<p>All there is to do is wait now… I take it on the 27th.</p>
<p>Ahh, now I’m nervous… People have already taken it. I’m taking it on the 31st. </p>
<p>I regularly get around a 14. I’m reaaaally hoping for some improvement. However, I’m less likely to leave questions blank on practice runs.</p>
<p>I don’t know if we’re allowed to post problems. I hope we are. This is question number 14 from 2010. </p>
<p><a href=“http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf[/url]”>http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf</a></p>
<p>A 5.0 kg block with a speed of 8.0 m/s travels 2.0 m along a horizontal surface where it makes a head-on, perfectly elastic collision with a 15.0 kg block which is at rest. The coecient of kinetic friction between both blocks and the surface is 0.35. How far does the 15.0 kg block travel before coming to rest?</p>
<p>(A) 0.76 m
(B) 1.79 m CORRECT
(C) 2.29 m
(D) 3.04 m
(E) 9.14 m</p>
<p>I hope this year’s test is easier than 2013…</p>
<p>Here’s how you do it. Use kinematic equations to find the velocity of the 5kg block the moment it hits 15kg block. The moment they collide, momentum is conserved due to no external forces. Then after that, find the velocity of the 15kg block the moment after the collision. Then use kinematic equations to figure out the distance the block travels.</p>
<p>I calculated the deceleration due to friction with (mu)g = 3.5 m/s^2. With kinematics laws, the final velocity of 5.0 kg is √50. Therefore the momentum is 5√50. The momentum is transferred to the heavier block, which goes off at a velocity of (1/3)√50. The deceleration is again 3.5 m/s^2. With kinematics, the distance it achieves is .794…</p>
<p>I’ve looked over my math, so nothing is wrong with my application of kinematics and stuff I believe. I must be missing some important concept or key step.</p>
<p>When determining the velocity of the heavier block, go through all the math.
Your conservation of energy equation at the MOMENT of the collision (simplified down) is:</p>
<p>49.9999=3V^2+v^2.</p>
<p>V=velocity of large block, v=velocity of little block</p>
<p>The conservation of momentum equation will read:</p>
<p>35.35533=15V-5v</p>
<p>Use those two equation to solve for V.</p>
<p>You should get that V is 3.535533906.</p>
<p>Don’t worry, it took me two times to get this right. I think the first time I made some careless mistake with my math that I have yet to find…
Lol no nvm, I found the issue</p>
<p>The problem’s interesting in that the speeds of the blocks are the same after the problem. By the way, where did you come up with the reasoning that the velocity of the big block would initially be (1/3)√50?</p>
<p>—what? I always thought that in a perfectly elastic collision, the objects trade momentums.</p>
<p>m1v1 = m2v2 —> v2 = (5)(√50)/(15) = (1/3)√50</p>
<p>I’ll be back to look more closely at the problem in a bit.</p>
<p>Whoaa buddy
The equation for the momentum in this problem is</p>
<p>m1v0=m2v2-m1v1</p>
<p>v0=initial velocity of the small block right before collision
v1=new velocity of small block after collision
v2=new velocity of big block immediately after collision</p>
<p>To be honest, I wouldn’t assume that the objects trade momentum unless you can prove it.</p>
<p>Oh by the way, I think they only trade momentums if the mass is the same.</p>
<p>Dang, you’re right. In fact, since a smaller mass is hitting a larger one, the smaller mass bounces back. Now that I think about it, that only makes sense. And I left out the v1 (velocity after collision of smaller mass) because I thought they trade momentums (and the second object had a momentum of zero).</p>
<p>Don’t you hate careless mistakes?</p>
<p>God, they really irritate me</p>
<p>Just a little tip for elastic collisions. This equation is derived from conservation of energy: </p>
<p>(v1) - (v2) = (v2)’ - (v1)’ for any pair of masses in a head-on elastic collision. </p>
<p>The difference in the objects’ velocities is the same before and after collision (same in magnitude, opposite direction). With the thinking that objects of different masses trade momenta, the above equation would be violated. </p>
<p>The problem using that equation: the velocity before collision of m1 is √50. Equating the momenta before and after the collision, 5√50 = 5(v1)’ + 15 (v2)’. There are two variables, so I will apply the above forumula: √50 = (v2)’ - (v1)’. Solving that equation for (v1)’ and plugging it into the first, you get (v2)’ = (1/2)√50. Then, using kinematics, you get the correct answer for the distance attained.</p>
<p>Yeah, I found that in my Schuam’s book, but I don’t always use it, because for some reason, it sometimes doesn’t work for me if it’s a complicated problem. Probably just because I make some math error, so I don’t really trust myself with it</p>
<p>Did you think that the 2010 exam was harder than usual? Anyone know the cutoff? I’ve taken 2007, 2008, 2009, and 2010 so far and I thought 2010 was the hardest. </p>
<p>I have a few questions from it. @Radbg74, do you know the worked out solutions to all of them? How do you find the worked out solutions? Would you mind if I asked a question or two on the thread?</p>
<p>Go ahead. And no, I reviewed them and figured out how to do them. 2013 was the hardest in my opinion. I thought 2010 was not bad, probably on par with 2009. 2007 was the easiest in my opinion (I don’t know if this is accurate, becuase I took 2008-2013 about a month ago and I took 2007 a few days ago and I got my highest score on 2007).</p>
<p>How have you done on the past tests?</p>
<p><a href=“http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf[/url]”>http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf</a></p>
<p>15 and 16. Let me know if you want me to copy and paste the problem. I’m sorry, but I can’t help you with the solution because I didn’t get very far at all. </p>
<p>For 15, I figured that it was an inelastic collision, so v’ of both masses is (mv / m+M). That’s as far as I got. I figured that I would set up an equation that equated the total kinetic energy right when the block mounts the ramp and the total kinetic + potential energy right when the block reaches its highest point. </p>
<p>And 16 depends on 15.</p>
<p>Never gotten above a 15, but I’m a reckless guesser.</p>