3.2: Momentum

Kinetic energy is a quantity that’s associated with motion. However, kinetic energy itself is not always conserved. If a cue pool ball runs into another ball, and the cue ball stops dead, the other ball goes off with the same speed that the cue ball came in at. In this case, the two balls have the same mass, so \(\ \frac{1}{2} m v^{2}\) is the same both before and after the collision; kinetic energy is conserved in the collision. However, if two cars hit each other in a head-on collision, and the tangled wreck of the two cars stops dead at the point of impact, kinetic energy is clearly not conserved, as the \(\ v\) of everything after the collision is zero. It’s not kinetic energy that’s conserved, but total energy. The kinetic energy that the cars had before the collision is, during the collision, converted into other forms of energy: heat, noise, and possibly some potential energy as the structure of the car is rearranged. So, sometimes, in some collisions, kinetic energy is conserved. However, in other collisions, kinetic energy is not conserved. Note that total energy is always conserved; it’s just that there are forms of energy other than kinetic energy, and sometimes kinetic energy can be converted to or from those other forms.

However, there is a quantity of motion that is conserved in every collision. If it is to be conserved in both the examples above, it can’t just be based on the speeds of the particles. While the speed would seem to be enough in the example of the pool balls, in the example of the cars there was a lot of speed to start with, but no speed after the collision. To work in both of these examples (and in general), this conserved quantity has to be something that takes into account both speed and direction. That quantity is momentum. It is traditional to use the letter \(\ p\) to represent momentum. The momentum of a particle is defined by:

\(\ \vec{p}=m \vec{v}\)

where \(\ m\) is the mass of the particle, and \(\ \vec{v}\) is the velocity of the particle. The magnitude of \(\ \vec{v}\) is traditionally written \(\ |\vec{v}|\), but is often just abbreviated as \(\ v\) without the arrow. Magnitudes of 3-vectors are always positive or zero; it does not make sense to say a 3-vector has a negative magnitude. The magnitude of velocity is what we call speed. You can’t have a speed of \(\ -50 \mathrm{~km} / \mathrm{h}\), but you can be moving at \(\ 50 \mathrm{~km} / \mathrm{h}\) in the negative-x direction.

One way of dealing with 3-vectors is to break them into components— an x-component \(\ v_{x}\), a y-component \(\ v_{y}\), and a z-component \(\ v_{z}\). For now, to keep things simple, we’ll only consider motion in one dimension, so that particles will not have any component of velocity in the y or z directions. Therefore, we can say that the particle’s velocity is \(\ v_{x}\) in the \(\ +x\) direction. If \(\ v_{x}\) is negative, it means that the particle is moving to the left. The speed, however, the magnitude of the velocity, is still positive; that’s just how fast it’s going, without reference to direction.

Just like total energy, it turns out that momentum is a conserved quantity. If you take everything into account (which is occasionally tricky), the total momentum before and after a collision or interaction must be the same. Consider the example of the two pool balls above. If the pool balls have a mass \(\ m\) and an x-velocity \(\ v_{x}\), then the initial momentum is just \(\ m v_{x}\). After the collision, it’s the other ball that’s moving, but the speed is the same, so the final momentum is \(\ m v_{x}\). The total momentum is conserved.

In the case of the two cars colliding, suppose that both cars have the same mass \(\ m\) and are approaching each other with speed \(\ v\). The car that is moving to the right has x-momentum \(\ m v_{x 1}=m v\), and the car that’s moving to the left has x-momentum \(\ m v_{x 2}=-m v\). Notice that the x-momentum of the car moving to the left is negative! In contrast, the kinetic energy of both cars is positive, and is the same: \(\ \frac{1}{2} m v^{2}\). The total momentum in the system is the sum of the momentum of the individual particles. Thus, the total x-momentum is \(\ m v_{x}+m\left(-v_{x}\right)=0\). After the collision, the velocity is zero, so the total momentum is still zero. Momentum is, in fact, conserved in the collision.