# Collisions

## Example 3: A Real Collision 2: $\mathbf{ a}+\mathbf{B}\to \mathbf{a}+\mathbf{B}$

Angus (we’ll call him $A$, rather than $a$) is a slight, but quick football safety trying to tackle Buster ($B$), the huge running back. Angus bounces off the lumbering Buster. (It could also be a small car and a large truck, but that's too scary.)

This is not a tackle (because they’d become a single entity in a tackle). Rather this is a bounce in which Angus and Buster will be separate.

<img align="center" src="./biglittle.png">

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Here are the data for this collision, again with fake units:

- $m(A)= 2$ and $v_0(A)=6$
- $m(B)= 4$ and $v_0(B)=2$ the other way... so if we take $A$'s direction to be positive, then $\vec{v}_0(B)=-2$, or since we're in one dimension, $v_0(B)=-2$
- So, $p_0(A)= 2\times 6 = 12$
- and $p_0(B)= 4\times (-2) = -8$

The final piece of data that I'll tell you that Angus' 

-  *final* momentum is $p(A) =-9.3$: so, he bounces back, off of Buster.

The operative football question for the defense, is whether Angus pushed Buster backwards for a loss or does he still carry his momentum forward for a gain.

**Questions:** 

1. What's Angus's final velocity?
2. What is Buster's final momentum after colliding with Angus? 

**Answers:**

1. What is $A$'s final velocity?

Our intrepid (courageous!) Angus's final velocity is easy. We know his mass ($m(A)=2$ and his momentum $p(A)=-9.3$, so we can get his velocity one of two ways:

- calculate it: 

  $$
  \begin{align*} p(A) &=-9.3 = m(A)v(A) \\
  v(A) &= \frac{p(A)}{m(A)} \\
  v(A) &= \frac{-9.3}{2} = -4.67
  \end{align*}
  $$
  
- or use an area plot. Set up an $m$ versus $v$ set of axes…draw a horizontal line at the known value of $m=2$ and construct the rectangular area to be the known momentum of $9.3$:

<img align="center" src="./angus_p.png">

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2. What's $B$'s momentum after the collision?

We could create areas and solve this but I think that more insight comes quickest if we use the thermometer approach. Here's our construction:

<img align="center" src="./scatter_example_therm.png">

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And, here's my reasoning:

* In (a) I've plotted the initial momenta for each object. Angus with a positive momentum of 12 and Buster with his negative momentum of $-8$. 
* The total momentum of the Angus-Buster system is the sum, which I can represent as the sum of the two thermometer lines, here the *difference* since one is negative.
* I can quickly see that the difference of those momenta leaves $4$, which is then the absolute total, $p(T)=4$. So the combination of the final momenta also have to sum to be that same value–momentum conservation.
* In (a) then, on the right hand side, I’ve put in the thermometer for that difference: $p(T)=4$.

That’s our initial situation and the key is creating that $T$ thermometer as the total…which has to be the same total in the final state, which is in the middle diagram, (b):

* I know the final $A$ momentum, and that's added in (b) and I brought along $T$ on the right: because THAT's the statement of momentum conservation!
* So we can construct $B$'s momentum which, when combined with that of $A$ will give us $T$ again.
* That's shown in (c). Count up the squares for $B$ and you'll see that $P(B)=\text{about }13.3.$

With pictures I'm solving the momentum conservation equations:


$$
\begin{align*}
p_0(A) + p_0(B) &= p(A) + p(B) \\
p(B) &= p_0(A) + p_0(B) - p(A) \\
p(B) &= (2)(6) + (4)(-2) - (-9.3) = 12-8+9.3 = 13.3 \end{align*}
$$

Either approach is fine!