Having seen Newton's laws of motion, it is a good time to introduce the concept of momentum. Momentum is often defined as the "quantity" of motion. It considers both how much mass is moving and with what velocity. The lower case 'p' is used for momentum since unfortunately 'm' is already used for mass. In mathematical terms momentum is a vector defined as[br][center][math]\vec{p}=m\vec{v}.[/math][/center]There is a close relationship between momentum and Newton's second law because acceleration is the time derivative of velocity. Consider taking the time derivative of momentum while mass is assumed constant. What do you get? Let's do it together: [br][center][math]\tfrac{d\vec{p}}{dt}=m\tfrac{d\vec{v}}{dt} \\[br]\tfrac{d\vec{p}}{dt}=m\vec{a} \\[br]\vec{F}_{net}=\tfrac{d\vec{p}}{dt}. [/math][/center][br]What this restatement of the second law tells us is that [b]the net force applied on an object equals the rate of change of the object's momentum[/b]. A net force will cause the momentum of an object to change. The larger the force the quicker the change in momentum. [br][br]In this equation, we can also see the first law. According to this expression, an object's momentum cannot change without a net force. If momentum can't change, then neither can velocity since momentum is proportional to velocity. The first law stated precisely this; that in the absence of a net force applied to an object, its velocity will be constant - both speed and direction of travel will be constant.

Since Newton's third law tells us that forces always come in pairs of equal and opposite vectors, we can learn something instructive about the momentum of any two interacting bodies - that it's conserved, or unchanging during the interaction. [br][br]Suppose we have two objects labeled 1 and 2 that exert forces on one another. It might be via collision, or what we think of as physical contact, or perhaps via gravitation from a distance. It makes no difference.[br][br]If we look at the force on 1 by 2 and the force on 2 by 1, Newton's 3rd law tells us they are equal and opposite. With that as a starting point, here is a simple way to show that momentum for the two-body system remains unchanged. We call this [i]momentum conservation[/i]: [br][br][center][math]\vec{F}_{1,2}=-\vec{F}_{2,1} \\[br]\int\vec{F}_{1,2}\;dt=-\int\vec{F}_{2,1}\;dt \\[br]m_1\int\vec{a}_{1}\;dt=-m_2\int\vec{a}_{2}\;dt \\[br]m_1\Delta\vec{v}_1=-m_2\Delta\vec{v}_2 \\[br]\Delta\vec{p}_1=-\Delta\vec{p}_2 \\[br]\Delta\vec{p}_1+\Delta\vec{p}_2=\vec{0} \\[br]\Delta\vec{p}_{system} = \vec{0} \\[br]\vec{p}_{system}=\text{constant}[br][/math][/center][br][br]

Really the only limitations on this derivation are that it "only" works up to around 30 million meters per second (v=c/10) and that it is only between two objects. I told you before that we will lift the first limitation in a later course by defining momentum a little differently. Regarding the two object limitation, recall from our discussions about fundamental forces that ALL interactions are only between two entities. In other words, if there is an object 3 that pulls on object 1 and on object 2, then we just bring object 3 into our system and once again the system (now of three particles) will conserve momentum. So it really is quite robust. That three-body system will really be three two-body systems that all interact: objects 1&2, 2&3, and 1&3. Each of those interactions will conserve momentum, and therefore so will the three-body system conserve momentum.[br][br]Another way to describe the validity is that [b]for any system comprising N>0 particles, momentum will be conserved (constant) in the absence of external forces[/b]. If there is some external force due to some external object, we can restore momentum conservation by bringing that object into our system when we do the mathematics. Realize that "bringing it into our system" is an intellectual and not a physical procedure. It simply means that we need to include another term for momentum in our accounting.

We often assume in situations like car accidents that momentum is conserved. We just mentioned above that momentum is only conserved in the absence of external forces. The trouble with car accidents involving two cars is that the friction force on the tires is often quite large. That friction force comes from the external force of earth on the cars. It is not practical to bring earth into our system since we want it to be our assumed fixed coordinate system. Therefore we should expect that using momentum conservation for car wrecks is approximate at best. [br][br][color=#1e84cc]EXAMPLE: Let's try it out nonetheless. Suppose the red car above has a mass of 1200kg and an initial velocity vector of [math]\vec{v}_{1,i}=(5\hat{i}-20\hat{j})m/s[/math] and the orange car has a mass of 1400kg and an initial velocity of [math]\vec{v}_{2,i}=(-10\hat{i}-10\hat{j})m/s.[/math] If the cars get tangled together during the wreck and slide as one mass, what will the velocity of that mass be right after the collision?[br][br]SOLUTION: We assume that momentum is conserved, or [math]\Delta\vec{p}_1+\Delta\vec{p}_2=\vec{0}.[/math] Expanding the deltas and separating terms gives us[br][center][math][br]\vec{p}_{1,f}-\vec{p}_{1,i}=-\vec{p}_{2,f}+\vec{p}_{2,i} \\[br]\vec{p}_{1,f}+\vec{p}_{2,f}=\vec{p}_{1,i}+\vec{p}_{2,i} \\[br]\text{Notice that this just means that momentum for the system is conserved. } \\[br]m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}=m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i} \\[br]\text{Since the cars stick together $\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}_f$, so } \\[br](m_1+m_2)\vec{v}_f = m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i} \\[br]\text{Plugging in and solving gives } \\[br]\vec{v}_f = (-3.1\hat{i}-14.6\hat{j})m/s.[br][/math][/center][/color]