WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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Ernest Rutherford showed that the
nucleus of a gold atom is very dense
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because he showed that helium nuclei
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which are incident on the gold nucleus
will bounce off and go backwards
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which suggests that this nucleus
of gold must be really heavy
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in order for it to make the helium
bounce all the way back.
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Well, so the scenario we have here is the
helium nucleus is incident with an energy
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of 8.00 times ten to the
minus thirteen joules
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and we're told that it bounces
back at an angle of 120 degrees
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and our job is to figure out what is the
angle of recoil of the gold nucleus
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and its speed and also to figure out the
speed of the helium's nucleus recoil.
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Now we're given the masses of
the helium and the gold nuclei
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and we have three things
that we want to find.
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So what that means is
that we will probably have
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to use three equations to figure it out.
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Let's put this arrow back here.
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Okay.
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So the first equation we get by considering
the x direction, conservation of momentum.
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So initially, we have all of our momentum
is included in the helium nucleus
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and we multiply the mass of the
helium nucleus by its initial velocity.
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We'll figure out what that
velocity is a little bit later
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using this kinetic energy that we're given.
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After the collision, we're going to have
the mass of the helium times its speed,
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multiplied by cosine of 180 minus
this *theta he* that's given.
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I like to use the reference angle in my
calculations, so I'm using this angle in here.
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This is 180 minus *theta he*
and because I am doing that
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I can put a negative sign
explicitly in the equation.
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You could instead have this be positive
and then go cosine of *theta he* or cosine
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of 120 in other words, that would be fine,
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but my preference is to make
the angle a reference angle
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so some number that's between zero and 90,
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and then put positives or negatives
in the equation explicitly.
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So, and then we'll add to that
this positive x component
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of the momentum of the gold nucleus.
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So that's mass of the gold nucleus
multiplied by its speed after the collision,
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multiplied by cosine of *theta au*.
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Okay.
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Well, there's not much
more we can do with that
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because there are three things we don't know.
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We don't know this angle here we don't know
this speed and we don't know this speed.
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So we need to consider the y
direction to find another equation.
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So initially there is no momentum
at all in the y direction
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because this helium nucleus is incident
straight along the x axis we'll say.
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Then after the collision the total
y component of the momenta
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must also be zero.
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So we have mass of the helium
nucleus times its speed,
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times sine of 180 minus
*theta he* and then minus,
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because this gold is going
downwards and to the right
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and so its y component of its
momentum is going to be negative
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so put a minus there
explicitly in the equation,
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and multiply that gold mass by its speed
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multiplied by sine of its
angle there, *theta au*.
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Okay.
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So that's two equations but
we have three unknowns
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so that means we need a third equation
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and that comes from
conservation of kinetic energy
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because we are told this
is an elastic collision.
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So we have the total
kinetic energy initially,
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which is one half mass of the helium times
its speed squared before the collision,
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equals one half mass of the helium
times its speed after the collision,
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squared plus one half mass of the gold
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times the speed of the gold
after collision, squared.
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So, we're going to --
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well, we have to look at
these equations first of all
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and think about what is our strategy
for solving for the unknowns.
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Now, one thing I'd like to
look at is these angles here
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because angles are
really tricky to solve for
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because they're inside a
trigonometric function.
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So this *theta au*,
we can't really get at it
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because itâ€™s contained within this sine
function, and likewise for this equation.
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In the x direction we have *theta au*
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is contained inside the cos
trigonometric function.
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So the first thing I like to
get rid of are these angles.
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So our goal is to --
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well, next thing to notice
is that the factors
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in front of these trig
functions are the same.
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We have *m au v au prime*
in front of the cos function
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and then we have *m au* *v au
prime* in front of the sine function.
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So that is a clue that if we
were to square this factor
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and add it to this factor squared,
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we'd have a common factor *m au v au*
all squared that we could factor out
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and then we'd have sine
squared plus cos squared.
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Sine squared plus cos squared
makes the number one.
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That's a nice trigonometric identity,
that's the Pythagorean identity.
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So we're going to do some rearrangement
here in order to make this possible.
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So we're going to rearrange things
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and then we're going to square
our rearranged y direction
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and square our rearranged x direction
formulae, and then we'll add them together
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and end up making the sine and
cosine each being squared,
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added together and that angle
will disappear, which is good.
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Okay.
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So version b of equation one,
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we'll isolate this *m au v au prime*
cos *theta au* term on one side,
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and so that equals *m he v he*
plus mass of the helium
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times speed of the helium after collision,
times cos 180 minus *theta he*.
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That's because I've just moved
this term to the left side
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and then switch the sides around.
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That's all that happened there
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and that brings this term isolated on
one side of the equation, which is nice
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because the next step is
we're going to square it.
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Then likewise for the y direction,
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we're going to isolate the sine
*theta au* term on one side
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by adding it to both sides or moving
it to the left so it becomes positive
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and then equals the other
stuff that was left over there.
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Then we'll square equation one b, so
this is equation one b squared now.
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So that's *m au* squared *v au prime*
squared times cos squared *theta au*,
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so there's nothing fancy
about the left side there.
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We just squared each of those factors.
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On the right side it becomes a little bit
more messy because we have a binomial
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and so we'll first of all factor out the
common factor, mass of the helium
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and we'll square that,
and then we'll square the
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*v he* plus *v he prime*
cos 180 minus *theta he*.
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We'll square that binomial and so
it's going to be the first term squared
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plus two times the first term
times the second term,
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then plus the second term squared.
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Okay.
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So now, we squared equation two b
as well which is straightforward
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because there is no binomials to consider
there, we just square each of the factors.
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Then we're going to add together
the equation one b squared
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to equation two b squared.
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So we're adding this equation to this
equation and that is shown on the next line.
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So we have *m au v au prime* squared as a
common factor among these two terms here
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and here so we can factor
it out leaving us with
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cos *theta au* squared plus
sine squared *theta au*.
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This is going to become the number
one and so mission accomplished,
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we got rid of the variable *theta au*.
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It doesn't appear anywhere else here.
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Then on the right hand side we have
a common factor * m he* squared
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and so we factor that out here and
then add everything else together.
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So that's *v he* squared plus two *v he prime*
cos 180 minus *theta he* plus this term
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and now this *v prime* helium
velocity after collision squared
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is a common factor between this and this
that remains, and so we'll factor that out.
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We have cos squared 180 minus *theta he*
plus sine squared 180 minus *theta he*
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which is another nice coincidence.
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This also becomes the number one
and this becomes the number one.
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So what we're left with is this line here.
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We'll call that equation four.
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So up to this line after we apply
the trigonometric identities
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to cos squared and sine squared here.
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Well that's a lot of work but
still not enough as it turns out
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because we have two unknowns remaining,
the velocity of the helium after collision
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and the velocity of the
gold after the collision.
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So, let's consider our kinetic energy formula.
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We're going to rearrange it to
solve for *v au prime* squared
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because if we can do that we can
substitute for it here in equation four.
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So we rearrange this kinetic energy
formula to solve for *v au prime* squared
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and multiply everything by two and
that gets rid of the fraction there.
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Then take this term to the left
side so it becomes minus
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and then divide everything by mass
of gold and you solve for this.
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So the velocity of the gold after collision
squared is going to be mass of the helium
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times the initial velocity of the helium
squared minus mass of the helium
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times the after collision
velocity of the helium squared,
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all divided by mass of the gold nucleus.
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So we'll substitute that in for *v au prime*
squared in equation four in this line now.
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So you see this is written here in red
because it replaced the *v au prime* squared.
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Otherwise everything else is
the same as in equation four.
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Then we have some work to do to solve for
the velocity of the helium after collision
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because this is going to end up
being a quadratic equation
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and we can use our quadratic
formula to solve it.
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But we have to change it to
the form that we're used to.
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We need to make it look like a x
squared plus b x plus c equals zero.
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Right now we have things on both sides of
the equation and we have brackets and so on.
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So we have some work to do.
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So first we'll expand the brackets
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by multiplying through by the
factors outside the brackets
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and that makes this *m au*
become power of one
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when it cancels with one of the *m au*'s there.
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So we have *m*, mass of the
gold times mass of the helium
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times initial velocity helium squared, minus
mass of the gold times mass of the helium
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imes after collision velocity
of the helium squared
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equals *m he* squared multiplied
by each of the terms in there.
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Then we collect everything on
the left side of the equation.
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Well what I actually did is I moved
everything to the right side
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and then switch the sides around.
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So we want to collect
the like terms together
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and so we have the *v he
prime* squared terms,
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there's going to be two of them, there's this
one and then there's going to be this one
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which is going to become positive
when it moves to the right side.
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So we have *v he prime* squared
times mass of the helium squared
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plus mass of the gold
times mass of the helium,
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plus two times mass of the helium squared
times velocity of the helium initially
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times velocity of the
helium after collision.
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So this is the *v he prime*
to the power of one term.
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I didn't do anything to
that, that's just copied.
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Then we have mass of the helium
times velocity of the helium squared,
00:12:58.816 --> 00:13:02.464
multiplied by mass of the helium
minus mass of the mass of the gold.
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That comes from combining
this term with this term
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and this term becomes minus
when it goes to the right side.
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These two terms have a common
factor of *m he* times *v he* squared
00:13:16.042 --> 00:13:20.128
that I've got factored out, leaving *m
he* to the power of one from this term
00:13:20.544 --> 00:13:24.618
and leaving *m au* behind from that term.
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All that equals zero.
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Now we have to plug in some numbers there
to figure out what our coefficients are
00:13:33.552 --> 00:13:38.912
in order to make it look like this
and to use our quadratic formula.
00:13:39.712 --> 00:13:41.829
So we need to know what the speed is
00:13:41.829 --> 00:13:45.134
first of all of the helium
nucleus before the collision,
00:13:45.460 --> 00:13:46.917
and so we know that its kinetic energy
00:13:46.917 --> 00:13:49.084
we're told is eight times ten
to the minus thirteen joules.
00:13:49.370 --> 00:13:53.205
So we'll solve for *v he* then by saying
that's the square root of two times that
00:13:53.482 --> 00:13:54.965
divided by the mass of the helium,
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6.68 times ten to the minus
twenty-seven kilograms.
00:13:57.722 --> 00:14:01.536
This gives 1.54765 times ten to
the seven meters per second.
00:14:02.517 --> 00:14:06.069
So, now we find all of our coefficients.
00:14:06.293 --> 00:14:08.405
So we plug in the mass
of the helium, squared,
00:14:08.768 --> 00:14:12.176
add that to the mass of the gold nucleus
times the mass of the helium nucleus.
00:14:14.384 --> 00:14:16.965
That's going to give this coefficient here,
00:14:18.362 --> 00:14:21.898
and then we have two times mass
of the helium nucleus squared
00:14:21.898 --> 00:14:26.120
times the initial velocity
of the helium nucleus.
00:14:26.730 --> 00:14:33.802
That gives us coefficient here and then
we add to that the mass of the helium
00:14:33.802 --> 00:14:36.087
times the initial speed
of the helium squared,
00:14:36.330 --> 00:14:40.453
times the difference between the
helium and the gold nuclei masses.
00:14:41.194 --> 00:14:44.949
That gives this coefficient
here of negative 5.15714
00:14:44.949 --> 00:14:47.630
times ten to the minus thirty-seven,
and all that equals zero.
00:14:47.950 --> 00:14:50.250
So then we plug into the quadratic formula
00:14:50.661 --> 00:14:54.650
to figure out what the velocity of
the helium nucleus is after collision.
00:14:55.157 --> 00:14:59.280
All of this works out to 1.49 times ten
to the seven meters per second.
00:15:01.408 --> 00:15:07.712
Then, we need to figure out the velocity
of the gold nucleus after the collision.
00:15:08.954 --> 00:15:12.549
So we look at equation
three b which is here,
00:15:12.768 --> 00:15:16.368
and we can take the square root of
both sides to solve for *v au prime*.
00:15:17.104 --> 00:15:22.565
So that's what I did here, and so
we've plugged in mass of the helium
00:15:22.565 --> 00:15:24.757
times initial speed of the helium squared,
00:15:25.008 --> 00:15:29.477
minus mass of the helium times the
final velocity of the helium squared,
00:15:29.856 --> 00:15:32.693
all divided by the mass of gold nucleus.
00:15:33.344 --> 00:15:35.989
We get 6.16 times ten to
the five meters per second.
00:15:37.445 --> 00:15:41.952
Then we need to figure out the
direction of the gold nucleus
00:15:42.890 --> 00:15:46.752
and so we look at equation two
b and one b and we divide them
00:15:49.104 --> 00:15:50.773
and we can get a --
00:15:53.674 --> 00:15:59.178
this will end up being tangent of *theta
au* because these factors will cancel
00:15:59.584 --> 00:16:04.565
and we'll have sine *theta au* divided by
cos *theta au* which is tangent *theta au*.
00:16:06.293 --> 00:16:10.821
I chose to do it this way because we'll
end up having tangent of this angle
00:16:11.130 --> 00:16:15.898
equals some stuff containing only one
of the numbers that we calculated.
00:16:16.528 --> 00:16:18.341
I wanted to minimize the number of
00:16:19.242 --> 00:16:22.150
of quantities that we
calculated in our equation,
00:16:22.150 --> 00:16:27.168
so doing it this way results in only using
the velocity of the helium after collision.
00:16:28.981 --> 00:16:32.736
So we have this is going
to be our numerator
00:16:33.056 --> 00:16:34.645
and we're going to divide
it by this denominator
00:16:34.645 --> 00:16:37.354
and we can see there's a common
factor *m he* everywhere
00:16:37.354 --> 00:16:43.539
so that's going to disappear from down here.
00:16:45.162 --> 00:16:47.893
So, that worked out to this.
00:16:48.645 --> 00:16:52.608
So we have velocity of the helium after
collision times sine 180 minus *theta he*,
00:16:52.860 --> 00:16:57.642
all divided by *v he* plus *v he prime*
times cos 180 minus *theta he*.
00:16:58.528 --> 00:17:02.442
So the angle is going to be the inverse
tangent of the velocity of the helium
00:17:02.442 --> 00:17:07.890
after collision times sine of 60, divided
by initial velocity of the helium
00:17:07.890 --> 00:17:12.084
plus final velocity of the
helium times cos of 60.
00:17:12.810 --> 00:17:14.954
This ends up being 29.3 degrees.
00:17:15.525 --> 00:17:19.637
So the velocity of the gold
nucleus after collision is 6.16
00:17:19.637 --> 00:17:23.831
times ten to the five meters per second,
29.3 degrees below the positive x axis.
00:17:25.178 --> 00:17:28.645
There! If you got this far,
congratulate yourself
00:17:28.645 --> 00:17:30.842
because that was a lot of work to follow.
00:17:32.570 --> 00:17:35.296
If you got the answer correct
on your own then awesome!
00:17:35.296 --> 00:17:39.099
Because you should be trying these problems
before you look at the solutions of course.
00:17:39.914 --> 00:17:43.525
Now the final kinetic energy of the
helium nucleus is going to be one half
00:17:43.525 --> 00:17:45.541
times its mass times its
final velocity squared
00:17:45.941 --> 00:17:49.872
and that works out to 7.38 times
ten to the minus thirteen joules.