## Questions/comments about classical mechanics

*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.

*Sun, Dec 10, 2:48 p.m.*-

**How to answer the multi-choice question number 93 in chapter 6 in Wolfson? Thanks**

I don't think that we ever assigned this question. I would encourage you not to waste your time on problems that aren't assigned -- they often deal with issues that we decided not to emphasize in this course, which means that you would be studying stuff that won't be relevant for the final exam.

Having said that, power is force/time -- think about it that way.

*Sun, Dec 10, 11:47 a.m.*-

**I am having some trouble on video example #1 for lecture 4. I keep getting an answer of g=a. I can't get the video to work, but do you know of an important step that I am missing if this is the answer I keep getting?**

The statement of the problem already has the two free body diagrams that you need. The direction of the acceleration is clearly UP for the mass on the left and DOWN for the mass on the right. So, I define the +x-direction to be up for the left mass and down for the right mass. $\Sigma F_x = ma_x$ becomes $T - mg = ma$ for the left mass, which gives $T = mg+ma$. $\Sigma F_x = ma_x$ becomes $2mg - T = 2ma$ for the right mass. Substitute in $m(g+a)$ for $T$ to get $2mg - mg - ma = 2ma$ which becomes $mg = 3ma$ so $a = g/3$.

*Sat, Dec 9, 8:13 p.m.*-

**For GOT IT? 3.3 in Wolfson Why does the perpendicular wind direction (no. 2) need higher velocity of the plane relative to the air (This is the case for no. 1 only because the wind has an opposite vertical component in this case)? Thanks**

AFirst comment: I would encourage you **not** to study for the exam by going through the reading assignments. There is a lot in the readings that you do not need for the final -- you'll be wasting a lot of your time going through the reading again. Everything that you need from the readings was covered in the lectures and problem sessions.

As for your question, you must draw a vector diagram showing the three velocity vectors to answer this question. $\vec{v_{total}} = \vec{v_{air}} + \vec{v_{plane relative to air}}$. $\vec{v_{total}}$ is 500 km/h $\hat{j}$, so if the velocity of the air is 100 km/h to the right (East), then the velocity of the plane relative to the air needed (which is the hypotenuse of the triangle that you drew) is larger than either of the legs of the triangle, which means it is larger than 500 km/h.

*Sun, Sep 24, 1:41 p.m.*-

**For the Fext=0 with gravity acting on an object question: is it because the normal force makes the net force on the object 0? Sorry about that!**

No. It is because the gravity doesn't have enough time to change momentum significantly. See response to earlier question.

*Sun, Sep 24, 1:33 p.m.*-

**If Fext=0, then Ptotal is conserved. Right? Then, why is Ptotal conserved if gravity acts on objects?**

Good question. Momentum conservation also works if the time elapsed is short enough so that the external force doesn't have enough time to change the momentum, e.g. immediately after something happens.

*Mon, Sep 18, 6:34 p.m.*-

**Why is it that for example, at the top of a loop, or when the yo-yo is at the top of its path, that we can say the force of tension or the normal force is zero. As in the velocity would be the square root of (g*R)?**

In most cases, we **CAN'T** say that the tension is zero for a yo-yo at the top of the loop. When I am doing an around-the-world with my yo-yo, for instance, the tension is usually not zero. Any time that the string is taut (i.e., tight), the tension is not zero and, in fact, the tension -- if the yo-yo is going fast enough must be non-zero to keep the yo-yo moving in the circle.

The **ONLY** time that you can say that the tension is zero is if the yo-yo is moving slowly enough so that it falls out of the loop. If you are looking for the minimum speed to keep the yo-yo in the loop, well, that corresponds to the case where the tension is **just barely** non-zero, i.e., a really, really small but non-zero number. So, if you want to find the minimum speed to keep the yo-yo from falling out, set the tension to zero.

*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, 10:22 p.m.*-

**Can you explain why normal force is either conservative or non-conservative?**

Normal forces are **non**-conservative. The work done by a normal force **does** depend on the path followed, and normal force **can** change the mechanical energy of an object. If you doubt that, pick up a book and raise it up. Your hand is doing work on the book via the normal force, and there is clearly a larger mechanical energy E after you have raised the book. $W_{nc} = \Delta E$, so if $E$ is larger, then $W_{nc} \ne 0$; i.e., the normal force is non-conservative.