Usapho 2014

Pre-College Issues High School Life
41

<p>Ok, here’s what you do. The moment the m reaches its max height, it will be moving in unison with M. That is the key realization you need. </p>

<p>So for 15, you set up your equations like this.</p>

<p>Conservation of momentum-inelastic collision.</p>

<p>m*v0=(m+M)v1</p>

<p>Conservation of energy- h represents the height </p>

<p>(1/2)m*v0^2=mgh+(1/2)(m+M)v1^2</p>

<p>You have two equations, two unknown variables (v1, and h). Solve for h.</p>

<p>16.</p>

<p>Since energy and momentum are going to be conserved throughout the entire thing,
you can set the initial energy of the system equal to the final energy of the system. Again, we have another inelastic collision.</p>

<p>Conservation of momentum:</p>

<p>v<em>f=velocity of small block after they separate
V</em>f=velocity of big mass after they separate </p>

<p>(m+M)v1= -mv<em>f+MV</em>f</p>

<p>Also note that: mv0=(m+M)v1</p>

<p>Conservation of energy-use the initial energy and set it equal to the final energy.
(1/2)m*v0^2= (1/2)m(v<em>f)^2+(1/2)M(V</em>f)^2</p>

<p>Three equations, three variables (v1, v<em>f, V</em>f)
Solve.
Note that when you solve for v<em>f, you will get it as a positive term since you already took into account that it would be negative by saying in the conservation of momentum “-mv</em>f” so don’t forget to take that into account when doing the quadratic formula for 16.</p>

<p>Also, 16 is not reliant on 15. You can solve 16 without the answer from 15.</p>

42

<p>How many do you guess on…?</p>

43

<p>It is crazy how much I over think these problems. Before I gave up, crazy ideas on solving those problems were going through my head (like mgsin(theta) and stuff to account for the forces on the incline and all that stuff). </p>

<p>Thank you very much. </p>

<p>2010, which I took a few days ago, was bad for me… 13 right, guessed on all the other ones but 2. I’ve looked over it all and found the solutions, with a few exceptions.</p>

<p>In the same exam, number 25 was crazy… I finally solved it just now after a lot of time on Wikipedia. It requires the Vis-Viva equation (at least that’s how I solved it in the end).</p>

<p>Also, I was mindblown by number 22. It seemed so simple. Apparently, however, if there is a container undergoing circular motion, and the container is full of a fluid and something that floats is suspended in the fluid, the thing suspended will be pulled IN because of the fluid, not pushed OUT like people in a car.</p>

44

<p>Yeah, I didn’t get 22 right on the first try and I too tend to overthink many problems. My god it’s unbelievable. </p>

<p>For 25, I found a much more basic way to solve it.</p>

<p>Use conservation of energy and momentum to find the velocity at the furthest point in terms of G, the mass of the planet, and R. </p>

<p>Then use mv^2/r=GMm/r^2 for the uniform circular motion. Since velocity is expressed in terms of G, the mass of the planet, and R it must be constant since none of those values change. Rearrange the equation to get v^2=GM/r and substitute for GM/r to get the answer.</p>

45

<p>Oh man, I know what you mean about overthinking. On F=ma 2013, problem 16, I had all these crazy ideas running through my head. I even tried to do the derivation myself… I lost five minutes and I got the answer wrong to boot.</p>

46

<p>good luck everyone…oh I want to say some of the problems and rage at them so much lol</p>

47

<p>Please do</p>

<p>jkjkjkjkjkjk</p>

48

<p>We’ll rage together after we’ve all taken it? </p>

<p>@Radbg, sorry but I couldn’t grasp how you solved 2010 #25. "Use conservation of energy and momentum to find the velocity at the furthest point in terms of G, the mass of the planet, and R. " I don’t understand. Sorry… could you clarify?</p>

<p>And #25 in 2008 was similar. It dealt with an elliptical orbit… Seems to be a a reoccurring theme. It would be very handy to have a reliable method for tackling these problems. Any ideas? </p>

<p><a href=“http://www.aapt.org/Programs/contests/upload/olympiad_2008_fnet_ma.pdf[/url]”>http://www.aapt.org/Programs/contests/upload/olympiad_2008_fnet_ma.pdf&lt;/a&gt;&lt;/p&gt;

<p>Answer is E. This stumped even my physics teacher. I quote “Where the hell does the 7 come from??”</p>

49

<p>Ok for 25 on 2010, what I meant that, if you write out the equations for energy and momentum, you should be able to express the velocity at the furthest point by only using the terms G, M (mass of planet), and R. For example, v=GM^2/R^3 (not the answer) Just do the math and it’ll work.</p>

50

<p>Okay Radbg, I spent a while trying. Here’s what I got. Not very encouraging…</p>

<p>v of circular orbit = √(GM / 2R)
v of elliptical orbit at the farthest point was where I got stuckded. I tried equating the PE + KE at the farthest point with the the PE + KE at the closest point along the elliptical orbit. But that introduced a new variable, v_closest point, that I had no idea how to solve for. As for momentum, how would momentum help here? The momentum is always changing in the elliptical orbit because the gravitational force acts at changing angles all the time. </p>

<p>Please be patient with me lol… I’d really like to know how you solve this type of problems. </p>

51

<p>Aha I googled some stuff. Were you talking about conservation of specific orbital energy? And with momentum were you talking about conservation of angular momentum (r1v1 = r2v2 at the apogee and perigee of an ellipse)? </p>

52

<p>You need to use conservation of energy and angular momentum for that one. As well as the acceleration for uniform circular motion. When I did it I also ended up solving a quadratic. It’s not that bad, it’s more like “here’s-a-tedious-but-fairly-simple-problem don’t-make-mistakes-go”</p>

53

<p>No, I’m just talkign about regular angular momentum and energy. There’s no need to know super specific stuff like specific orbital energy. ifailedusamo is right. It’s just are regular question that requires you to think a little bit after you write down the standard stuff when approaching a problem like this. </p>

54

<p>Okay, here it goes. Solving for circular velocity (V) in terms of v-nought (v)</p>

<p>1) Velocity around circular orbit: GM / (R^2) = V^2 / 2R -> V = √(GM / 2R)</p>

<p>2) v-nought in terms of velocity at perigee (v<em>p) of ellipse using conservation of angular momentum (h). h = rmv: 2Rmv = Rm(v</em>p) → v_p = 2v</p>

<p>3) v-nought in terms of velocity at perigee (v<em>p) of ellipse using conservation of energy. The gravitational potential energy plus the kinetic energy is equal at the perigee and apogee: (GmM) / (2R) + (mv^2) / 2 = (GmM) / (R) + (m(v</em>p)^2) / 2 —> substituting v_p = 2v → GM + Rv^2 = 2GM + 4Rv^2 → v = √(-GM / 3R)</p>

<p>4) The ratio of V to v is √(-3/2)… obviously there is a tiny error somewhere in my setting up of equations (not my arithmetic, I double checked)… But I got hella close! Thanks guys. </p>

55

<p>Gravitational potential energy is always negative.</p>

56

<p>Okay, thanks, it works now. </p>

<p>I just took the 2011 practice F=ma exam and got a 16. I was much less reckless with the guessing. Could anyone tell me the cutoff for that year? </p>

57

<p>over 15 is probably is safe passing bet.</p>

58

<p>oh lord, I take it tomorrow. wish me luck</p>

59

<p>Good luck and thanks for everything! May the Korean Jesus see that you do well.</p>

60

<p>A spaceborne energy storage device consists of two equal masses connected by a tether and rotating
about their center of mass. Additional energy is stored by reeling in the tether; no external forces
are applied. Initially the device has kinetic energy E and rotates at angular velocity ω. Energy is
added until the device rotates at angular velocity 2ω. What is the new kinetic energy of the device?
(a) √2E
(b) 2E
(c) 2√2E
(d) 4E
(e) 8E</p>

<p>Why is the answer not D? I thought KE was .5Iω^2, so shouldn’t doubling the angular velocity quadruple the KE? The answer is B by the way. This is from the 2008 F=ma.</p>