Fundamentals of Physics/Linear Momentum and Collisions

In the previous chapters, we concerned ourselves only with isolated objects. In this chapter, we'll see what happens when two (or more) objects interact which each other, in the form of contact between the two bodies (as in a collision, or in the sudden motion of two or more objects due to a need for them to separate due to an explosion).

Forces between two colliding objects

When two objects come in contact with each other, each exerts a pushing force on the other (think of the last time you bumped into someone in a crowded place: you felt a push, and so did they). Further, the force that one exerts on the other is always exactly the same. This is Newton's third law of "equal and opposite reaction forces." For example, if two cars collide, during the collision, car A will exert a force on car B ($F_{AB}$), and car B will exert the exact same force on car A ($F_{BA}$). The two forces will have the same strength, but be in exactly opposite directions to one another. In other words, $F_{AB}=-F_{BA}$. It doesn't matter if one car is heavier (more massive) than the other. The push force from one car will equal the push force from the other. What if one car is a small Honda and the other car a huge SUV? If so, when in contact, the force the Honda exerts on the SUV will be equal to the force the SUV exerts on the Honda, only in the opposite direction. What about a bug hitting a car windshield? The force of the bug on the windshield is equal to the force of the windshield on the bug, only in the opposite direction. Why then does the bug get crushed and the SUV doesn't even feel the collision? Because the resulting motion \emph{after the collision} is driven by the acceleration the body takes from the collision, while in contact with the other object. Suppose the equal and opposite force of the bug-windshield collision is 0.1 N. The bug has a mass of 0.001 gram, or 1x 10^-6 kg. Its resulting acceleration will be a=F/m=0.1 N/1 x 10⁻⁶ kg=100,000 m/s². The SUV, with a mass of about 4,000 kg gets an acceleration of a=0.1 N/4000 kg =0.000025 m/s².

Impulse

Collisions are typically very brief, say $1$ ms, or $0.001$ s. During this time, a parameter called "impulse" exists, defined as $J=\Delta p$, which is the change in an object's momentum. How far does the bug and SUV move in this time? The bug from above will move $5$ cm, the SUV will "move" about the diameter of an atom making up the windshield. The bug gets crushed because its internal structure cannot sustain an acceleration of about 10,000g.

Momentum

This equal and opposite force idea leads to momentum, which is defined as $p=mv$ or more correctly, ${\vec {p}}=m{\vec {v}}$. Notice ${\vec {p}}$ involves velocity directly. We also have "conservation of momentum" that says that

${\vec {p}}_{before}={\vec {p}}_{after}$.

The "before" and "after" refer to before and after a collision. This law itself allows us to ignore the physics \emph{of the collision} and instead focus on the physics \emph{just before and just after the collision}. More correctly, the law is

$\Sigma {\vec {p}}_{before}=\Sigma {\vec {p}}_{after}$,

indicating that the law means add all objects carrying momentum before a collision and set equal to the sum of all momenta carrying objects after the collision. Since $p$ is a vector, so you must sum the momenta of all objects in the $x$ direction, then in the $y$ direction, both before and after the collision in order the conservation law to be helpful. Lastly, there are two types of collisions, elastic and inelastic. In elastic collisions, the colliding objects bounce off of each other, while in inelastic, they all stick together creating a new "conglomerate mass" which is the sum of the individual masses that stuck together. In applying the conservation law for an inelastic collision, you typically have something like

$m_{1}{{\vec {v}}_{1before}}+m_{2}{{\vec {v}}_{2before}}+m_{3}{{\vec {v}}_{3before}}+...=(m_{1}+m_{2}+m_{3}+...){\vec {v}}_{after}$.

Notice that there's only one velocity after the collision

(${\vec {v}}_{after}$) because only the "big blob" is moving after they all collided and stuck together. For an elastic collisions, where two objects (1 and 2) collide along a single axis, we'll have

$v_{1final}={\frac {m_{1}-m_{2}}{m_{1}+m_{2}}}v_{1before}+{\frac {2m_{2}}{m_{1}+m_{2}}}v_{2before}$

and

$v_{2after}={\frac {2m_{1}}{m_{1}+m_{2}}}v_{1before}+{\frac {m_{2}-m_{1}}{m_{1}+m_{2}}}v_{2before}$.