OPENING QUESTION: The photoelectric effect happens when a photon of light strikes an electron in orbit around an atom of a semiconducting material and knocks that electron away from the atom which allows the electron to flow freely down an electric wire.

There's a problem here... photons don't have mass. If they don't have mass, how can they interact with an electron? <Hint: see chapter title above>

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OBJECTIVE: I will be able to do basic momentum problems during today's class

FORMULAE OBJECTUS:

p = mv

J = F∆t = ∆mv = ∆p

∫Fdt = ∆p

WORDS O' THE DAY:

momentum (p)

impulse (J)

conservation of linear momentum

WORK O' THE DAY:

Momentum is a characteristic that EVERY MOVING OBJECT has. Back in Newton's time, we thought that an object had to have mass in order to have momentum....that was until Eistein came around and said 'not so fast' (so to speak)

However, since this isn't a class that uses relativity we'll (sadly) have to stick with the Newtonian universe in which an object MUST HAVE mass and velocity to have momentum.

In fact, LINEAR (we'll talk about angular momentum next unit) momentum is defined to be:

p = mv

(and no, I dunno why they chose "p", but someone did)

Furthermore, momentum is ALWAYS conserved...which comes in really handy when we talk about collisions between objects.

Consider this:

A billiards player shoots the cue ball across a table top. The ball has a mass (m) and velocity (v).

Let's look at two different cases in what happens next:

1) The cue strikes the 8 ball and then stops moving...transferring 100% of it's momentum into the 8 ball. In that instance:

m_cuev_cue = m₈v₈

 

2) Now let's look at the instance where the cue ball strikes the 8 ball but continues moving:

m_cueInitialv_cueInitial = m_cueFinalv_cueFinal +m₈v₈

 

From that situation we can generalize a basic idea in LINEAR momentum:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

 

generally speaking then:

p_1i + p_2i = p_1f + p_2f

Or

So... in an isolated system where no energy is input from outside:

(P_1f - P_1i ) + (P_2f - P_2i) = 0

or

∆P₁ + ∆P₂ = 0

or

∆m₁v₁ + ∆m₂v₂ = 0

In otherwords: The total momentum of two objects before a collision is equal to the momentum of those two objects AFTER they collide.

Linear Momentum is ALWAYS conserved (that's a big one folks!)

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Keep in mind that unlike energy and work, momentum is most definitely a vector... the direction in which two objects collide AND the direction they take after a collision are important considerations.

So.... be KEENLY aware of whether two objects are colliding in one dimension AND then rebounding in one dimension, or whether a second dimension is involved either before or after the collision

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One of the few times I'll have you actually look at a derivation--

Check out the derivation of Newton's 2nd using calc on page 249:

 

Let's take a gander at Newton's 2nd:

∑F = ma

However, we know we can rewrite acceleration as dv/dt so:

∑F = m(dv/dt)

However, in most situations mass is constant so we can include that inside the derivative (as long as mass is constant):

∑F = (dmv/dt)

∑F = (dp/dt)

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Do the worked example on page 251

Now let's take a look at what constitutes a NON-ISOLATED system. Keep in mind that when we discussed energy, we had to be concerned with energy crossing from outside the sytem to inside the system.

In very much the same way, we now have to be concerned about outside FORCES acting on a system.

If an outside force acts upon a system, it MUST change the momentum of that system.

We define that interaction in terms of the IMPULSE imparted on that system

When we are dealing with constant, non-varrying forces we can use simple algebra:

J = F∆t

However, when we are dealing with variable forces, we gotta bust out the calc:

J = ∫Fdt evaluated between t_i and t_f

Impulse is a measure of how much an outside force changes the momentum of the member(s) of the system acted on by that force

So

J = ∆p = F∆t

Work through example on page 255

Like momentum, Impulse is a vector

Let's talk about the Martian Landing lab coming up in a week or so.