Questions/comments about homework (assigned or hand-in)

Sun, Dec 10, 6:41 p.m. - For question 8 in Chapter 9 in Wolfson, I said that because momentum has to be conserved, there is even a very small velocity maintaining the momentum in the after picture, which conserves a part of kinetic energy. So, my answer was NO. Is that correct?

No, that isn't correct. It is possible to have an inelastic collision where all the kinetic energy is lost. A 1-kg ball of ground beef is moving to the right with a speed of 2 m/s. Another 1-kg ball of ground beef is moving to the left with a speed of 2 m/s. They collide and stick together. If you conserve momentum you will find that total momentum is zero in the before picture, so the ground beef must be motionless in the after picture since it is now a 2-kg, stuck together mass of meat. So, all of the initial kinetic energy is gone.

Mon, Dec 4, 2:18 a.m. - Ch. 11 Wolfson #58. Potential energy of the spring equals the kinetic energy of the mass... and the rotational energy of the turntable or no? <p>How does the turntable have any rotational energy? Let's think about this. There's no friction. The mass can't exert a force on the table, then. How, then, is the table moving? Isn't the mass basically not in contact with the table? Please explain!

No, it is not correct that potential energy of the spring equal kinetic energy of the mass. Please, please, please, please ... don't ever start a problem with ``potential energy of (something) equals kinetic energy of (whatever)'' -- I can promise you that we will ALWAYS deduct at least 2 or 3 points if you start a problem like that. The starting principle is that since $W_{NC} = 0$, that means that mechanical energy is conserved, i.e., $E_{before} = E_{after}$.

It is true that $E_{before} = K_b+U_b = U_{spring}$ since everything is motionless in the before picture (which I know that you have drawn because it is the first step in the step-by-step approach and I trust that you now know never to leave out any steps). $E_{after} = K_a+U_a = K_{trans} + K_{rot} + 0$, so, yes, the turntable does have rotational energy.

By the way, don't forget conservation of angular momentum as well -- that plays an important role.

Answering your question -- it is correct that the mass is not in contact with the table. It is flying off in one direction, and the table is then rotating. Total angular momentum is clearly zero in the before picture that you have sketched, so with the mass flying off and clearly carrying angular momentum (calculate it using $\vec{L} = \vec{r} \times \vec{p}$), the turntable must be carrying the opposite angular momentum (it must be spinning) to keep the total at zero.

By the way, conceptually, when the spring is pushing the mass, it is also pushing back on the table (by Newton's 3rd Law), so the table gets a torque while the spring is pushing that causes it to rotate.

Sun, Dec 3, 4:23 p.m. - For Question A44c, why in the world does the yoyo move towards the string, let alone accelrate so quickly? My brain is litterally on the verge of melting; this is trully the first time in this course where I cannot fathom a reason why something is happening. But I'm sure it will be a simple answer.

Yeah, it's a great puzzle, isn't it? I love this particular question, because almost no one ever expects it to roll the way that it does.

The answer, of course, can be found in the standard step-by-step approach that we have discussed. Draw a free-body diagram for the yo-yo, and write down Newton's Second Law in component form (you only need the component in the direction of acceleration). Draw a torque diagram and apply $\tau_{net} = I\alpha$ (warning: you'll need two different radii, one -- let's call it $r_1$ -- for the small radius of the spool around which the string is wrapped, and a second -- let's call it $r_2$ -- for the outer radius of the yo-yo. And then $a = r_1\alpha$. Using those three equations and solving for $a$, you'll find that it will accelerate in the direction of the tension in any case where $r_2 \gt r_1$.

By the way, the result is quite interesting, because if you use the center of the yo-yo as the origin for torque calculations, the torque from the tension causes the yo-yo to twist one way, and the torque from the friction causes the yo-yo to twist the other way. The friction torque wins this battle, resulting in the yo-yo accelerating the way you found; however the tension force has to be stronger than the friction force, resulting in the yo-yo accelerating in the direction of the tension. These statements are consistent because of the fact that the radius for the friction force is larger than the radius for the tension force, so the friction torque can be (and is) larger than the tension torque, even though the friction force is smaller than the tension force.

By the way, there is a quicker way to argue that the yo-yo has to accelerate in the direction that you found: draw the torque diagram, but use the contact point with the table as the observation point when determining the net torque. When you do that, there will only be one force that causes a torque, and it should be easy to figure out the direction of that torque. And then consider that your observation point is the contact point, and then you'll be able to see which way the yo-yo will rotate. This is a quick approach, but a bit subtle.

Sun, Nov 5, 11:41 a.m. - For SUPP CH 9 #16b, I thought that delta-S total should be conserved. Why, then, is it 0.6 J/K if no energy is being added to the system as a whole?

Ugh. Here is again the fallacy that ``conservation'' has anything to do with entropy. It doesn't. Something is conserved if its value doesn't change. That's not the case with entropy. Entropy can and often does increase during a process.

In 9-16, you are taking a high temperature object and putting it in contact with a low temperature object, in which case heat will flow from the hot to the cold object, resulting in a decrease in the entropy of the hot object, and even larger increase in the entropy of the cold object, and an overall increase in entropy of the entire system. An increase in entropy means that the system is going to a more likely state. THAT'S THE WHOLE POINT!!! Heat flows from hot to cold because it is overwhelmingly, mind-bogglingly more likely that it will do so, rather than having no energy flow or (even less likely) a flow from cold to hot.

Sun, Nov 5, 11:34 a.m. - For SUPP CH 9 #14, the relationship is given to us in 9.10. Where might we want to look to begin deriving that relationship? Does it have to do something with how thermal energy is conserved, perhaps?

Please don't answer 9-14 by saying, ``$S_{AB} = S_A + S_B$ because equation 9.10 says so.'' You can prove this by (a) recognizing that the total multiplicity $\Omega_{AB} = \Omega_A \Omega_B$, from Boltzmann's relation for entropy $S = k_Bln\Omega$, and from the mathematics of natural logs. Specifically, remember that $ln(ab) = ln(a)+ln(b)$.

Sun, Nov 5, 11:27 a.m. - For SUPP CH 9 #7c, should the explanation be something along the lines of the fact that 7b has more energy units than 7a and an incremental unit of energy added to 7b yields a lower increase in entropy than an incremental unit of energy added to 7a? Do we have to explain anything on a molecular level? Even if we don't, though, how could one explain this phenomenon on a molecular and probabilistic level (rather than just a computational one)?

Yes to your first question. As for your second question, any explanation should involve probabilities/multiplicities and that is what you are doing in parts a and b. So, no -- once you have completely explained something, you do not have to come up with another explanation.

As for your third question: you are explaining this on a probabilistic level. That's what the computation in (a) and (b) are doing. Remember: entropy is a measure of probability (actually, likelihood, and the natural log of that).

Sun, Nov 5, 11:13 a.m. - For SUPP CH 9 #5, would the approach to answering this question be that the temperature of both can't go up (one has to go up and one has to go down per dS/dE = 1/T)? How do we prove entropy is conserved, though?

Answer to your first question: no. Answer to your second question: why would you want to prove the entropy is conserved? There is no ``conservation of entropy'' principle and if you ever try to prove that there is one, you will be wasting your time.

In class last week, we said that for dice, there is only one microstate that gives a total of 2 for a roll of two dice, but there are 6 microstates for a total of 7. So, rolling a 7 is more likely than rolling a 2, which means that ``7'' has a higher entropy. Does that mean that you will never roll at 2? Why is it that you can end up with smaller entropy states when rolling 2 dice, but you never end up with lower entropy states in thermodynamic systems?

Hint number 1: if you were working with an Einstein solid that had only two particles in it, you would often see the system entropy decreasing. Why don't you ever see an entropy decrease for a solid with, say, a mole of atoms?

Hint number 2: you might want to use the words, ``really'', ``unbelievably'' and ``mind-bogglingly'' in your answer when discussing probabilities/likelihoods when discussing systems with large numbers of molecules in the Einstein solid.

Sun, Nov 5, 10:56 a.m. - For SUPP CH 9 #3c, why can we not simply use the multiplicity formula (subbing q=6 and N=3)? What may be a better approach to solving this problem instead (to get 10/216 instead of the answer that the multiplicity formula gives with the approach I outlined above: 28).

The multiplicity formula that we use $\Omega = \frac{(q+N-1)!}{q!(N-1)!}$ is only valid for the Einstein solid. Please make sure that you have that on your 3x5 card and that you only use this formula if the problem specifically states that you are dealing with an Einstein solid.

For this problem, by far the easiest way to do it is to simply write down any combination of the three dice that will give you a sum of 6. I'll give you the first one: 1 on the first die, 1 on the second and 4 on the third. Keep going.

Sun, Nov 5, 10:55 a.m. - For SUPP CH 7 #15e, why do you have to use force = (NkT)/L instead of density * volume * average force per molecule (derived in part 15d)?

You aren't using NkT/L -- I'm not sure where that came from. I think that you are overthinking part (e). Once you have figured out in part (d) how much force a single molecule is exerting on the wall, simply multiply by the number of molecules in the box that are hitting the wall. Note the second assumption at the top of the problem that says pretend that a third of the particles are moving in each direction.

And the good news is that you will know if you have done this correctly because when you take the force determined in part (e) and divide by the area of the wall (in part f), you should get a pressure of 1 atm (i.e., 101 kPa).

Sun, Nov 5, 10:53 a.m. - For SUPP CH 7 #12, is the answer that you are adding thermal energy by dragging the molecule? What I thought is that if you add enough thermal energy to a few atoms/molecules, the thermal energy is dissipated across all neighboring molecules.

This is something that you should be noticing when you actually do the simulation. (By the way, if your response to this question doesn't explicitly discuss what you did and what you saw in the simulation, then don't expect to get full credit -- you actually have to do the simulation.) Yes, the energy that you put into dragging a single molecule gets transferred to other molecules (not just the neighboring ones). Can you explain why? And why does that result in melting of the solid? For that, refer back to 7-1(c).

Sun, Nov 5, 10:52 a.m. - For SUPP CH 7 #1c, is the answer that if the thermal energy is high enough that the molecules are too far from their minimum of their potential well, the pairs of molecules are stretched out far enough (at least d/10 per the Lindemann criteria) that causes the interaction of molecules to be weaker (the model of the spring goes away)?

It isn't that the interaction between the molecules goes away -- that's what happens when the stuff turns into a gas. It's only in the gas state that the molecules are far enough so that the interaction is negligible. See Fig. 6.1 for more about that.

A solid doesn't melt when the interactions become negligible. Rather, a solid melts when the atoms in the solid can move around (due to thermal jiggling) so that they can start exchanging places, rather than basically jiggling in place. That's the key difference between a solid and a liquid -- in both of them, the atoms are still bound together, but in a liquid, the atoms are still able to move around.

And that is the key to understanding why comparing $x_{therm}$ to the equilibrium spacing $d$ for a solid is the key. I'll let you figure out the rest of the argument.

Sat, Nov 4, 4:57 p.m. - Silly question, but seriously not sure about this: Are we allowed to do use blank paper instead of lined paper for the hand-in set?

I don't see any harm in that. Go for it.

Sat, Nov 4, 3:59 p.m. - For Supp CH 7 #12, where can we find the molecular dynamics applet?

It can be found on the calendar page for Lecture 16 or 17. You can also get to it by clicking here:

Sun, Oct 29, 2:36 p.m. - I noticed that there are no answers to CH8 on the back of the supplementary reading. Would you please post it on the website?

Thanks for catching this. We have posted the answers for Ch 8 on the calendar page for the October 31 lecture.

Sun, Oct 8, 4:32 p.m. - How do you do A38 and A39. I can't figure out a method even after reviewing the lecture example, and the strategy. The video examples also don't work on my computer. Thank you in advance.

The method discussed in lecture and on the ``lecture materials'' for this past Thursday's lecture works perfectly for both of these problems, but you can't skip any steps. For each problem:

  1. Draw ``before'' and ``after'' sketches. For A38, the Before sketch should show the motionless particle and the photon that is about to strike it. The After sketch should show just a single particle that is moving. For A39, the Before sketch should show a moving particle and a stationary particle, and the After sketch should show a motionless particle and a photon.
  2. Write down p and E for each particle in your sketches. Some of this information is given in the statement of the problems. For some of it, you will need to use $E^2 = p^2c^2+m^2c^4$. Count how many unknowns you have. If more than two, then use $E^2 = p^2c^2+m^2c^4$ to write some of the unknowns in terms of other unknowns.
  3. Once down to 2 unknowns, then $\Sigma p_{x,before} = \Sigma p_{x,after}$ and $\Sigma E_{before} = \Sigma E_{after}$.
  4. You now have two equations and two unknowns. Solve.

Sun, Sep 24, 2:10 p.m. - To add on, the problem for A74 is that if you try to draw the FBD of the ball, there is mg acting down and no normal force. I guess you'll have to assume that the ball is rolled on the ice instead...

Everything for the handin set due tomorrow deals with conservation of momentum. You don't need free body diagrams. Before and after sketches, momentum before - in components if necessary - momentum after, set befire equaks aftet, solve. If elastic, also use speed of approach ewuals soeed if recession.

(By the way, the fact that there is acceleration in the y-direction doesn't have any effect on its x-motion. And besides, the problem specifically says that the ball is moving horizontally when the child catches it. So, you don't need to worry about what has been happening with its vertical motion before she catches it.)

Sun, Sep 24, 2:06 p.m. - For A74 on Hand-In Set #5, the problem states that the girl "catches a 1.1 kg ball moving horizontally at 9.5 m/s." Can we assume that from the time the ball is thrown horizontally to the time the girl catches the ball, the ball gains no vertical velocity/momentum?

You don't have to worry about vertical momentum. Horizontal is all you need.

Mon, Sep 18, 3:28 p.m. - For number A22. Extreme Skiing, how do you begin to solve part f? Is it asking about the maximum magnitude of the normal force at point B?

(1) Draw a sketch of the skier at the point B. (Optional)

(2) Draw a free-body diagram for her at point B.

(3) Write down $\vec{F_{net}} = m\vec{a}$.

(4) Choose coordinate axes.

(5) Write Newton's law in component form. You'll need only one direction in this case (the direction of the acceleration, which is toward the center of the circle).

(6) Solve for the normal force.

Now, you will need the speed of the skier at point B to finish this off. To get her speed at point B, use the standard approach for mechanical energy problems:

(1) Draw before and after sketches (with before when she is starting up the hill and after at point B)

(2) Write down an expression for $E_{before}$.

(3) Write down an expression for $E_{after}$.

(4) $W_{nc} = 0$ so $E_{before} = E_{after}$.

Solve for the speed (which should be in $E_{after}$), pop it into your expression for $N$, solve, and pat yourself on the back.

Sun, Sep 17, 10:24 p.m. - Can you explain the clock problem from one of the homework problems? I am confused about how to get the acceleration. On one hand, you have displacement over time is the average velocity. Then, you also have 2*pi*R/time, which gives the magnitude of the velocity at any point, which is constant. Please explain in full if possible--thanks!

The problem asks for the average acceleration with is defined as $\Delta \vec{v}/\Delta t$. $\Delta \vec{v} = \vec{v_f} - \vec{v_i}$. The velocity $\vec{v_f}$ is the velocity of the hand when it is at the bottom (i.e., 6:00) and $\vec{v_i}$ is the velocity of the hand when it is at the top (i.e., 12:00). In both cases, the velocity has a magnitude $v = (circumference)/(rotation period) = 2\pi r/(12 hours)$. Once you have this numerical value for the speed, then $\vec{v_i} =$ (this numerical value)$\times \hat{i}$ since the hand is moving in the positive x-direction at the top, and $\vec{v_f} =$ (this numerical value)$\times (-\hat{i})$ since the hand is moving in the negative x-direction at 6:00.

Do the subtraction $\Delta \vec{v} = -(something)\hat{i} - (something)\hat{i} = -2(something)\hat{i}$, divide by 6 hours (which is $\Delta t$ between 12:00 and 6:00) and convert units to meters and seconds.

Sat, Sep 16, 1:29 p.m. - Are there Assigned Problem for Wednesday (9/20)'s problem session?

Nope. But, as I mentioned in lecture Thursday, there IS problem session on Wednesday and you will be receiving a 5-point grade based on the activities in that problem session. But no assigned problems for Wednesday.

Thu, Sep 14, 5:33 p.m. - In Hand-In Set #4, can we assume the drag force to be negligible for A25 and A78?

Well, if you assume that the drag force (friction) is negligible for A25, that would certainly make the problem easier. ``Determine the magnitude of the friction force ...'' Answer: ``We are going to assume that friction is negligible. Therefore, the answer is zero!''

In other words, no, you can't assume drag force to be negligible for A25, since that is what the problem is asking you to measure. And it clearly isn't negligible because the block is stopping.

(Did you mean to ask, ``Can we assume that air resistance is negligible?'' If so, the answer is yes -- don't worry about air resistance, which really is negligible in this case.)

As for A78: Yes, neglect air resistance. You can't do the problem unless you neglect air resistance here which is actually a reasonable approximation -- air resistance really wouldn't have a huge effect here.

Thu, Sep 14, 5:19 p.m. - For A24 on Hand-In Set #4, can we assume that drag force is negligible since the question asks us to 'energy' potential energy?

That is an assumption that you will have to make, yes, but say explicitly that you are, in fact, neglected drag forces here.

Wed, Sep 13, 8:04 p.m. - For A25, I understand that you must first find the potential energy of the spring using U= .5(k)(x)^2 and plug in, then use fk=(mu)(m)(g) to find that mu=U/((m)(g)(d) but there is no mass value given for the block? How would you go about solving this provlem without a mass value given?

A few comments here: first, it took me quite a while to figure out how you are going about this problem here. If you are doing this on an exam, please make sure that you start with some general principle or a sentence explaining how you are going about this. And don't make statements like $f_k = \mu mg$ or $\mu = \frac{U}{mgd}$ without explaining where that came from.

Actually, now that I look at this more ... where did $\mu = \frac{U}{mgd}$ come from? And why are you looking for $\mu$ when the problem is asking for the magnitude $f$ of the friction force?

There is a very straightforward way of doing this problem: use either $W_{net} = \Delta K$ or $W_{nc} = \Delta E$. Either approach will work fine. The key is that $W_{net} = W_{spring} + W_{friction}$ and $\Delta K = 0$ since it is motionless at the beginning at at the end. $W_{spring} = \frac{1}{2}kx^2$ and $W_{friction} = f\Delta rcos(angle)$. Throw into work-KE and solve for $f$.

OR if you use $W_{nc} = \Delta E$, then $W_{nc}$ is simply the work done by friction and $\Delta E$ is the difference between the mechanical energy at the end (rhymes with ``hero'') and beginning (just spring potential energy).

Note that you don't need the mass of the object for either of these approaches.

Sun, Sep 10, 7:36 p.m. - For A18, if I were to find the initial velocityof my dart, how would I do that? Or are there any other ways to find the friction? Thank you!

You already have found the initial speed of your blow dart. Back for the first hand-in set of the semester, for problem A4 you fired a blow dart straight up into the air, timed how long it was in the air, and used that timing to determine how fast the blow dart is moving when it leaves the blow dart gun.

You can use the same speed for A18 -- the fact that you are firing the blow dart horizontally doesn't really affect the speed when it leaves the blow dart gun.

And, yes, you are correct that you do need this information to find the friction.

Sun, Sep 10, 12:00 a.m. - For the Chapter 1 question 2 on the supplementary reading, it asks for a comparison between the results for question #1 and question #2. Does this mean that we have to do question # 1 as well or not?

Of course you have to do question 1 as well -- that is one of the assigned homework problems from this past Wednesday. Every assigned problem is required in this course. The good news is that you have already done it.

You don't have to show all the work for question 1 -- you can refer to the answers that you got for number 1 and write a couple of sentences about the similarities and/or differences between the results for number 2, including a brief discussion explaining why there are any differences and if they are what you would expect them to be.

Thu, Sep 7, 7:08 p.m. - For question 21 Chap 6, do you have to prove that acceleration is zero before using the W=Frcos equation? Thanks friendo

There isn't anything in this problem to indicate that the acceleration is zero. It could be zero, or it might not be zero. It actually doesn't matter at all what the acceleration is. The relation W = Frcos(theta) only requires that the force be constant during the process, which is the case here.

Sun, Aug 27, 1:10 p.m. - What is the expectation for hand-in sets in terms of formatting?

At the top, put your name, your PROBLEM SESSION instructor's name, and the time of your problem session (NOT the lecture section). For each problem, the issue of ``formatting'' is answered by the ``Show All Work'' policy that I discussed at the end of this past Thursday's lecture. This is spelled out point-by-point on the ``Lecture materials'' PDF that is posted on Thursday's calendar page, but the short summary:

(1) Every problem/question must start with either a generally-applicable equation or definition (i.e., something that is always valid); a special-case formula WITH A STATEMENT explaining why that formula is valid (e.g., ``a is constant, therefore ...''); or a sentence explain what you are doing.

(2) There should be sufficient steps that show how you get from your starting point to the answer; and

(3) The answer must have correct units.

Sat, Aug 26, 5:39 p.m. - I am sorry to disturb you on the weekend, but I wanted a clarification on one of the homework problems. Problem A83 tells that the balls in the second pit are half the diameter of the balls in the first pit. Would you please clarify if the previous statement is referring to the second ball's radius or diameter. This difference is turning out to be humongous in my final answer.

The diameter $D_2$ of the balls in the second pit is one-half the diameter $D_1$ of the balls in the first pit. I.e., $D_2 = \frac{1}{2}D_1$.

If you were considering the radii, then the radius of the second is 1/2 the radius of the first: $r_2 = \frac{1}{2}r_1$.