Physics Practice Problems (PH212)

Another student and I were quizzing each other using Google Talk I decided to post the questions, answers, and solutions here for everyone to use.

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#### By number (in numerical order):

A. Relativity_____1. Light______2. Sound______3. Sound and Wave Motion____4. Sound______5. Sound______________ 6. Sound and Wave Motion____7. Rotation____8. Fluids______9. Fluids______10. Oscillations________________________ 11. Fluids and Oscillations____12. Relativity__13. Rotation__14. Rotation___15. Light_______16. Light______________ 17. Oscillations

#### By topic (in alphabetical order):

Fluids: 8,9,11. Light: 1,15,16. Oscillations: 10,11,17. Relativity: A,12. Rotation: 7,13,14. Sound: 2,3,4,5,6. Wave Motion: 3,6.

A. Dennis didn't really solve the last question (9) on the first mid-term, so I did it here:

The question is that something happens on Earth and the Sun at the same time as seen from Earth, but aliens traveling toward Earth from the Sun say it's 5 seconds apart, so what's the velocity of the aliens? The distance from Sun to Earth is 1.496·10^{11}meters.

The Lorentz transformation for time (eq 37-5, page 934, discussed in section 37-8, page 932) is shown at left. Since two events occur, we'll need two times and two x-coordinates, let's call them t_{1} and t_{2}. The Lorentz transformations for them are at left and left too. Let the primes be from Earth's reference frame, so that the time seen by the aliens will be t_{1}-t_{2}=5 seconds, the time seen on Earth will t_{1}'-t_{2}'=0, and the Sun-Earth distance will be x_{1}'-x_{2}'=1.496·10^{11} meters, which I'm going to call d for convenience (who wants to type 1.496·10^{11} over and over again?). Now we just plug in the Lorentz transformations for t_{1} and t_{2} shown at left into t_{1}-t_{2}=5, which yields the equation shown below. Notice that γ is the same for both, because v is the same, so we can

just factor out γ, yielding the equation shown at right, where we substitute in t1'-t2'=0 and x1'-x2'=1.496e11=d. Now we get γ·v·d/c²=5, which is easily solved by bringing c² to the other side (5·c²=γ·v·d) and plugging in γ=1/√(1-v²/c²), bringing √(1-v²/c²) to the other side too (5·c²·√(1-v²/c²)=v·d), then squaring both sides (25·c^{4}-25·v²·c²=v²·d²). Now you bring 25·v²·c² to the other side and factor out v² (v²·(d²+25·c²)=25·c^{4}), and you (should) get **v=5·c²/√(d²+25·c²)=3·10 ^{6}m/s=.01c**.

1. How far below the surface of a swimming pool can you be, and still see light from a 2 m depth marker on the wall 4 m in front of you, be totally internally reflected from the surface? Assume the surface of the water is completely flat and level. The index of refraction of water is 1.33 and that of air is 1.

Answer: [1.5m=2m, depends on the sigfig] Solution: [The critical angle (theta-c) for total internal reflection is the arcsine of the index of refraction of air divided by that of water (if you don't know this already see 33-7, page 826), theta-c=arcsin(1/1.33). Light hits the surface a certain distance from the wall, a distance we'll call d. By standard trig, tan(theta-c)=d/2m (remember theta-c is the angle from the normal, for more info see 33-5, page 822). You are 4 m from the wall, so if the light hits the surface d from the wall, you're 4m-d from where the light hits the surface. The angle of reflection equals the angle of incidence, so trig again tells us that tan(theta-c)=(4m-d)/h, if we let h be the depth we're looking for. We have two equations and two unknowns, solve for h and we get 1.5m]

2. Annoyed at the incessant 1000Hz tone his broken computer was making, a technophobe kicked one of his wireless speakers away from him. He then heard a beat frequency of 1Hz. How fast did he kick the speaker? The speed of sound is 343m/s. <Quote from virtual equation sheet: f'=f( (v - vo)/(v + vs) ) when observer and source are moving away from each other.>

Answer: [0.3m/s] Solution: [The beat frequency (16-6, page 429) of 1Hz is means that due to Doppler shift (16-7, page 432), the speaker he kicked seemed to shift to a frequency 1Hz lower or higher than 1000Hz. Since the speaker was kicked away from him, the source and observer are moving away from each other, the vo=0, and vs is what we're trying to find. Since they are moving away from each other, f'<f=1000Hz, so f'=999Hz. Now we just solve for v-s, and get 0.3m/s.]

3. A physics student wrote a program to sound tones from 1000Hz to 2000Hz, and found that a 0.25m long open tube resonated at 1200Hz and 1800Hz. How fast does sound travel in the tube? <Quote(s) from virtual equation sheet: fn=v/lamda-n=n·v/(2·L)=n·f1 and L=n·lambda-n/2>

Answer: [300m/s=3.0e2m/s] Solution: [We know that the fundamental frequency f1=1800Hz-1200Hz=600Hz(6.0e2 Hz), since it's an open tube (16-4, page 424, and 15-9, page 406). The length of an open tube is always half the wavelength of the fundamental frequency (also in 16-4 and 15-9, pages 424 and 406, respectively), so lambda1=2·L=0.50m.v=f·lambda=600Hz·.50m=300m/s=3.0e2 m/s]

4. The next day, that student bet his friends that he could make a 45 degree cone blindfolded. He then used a homemade railgun to shoot paper clips (it can't shoot anything else) through the same tube as in above. How fast should he shoot the paper clip so it makes a shock wave cone with an edge 45 degrees from the direction of travel of the paper clip? Assume that the speed of sound in the tube is the same as you calculated in the last problem. <Quote from virtual equation sheet: sin theta=speed of sound/speed of object (16-8, page 435)>

Answer: [420m/s] Solution: [Speed of object=speed of sound/sin(45 degrees)=300m/s·2/sqrt(2)=420m/s]

5. The student cannot actually directly control the speed of the paper clip. Instead, he uses yet another computer program (imagine his electric bill) to use a high-speed webcam to take pictures of the paper clip and measure the angle of the shock wave cone outside the tube, then he adjusts a potentiometer on the railgun accordingly. What angle does he try to get the shock wave cone outside the tube to be? The speed of sound oustide the tube is 343 m/s. When asked why he didn't just make the 45 degree cone outside the tube in the open air, he explained that it was so complex this way if he made a mistake his friends wouldn't know anyway.

Answer: [55 degrees] Solution: [When the paper clip moves at 420m/s, the speed he wants, the angle of the cone outside the tube will be arcsin(343/420)=55 degrees]

6. A tuning fork of frequency f establishes a resonance with a closed tube of length L (closed on one end). The experimenter increases the frequency until another frequency, f', is found to establish a resonance with the tube. Find the speed of sound in the tube as a function of f, f', and L.

Answer: [2L·(f'-f)] Solution: [Because the tube is closed, the fundamental frequency is half the difference between any two adjacent frequencies (16-4, page 425), or f1=f'-f. The length of the tube L is twice the wavelength of sound at the fundamental frequency, lambda1=2L. v=f·lambda=2L·(f'-f)]

7. A uniform sphere of mass 1.0kg and radius 2.5cm rests on a piece of paper, 20cm (two sigfig, as in 2.0e1cm) from the edge of the paper. You yank on the paper with 1.0N of force in a direction perpendicular to the edge, so that the sphere rolls all of 20cm off the paper (from the reference frame of the paper). How long does it take the ball to roll off the paper? Assume that the 1.0N of force is constant across the time it takes the sphere to roll off the paper. Ignore the mass of the paper and any frictional forces on the paper from the surface the paper is resting on, ie assume the force on the sphere is 1.0N all the way (no slipping, either). <Quote from virtual equation sheet: Moment of inertia of uniform sphere=2/5·M·r^2>

Answer: [0.40 seconds] Solution: [Long time no see, eh? Remember the kinematics equations from PH211? Dennis doesn't have you memorize em but I think you should, and anyhow you should know how to integrate a=a. v=v0+v·t, and x=x0+v0·t+0.5·a·t^2, simple integration (if you have no idea what I'm talking about, better go back and review Chapter 2, or if you just don't remember how to get the equations, see 2-5, page 26). Then at the beginning of this term, we learned that under two conditions, those same equations apply to rotation if we substitute theta for x, omega for v, and alpha for a. Those two conditions are when the axis is fixed relative to any inertial reference frame or when the axis of rotation goes through the center of mass, which it does, so those kinematics equations are valid here (10-2, page 243, although you may want to reread the whole Chapter 10 starting from page 239). Also, torque=R cross F=I·alpha, remember? (If not, see 10-5, page 247, and you may want to keep reading til you finish the chapter.) So alpha=(R cross F)/I. Now, R cross F=r·F, if we let r be the radius of the sphere and F be the 1.0N force (10-5, page 243), and I=2/5·M·r^2, so after canceling the r's we get alpha=5·F/(2·M·r). Plugging this into the kinematics equations, and letting theta0 be zero and, since they start from rest, omega0 to be zero, theta=0.5·alpha·t^2=t^2·5·F/(4·M·r). What do we want theta to be? We want the sphere to roll 20cm, and it has a radius of 2.5cm and hence a circumference of 2·pi·r=5.0·pi·cm. 20cm is thus 4/pi revolutions, and each revolution is 2pi radians, so we want theta to be 8 radians. So 8=t^2·5·F/(4·M·r), we know F, M, and r, solve for t and we get t=0.40seconds]

8. A clean energy company wants to generate hydroelectricity not by damming an entire river, which would destroy the river's ecosystem, but by collecting rainwater in a giant tank. They construct a cube shaped tank with a length S to each side, and after it fills up with rainwater, they open a valve at the bottom that leads to a pipe of radius R that is level along it's entire length. Given that the pressure in the pipe is P, how fast is water in the pipe flowing? Ignore the compressibility of water and the fact that the water level at the top of the tank gradually drops. The density of water is rho. <Quote from virtual equation sheet: P+0.5·rho·v^2+rho·g·y=constant>

Answer: [sqrt(2·( (1atm-P)/rho + g·S ))] Solution: [That equation is Bernoulli's equation (eq 13-8, page 346, see section 13-8, page 345). It is typically written as P1+0.5·rho·v1^2+rho·g·y1=P2+0.5·rho·v2^2+rho·g·y2, but I wrote it the way I did above because it'll throw you off and it's shorter to write. Anyhow, lets have the 2's be at the top of the tank and the 1's be at the pipe, so that P1=P and P2=1atm=1.013256e5Pa, since the tank has to be open to the air to be able to collect rainwater, and we want to find v1. Let y1 be zero, so y2=S, and since we're ignoring the fact that the water level at the top of the tank gradually drops, v2=0. Now we have P+0.5·rho·v1^2=1atm+rho·g·S, and we just solve for v1.]

9. A 2.0cmx3.0cm business card weighing 4.0g is suspended on a thread with negligible mass. You blow right next to it with a velocity of 0.50m/s for 0.50s. How fast does the business card move (at first)? The density of air=1.29 kg/m^3. <Quote from virtual equation sheet: P+0.5·rho·v^2+rho·g·y=constant>

Answer: [0.048m/s] Solution: [During the half a second of blowing, the pressure on either side of the business card goes from equal (both 1atm) to unequal, one side still 1atm but the other side less (13-8 and 13-9, pages 345 and 347), so the card is pushed to the side. Since the air on either side is of equal height, the rho·g·y on either side cancels, so we have P1+0.5·rho·v1^2=P2+0.5·rho·v2^2. Let the two's be on the side we're blowing, so that v1=0 and v2=0.50m/s, and the pressure difference between the two sides is P1-P2=0.5·rho·v2^2=0.5·1.29·0.50^2=0.16 Pa. Now we're halfway there. Pressure=force/area, so force=area·pressure, and force=impulse·time, so impulse=force/time, and since impulse is change in momentum, when momentum starts at zero as it does here, impulse=resulting momentum. Let v=speed of card, m·v=momentum=impulse=force/time=area·pressure/time, so v=2.0cm·3.0cm·0.16Pa/(0.50s·4.0g)=0.048m/s]

10. As a continuation of the problem above (9), determine the angle theta the thread makes with the vertical as a function of time (or an approximation thereof) if the thread is 20cm (2.0e1cm) long, assuming it all started from rest (let theta=0 when t=0). <Quote from virtual equation sheet: x(t)=A·cos(omega·t+phi) and omega=sqrt(k/m) and omega=sqrt(g/L) and k=mg/L> Hint: Not all these equations are needed.

Answer: [0.0069·sin(7.0·t)] Solution: [There are several ways to do this, but the first that occurs to me is cons. of energy. You should know that total E=0.5·k·A^2=0.5·m·vmax^2 (14-3, page 369), but since k=mg/L, and the m's cancel, g·A^2=L·vmax^2. We know g, L, and vmax (the answer to number 9), so all we gotta do is solve for A=vmax·sqrt(L/g) and plug it into x(t)=A·cos(omega·t+phi). This would then be theta(t)=vmax·sqrt(L/g)·cos(sqrt(g/L)·t+phi), and since theta=0 when t=0, cos(phi)=0, so phi=pi/2, which would make theta(t)=vmax·sqrt(L/g)·sin((sqrt(g/L)·t)=0.048·sqrt(0.20/9.8)·sin(sqrt(g/L)·t)=0.0069·sin(7.0·t)]

11. A cylinder of radius r floats in a cup of perfectly still fluid in such a way that it's always upright. If you were to push it straight down a distance d, but not so much you don't push it down so much it's totally submerged or shoots all the way out of the fluid or touches the bottom or edges of the cup, it would come back up, right? And then because of inertia, it would come back *past* where it floats perfectly, right? And if there is no viscosity friction etc, it'll keep doing this, bobbing up and down, that is, oscillating back and forth around an equilibrium point, right? But is this motion *simple* harmonic motion? Explain why you do or do not think so and if you do think so, provide a value for k (the proportionality constant of the restoring force, not the wave number) in terms of properties of the fluid and the cylinder, such as density of the fluid (rho), gravitational acceleration (g), volume of cylinder (V), etc. Assume that the radius of the cup is so close to the radius of the cylinder that you can ignore complications resulting from waves at the surface. Hint: Review Chapter 14 for the definition of simple harmonic motion. Somehow this strikes my as just the type of obscure concept that Dennis would put on the test. And don't let the length of the solution and answer scare you, I just do a lot of talking.

Answer: [Yep, and k=rho·g·pi·r^2, where rho=density of fluid, g=gravitational acceleration, pi=irrationally annoying irrational number that is the circumference of a circle divided by it's diameter, and r=radius of cylinder.] Solution: [The key is that the definition of simple harmonic motion is that the restoring force to the equilibrium point is directly proportional to the displacement (F=k·x, F~m·g·theta), something I suspect that, in spite of slight emphasis on this when the pendulum was introduced, passed over everyone's heads (except the other guy who got 50). This is also why I bet Dennis will put this on the test, just like he wanted to make sure everyone was clear that constant velocity implies that net force equals zero, something he said used to pass over almost all physics student's heads, although I could be wrong since it wasn't covered on either midterm and I don't recall him ever mentioning it. So, now that we have the primary stumbling block out of the way, is this SHO? That is, is buoyancy proportional to the displacement? Well, gravity acts with a force mg, and buoyancy acts with a force rho·V·g, where rho is the density of the fluid and V is the volume displaced. What's the volume displaced? Let's say it floated originally so that it displaced a volume v0, and that V-v0=vd, with vd being the additional displaced volume as a result of pushing it into the fluid. rho·v0·g cancels out the weight, mg, so we're left with rho·g·vd. Since the cylinder is always upright, vd=d·pi·r^2, so the restoring force=rho·g·d·pi·r^2, and rho, g, pi, and r^2 are all constants, so yes, the restoring force is proportional to the displacement and the motion is simple harmonic and k=rho·g·pi·r^2.]

12. Two space stations want to synchronise their clocks. Heaven Station fires a laser pulse at exactly midnight to Haven Station (dumb names or what?), 540e6km away. Taking this distance into account, Haven Station sets all onboard clocks to the appropriate time as soon as the light pulse is detected. However, unbeknownst to either station, Heaven Station has begun to drift away from Haven Station at a speed of 0.01c, a considerable velocity, to be sure, but when you're so far apart with unsynchronised clocks, it's hard to tell. How much faster or slower will Haven Station's clocks be than Heaven Station's when the light pulse is detected? Ignore any change in distance resulting from the drift, consider only relativistic effects. <Quote from virtual equation sheet: L=L0/gamma and delta-t=delta-t0·gamma and x=gamma·(x'+v·t') and t=gamma·(t'+v·x'/c^2) and gamma=1/sqrt(1-v^2/c^2)> Note: I included all these equations just for completeness and to make it confusing like a real equation sheet.

Answer: [18 seconds slower.] Solution: [Only one way to do this one, though you might think time dilation and length contraction would work, too. Unfortunately, length contraction works only at one point in time and time dilation only works for one point in space, so neither apply here, since both the time and position is different from between the two events, namely, when the light pulse was fired and when it was detected. So you have to use the messy Lorentz transformations. Before you continue, I would advise you to make sure you understood how to do the last question on the first PH212 midterm, the one about the aliens and the DOD and solar flare, if not see problem number A. I will use the same strategy as I did in A, making the 1's be when the event of firing and 2's be the detection. Let the primes be in Haven Station's frame of reference, so that x2'-x1'=540e6km, because event 2, the detection, takes place 540e6km away from event 1, the firing of the light, and t2'-t1' = 1800s, because after event 1 occurs, namely the light pulse is fired, it takes 1800s for the light to cross the not-so-vast stretch of space and be detected (you should not need my help to calculate this). Now what we want to calculate is the time between events 1 and 2 in Heaven Station's reference frame, mathematically t2-t1. Now we merely plug in the Lorentz transformations (you may want to see problem number A) and then factor and rearrange so that we can easily plug in t2'-t1' = 1800s and x2'-x1'=540e6km and v=0.01c. You should get 1818 seconds. Now what does the 1818 seconds *mean*? It means that the time it takes for the light pulse to travel to Haven Station is 1818 seconds in Heaven Station's reference frame, so when Haven Station detects the light Heaven Station's clocks will read 12:30:18, but Haven Station *thinks* that Heaven Station's clocks read 12:30:00, because 30mins is the time it takes the light to travel across and be detected from Haven Station's frame of reference, and Haven Station thinks that Heaven Station is in the same reference frame. So Haven Station's clocks will be 18 seconds slower than Heaven Station's (probably bigger than you imagined, huh?).]

13. A ball weighing 4g with a 2cm radius rolls down a ski jump-like ramp. The bottom of the ramp is 1m below where the ball started rolling. The place where the ball flies off the ramp is 20cm above that. If the ball were a unifom sphere and the only friction present was static friction with the ramp, how high would the ball go? <Quote from virtual equation sheet: moment of inertia of uniform sphere=2/5·M·r^2>

Answer: [77cm] Solution: [Your conservation of energy radar should have been beeping like crazy when you read that question (if it wasn't go reread chapter 8). The ball started out with mg·1m of energy at the top of the ramp, then when it left the ramp, it had mg·20cm+0.5·m·v^2+0.5·I·omega^2, and when it reaches the maximal height we'll call h, it has mgh+0.5·I·omega^2. Now we recall what we learned in chapters 10 and 11 (or go reread em on pages 239 and 279, respectively) and plug in I=2/5·m·r^2 and omega=v/r, so that we now have two equations, mg·1m=mg·20cm+0.5·m·v^2+0.5·I·omega^2 and mg·1m=mgh+0.5·I·omega^2, and two unknowns, h and v. We solve for v in the first equation (v^2=10/7·g·80cm) and plug that into the second equation and solve for h, to get h=1m-v^2/5g=1m-2/7·80cm=77cm. Notice that like so much in classical mechanics, the mass is irrelevant.]

14. Sasha, the substitute physics professor and brown belt in aikido, is doing an experiment on rotation. She is standing on a rotating platform and does an perfect aiki roll from the axis of rotation out exactly (infinite significant figures) one meter that causes the rotational velocity to halve from 0.2rev/s to 0.1rev/s. If she weighs 60kg and is small enough as compared to the platform that she can be treated as a point mass, what is the moment of inertia of the platform? Ignore Coriolis (in an aiki roll one is always in contact with the ground/platform, so friction counterbalances the Coriolis) and frictional forces on the platform from the ball bearings supporting the platform.

Answer: [60kg·m^2] Solution: [Your conservation of angular momentum radar should have been beeping like crazy when you read that question (if it wasn't go reread chapter 11). Let I be the rotational inertia of the platform, m be Sasha's mass, d be that 1m she rolls, omega2 be the initial omega (0.2rev/s), and omega1 be the final omega (0.1rev/s). The initial angular momentum is the sum of that of the platform, I·omega2, plus that of her, which is zero since she's at the center. The final angular momentum is I·omega1+omega1·m·d^2. Because there is no friction or anything and hence no net torque, those two are equal and the only thing we don't know is I, which we solve for to get I=omega1·m·d^2/(omega2-omega1)=0.1rev/s·60kg·(1m)^2/(0.2rev/s-0.1rev/s)=60kg·m^2. Notice there is no need to convert the rotational velocities to rad/s, because the 2pi would have canceled anyway.]

15. You observe that the plane polarization of light from your brand new LCD monitor is set at an angle rather than vertical or horizontal, but it is hard to accurately tell the angle of the polarization. After night falls, you close all your windows and lights etc until the LCD monitor is the only source of light in the room. A handheld scanner tells you that a Polaroid held so light passing through is plane polarized vertically lets through 90.5% of the original light. Calculate the angle of plane polarization of the LCD. <Quote from virtual equation sheet: I=I0·cos^2(theta)>

Answer: [18 degrees] Solution: [Just reverse engineer the formula (it's from 36-11, page 907). I=I0·90.5%, and I=I0·cos^2(theta), therefore cos^2(theta)=0.905 and cos(theta)=sqrt(0.905) and theta=arccos(sqrt(0.905))=18 degrees.]

16. You observe that when monochromatic light from a particular light source reflects off a soap film at normal incidence, there are many alternating bright and dark fringes. When directed through a diffraction grating with 10,000 lines per cm, the first order maximum on a screen 1.0m away is 66cm from the central bright spot. How thick is the soap film at the first bright fringe? The index of refraction of water=1.33<Quote from virtual equation sheet: sin(theta)=m·lambda/d>

Answer: [100nm] Solution: [It's all in the trig. tan(theta)=0.66/1.0=0.66, so theta=arctan(0.66) and sin(arctan(0.66))=m·lambda/d (this is from 36-7, page 900). m=1, and if there are 10,000 lines in every cm, the spacing between any two lines d=1cm/10,000, so lambda=sin(arctan(0.66)·1cm/10,000. At the first bright fringe, where there is constructive interference, light reflected off the first boundary is 180 degrees out of phase with the incoming light (fixed boundary condition), and light reflected off the second boundary is in phase with the incoming light, so the two light rays are out of phase with each other(if you have no idea what this is about, see 35-6, page 877). But one of them is also displaced twice the thickness of the soap film, so if twice the thickness is equal to half the wavelength of the light in water, they'll be in phase with each other and constructively interfere. This is expressed mathematically as the equation 2·t=lambda-in-water/2, or t=lambda-in-water/4. You shouldn't need the equation sheet to tell you that v=f·lambda (although it will be on there I think), nor should you need it to tell you that v=c/n. So c/lambda=f, and c/(n·lambda-in-water)=f, so lambda-in-water=lambda/n=sin(arctan(0.66)·1cm/(10,000·1.33) and the thickness, which is one-quarter that, will be t=sin(arctan(0.66)·1cm/(4·10,000·1.33)=1.0e-7m=100nm.]

17. A 5g, radius 1.5cm ball rolls down a ramp onto a level table and smashes into a 5g piece of clay at rest attached to a massless level spring. The ball and clay stick together and oscillate with a period 3s and an amplitude of 10cm. If the ball were a uniform sphere and the ramp and table had enough static friction that the ball rolls with no slipping, from how high on the ramp must the ball have started rolling? <Quote from virtual equation sheet: rotational inertia of a uniform sphere=^{2}/_{5}·M·r² and T=2π/ω and ω=√^{k}/_{m} and vmax=ω·A and ω=v/r and K=0.5·I·omega^20>

Answer: [12.5cm] Solution: [This one isn't quite as bad as it looks, you just gotta work backwards. So, obviously, there are two parts. Working backwards as we are, we start with the oscillation. Looking ahead, what do we want to find? The clay was at rest before impact, so the clay must have been in equilibrium position. The ball and clay stuck together, so though cons. of energy is invalid, cons. of momentum (linear) can still be used to find the speed of the ball before impact. So, starting from the end (remember we're working backwards), if T=2π/ω, then ω=2π/T, which we plug into v_{max}=ω·A to get v_{max}=2πA/T (bits and pieces from Chapter 14). Since the clay was in equilibrium position at impact, vmax was the speed of the ball+clay after impact (14-2, page 364). Using cons. of momentum, 2m·vmax=m·v, where v is the speed of the ball before impact and m is the mass of the ball or clay (they're the same, right?). Cancel the m's--notice that the m's cancel only because the clay and the ball are of the same mass--and you get v=4πA/T. Now the rotation. Cons. of energy tells us that the ball must have been a height h up the ramp such that mgh=½·m·v²+½·I·omega², where the omega, different from the omega before, is the angular velocity of the ball itself, rather than a property of the oscillation, and I is the moment of inertia of the ball, given by the formula I=^{2}/_{5}·m·r² (10-6, page 249). When a ball rolls without slipping, omega=v/r (10-3, page 244), and we know that v=4pi·A/T, so that cons. of energy equation above becomes, after doubling both sides, 2mgh = m·v² + 2/5·m·r²·v²/r². Now we cancel the m's and plug in v and solve for h and we get h=12.5cm.]