Section 8.4 Collisions
“-----”―------
We will be using the principle of conservation of momentum to analyze what happens in collisions. This topic is usually presented with a discussion of the energy of the system, which breaks the world of collisions into two categories. If no energy is lost, then the collision is said to be elastic. These types of collisions basically do not exist in reality, and are considered to be a theoretical ideal. A collision in which energy is lost is said to be inelastic. The ways that energy would typically be lost would be through heat and sound, and while we say that the energy is lost, it’s more correct to say that the energy has been converted into another form that cannot be accounted for in the motion of the particles.
The algebra of these calculations can get a little bit difficult, and it’s easy to get lost in the details. So we’re going to focus on setting up the equations and laying out the trajectory of the algebra, but we won’t get bogged down in those details.
A Moving Mass Hits and Sticks to a Stationary Mass.
In the first scenario, we’re going to imagine a mass \(m_1\) is traveling with velocity \(v_1 \gt 0\) and collides with another mass \(m_2\text{,}\) which is at rest. After the two masses collide, they stick together and act as a single object. We are interested in how fast the combined object is moving. We begin by creating diagrams that represent the motion before and after the collision. We will label the final velocity of the combined object as \(v_\text{f}\text{.}\)
INSERT DIAGRAM
Before the collision, the momentum of the system is \(m_1 v_1\text{.}\) After the collision, the momentum of the system is \((m_1 + m_2) v_f\text{.}\) Since momentum is conserved, we know that these two quantities are equal to each other. So we can set up an equation to solve for \(v_\text{f}\text{.}\)
\begin{align*}
m_1 v_1 \amp = (m_1 + m_2) v_\text{f}\\
v_\text{f} \amp = \frac{m_1}{m_1 + m_2} v_1
\end{align*}
The important observation about this equation is that the two objects together will always be moving slower than the object by itself. This should make intuitive sense because if it were moving faster, there would have to be something that propelled it forward after the two pieces collided. If \(m_1\) is large compared to \(m_2\text{,}\) the result would be that the combined object is moving close to the initial speed of the mass. If \(m_1\) is small compared to \(m_2\text{,}\) then the combined object will be moving very slowly. Both of these scenarios should feel fairly intuitive.
Note that the kinetic energy after the collision is
\begin{align*}
KE \amp = \frac{1}{2} (m_1 + m_2) (v_\text{f})^2\\
\amp = \frac{1}{2} (m_1 + m_2) \left( \frac{m_1}{m_1 + m_2} v_1 \right)^2\\
\amp = \frac{m_1}{m_1 + m_2} \cdot \frac{1}{2} m_1 v_1^2.
\end{align*}
In other words, we can directly calculate the proportion of the energy that is lost in the interaction based on the masses, even though we have no information about the interaction. This means that the collision is inelastic.
A Moving Mass Bounces off of a Stationary Mass Elastically.
We will now consider the scenario with the same initial condition, but a different outcome. The mass \(m_1\) travels with velocity \(v_1 \gt 0\) and collides with another mass \(m_2\text{,}\) which is at rest. This time, we will assume that they remain two separate entities after the collision and that the collision is purely elastic, meaning that no energy is lost. We will assume that the final state has mass \(m_1\) traveling with velocity \(v_1'\) (which may be positive or negative) and mass \(m_2\) traveling with velocity \(v_2'\) (which will always be positive).
Conservation of momentum leads to the equation
\begin{equation*}
m_1 v_1 = m_1 v_1' + m_2 v_2'.
\end{equation*}
Conservation of energy leads to the equation
\begin{equation*}
\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2.
\end{equation*}
There is a bit of algebra to do here. It’s not particularly difficult, but there are just a lot of steps. In the end, you get the following results:
\begin{align*}
v_1' \amp = \frac{m_1 - m_2}{m_1 + m_2} v_1 \\
v_2' \amp = \frac{2m_1}{m_1 + m_2} v_1
\end{align*}
Notice that the direction of the mass \(m_1\) depends on whether it has more or less mass than mass \(m_2\text{.}\) If the mass is smaller, it bounces back, and if the mass is larger, then it continues forward. And if the masses are identical, then the mass comes to a stop.
Two-Dimensional Collisions.
We can use the same techniques to solve for collisions in two dimensions. This would be the classic setup for billiard balls, where instead of hitting the ball straight-on, it is hit to one side or the other. This requires us to introduce two more variables \(\theta_1\) and \(\theta_2\text{,}\) which represent the angles at which the balls leaves the collision relative to the initial velocity. Solving these equations requires a bit of trigonometry and a lot of algebra, so we’re going to skip all of the details.
The important fact when thinking about this problem is that momentum is conserved in all directions. If we have the initial particle traveling in the x-direction, then the momentum in the y-direction after the collision will be zero.
