For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

### AP Practice Exam: Multiple Choice, Part 2

### AP Practice Exam: Multiple Choice, Part 2

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Problem 36
- Problem 37
- Problem 38
- Problem 39
- Problem 40
- Problem 41
- Problem 42
- Problem 43
- Problem 44
- Problem 45
- Problem 46
- Problem 47
- Problem 48
- Problem 49
- Problem 50
- Problem 51
- Problem 52
- Problem 53
- Problem 54
- Problem 55
- Problem 56
- Problem 57
- Problem 58
- Problem 59
- Problem 60
- Problem 61
- Problem 62
- Problem 63
- Problem 64
- Problem 65
- Problem 66
- Problem 67
- Problem 68
- Problem 69
- Problem 70

- Intro 0:00
- Problem 36 0:18
- Problem 37 0:42
- Problem 38 2:13
- Problem 39 4:10
- Problem 40 4:47
- Problem 41 5:52
- Problem 42 7:22
- Problem 43 8:16
- Problem 44 9:11
- Problem 45 9:42
- Problem 46 10:56
- Problem 47 12:03
- Problem 48 13:58
- Problem 49 14:49
- Problem 50 15:36
- Problem 51 15:51
- Problem 52 17:18
- Problem 53 17:59
- Problem 54 19:10
- Problem 55 21:27
- Problem 56 22:40
- Problem 57 23:19
- Problem 58 23:50
- Problem 59 25:35
- Problem 60 26:45
- Problem 61 27:57
- Problem 62 28:32
- Problem 63 29:52
- Problem 64 30:27
- Problem 65 31:27
- Problem 66 32:22
- Problem 67 33:18
- Problem 68 35:21
- Problem 69 36:27
- Problem 70 36:46

### AP Physics 1 & 2 Exam Online Course

### Transcription: AP Practice Exam: Multiple Choice, Part 2

*Hi everyone! I am Dan Fullerton and I would like to welcome you back to Educator.com.*0000

*In this lesson we are going to continue our work through the 1998 AP Physics B examination for practice.*0004

*Starting at multiple choice question Number 36, we will continue through the end of the multiple choice section.*0011

*Let us dive right in.*0017

*As I look at Number 36 here, we are talking about number of un-decayed atoms as a function of time, and what we can see here is if we are looking for the half-life -- well that is the point where the number of un-decayed atoms has cut in half...*0019

*...so on the graph, that should be pretty obvious to see that that happens right at point A, which is choice A.*0033

*Moving on to 37, we are talking about photons of light that have a frequency and momentum, but what about the momentum if they have a frequency of 2 times their initial frequency?*0042

*Well to do that, let us start by looking at some of the relationships involved. *0055

*We will start with our wave equation, V = F(λ), but we also know that wavelength = H/momentum.*0058

*I could write this as V = FH/P, and now what I am going to do is I am going to try and get all the constants on one side.*0068

*Since it is a photon, (V) is going to be constant, and that is a speed and we also have this H constant, so I am going to write that V/H = F/P.*0077

*Knowing that this is constant, we know that F/P must remain the same, so if V/H = F/P -- well let us take a look here -- V/H which is C/H (the speed of light) must equal F/P.*0090

*You can ignore this part now, C/H = F/P.*0109

*What happens then if we have a frequency of 2F?*0112

*Well if we put a frequency of 2 here and we want this to remain constant, we must have 2 down there.*0117

*So what do we do? We doubled the momentum -- answer choice A.*0123

*All right. Number 38 -- This problem is a little more involved, there is a little more to it than it looks like at first glance.*0133

*So we have the block that has a 3 kg mass and we know what its stretch is when it is at equilibrium.*0139

*First thing, let us use that to find its spring constant.*0145

*For the 3 kg mass that is on there, we know that the spring constant is the force divided by the displacement, so that is going to be mg/displacement, or 3 kg × 10 m/s ^{2} over its displacement which is...*0149

*...let us leave it in centimeters to make life easy -- 12 cm, so that will be 30/12 N/cm.*0166

*Now we are going to replace that by a 4 kg block and when we do that, we release it from its highest position, and it is going to fall and they want to know how far it will fall before it reverses, so it is going to fall to the equilibrium position and then another same distance past that until it stops.*0175

*To figure that out, let us first figure out that equilibrium position -- X = F/K, so that is going to be our force (4 kg) × 10 m/s ^{2}/K (30/12), so I could re-write that as 4 × 10 = 40/30...*0195

*...and I will put the 12 in the numerator and with just a little bit of math here, we have 4 × 12 = 48/3, and that is going to be 16 cm, so that is the new equilibrium position.*0216

*If it starts at the highest point, it falls 16 cm to the equilibrium position and then another 16 cm before it stops and will pass back up, its total must be...*0228

*...How far does it fall? 32 cm and that answer would be choice D.*0239

*Number 39 -- We know an object has a weight (W) when it is on the surface of the planet at a radius (r), so it starts with weight (W).*0250

*What happens to the gravitational force when you quadruple that distance, when it goes from r to 4r?*0258

*Well remember, gravity is an inverse square law. *0265

*If you increase the distance, the force must be going down, and if you quadruple the distance, multiply it by 4 -- well it has squared the effect so that is 16, so you will get 1/16 the initial weight -- Answer E.*0268

*Number 40 -- We are trying to find the kinetic energy of a satellite that orbits the earth in a circular orbit of radius (r).*0286

*Well if it is moving in a circular orbit, that must be caused by a centripetal force which is mv ^{2}/R and what is causing that centripetal force? *0296

*Well the force of gravity, which is G times the mass of our satellite and the mass of our planet divided by r ^{2}.*0307

*I am going to solve to get mv ^{2} by itself, multiply both sides by R and I get that mv^{2} is going to be equal to GmM/R.*0316

*Why did I solve for mv ^{2}? Well I know kinetic energy is just 1/2 mv^{2}.*0328

*So it is going to be 1/2 × mv ^{2} (GmM/R), so our correct answer there must be choice B.*0336

*Number 41 -- We have a couple of objects that are moving parallel to the x-axis and then they are going to collide and we are trying to find out the y-component of the velocity of one of these objects.*0354

*Well first thing I think of when I hear collision is conservation of momentum, and in this case we only have to worry about the conservation of momentum in the y-direction.*0366

*We do not have to worry about the (x) at all because all we are asked about is the y-component of the velocity. *0375

*We know that the initial momentum in the y-direction must equal the final momentum in the y-direction, and since they are both traveling purely horizontally to start the problem, the initial momentum in the y-direction must be 0.*0380

*Therefore 0 must be equal to -- well the final momentum in the y-direction, we have the mass of the .2 kg object and it is traveling with a velocity of 1 up, so that must be added to the momentum of the second object, which has a mass of 0.1 kg and some unknown velocity.*0393

*That is a pretty easy equation to solve: 0 = .2 + .1 V, which implies then that .1 V = -.2 or therefore V = -2 m/s, where all that negative means is it is downward, so the correct answer must be A, 2 m/s downward.*0416

*Number 42 -- We have a beam of white light incident on a prism and it talks about what is going on here, producing a spectrum and as that is filled with air, you get one effect. *0442

*Then what happens -- if you fill it in with water, which has an index of refraction of 1.3 -- well you are still going to get a spectrum. *0454

*You are still going from a lower index to a higher index into the prism and a higher to a lower index as you leave the prism.*0462

*But as you have that smaller change in the index of refraction, you are going to get less refraction; you are going to get less dispersion, and therefore you are going to get a compressed spectrum.*0469

*It is going to take a lot less space than it did before in order to show all of those colors. *0478

*You are going to get less separation between the red end of the spectrum and the violet end of the spectrum, so our correct answer must be E.*0484

*Numbers 43 and 44 have to do with these distance-time graphs and as I look at these distance-time graphs, the first thing I notice is the first graph has a constant slope.*0496

*That means it must have a constant velocity, so constant velocity for graph 1.*0507

*Graph 2 has 0 slope, therefore that must have a 0 velocity.*0514

*Graph 3 has an increasing slope, therefore it must have an increasing velocity.*0520

*In this case you have some acceleration and acceleration does not equal 0.*0526

*Now let us look at the questions.*0531

*Talking about the magnitude of the momentum is increasing in which case?*0533

*Well the mass is not increasing so the only way momentum could increase is if velocity is increasing, so that rules out 1 and 2 and the answer must be 3 only, so for answer 43, the correct answer is B.*0537

*If we go to 44, the sum of the forces is 0, well when the sum of the forces is 0, that means the acceleration must be 0 by Newton's Second Law.*0551

*The only place where the acceleration is 0, is 1, constant velocity, and 2, no velocity, so the answer there must be 1 and 2, or answer C.*0561

*Number 45 -- You have a metal spring inside a uniform magnetic field, and we are looking for a condition that will not cause a current to be induced and if you do not want a current, we cannot have any induced EMF.*0581

*So which of those is going to have some effect where you are not going to have a change in flux because remember the EMF is equal to the opposite of the change in magnetic flux divided by the time interval. *0594

*If you were to change the magnitude of the magnetic field, you are going to change the flux -- it cannot be A.*0606

*Increase the diameter of the circle? -- Well if you do that, you are changing the flux, you have more area.*0611

*Rotate the spring about the diameter? -- You are going to change the flux as you rotate that.*0616

*Moving the spring parallel to the magnetic field? -- Just moving the whole thing parallel to the magnetic field is not going to do anything to the flux.*0622

*I am thinking D must be our best answer, but let us check.*0629

*E, Moving the spring in and out of the magnetic field? -- No, as you do that you are going to change the flux, so if you want to have no current induced, you cannot have any change in flux, therefore that must be 45 (D).*0633

*All right. Let us take a look at Number 46 here.*0653

*We are looking again for 46 and 47 at a magnetic field on a proton beam.*0660

*It says "Of the following, what is the best estimate of the work done?"*0667

*Well let us draw a diagram quickly.*0671

*If there is our magnetic field and we are going to have protons moving in a circle -- let us just draw our circle there and as they do that, the force required for it to move in a circle must be toward the center of the circle.*0673

*But its velocity is at a right angle to that, and if you recall work is F(δ)R cos(θ), but in this case, θ equals 90 degrees so cos(θ) equals 0, therefore the work done must be 0 -- answer A.*0695

*A magnetic field cannot do any work on a moving charge.*0714

*It can change its direction; it can accelerate it but it cannot do work on it.*0718

*Number 47 -- What is the best estimate of the speed of the proton as it is moving in that circle?*0723

*Well the centripetal force must be caused by the magnetic force which is QVB sin(θ) and that must be equal to (M) times the acceleration (mv ^{2}/R), because it is moving in a circle. *0730

*Therefore, we know θ = 90 degrees, so sin(θ) = 1, so we could say that QVB = mv ^{2}/R.*0746

*A little simplification here -- we can divide velocity out of both sides -- solving for V then, I could say that V = QBr/M and substituting in my values, Q (1.6 × 10 ^{-19}), B (0.1 tesla), radius here (0.1 m) and the mass of our proton (1.67 × 10^{-27} kg).*0760

*That looks awfully tough given that you do not have a calculator for the multiple choice part of the question.*0794

*Let us just estimate the order of magnitude.*0799

*Right away let us say 1.6/1.67 and that is pretty close to 1.*0801

*Now we have 10 ^{-19} × 10^{-2}, and that is going to be 10^{-21}/10^{-27}, so I would say that that is going to be around 10^{6}m/s then which will give us answer C.*0807

*You do not really have anything else that is really close to that value, so you could figure that one out without having to go to the calculator.*0826

*Checking out number 48 here -- We have a loop of wire in the plane of a page perpendicular to a magnetic field (B) out of the page.*0838

*And it looks like this is a Lenz's Law problem -- trying to figure out the direction of the induced current in the wire. *0847

*If the magnitude of the magnetic field is decreasing and it is coming out, the induced current wants to create a magnetic field to oppose that, so it is also going out, because that is decreasing; it wants to keep it the same.*0855

*The right-hand rule, as I then wrap the fingers of my right hand around to the direction of that induced magnetic field, I am going to find that this is counterclockwise around the loop or choice D -- a Lenz's Law right-handed rule problem.*0869

*Taking a look at Number 49 -- We have an object making ripples or waves of frequency 20 Hz and it tells us the speed and if the source is moving to the right with some speed, at which of those labeled points will you get a highest possible frequency?*0889

*Well a shift in frequency -- we are talking about the Doppler Effect -- and you get the highest shift in frequency when the source and observer are moving toward each other.*0907

*That must be point C, which you could also see as you look at the diagram -- you can see that C is going to pick up those wave fronts at a higher frequency than the place behind the point.*0914

*(A) would be a lower frequency, for example, but C must be our best answer for the highest frequency.*0926

*And Number 50 -- We have an object in front of a plane mirror.*0936

*Well what do you see when you have a plane mirror? You see the exact inverse image, so what are you going to have there by the Law of Reflection? -- D.*0940

*All right. Number 51 -- I have sound waves of some wavelength incident on two slits in a box with non-reflecting walls and we are going to try and find -- well it tells us the first order maximum, where it is from the central maximum, and we want to find the distance between the slits.*0952

*It is a double slit diffraction problem, therefore, D sin(θ) = M(λ), and what we are really looking for here is D.*0971

*So let us solve that for D = M(λ)/sin(θ), but as you look at your diagram there, sin(θ) is opposite/hypotenuse.*0981

*The opposite side is 3 m, the hypotenuse is 5 m, so sin(θ) is 3/5, so as I plug in my values here I would say that D is equal to -- well, M = 1, our wavelength = .12 and sin(θ) 3/5...*0996

*...so 3/5 down here or I can put that 5 up there -- 5 × .12 = .6/3 or .2 m -- Choice D.*1018

*Number 52 is another PV diagram where we have an ideal gas that corresponds to .1 on the graph and what we are looking for is which of the following relationships are true?*1037

*Is T1 less than T3? -- Well that cannot be the case because it is isothermal, constant temperature, so it is not choice A.*1049

*B -- T1 is less than T2 -- absolutely, as you go up and to the right, you see higher temperatures. *1057

*T2 is greater than T1, so that must be our answer, answer B because T2 is up and to the right, the furthest most up and to the right point on our graph.*1063

*Number 53 is another ideal gas problem. The absolute temperature of a sample of an ideal gas is doubled, but the volume is held constant.*1080

*We know that PV = NRT and we want to know what effect that it is going to have on the pressure and density.*1089

*So I am going to get pressure on a side by itself and pressure = NRT/V, and in this problem itself -- well N is constant, R is constant and V is constant, and what we are going to do is we are going to double the temperature.*1096

*If I double the temperature, multiply the right side by 2, I have to double the left side.*1112

*We must have twice the pressure, which means P doubles.*1117

*How about density? Well density remember is mass divided by volume. *1123

*Mass is not changing and the volume in this problem is constant, that is a given, therefore, density must be constant in this problem, so the choice where we have double the pressure and density remains the same is choice C.*1129

*Number 54 -- This is just a nasty question. *1150

*I am really not sure why they would include something like this on an AP exam, but now that it is here let us just take a quick look at how you would solve something like this without having any background in it.*1154

*We are trying to find the number of atoms that you could stick on the top of a pinhead, which is really an order of magnitude estimation problem.*1166

*Well the radius of an atom is about 1 angstrom or 10 ^{-10} m, just as a very, very, very rough estimate.*1174

*Therefore the area of the atom -- we will call πr ^{2}, which is going to be π × (10^{-10})^{2} or π × 10^{-20} square meters -- rough, rough, rough.*1185

*Let us do the same thing for the pinhead. *1201

*If we want the area of the pinhead, then that is going to be πr ^{2} and if the diameter is 1 mm, half a millimeter is its radius...*1204

*...so 0.5 × 10 ^{-3} m^{2}, is going to give us 25 π × 10^{-8} square meters.*1217

*If we want the number of atoms, I will take the area of the pinhead, which is 25 π × 10 ^{-8} m^{2} divided by one single atom's area (π times 10^{-20})...*1231

*...and π's will cancel out and we have 10 ^{-8} up here, so that will be 10^{-12} down here, so I get something like 25/10^{-12}, which will be 25 × 10^{12}... *1253

*...or about 2.5 × 10 ^{13}, which the closest answer or the only thing even remotely close to that is choice B -- 10^{14}.*1267

*That is how you could estimate something like this, but if this is the problem that gave you trouble, you are doing just fine.*1279

*Let us take a look here at Number 55 -- In an experiment we have light of a particular wavelength incident on a metal surface and we have electrons being emitted.*1289

*This sounds like the photoelectric effect.*1298

*If we want more electrons per unit time, but we want less kinetic energy with each electron, what should we do?*1301

*Well if we want more electrons, that means we need more light waves, more intensity.*1309

*But we also want them to have less kinetic energy, so if we want less kinetic energy that means that the light waves that hit them have to have less energy; those photons have to have less energy, they have to have a lower frequency and if they have to have a lower frequency that means they must have a longer wavelength.*1319

*The correct answer here would be B, increase the intensity and the wavelength of the light.*1339

*Now notice if you increase the wavelengths too much, those photons will not have enough energy to free any photoelectrons, so at a certain point, you keep reducing that wavelength, and you are not going to get any emitted photoelectrons. *1346

*Number 56 -- All right, we have an object moving up and down with an acceleration given as a function of time by A = A sin(ω)T, where A and ω are constants, so what is the period?*1361

*Well, period -- just right back from simple harmonic motion -- is just 2 π/ω -- one of our formulas, so there is our choice right there.*1381

*All we have to really do is know the formula for 56.*1391

*Number 57 -- we have a ball at rest and is then kicked.*1397

*It is given some speed and a direction and we want to find the impulse imparted to the ball by the foot.*1401

*Well, impulse is change in momentum, and change in momentum is going to be mass × change in velocity.*1406

*Our mass (0.4 kg) and our change in velocity from 0 to 5 m/s is just going to be 5 m/s or 2 N-s, so the answer is C.*1414

*Number 58 -- We have a wheel with a couple masses on it and we want to know what the masses are so it remains in static equilibrium, so that it is not rotating about that center point.*1430

*That sounds like a net torque equals zero problem, a rotational equilibrium problem.*1440

*If our net torque = 0 -- well our net torque we have from the left-hand side, that (m) -- we have a torque of mgR, from the (M) on the upper left...*1445

*...that torque is going to be MgR × cos(60 degrees) and causing the clockwise rotation -- so a negative torque -- on the right-hand side we have 2MgR.*1461

*And again all of that must be equal to 0, so really what we are doing here is we are solving for (m).*1477

*This implies then that -- let us say -- we can do some cancellations here.*1482

*We have (g) and (R) in all of these so gR, gR, gR -- so this implies then that m + M cos(60 degrees) - 2M = 0 or m = 2M - M cos(60 degrees).*1489

*But we know the cos(60 degrees) is 1/2, so m = 2M - M/2 or 3/2(M) -- answer C.*1514

*Number 59 -- We have a rock thrown horizontally off a building from some height (H).*1535

*The speed of the rock as it leaves the hand horizontally is V0.*1542

*How much time does it take to travel to the ground?*1546

*To travel to the ground -- we do not care how fast it is going horizontally, that is a vertical kinematics sort of problem.*1550

*Vertically we know that the initial velocity in the y-direction is 0, we know the change in displacement, the change in distance, or the distance traveled is (H), and its acceleration is going to be G, assuming we are calling down the positive y-direction.*1556

*In order to find the time, I would use the equation δy = V0t +1/2at ^{2}, and since V0 is 0, that term goes away...*1573

*...and I can rearrange this to say that T = 2H/G square root where H was my δy and (a) was my G, so the answer there must be E.*1586

*Now Number 60 -- a connected problem, says what is its kinetic energy just before it hits the ground?*1605

*All right, well the kinetic energy -- we are going to have kinetic energy from its horizontal velocity and its vertical velocity.*1612

*We could do this by conservation of energy if we wanted to make it really easy, because its horizontal velocity is not changing, so the contribution horizontally is going to be 1/2 mv0 ^{2}.*1619

*Now vertically, it initially starts with potential energy mgh, right?*1632

*Well when it gets down to the ground, what is its kinetic energy going to be?*1637

*All that potential has become kinetic, so vertically that has become (mgh), so kinetic energy is the horizontal contribution plus the vertical contribution.*1641

*And of course you could go through and solve for the velocity vertically, plug it into 1/2 mv ^{2} and work through it that way, but the easy way -- we have kinetic energy horizontally and you have this contribution from the vertical component.*1653

*By conservation of energy, you are all set right there, answer D.*1667

*Number 61 -- We are looking for which of the statements is not the correct assumption about ideal gases.*1676

*Well molecules are in random motion -- Yes, that is one of our assumptions.*1682

*The volume of the molecules is negligible compared to the volume occupied by the gas, or you have a lot of free space in there -- Yes.*1685

*The molecules obey Newton's Laws of Motion -- I certainly hope so. Yes.*1694

*The collision between molecules are inelastic -- Absolutely not. *1698

*We have to assume that all the collisions are elastic, so the correct answer here, the one that is not an assumption that we need to be in ideal gas is D.*1702

*Number 62 is another ideal gas problem. *1715

*Here we have an ideal gas in a tank at constant volume; it is going to absorb heat energy so that its temperature doubles and we know V1 is its initial average gas molecule speed and V2 is the speed after it absorbs that heat.*1717

*Let us find the ratio of V2/V1. *1733

*First thing is let us figure out what happens to the kinetic energy when we double the temperature.*1736

*We know that the average kinetic energy is 3/2 times Boltzmann's constant times the temperature. *1740

*If we double the temperature, we have to double the kinetic energy, so K average is doubling.*1748

*Now we also know kinetic energy is 1/2 mv ^{2} and if we are looking for velocity, we could rearrange that to say that velocity is 2K/m square root.*1757

*So what do you actually do if you want to put a 2 in here if you double the kinetic energy?*1770

*Well you have multiplied the right side by square root of 2, so you have to multiply the left side by square root of 2, so really what do we do? *1775

*We have square root of 2V, so the correct answer must be C.*1783

*All right, moving on to Number 63 -- We have two people, unequal mass, they are still on ice, they push each other and they go off in their separate directions.*1791

*What is true after they pushed each other?*1805

*The kinetic energies are equal? -- No, it does not have to be.*1808

*The speeds of the people are equal? -- Not necessarily.*1811

*The momenta of the two people are of equal magnitude? -- Yes, that is the Law of Conservation of Momentum, so C must be correct.*1815

*Let us take a look at Number 64 -- We have two parallel conducting plates separated by a distance (d) and they are connected to a battery of some electromotive force.*1826

*What is correct if the plate separation is double while the battery remains constant?*1837

*We are looking for what happens to the electric charge or potential difference or capacitance.*1843

*Well let us start -- if they are parallel plates, the capacitance is ε0 A/d, and if we are doubling the distance, we have to cut the capacitance in half, so we know that C is halved.*1848

*Now if that is the case, we also know C = Q/V, therefore the charge equals CV.*1863

*If we cut C in half, what happens to Q if we are holding voltage constant?*1870

*We must have cut Q in half or we halved the charge, so the correct answer for 64 would be B, the electric charge on the plates is halved.*1877

*Number 65 -- We have some concentric loops that have different radii, B and 2B, so one has twice the radius of the other.*1891

*What is the resistance of the wire loop with the bigger radius if the resistance of the one with the initial radius is R?*1901

*Well if you double R and circumference = 2 π × R -- if you double R, you double the circumference, therefore what you are doing is you are doubling the length of your wire.*1909

*Now resistance, if you recall, is resistivity times length divided by area.*1924

*If we have doubled the length, we must have doubled the resistance, therefore our answer is D.*1931

*Number 66 -- We have a uniform magnetic field perpendicular to a plane that is passing through a loop and we have two different radii of wires and the induced EMF in the wire loop of radius B is given.*1942

*What is the induced EMF if we have twice the radius?*1956

*Well remember, EMF is the opposite of the change in magnetic flux divided by time, but as I look at this problem, the changing flux is the same in both loops.*1969

*As you go from the loop of radius B to the radius of 2B, there is no difference in the amount of flux or in the change of magnetic flux, therefore, it is going to be the same.*1973

*The correct answer is going to be C. *1980

*There is your induced EMF in the wire loop of radius 2B, the same as you have in the wire of loop radius B.*1988

*Number 67 -- We have a stationary object exploding into three pieces of different masses, and we know when they explode that we have a couple pieces moving off at right angles to each other and it gives us their momentum.*1998

*We want to find the magnitude and direction of the piece that has the bigger mass.*2011

*Well if they started at rest and you had one piece that went that way, one piece that went that way, by conservation of momentum, your other piece must have gone that way so that their momentum all tally up.*2015

*So our direction is figured out.*2029

*Now to get its magnitude, let us use conservation of momentum.*2031

*In the x-direction, we know that the initial momentum 0 must equal the momentum of the pieces which is (mv) to the right plus 3m times our VX prime...*2034

*...the x-component of the velocity of our larger piece, so therefore VX prime must be equal to -V/3.*2051

*And if we did this in the y-direction, we are going to get the same math, only to find out that V prime Y equals -V/3 and we could do that by symmetry as well.*2064

*So what is the total V prime? *2072

*Well to do that, we have to realize that our x-component is -V/3 and our y-component is -V/3 so we can use the Pythagorean Theorem to figure out the total.*2075

*That is going to be the square root of -V/3 ^{2} + -V/3^{2}, which will give us the square root of V^{2}/9 + V^{2}/9...*2088

*...which is going to be the square root of 2V ^{2}/9 or square root of 2/3V.*2102

*The correct answer must be D.*2113

*Number 68 -- We have a rod on a table top pivoted at one end and it is free to rotate without friction.*2121

*We are applying a force at the end at some angle and what we want to do is see where would we have to apply that force if it were perpendicular?*2129

*That means that what we really want is we want the same torque.*2137

*All right, well, our torque is FR sin(θ), which implies -- well the initial torque when we apply (F) at the end of the rod, and that is our force (F), our arm is (L) times the sine of that angle θ.*2140

*And that is going to be equal to the torque if we move that force in, so if we apply the same force perpendicularly, but at some distance (x), (x) is going to tell us at what position we have to apply that.*2161

*If I solve for (x) and if I divide both sides by (F), I find that (x) is going to be equal to L sin(θ), so the correct answer is A.*2172

*Number 69 -- Which of the following imposes a limit on the number of electron in an energy state of an atom?*2187

*Well this one you just pretty much have to know. That is going to be B, the Pauli exclusion principle.*2193

*It says you can only have a certain number of electrons at each energy level.*2200

*And Number 70 -- It gives us a capacitor and its charged to a set potential difference. *2205

*Find the energy stored in a capacitor.*2212

*This looks pretty straightforward. The energy in a capacitor is 1/2 CV ^{2}.*2214

*That will be 1/2 × 4 × 10 ^{-6} (our capacitance in farads) × 100 volts^{2} or 1/2 × 4 = 2 × 10^{-6} × 100^{2} = 10^{4}...*2221

*...so I am going to get 2 × 10 ^{-2} and the units of energy are joules, so the correct answer there is E.*2242

*All right, so we have gone through the multiple choice portion of the test.*2253

*Check your answers, see how you did and if there is anything you need to review, go back, take some time, check it out and then in our next session, we will start on the free response portion of the test.*2258

*Thanks so much for your time everyone and make it a great day.*2267

3 answers

Last reply by: Professor Dan Fullerton

Sat Jul 9, 2016 6:15 AM

Post by Peter Ke on July 7, 2016

For problem 45, how does rotating the spring about a diameter causes a change in flux?

1 answer

Last reply by: Professor Dan Fullerton

Sun May 5, 2013 5:40 AM

Post by Saki Amagai on May 4, 2013

For #65, shouldn't the area of the wire also increase, if the radius increases? cuz a=pi r^2