WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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I am Dan Fullerton and today we are going to finish off the free response section of the 1998 AP Physics B examination.
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We will start with problem 5 and we will work right through to the end, so let us dive right in to question number 5.
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Now number 5 has to do with this apparatus, where we have a tuning fork, a string, a pulley, and a hanging mass.
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For Part A, it wants us to determine the wavelength of the standing wave and you could probably just see that by looking at the diagram that that is going to be 0.6 m, but it is usually a good idea to try and justify your answer on these exams.
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I would write something like wavelength is equal to the total length divided by 2 because we have two full waves in that region, so that is going to be 1.2 m/2 or 0.6 m, and that should be plenty sufficient for that part of the problem.
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Now for B, it says determine the speed of the transverse wave along the string.
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Well, we are experts by now using the wave equation, V = F(λ).
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The frequency it gives us is 120 Hz and we just determined the wavelength is 0.6 m, so that is just going to be 72 m/s.
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For C, it tells us that the speed of the wave increases with increasing tension in the string, so what should we do to the value of that mass -- increase it or decrease it if we want to have double the number of loops in the standing wave pattern?
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To start that, I realize that if we want double the number of loops, we need the wavelength to go way down and frequency remains the same, so if wavelength is going way down, we need velocity to decrease.
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Velocity must decrease for wavelength to decrease...
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...which implies then that the mass must decrease in order to reduce the tension and our key answer here is mass must decrease.
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Part D -- If a point on a string and an anti-node moves a total vertical distance of 4 cm during a cycle, what is the amplitude of the standing wave?
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This is a very tricky question just because of the wording, not usually the concept behind it, but here is what they are saying.
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If there is our equilibrium point, here is our standing wave at one point in time -- that portion of it and the other portion down here.
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What they are really saying is as this point, the anti-node, travels here and down there, back up to the point where it started -- that entire distance it has traveled is 4 cm, but the amplitude is just that amount, that displacement.
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To solve this, I would say that that 4 cm is 4 amplitudes, therefore your amplitude must be 1 cm.
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The tricky part there is reading that question carefully.
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Moving on to the next problem, Number 6 -- We have a heavy ball at the end of a light string making a pendulum and we are talking about point (P) when it is at its lowest point and (Q) when it is at its highest point in its path as a pendulum.
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For Part A1, it wants us to draw on the diagram the direction of the acceleration and velocity vectors when it is a point (P), when it is at its lowest point.
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When it is at its lowest point here in its path, what does that diagram look like?
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If there is our object -- I would say since it is moving in a circular path, at this point, the acceleration must point toward the center of the circle.
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The velocity, assuming it is moving to the right across the screen -- if that is the case -- at that specific instant in point, its velocity vector would probably look like that.
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Now for A2, we are going to do the same thing for point (Q), but at point (Q) now, we are looking at our pendulum when it is at its highest point over here.
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Here is our point (Q), so now over here -- well at its highest point there for a split second it stops, so velocity is easy because velocity is 0.
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The acceleration -- well the direction it is going to move next is going to be in that direction.
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If you remember when we started talking about pendulums in earlier videos, we even calculated what that acceleration would be, what the force would be that would cause it.
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So that gets us through Part A.
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Now for Part B, it says after several swings, the string breaks.
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Assuming the mass of the string is negligible and the air resistance is negligible, sketch the path of the ball if it breaks at point (P) or (Q).
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If it breaks at point (P) here when it is at its lowest point, what is it going to do from there?
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It is going to become a projectile; it is going to follow a projectile path -- projectile motion or parabolic path.
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But what happens if the string breaks when it is over here at position (Q) when it is at its highest point?
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That will be (B2) and if that happens when it is at its highest point, it is completely stopped.
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If the string breaks, it is just going to fall; it becomes a free-fall object, so in that case there is our object and it is going to enjoy a little bit of a trip downwards or free-fall.
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That is all there is to that one. Excellent!
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Number 7 -- Now we are taking a look at an optics problem where we have a diffraction grading with 600 lines/mm and we are studying the line spectrum of light produced by a discharged tube with that setup.
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It tells us the distance of the grading to the screen and it also says where the observer sees that first order of red line at a distance Yr, which is 428 mm or 0.428 m from the hole.
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It wants us to calculate the wavelength of the red line in the spectrum.
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The first thing I am going to do is I am going to figure out the separation between those slits (d).
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If we have 600 lines/mm, then (d) is going to be -- well we have 1 mm over those 600 lines, which is 0.001 m/600 or 1.67 × 10^-6 is going to be our separation distance.
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If we have (d)sin(θ) equal to m(λ) times m(λ) then our wavelength (λ) is going to be (d)sin(θ)/m...
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...but sin(θ) here, we do not know off hand, but sin(θ) is the opposite of the hypotenuse.
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The opposite side is Yr and our hypotenuse, we will write there.
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As I look at our hypotenuse, I can get that from the Pythagorean Theorem and our hypotenuse is going to be the square root of...
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...Well it is our length squared plus Yr², so then I could write that our wavelength is going to be equal to d/m × sin(θ), which I am going to write as Yr/L² + Yr² square root.
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Now I can start to substitute in my values and say that λ is going to be equal to -- well our (d) we figured out is 1.67 × 10^-6 m from right there...
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...divided by m, and since we want the first order, (m) is 1 × Yr, which is 0.428 m, divided by the square root of...
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...Well, L² is going to be 1 m² + Yr², which is 0.428 m² and with a little bit of calculator work, I can show then that the wavelength is 6.57 × 10^-7 m or 657 nanometers.
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We have our first answers there for Part A.
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Moving on to Part 7B -- It talks about the Bohr model and energy levels, but what we are trying to find now is the value for (n), for the initial level of the transition.
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To start I am going to figure out what energy this is going to correspond to, so our energy is (hf) -- but we also know that V = f(λ), which is also equal to (c), so I could replace frequency with c/(λ) to say that E is hc/λ...
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...which is Planck's constant 6.63 × 10^-34 × 3 × 10^8 m/s (c)/6.57 × 10^-7 m (wavelength), so that my energy is 3.03 × 10^-19 J.
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Let us convert that to electron-volts so that we are consistent.
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We have 3.03 × 10^-19 J and we want to convert that to electron-volts.
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We know 1 eV is 1.6 × 10^-19 J and my joules will cancel out and I am left with about 1.89 eV.
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Now to find the energy of that photon is the initial energy minus the final energy, which is going to be -13.6 eV over our initial (n)² minus...
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...and our final energy is -13.6/2², since we are dropping to the n = 2 level.
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So we have to find what ni² is equal to because the energy of our photon is 1.89 eV, so now it becomes an exercise in Algebra to solve for ni.
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First I would write -13.6/ni² and I have two negatives, so that is going to be +13.6/2², which turns out to be 3.4 must equal 1.89...
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...therefore -13.6/ni² equals -- subtract 3.4 from both sides and I get -1.51 eV on the right hand side or with a little bit of cross-multiplication, ni² = -13.6/-1.51...
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...which is right about 9, so if ni² = 9, then that means ni must be equal to 3.
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We must be dropping from the n = 3 level to the n = 2 level.
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Finally, Part C -- Qualitatively we want to describe what is going to happen to that first order red line if the diffraction grading is switched to one that has 800 lines/mm instead of 600 lines/mm.
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If we do that, we are going to have a smaller separation between those lines and we are going to end up with more diffraction, so Yr is going to increase.
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Last question on the test -- Number 8 -- Here we have a long straight wire in the plane of the page and it is carrying a current (i) and point (P) is in the plane of the page and above it some distance (d).
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We are going to ignore gravity, but with reference to that coordinate system, what is the direction of the magnetic field at (P) due to the current in the wire?
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That is a right-hand rule problem.
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Take your right hand, point the thumb of your right hand in the direction of the current flow through the wire and wrap your fingers around it, and up above that wire, you should see that you are going into the page...
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...or from your perspective, current to the right, wrap the fingers around and up above that wire, so that your fingers point into the page, so your direction must be into the page, which is also -Z by the coordinate system given here, due to our right-hand rule.
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For Part B -- Now we have a particle of mass (m) and the positive charge (q) is traveling as well, with some speed V0, when it is at point (P).
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With reference to the coordinate system, what is the direction of the magnetic force?
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Another right-hand rule problem, so take the fingers of your right hand, point it into the direction of the velocity of the particle and then bend them into the direction of the magnetic field into the page.
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When you do that, your thumb should point down.
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Try it out again -- right hand in the direction of positive particles moving, bend your fingers into the page in the direction of the magnetic field and when you do that from your perspective, your thumb should point down.
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The direction of the force on that is going to be down or from the coordinate system given here, that is -Y, again due to right-hand rules.
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Now for Part C -- Find the magnitude of the magnetic force acting on the particle at (P) in terms of the given quantities.
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The magnetic force is (qvB)sin(θ), but as we see in this problem, θ equals 90 degrees, so sin(θ), which is sin(90) is going to equal 1, so we can just write that the magnetic force is (qvB).
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But what is (B)? Well the magnetic field strength due to a wire (B) is μ0I/2πr.
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In this problem, we also know that v is v0, which is a given and our (r), or our distance from the wire is (d).
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I can put that all together to say then that the magnetic force (FB) is going to be...
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...we have (q) and instead of (v), I will put v0 and instead of (B) we will have μ0I/2π, and instead of (r), we have (d).
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So the magnetic force is qv0μ0I/2πd.
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On to Part D -- An electric field is applied that causes the net force on the particle to be 0 at that point.
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In that case, if we have the magnetic force down, the electric field must be providing a force up.
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So what is the direction of the electric field that could accomplish that?
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Well, if you want the force up on a positive particle, you must have the magnetic field go up, so up would be our direction or we could write +Y, because it must oppose the magnetic force.
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Part E -- Find the magnitude of the electric field now.
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Well, remember the electric field must balance or match the magnetic field in terms of magnitude, so that is going to be in magnitude 2v0μ0I/2πd, but we do not want the electric force, we want the electric field.
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The electric field is the electric force divided by the charge, so that is going to be our electric force, qv0μ0I/2πd...
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...and we still have to divide out that (q), so those will make a ratio of 1 and that is going to leave me with my answer of V0-μ0I/2πd.
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That concludes our examination of the 1998 AP Physics B test.
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Hopefully you got a great start and these are all starting to come together and make sense.
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Thank you so much for your time and make it a great day everyone!