7.1: Momentum Conservation
The universe has many remarkable qualities, among them a rather beautiful symmetry: the total amount of motion in the universe is constant. This law only makes sense if we measure “motion” in a specific way: as the product of mass and velocity. This product, called momentum, can be transferred from one object to another in a collision. The rapidity with which momentum is exchanged over time is determined by the forces involved in the collision. This is the second of the five fundamental conservation laws in physics. The other four are conservation of energy, angular momentum, charge and CPT. (See Feynman's Diagrams for an explanation of CPT.)
Key Equations and Definitions
\begin{align*}\text{Uses of Momentum}
\begin{cases}
p = m v & \text{Momentum vector}\\
F = m a & \text{Newton's Second Law}\\
m a= \frac{m \Delta v}{ \Delta t}=\frac{\Delta p}{\Delta t} & \text{Newton's Second Law in terms of momentum}\\
\sum p_{\text{initial}} = \sum p_{\text{final}} & \text{Momentum is constant in closed systems}\end{cases}\end{align*}
Key Concepts
 Momentum is a vector that points in the direction of the velocity vector. The magnitude of this vector is the product of mass and speed.
 The total momentum of the universe is always the same and is equal to zero. The total momentum of an isolated system never changes.
 Momentum can be transferred from one body to another. In an isolated system in which momentum is transferred internally, the total initial momentum is the same as the total final momentum.
 Momentum conservation is especially important in collisions, where the total momentum just before the collision is the same as the total momentum after the collision.
 The force imparted on an object is equal to the change in momentum divided by the time interval over which the objects are in contact.
 Impulse is the change in momentum on one of the objects in a collision. Impulse force is the force imparted on one of the objects as defined above.
 Impulse is how momentum is transferred from one system to another. You can always determine the impulse by finding the changes in momentum, which are done by forces acting over a period of time. If you graph force vs. time of impact the area under the curve is the impulse.
 When calculating impulse the time to use is when the force is in contact with the body.
 Centerofmass is the 'balancing point' of an object. For uniform density ball, the center of mass is at the exact center. All collisions in this chapter are treated as center of mass collisions.

Internal forces are forces for which both Newton’s \begin{align*}3^{rd}\end{align*}
3rd Law force pairs are contained within the system. For example, consider a twocar headon collision. Define the system as just the two cars. In this case, internal forces include that of the fenders pushing on each other, the contact forces between the bolts, washers, and nuts in the engines, etc.  External forces are forces that act on the system from outside. In our previous example, external forces include the force of gravity acting on both cars (because the other part of the force pair, the pull of gravity the Earth experiences coming from the cars, is not included in the system) and the forces of friction between the tires and the road.
 If there are no external forces acting on a system of objects, the initial momentum of the system will be the same as the final momentum of the system. Otherwise, the final momentum will change by \begin{align*}\triangle p= F \triangle \mathrm{t}\end{align*}
△p=F△t . We call a change in momentum, \begin{align*}\triangle p\end{align*}△p , an impulse.
Key Applications
 Two cars collide headon...two subatomic particles collide in an accelerator...a bird slams into a glass office building: all of these are examples of onedimensional (straight line) collisions. For these, pay extra attention to direction: define one direction as positive and the other as negative, and be sure everybody gets the right sign.
 A firecracker in midair explodes...two children push off each other on roller skates...an atomic nucleus breaks apart during a radioactive decay: all of these are examples of disintegration problems. The initial momentum beforehand is zero, so the final momentum afterwards must also be zero.
 A spacecraft burns off momentum by colliding with air molecules as it descends...hail stones pummel the top of your car...a wet rag is thrown at and sticks to the wall: all of these are examples of impulse problems, where the change in momentum of one object and the reaction to the applied force are considered. What is important here is the rate: you need to come up with an average time Dt that the collision(s) last so that you can figure out the force \begin{align*}F = \triangle p / \triangle \mathrm{t}\end{align*}
F=△p/△t . Remember as well that if a particle has momentum \begin{align*}p\end{align*}p , and it experiences an impulse that turns it around completely, with new momentum \begin{align*}p\end{align*}−p , then the total change in momentum has magnitude \begin{align*}2\mathrm{p}\end{align*}2p . It is harder to turn something totally around than just to stop it!  A car going south collides with a second car going east … an inflatable ball is thrown into the flow of a waterfall … a billiard ball strikes two others, sending all three off in new directions: these are all examples of twodimensional (planar) collisions. For these, you get a break: the sum of all the momenta in the \begin{align*}x\end{align*}
x direction have to remain unchanged before and after the collision — independent of any \begin{align*}y\end{align*}y momenta, and viceversa. This is a similar concept to the one we used in projectile motion. Motions in different directions are independent of each other.  Momenta vectors add just like any other vectors. Refer to the addition of vectors material in Chapter 1.
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