View source for Physics with Calculus/Mechanics/Momentum and Conservation of Momentum

== Momentum and Energy ==

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[[w:Momentum|Momentum]] is a physical quantity that is equal to the [[w:mass|mass]] of an object multiplied by its [[w:velocity|velocity]]. Momentum is related to energy, and like [[w:energy|energy]] it remains conserved in a [[w:closed system|closed system]] (a system where no energy enters or leaves). Classically, momentum is defined as such:

<math>\mathbf{p}=m\mathbf{v}</math>

where the bold P and v indicate [[w:vector|vector]]s.

Where [[w:kinetic energy]] is defined as:

<math>E=\left(\frac{1}{2}\right)mv^2</math>

Therefore, 

<math>2E/\mathbf{v}=\mathbf{p}</math>

Also note

<math>\frac{dE}{dv}=p</math>

That is, momentum can be used to measure the change in kinetic energy as an object changes velocity.

== Force and Momentum ==

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[[w:Force|Force]] is defined as the mass of an object multiplied by its [[w:acceleration|acceleration]]:

<math>\mathbf{F}=m\mathbf{a}</math>

Acceleration is the [[w:derivative|derivative]] of velocity, which is the vector form of distance over time. Thus, the integral of acceleration is velocity:

<math>\int adt=v</math>

Using this, we can easily see how the integral of Force is momentum:

<math>\int Fdt = m\int adt = mv = p</math>


== Conservation of Momentum ==

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Recall the main result from [[Physics with Calculus/Mechanics/Center of Mass|Center of Mass]] that for a system of particles

<math> F = M \frac{dV}{dt} = \frac{d MV}{dt} </math>

where F is the sum of the external forces, M is the total mass, and V is the velocity of the center of mass (that is, the time derivative of the center of mass).  If F = 0, we have <math>MV = \sum_i m_i v_i = const</math>.  If we define <math>m_i v_i</math> to be the momentum of a particle then we have the law of conservation of momentum -- that the total momentum is conserved when there are no external forces.  In other words, the momentum you start out with will be the momentum you end up with. Say you're playing pool, and you go to smack the 8-ball in the right corner pocket. As a challenge, however, your opponent rolls the 8-ball across the table. The conservation of momentum can be summed up with a simple equation:

<math>\sum p_i = \sum p_f</math>

As you smack the cue ball toward the 8-ball and your opponent rolls the 8-ball across the table, you give the cue ball some velocity v_1 and the 8-ball a velocity v_2. The initial momentum for the entire system is therefore:

<math>m_1v_1 + m_2v_2 = \sum p_i</math>

Being the pool player with some sweet skills that you are, the cue ball smacks the 8-ball and it goes flying into the pocket. Right after the collision, the momentum of the system is:

<math>m_1v_1' + m_2v_2' = \sum p_f</math>

Can you guess what we can do next? Since we know that the momentum of the system is conserved, we can set these two equations equal to each other:

<math>m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' = \sum p_i = \sum p_f</math>

Now, there are a plethora of ways to solve these equations, depending on what values are given or determined through experiment. If we determine the masses and the initial velocities plus one of the final velocities, we can solve the equation by simple algebra this way:

<math>(m_1v_1 + m_2v_2 - m_1v_1')/m_2 = v_2' \Rightarrow (m_1v_1 + m_2v_2 - m_2v_2')/m_1 = v_1'</math>

The "<font style="vertical-align:-2%;"><math>\Rightarrow</math></font>" symbol here means "implies". Obviously, if you can solve for v1' here you can solve for v2'.

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Now let's try something a bit harder. Say that we're poor college students and can't afford to purchase machines that measure velocity. Instead, however, we can measure forces and masses. We intentionally hit the cue ball with a force of 3N and roll the 8-ball with a force of 1N. If we want to find momentums now, we'll have to calculate them using our given information. Well, we can use the nifty equation

<math>\int Fdt = \int madt = mv = p</math>

(I just integrated F = ma over time). Our momenta are therefore easily calculated:

<math>\int_{t_1}^{t_2} 3\, dt = p_1\quad and\quad \int_{t_1}^{t_2} 1\, dt = p_2</math>

Remember that because the integral of force is momentum, momentum is equal to the area under the curve of force (by time). In other words, when you smack a cue ball you exert some force over some time interval. If you were to plot this, the momentum of the cue ball at the end of the time interval would be the total area under that curve to that point. 


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The principle of conservation of momentum can, of course, be applied to a situation where two colliding objects creates one big object. Say that your billiard balls were made of a clay, and just as your cue ball hit the 8-ball, they stuck together to form one large mass of clay. the conservation of momentum equation would now look like

<math>\sum p_i = m_1v_1 + m_2v_2 = \sum p_f = (m_1 + m_2)/v'</math>


--[[User:Mattciv|Mattciv]] 18:21, 7 August 2005 (UTC)

==Continuous Systems==

Now is a nice time to look at an interesting class of problems involving continuous mass distribution.  The classic example is a rocket that burns fuel and ejects exhaust at a velocity v ''relative to the rocket.''  If the rocket starts out with mass <math>M_0</math> and burns fuel at a rate of b, what is the velocity of the rocket as a function of mass burned (or equivalently time since the mass burned is bt).

Consider the rocket at two times t and t + dt.  At t it has a velocity forwards of u, and mass M.  At t + dt, it has velocity u + du and mass M - dm and has expelled a small mass dm at velocity u - v forward (so that it is traveling at -v relative to the rocket).  Using conservation of momentum,

<math> Mu = (M - dm) (u + du) + (u-v)dm</math>.

This simplifies to

<math> du = v \frac{dm}{M}</math>

noting that the product of two small things is really, really small and we can safely take <math>du dm</math> to be zero.

Integrating gives

<math> \Delta u = v \ln(\frac{M}{M_0})</math>.

Or, if we want to find u as a function of time, just substitute <math> M = M_0 - bt</math>.

As you can see, the conservation of momentum provides a powerful tool for problems such as these.  If we wanted to find the velocity of a rocket that's going away from earth, we have to note that the conservation of momentum does not quite hold, but that <math> p(t + dt) = p(t) + Fdt</math> which is the same thing as F = ma.

Using differentials (or deltas if you're a stickler for that kind of thing) is a very useful technique that will allow you to write a differential equation for most systems (all if you're very clever), so it is a trick well worth remembering.

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