Conservation of Momentum

The bench starts with two trolleys at rest in the middle of a frictionless track, A with 1 kg on the left and B with 3 kg on the right, and a spring squeezed and latched between them, holding 6 J. The Release the spring button opens the latch. For 0.4 s the spring pushes the two in opposite directions, and from then on it touches neither of them; A carries on at 3 m/s to the left and B at 1 m/s to the right, each with the velocity it had when the push ended.

The marks the trolleys leave on the track every second make the difference easy to see. A’s marks lie 3 m apart and B’s 1 m apart, and the ratio between the two spacings holds for as long as there is track. While the spring sits between them it pushes A to the left and B to the right, and one may think of this as A pushing B through the spring and B pushing A back. Newton’s third law says that these two forces have, at every instant, the same size and opposite directions, so the two red arrows on the bench are equally long, although one of them acts on a trolley with three times the mass of the other.

The two forces also start and stop together, so the impulses they deliver are equal and opposite. In step 1 of the chapter 9 lesson, the impulse appeared as the change in momentum; here it takes A’s momentum and B’s away from zero by the same amount, with opposite signs. A finishes with 1 kg times −3 m/s, B with 3 kg times 1 m/s, and the sum is zero, as it was before the latch opened.

spring force, while it acts   velocity

With equal masses on the controls, the two leave equally fast in opposite directions. That could be predicted with no arithmetic at all, from symmetry alone, since the set-up looks the same in a mirror. With different masses, the ratio of the velocities is the inverse of the ratio of the masses, with the sign reversed, and the trolley with three times the mass leaves with a third of the speed.

The spring’s energy is shared in another way. Since the two momenta have the same size \(p\), the kinetic energy of each trolley, \(p^2/2m\), is larger for the one with less mass. Of the spring’s 6 J, A takes 4.5 J and B only 1.5 J, in the inverse ratio of the masses. When an adult and a child on roller skates push off from each other, the child moves away faster and with most of the energy, even though the forces are equal, as in the secondary-school lesson on action and reaction.

Feynman remarks that the experiment could even serve to compare masses, by calling two masses equal when they fly off equally fast, and that this seeming convention already commits us to laws that only experiment can confirm.

In Feynman: §10-1 Newton’s Third Law ↗ · §10-2 Conservation of momentum ↗

Now the trolleys come towards each other and stick together when they touch, as if their bumpers were lined with Velcro. With the starting values, A has 3 kg and moves at 2 m/s to the right, B has 1 kg and comes at 2 m/s to the left, and after the collision the pair moves on together at 1 m/s. The momenta beforehand, 6 kg·m/s for A and −2 kg·m/s for B, add up to 4 kg·m/s, and the 4 kg of the pair at 1 m/s give the same sum. During the collision each trolley pushes the other with equal and opposite forces for the same time, and the argument of step 1 applies again: the momentum one of them loses, the other gains.

The velocity at which the pair moves on has a meaning of its own. It is the velocity of the centre of mass, the average of the velocities weighted by the masses, which the bench marks with a triangle under the track. The triangle moves at 1 m/s before, during and after the collision, as if nothing had happened, and the stuck trolleys simply fall in with it. The secondary-school lesson on collisions reaches the same formula through conservation; what matters here is what changes when the observer changes.

trolley A   trolley B   total momentum   centre of mass

The frame buttons show the same collision as seen by someone travelling beside the track at a constant velocity \(u\), and all that takes is to subtract \(u\) from every velocity. For someone keeping pace with A before the collision, at 2 m/s, A is at rest, B arrives at −4 m/s and the pair leaves at −1 m/s. Every number has changed, and the total momentum is −4 kg·m/s before and after. For someone moving with the centre of mass, the two arrive with opposite momenta, 3 and −3 kg·m/s, and the pair stays put. Being at rest or in motion depends on who is watching, and no experiment in mechanics tells apart two observers who each move in a straight line at constant velocity; this is Galilean relativity.

Feynman uses this idea the other way round, to reach the conservation of momentum without speaking of forces. Two equal masses that meet head-on at equal speeds and stick can only end up at rest, since symmetry favours neither side. With 2 kg on both trolleys, 1.5 m/s for A and −1.5 m/s for B, the Trolley A button shows the same collision as a trolley at −3 m/s running into one at rest, and the pair leaves with half of that, −1.5 m/s. By combining symmetric cases with changes of frame he extends the rule to unequal masses, and in every case the sum of the masses times the velocities is the same before and after.

Kinetic energy does not fare so well. With the starting values, 8 J before the collision become 2 J after it, and the missing 6 J went into deforming the bumpers and into heat. The energies before and after change from one frame to another, but the difference between them is the same in all three; seen from the centre of mass, all the energy there was is lost. Step 3 returns to this point with collisions in which the trolleys do not stick.

In Feynman: §10-3 Momentum is conserved! ↗

With the bench’s starting values, A, of 1 kg, comes at 4 m/s towards B, of 3 kg, at rest, and the coefficient of restitution is set to 0.5. After the collision A comes back at −0.5 m/s and B moves on at 1.5 m/s. The momenta add up to 4 kg·m/s before and after, as in the two previous steps, while the kinetic energy drops from 8 J to 3.5 J. In the bars, the black mark for the total momentum stays in the same place on both rows, and the energy bar shrinks, leaving the part that was lost hatched.

Momentum is conserved in any collision because the contact forces are internal to the pair, equal and opposite, whether the bumpers are made of rubber or of steel. Kinetic energy has no such guarantee. During contact the bumpers deform, and whatever part of the deformation does not undo itself remains as heat, sound and vibration in the material. The coefficient \(e\) measures how much comes back, comparing the speed at which the trolleys separate after the collision with the speed at which they were closing in before it. With \(e = 1\) the collision is elastic and kinetic energy is conserved as well; with \(e = 0\) the trolleys do not separate, they stay stuck, and we are back in step 2.

With equal masses, B at rest and \(e = 1\), A halts and B takes over the velocity A had; the velocity passes whole from one trolley to the other. This is what happens in Newton’s cradle, the desk ornament with steel balls hanging side by side, where the ball that strikes stays still and the one at the far end swings out. With different masses there is no such hand-over, and a light trolley that hits a heavy one elastically bounces back.

trolley A   trolley B   total momentum   energy lost

The View from the centre of mass button makes the arithmetic simpler. For someone travelling with the centre of mass, at 1 m/s with the starting values, the total momentum is zero before and after; A arrives at 3 m/s, B at −1 m/s, and after the collision they leave at −1.5 and 0.5 m/s. Each has reversed its direction and kept the fraction \(e\) of the velocity it had. For the momentum to stay at zero, the two velocities must keep the inverse ratio of the masses, which is why a single number is enough to describe the whole collision.

The energy also splits into two parts. One belongs to the motion of the centre of mass, \(\tfrac12 (m_1+m_2)\,v_{\text{CM}}^2\), 2 J with the starting values, which no internal force can alter because it depends only on the total momentum. The other belongs to the relative motion, 6 J with the same values, and the collision can take energy only from this part, which is multiplied by \(e^2\) and falls to 1.5 J. The 4.5 J lost are the same seen from the track or from the centre of mass, as the second equation below states, where \(m_1 m_2/(m_1+m_2)\) is known as the reduced mass.

Feynman notes that bodies with no inner parts, such as the atoms of a gas, have almost no way of keeping the energy a collision takes from them, so they bounce off one another almost elastically; even for them, \(e\) falls a little short of 1.

In Feynman: §10-4 Momentum and energy ↗

The rocket on the bench starts at rest, full, with ten times the mass it will have when empty, and throws its fuel backwards at 3,000 m/s relative to itself. With a single portion, the 90% of the mass that is fuel leaves all at once and the rocket ends up at 2,700 m/s. With the same fuel split into 1,000 portions it reaches 6,896 m/s, more than twice as much, without spending a single gram more.

There is nothing outside the rocket for it to lean on, and nothing is needed. The system is the rocket together with its fuel, and the total momentum, zero at the start, stays zero. Each portion that leaves backwards carries off negative momentum, and what is left of the rocket gains the same amount forwards, like the trolleys of step 1 pushed by the spring. The secondary-school lesson on action and reaction treats the rocket through the pair of forces between it and the gases; what matters here is how much velocity each portion is worth.

If the portion of mass \(\Delta m\) leaves at \(u\) relative to the rocket, measured once the push is over, and \(m\) is the mass of the rocket with the portion still on board, conservation of momentum gives exactly \(m\,\Delta v = \Delta m\,u\). With a single portion, \(\Delta m\) is 90% of \(m_0\), and \(\Delta v\) comes out at 90% of \(u\), the 2,700 m/s on the bench. The animation shows everything from the point of view of someone who stayed at the starting point, which is why the last portions, thrown out when the rocket already moves faster than \(u\), travel forwards, only more slowly than the rocket.

Splitting helps because \(m\) keeps falling. The first portion pushes the full rocket, while the last one pushes little more than the empty shell, and the same mass of fuel is worth much more velocity there. As the portions become smaller and smaller, the sum of the \(\Delta m/m\) turns into an integral, and the result is \(\Delta v = u\ln(m_0/m_f)\), the equation Konstantin Tsiolkovsky published in 1903. On the graph the dots for each split climb towards the dashed line, always from below, and with a ratio of 10, ten portions give 85% of the limit and a hundred give 98%.

The logarithm drives a hard bargain. To double \(\Delta v\) the mass ratio has to be squared, and a rocket bound for a low orbit needs a \(\Delta v\) of around 9 km/s, counting the losses to gravity and to the air. With \(u\) = 3,000 m/s not even the control’s ratio of 20 gets there. That is why most of the mass of a rocket on the launch pad is fuel, and why it climbs in stages; each empty stage is dropped along the way, and the next one no longer has to accelerate that dead weight.

The one-portion case is the kick of a firearm, a comparison Feynman himself makes. The bullet leaves forwards and the gun moves back, the more slowly the heavier it is compared with the bullet.

In Feynman: §10-2 Conservation of momentum ↗ · §10-4 Momentum and energy ↗

The bench pushes an electron, at rest to begin with, with a constant force of 1.6·10⁻¹³ N, the force an electric field of a million volts per metre exerts on its charge. It all happens within nanoseconds, and the animation shows 10 ns in 8 s, in slow motion. By the end of that time the momentum has reached 5.86 times \(mc\), which for the electron is 2.73·10⁻²² kg·m/s, and the velocity 0.986 of the speed of light. By Newton’s arithmetic, the same momentum divided by the mass would give 5.86 times \(c\).

The momentum graph is a straight line because force is still the rate of change of momentum, and a constant force adds the same amount of momentum every nanosecond, without limit. What changes is the link between momentum and velocity. With \(p = \gamma m v\), the velocity comes from the second equation below, and when \(p\) grows much larger than \(mc\) the square root in the denominator approaches \(p/mc\) and \(v\) approaches \(c\) without getting there. On the velocity graph, Newton’s dashed line crosses the line of \(c\) at 1.71 ns and enters the shaded band, where no body with mass can be.

Why \(\gamma m v\) and not \(mv\)? Step 2 showed that changing frame alters the numbers of a collision without spoiling the conservation, provided velocities add as in Galilean relativity. In Einstein’s relativity they combine in another way, and a sum of \(mv\) that is conserved for one observer stops being conserved for another who passes by at high speed. The quantity conserved for everyone is \(\gamma m v\). At everyday speeds the difference vanishes: at the 3,000 m/s of the rocket in step 4, \(\gamma\) exceeds 1 by 5·10⁻¹¹, and only at 0.6 \(c\) does it reach 1.25.

momentum   velocity   Newton’s prediction, \(p/m\)

In the second half of the run, from 5 to 10 ns, the electron gains as much momentum as in the first, 2.93 \(mc\), while its velocity rises only from 0.946 to 0.986 \(c\). In an accelerator the effect is more extreme still. CERN’s LHC takes in protons at 450 GeV, when they travel about 650 m/s below \(c\), and brings them to 6.8 TeV, with about 15 times as much momentum, leaving them some 3 m/s short of \(c\). Each lap of the ring gives one more push, and what grows is the momentum, while the velocity barely moves.

Kinetic energy follows momentum, not velocity. It is \((\gamma - 1)\,mc^2\), as in the chapter 4 lesson, and it too grows without limit as \(v\) approaches \(c\). What the constant force delivers every nanosecond, in momentum and in energy, keeps reaching the electron; it simply no longer shows up as velocity.

Feynman writes the same law another way, keeping \(mv\) and letting the mass itself carry the factor \(\gamma\); most texts today prefer to keep the mass fixed and put the \(\gamma\) into the momentum, and the numbers are the same. He also points out that momentum, unlike energy, cannot slip away unseen into the disordered motion of atoms, and that it can also be carried by the electromagnetic field, as light shows when it pushes on whatever it falls upon.

In Feynman: §10-5 Relativistic momentum ↗