Kinetic energy, bullets and collisions
Energy is a fundamental notion in physics and appears in a variety of forms, such as: kinetic energy, potential energy, heat energy, and electrical energy. Kinetic energy is particularly important in our daily life, and is an excellent example of a quadratic relation.
In this step we will

define the kinetic energy of a mass \(\normalsize{m}\) moving with velocity \(\normalsize{v}\)

learn about the Law of Conservation of Energy and its implications for bullets

see how physicists analyse elastic particle collisions.
Definition of kinetic energy
Physicists regard kinetic energy as energy due to motion, and sometimes also refer to the amount of work that a moving particle can do. For a single particle of mass \(\normalsize{m}\) moves with velocity \(\normalsize{v}\), its kinetic energy is officially defined as
\[\Large{T=\frac{1}{2}mv^2}.\]The dependence of the kinetic energy \(\normalsize{T}\) on the velocity \(\normalsize{v}\) is quadratic. This is a marked difference from what we saw with the momentum \(\normalsize{p=mv}\), which depends linearly on velocity. So if we double the speed of a particle, its momentum doubles, but its kinetic energy is multiplied by four. If we triple the speed of a particle, its momentum triples, but its kinetic energy is multiplied by nine.
In general energy may shift from one form to another during interactions, but the Law of Conservation of Energy states that the total amount of energy is conserved. With a swinging pendulum for example, the mass has at the top of its swing no kinetic energy, but only potential energy, while at the bottom of its swing it has the most kinetic energy and the least potential energy. In real life, most moving particles lose energy through friction, so that their kinetic energies eventually are transferred to heat energy.
The difference between momentum and energy
Somewhat simplistically, the momentum of a moving particle is a measure of how much push it has; whereas the kinetic energy is a measure of how much damage it can do. This explains why martial artists practice to punch fast. With a small increase in the speed of a punch, the power to damage increases quadratically.
This is also the reason why bullets are so deadly, even if they are relatively light; they have a high speed. Rifle bullets typically have two or three times the speed of pistol bullets, so that even with the same size, the energies of rifle bullets are between four and nine times the energy of pistol bullets, making them correspondingly more deadly.
Here are some figures: the caliber of a bullet is usually measured in inches, unless explicitly stated in millimetres, and refers to the diameter of the circular crosssection.
Firearm  Caliber  Muzzle energy (joules) 

air gun spring  .177  20 
air gun PCP  .22  40+ 
pistol  .177  159 
pistol .357 Magnum  .177  873 
rifle  .30  2,000 
Q1 (E): The Lone Ranger’s rifle shoots bullets at 3000 ft/sec, while Butch Cavendish’s rifle shoots the same size bullet at 4000 ft/sec. How much more energy do Butch’s bullets have?
Q2 (M): If the Lone Ranger’s rifle shoots .30 caliber bullets at 3000 ft/sec, and his sidekick Tonto’s pistol shoots .20 caliber bullets at 1000 ft/sec, then how much more energy do the Lone Ranger’s bullets have? [Let’s assume that the bullets have the same lengths.]
The quadratic dependence of energy on velocity also explains the scenes in the delightful movie Gravity, starring Sandra Bullock, where even light particles can do significant damage if they are moving swiftly enough, (even if this might stretch the bounds of what we can expect orbiting satellites to experience).
What happens when things bounce, or go bang
The Law of Conservation of Energy is a sister to the Law of Conservation of Momentum, which implies that in a collision between a number of different particles of varying speeds and masses, the total momentum is also conserved. However there are some big differences: prominently that Momentum doesn’t change into something else, but energy can assume quite a lot of different forms, and momentum has a directed aspect, unlike energy.
If you fire a gun, the law of Conservation of Momentum says that the momentum of the bullet is exactly equal to, but opposite in direction, to the recoil of the gun. However the energy of the bullet is much more than the energy of the gun: as far as energy is concerned the energy of the chemical explosion is changed into the combined energy of bullet and gun, along with heat and noise.
An explicit physics computation (advanced)
Fortunately for physicists, having both laws at hand to study collisions often is enough to predict quite a lot about what happens. This is especially so in elastic collisions, where no energy is assumed lost due to friction, heat or noise. Let’s illustrate this by making an (advanced) computation: feel free to skim this if you like, but do have a look at the animations.
A particle of mass \(\normalsize{m_1=3}\) is traveling right (in the positive direction along a onedimensional axis) with velocity \(\normalsize{v_1=5}\) meters per second. It hits (elastically) a second particle of mass \(\normalsize{m_2=1}\) which is stationary, so we write \(\normalsize{v_2=0}\).
Suppose that after the collision the respective velocities are \(\normalsize{w_1}\) and \(\normalsize{w_2}\). Can we determine these two velocities? Can we predict what happens? This is a powerful example of how a bit of linear and quadratic mathematics solves an interesting realworld problem.
First, we gather our facts
before collision  after collision  

total momentum  \(5 \times 3 + 0 \times 1 = 15\)  \(3 w_1 + w_2\) 
total kinetic energy  \(\frac{1}{2} \times 3 \times 5^2 + \frac{1}{2} \times 1 \times 0^2= \frac{75}{2}\)  \(\frac{1}{2} \left( 3 w_1^2 + w_2^2\right)\) 
By Conservation of Momentum and Energy:
\[\Large 3w_1 + w_2 = 15,\quad \frac{1}{2} \left(3w_1^2 + w_2^2\right) = \frac{75}{2}.\]We can rewrite the first equation as \(w_2 = 153w_1\) and substitute it into the second equation to get
\[\Large \frac{1}{2}\left(3w_1^2 + \left(153w_1\right)^2 \right)= \frac{75}{2}.\]After some algebraic simplification, this equation becomes
\[\Large 2w_1^2  15w_1 + 25 = 0\]which has solutions \(w_1 = 2.5\) or \(5\). We can use the Conservation of Momentum equation \(w_2 = 153w_1\) to calculate the corresponding values of \(w_2\), which are \(7.5\) or \(0\). Notice we have the precollision scenario where the velocities are \(v_1= 5\) and \(v_2=0\) and a postcollision scenario with velocities \(w_1=2.5\) and \(w_2 = 7.5\) meters per second.
Q3 (M): What happens if we change \(\normalsize{m_2}\) to \(\normalsize{5}\)?
Q4 (M): What happens if we change \(\normalsize{m_2}\) to \(\normalsize{8}\)?
If you like this kind of calculation, you might like Norman’s approach using some linear algebra, and even geometry: Norman’s video on collisions via linear algebra.
Answers
A1. The ratio is \(\normalsize{4000^2/3000^2}\), or \(\normalsize{16/9}\). Cavendish’s bullets have almost twice as much energy.
A2. In this situation things are more complicated since not only are the velocities different, the masses are too. Let’s remember that the mass of a cylindrical bullet of a fixed length is proportional to the crosssectional area, which is proportional to the square of the radius, and so also to the square of the diameter. Thus the ratio of the energies of the bullets is the ratio of \(\normalsize{\frac{1}{2}(.3)^2(3000)^2}\) to \(\normalsize{\frac{1}{2}(.2)^2(1000)^2}\). This is the ratio of \(\normalsize{81/4}\), or more than \(\normalsize{20}\) to \(\normalsize{1}\).
A3. After an analysis much the same as the previous one, we find that \(w_1 = \frac{5}{4}\) and \(w_2 = \frac{30}{8}\) metres per second.
A4. In this case \(w_1 = \frac{25}{11}\) and \(w_2 = \frac{30}{11}\) metres per second.
© UNSW Australia 2015