it flies apart the fragments' momenta still have to sum to zero. Fire a bullet one way and the rifle recoils the other, because zero must stay zero.

Momentum in two dimensions

Momentum is a vector, so in an oblique collision it is conserved along each axis independently. Resolve every momentum into components before and after, then demand that the x-components balance and the y-components balance: two equations from one law. A snooker ball glancing off another leaves at an angle precisely so that the sideways momenta cancel. The method is the one used for projectiles: pick axes, resolve, then treat each axis on its own.

Elastic and inelastic

Momentum is conserved in every collision; kinetic energy is not. An elastic collision conserves kinetic energy too, and it is rare outside colliding gas molecules and good billiard balls. An inelastic collision hands some kinetic energy over to internal energy, sound and deformation. A perfectly inelastic one is where the objects stick together and travel on as one lump.

Total energy is always conserved. Kinetic energy is the quantity that need not be. Testing it is pure arithmetic, so compute ½mv² before and after and compare the two.

There is a second test. Compare the relative speed of approach with the relative speed of separation. In a perfectly elastic collision the two are equal. Two trolleys closing at 4.0 m s⁻¹ bounce apart at 4.0 m s⁻¹, whatever their masses. Anything less than that on the way out means kinetic energy was lost, and a separation speed of zero is the perfectly inelastic case where they move off together. Use whichever test the question asks for.

Kinetic energy written with momentum

Momentum and kinetic energy are two readings of the same motion, and one substitution converts between them. Both definitions carry the same velocity, p = mv$p = mv$p = mv and E_k = 12mv²$E_{k} = \tfrac{1}{2}mv^{2}$E_k = (1/2)mv^2, so eliminating it takes two lines. Rearrange the first to v = p/m, feed that into the second, and ½m × p²/m² collapses to

with a single power of the mass left in the denominator. Both starting equations are given, so the result can be derived rather than recalled. It also inherits the ½mv² it was built from, so it is a non-relativistic result and holds only while the speed stays well below the speed of light.

Read it as a statement about a fixed momentum. Share one value of p between a light object and a heavy one and the heavy one ends up with less kinetic energy, in inverse proportion to its mass. That is the skater and the ball above, quantified. Their momenta match in magnitude, and the 4.0 kg ball leaves with 50 J while the 60 kg skater has only 3.3 J, fifteen times less for fifteen times the mass. Turned the other way about, p is the square root of 2mE_k, which is the quickest route from a kinetic energy to a momentum and the usual opening move of a de Broglie calculation.

Impulse, and why crumple zones work

Rearranging the second law over a constant force gives the impulse.

The same change in momentum can be produced by a large force acting for a short time or by a smaller force acting for a longer time. Crumple zones and airbags increase the time.

Crumple zones, airbags, crash mats, bending your knees on landing. Not one of them alters Δ(mv), which was fixed the moment you knew how fast you were going. What they stretch is Δt, and with the product pinned, a longer Δt forces a smaller F. Present the argument in exactly that order: the fixed Δ(mv) first, then the stretched Δt, then the smaller F.

On a force-time graph the area beneath the curve is the impulse, which is how questions handle a force that varies through an impact and never settles on a single value of F. For an occupant brought from the crash speed to rest, the momentum change is fixed. Stretching the stopping time lowers the average force, and usually the peak force with it, and that is where the engineering goes.