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 ▼ Howard Owen Posting Freak Posts: 1,830 Threads: 113 Joined: Aug 2005 01-18-2013, 12:54 PM This is a joke, of course. What's the real answer? ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 01-18-2013, 02:11 PM ```t = 0.4503 + 2*sqrt(0.4)/9.81*(sqrt(1489.5) + sqrt(1488.5) + sqrt(1487.5) + sqrt(1486.5) + ... + sqrt(1.5) + sqrt(0.5)) seconds ``` This has to be checked though. ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 01-19-2013, 12:02 PM I would prefer the vertical shot, or at least letting the ball being dropped in free fall (which would not affect the result). Less probability of it being stopped by sticky dead organic matter :-)  Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-18-2013, 02:48 PM At the beginning, the ball has an energy of Wo = 2.5 x 20^2 + 5 x 9.8 x 10 = 1490 J. It needs a time of to to touch the ground the first time. Then it rebounces with an energy of W1 = 1489 J to a peak height of h1 = W1 / (m x g). Raising to h1 takes the same time as falling from h1, so t1 = 2 (2 x W1 / (m x g^2) )^(1/2) = 0.129 x SQRT(W1). Adding all the bounces results in T = to + 0.129 x SUM with SUM being the sum from 1 to W1 of all energy square roots. Now I've to find to and said sum ... d:-) ▼ Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-18-2013, 03:02 PM OK, to = - vo / g + SQRT((vo / g)^2 + 2 ho / g) = 0.450s. Adding all the square roots for the sum sounds like brute force ... d:-/ Edited: 18 Jan 2013, 3:03 p.m. ▼ Thomas Ritschel Junior Member Posts: 17 Threads: 1 Joined: Nov 2012 01-18-2013, 03:11 PM Quote: Adding all the square roots for the sum sounds like brute force ... I've got 933,66 seconds (after about 50 secs on a HP 50g). Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 01-18-2013, 05:09 PM Quote: Adding all the square roots for the sum sounds like brute force ... Integration gives a fair approximation: ```Sum(n=1,1489,sqrt(n)) = 38323.6742768 |x=1489 2/3*x*sqrt(x)| = 38303.9206363 |x=0 ``` Nigel J Dowrick Member Posts: 167 Threads: 13 Joined: Sep 2008 01-18-2013, 03:09 PM I think we need more information. Without knowing its direction of motion after each bounce I don't see how to calculate the height it reaches, nor the time it takes to return to the ground. We can assume that the horizontal component of velocity remains unchanged, or that both components are reduced in proportion so that the direction doesn't change, but I think some such assumption is needed. Nigel (UK) ▼ Thomas Ritschel Junior Member Posts: 17 Threads: 1 Joined: Nov 2012 01-18-2013, 03:14 PM I think we can separate the horizontal and the vertical motion. Then after each bounce it's maximum height is reduced by 1_J/(m*g). This results in 490.33 bounces. Edited: 18 Jan 2013, 3:16 p.m. ▼ Nigel J Dowrick Member Posts: 167 Threads: 13 Joined: Sep 2008 01-18-2013, 03:31 PM This is equivalent to assuming that the force from the ground is always vertical. In this case the horizontal motion makes no difference - I'm not sure that this is what the question intends. Or maybe it is! I agree that this is the most natural assumption to make. Nigel (UK) Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-18-2013, 04:59 PM Hmmh, f...g language! >:-( I've assumed the ball is shot vertically right to the ground and misinterpreted the "ball is shot directly right". Such a horizontal shot will not change Wo but the other calculations. So to = SQRT(2 ho / g) = 1.429s. And right after first touch down, the velocity corresponding to the kinetic energy of 1489 J must be split into a vertical and horizontal component according to the angle. Right before touch down, horizontal velocity is 20 m/s, and vertical is SQRT(2 g ho) = 14.007 m/s. This gives an angle of arctan(0.7) = 35.006°. The outgoing ball has a smaller total velocity, but is reflected conserving this angle. Now, that's looking like work now ... d:-I ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 01-18-2013, 05:11 PM Quote: I've assumed the ball is shot vertically right to the ground and misinterpreted the "ball is shot directly right". So have I :-( Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-18-2013, 05:27 PM Luckily, it's easier than it first looked: when total velocity is reduced by a factor of SQRT(Wo / W1), each of its components is reduced by the same factor due to geometric reasons. So the squared vertical velocity vv1^2 = vvo^2 x 1489 / 1490, vv2^2 = vv1^2 x 1488 / 1489, ... vv1490^2 = vv1489^2 x 1 / 2. And the time interval between the first two bounces is simply dt1,2 = 2 vv1 / g. So we end with T = 1.429s + 0.204 x (vv1 + vv2 + ... + vv1490). I frankly admit I'm too lazy to calculate that sum. d:-> Edited: 18 Jan 2013, 5:32 p.m. ▼ Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-19-2013, 04:00 AM As can be seen from the post above, T = 1.429s + 0.204 x (vv1 + vv2 + ... + vv1490) = 1.429s + 0.204 x (vvo / SQRT(1490)) x SUM = 1.429s + 0.07403 x SUM with SUM being the sum of all SQRT(i) from i=1 to 1489. So I eventually took the [Sigma] function of my WP 34S, wrote a micro-program ```LBL'S' SQRT RTN ``` , entered 1489.00001, called [Sigma]'S', and got 38323.7 after some 65s. Thus I return with T = 2838s. d:-) Edited to correct a calculation error found by doublechecking. Edited: 19 Jan 2013, 6:09 p.m. after one or more responses were posted ▼ fhub Posting Freak Posts: 1,216 Threads: 75 Joined: Jun 2011 01-19-2013, 08:30 AM Quote: Thus I return with T = 204s. Well, I would say that's almost impossible. ;-) If you only consider the vertical part (i.e. ignore any energy loss horizontally), then the total time is 933.82 sec (which is quite easy to calculate). But if you take the full (vertical+horizontal) movement, then the energy loss splits to both directions, so in the vertical direction there's definitely less than 1J lost at every bounce (I got a ratio of dEy :dEx = 0.3291 : 0.6709 from the speed ratio v0y:v0x), and this means that the total time must increase of course - I got a total time of 2837.47 sec in this case. My only problem is that I'm not yet sure which of both solutions are indeed correct, in other words: is really only the vertical direction affected by the 1J energy loss? Franz Edited: 19 Jan 2013, 8:35 a.m. Bill Carter Member Posts: 54 Threads: 0 Joined: Aug 2011 01-19-2013, 01:42 AM Walter- I would ignore the horizontal component. If there's no air friction, why would there be ground friction or rolling resistance? If there were either of those things, the horizontal component would cause the ball to spin on impact. In doing so, at least some the kinetic energy from the horizontal component would be converted to rotational kinetic energy rather than being "lost". On each bounce, the relative velocity of the surface of the ball contacting the Earth would decrease, so the constant angle approach would fall apart. The only "self consistent" approach is to assume that the horizontal component is constant. Thinking of it another way, suppose the ball rolled out of whatever it was "shot" from. While the center of the ball has a horizontal velocity of 20 m/s, the bottom of the ball has a horizontal velocity of 0. When the ball bounces, there is no difference in horizontal components between the ball's surface and the ground, so no energy is transferred/lost in the horizontal. -Bill John M (Ottawa) Junior Member Posts: 6 Threads: 2 Joined: Oct 2011 01-18-2013, 06:49 PM   hugh steers Senior Member Posts: 536 Threads: 56 Joined: Jul 2005 01-18-2013, 07:29 PM 933.8 seconds? ```mass = 5 g = 9.81 sqrt = math.sqrt function fall(h0) -- first drop local t = sqrt(2*h0/g) -- start energy - 1, first bounce local e = h0*mass*g - 1 local h local c = 0 -- each bounce while e > 0 do c = c + 1 -- height reached with remaining E h = e/(g*mass) print(c, "bounce height", h) -- time is TWICE for up and down t = t + sqrt(2*h/g)*2 -- up then down again -- lose energy e = e - 1 end print("total time", t) end ``` Chris Dreher Junior Member Posts: 18 Threads: 5 Joined: Aug 2012 01-19-2013, 01:26 AM If you go with the dead body scenario, the 1.4s answer is correct. :-) ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 01-19-2013, 11:55 AM The problem said "assume no air resistance", not "assume no air at all". D- would be fair enough, I think :-) Bunuel66 Member Posts: 59 Threads: 5 Joined: Jul 2011 01-19-2013, 02:59 PM The text says 'stop bouncing', not 'stop moving'... Assuming no air resistance and no friction (let say that the energy is dissipated only by mechanical deforming of the ball). Then the horizontal component doesn't affect the problem, the ball will keep its horizontal speed even after its last bounce. At start its potential energy is m*g*h0=5*9.81*10=490.5 J=E0 Falling from h0 to ground takes: sqrt(2*h0/g)~2.04s=t0 Then the following bounce is initiated with an energy of: E1=489.5 J The time needed by the ball to impact the ground again is t1=(2/g)sqrt(2*E1/m) (beware that t0 is only one way down...) By iteration: T=2.04+(2/9.81)sqrt(2/5)[sqrt(489.5)+sqrt(488.5)+..+sqrt(0.5)] The last term of the sum is <1 but we assume then a complete dissipation of the energy. I found T~931.3s My 2 cents.... Dave Shaffer (Arizona) Posting Freak Posts: 776 Threads: 25 Joined: Jun 2007 01-19-2013, 09:30 PM I used to teach this stuff - this is an exceedingly poorly-stated (but interesting) problem. As I see it, there is no obvious solution. You need to make assumptions not stated in the formulation of the problem. Various of the answers above address my 3 possibilities below. Take your choice! In fact, I rather doubt that it is a real physics problem (if so, and if I were Department Chair, I would sit the original poser of the question down and ask him to refine his problem statement!), and that it is only a set-up for the "dead bodies" answer. Otherwaise, there are three ways to attack it, all of which involve assumptions (which is why it is poorly stated). 1) You can chose to ignore Y effects - in which case only the X speed changes, by a kinetic energy change of 1 joule per bounce 2) You can chose to ignore X effects - in which case the X speed does not diminish, and the Y speed/height of bounce changes by 1 joule of kinetic energy per bounce or 3) You can chose that the 1 joule loss is apportioned (but you must still state your assumptions as to how much!) to both X and Y speeds. (there is a 4th mode, noted above, that the ball would start to rotate and some energy would go into spinning up the ball - but since we don't have a size for the ball, you can not really address this - i.e. you can not calculate the moment of inertia!) If I had to pick, I would go for Walter's equal-angle bounce version above - it is my mode 3. However, I see no logical reason, other than by stating your case, that the bounce take-off angle should match the incoming angle - this requires an assumption as to the apportionment of energy loss to the X and Y components. (Light from a mirror bounces this way, but real-world balls could do almost anything!) The question poser gets the failing grade!!!! ▼ Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-20-2013, 04:22 AM Usually such problems are stated in discussing so-called 'point mechanics'. So we assume a 5kg ball of neglegible radius ... (so it can't take rotation energy - but don't ask me how it can bounce then). And about the ways you call 1 and 2: How shall the ball know it may only dissipate x- *or* y-energy? That's the reason I chose the way you call way 3. As Albert stated: "Make everything as simple as possible but not simpler." d;-) Edited: 20 Jan 2013, 4:36 a.m. ▼ fhub Posting Freak Posts: 1,216 Threads: 75 Joined: Jun 2011 01-20-2013, 08:08 AM Quote: And about the ways you call 1 and 2: How shall the ball know it may only dissipate x- *or* y-energy? That's the reason I chose the way you call way 3. And chosing this way 3) is indeed certainly better (i.e. moré realistic) than 1) or 2). BUT - no matter how you distribute the energy loss of 1J at each bounce to the x- and y-direction, you can never get less than T=933.82 sec for the total time. And here's why: Method 2) (where the full energy loss of 1J is taken from the vertical movement) gives the shortest possible time, and this time is 933.82 sec (Bunuel66 has explained it quite detailled in message #22, only his value is a bit inaccurate). The y-energy is 490.5J at the beginning, losing 1J at every bounce, there are 490 bounces (stopping at the 491.). Now if you split this 1J energy-loss to both (x and y) directions - no matter in which ratio - this means that the energy-loss in vertical (y) direction is less than with method 2) above. This results in _more_ bounces (than 490) AND in _longer_ single times between each bounces (because of higher y-energy). So of course summing up _more_ bounces with _longer_ times can't result in shorter total time (than the lowest limit of 933.82 sec). The method I've used in my last answer (to your posting) above was to split this 1J energy-loss to Ex and Ey in the same ratio as both energies: dEy:dEx = Ey:Ex = (m*g*h): (m/2*v0^2) This together with dEx+dEy=1 gives dEy=0.3291J Since now this 1J loss is taken from the whole energy (which is 1490.5J at the beginning), there are now 1490 bounces and the complete formula is the following: T = t0+2*SUM(ti,i=1..1490) = sqrt(2*h/g)+2/g*sqrt(2/m)*SUM(sqrt(m*g*h-i*0.3291),i=1..1490) = 2837.47 sec Of course this (my) kind of energy splitting (dEy:dEx = Ey:Ex) is not the only possible way since the original problem is indeed very poorly stated (as Dave already mentioned), but my main point was that any total time less than about 933 sec is in fact impossible! Franz ▼ Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-20-2013, 08:17 AM You're pushing at an open door (see above) - you may save your kinetic energy for more valuable tasks. d:-) ▼ fhub Posting Freak Posts: 1,216 Threads: 75 Joined: Jun 2011 01-20-2013, 08:48 AM Quote: You're pushing at an open door (see above) - you may save your kinetic energy for more valuable tasks. Is that all you can say about your obvious false result? ;-) Having several different ways to split this energy-loss into x- and y-directions doesn't change anything in my statement that the absolute minimum total time is 933.82 sec, and that 204 sec is definitely wrong, point. And BTW, YOU asked me to show my calculation! Nevertheless have a nice Sunday, ;-) Franz Edited: 20 Jan 2013, 8:49 a.m. ▼ Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-20-2013, 09:44 AM Just for the record: I did my (necessary) correction job before you unveiled anything. So I don't understand your post of today 8:08 a.m. nor the one right now. The IMHO aggressive attitude shown seems having causes which might be interesting for specialists in other fields - but, alas, that's an old topic. d:-I ▼ fhub Posting Freak Posts: 1,216 Threads: 75 Joined: Jun 2011 01-20-2013, 10:03 AM Quote: Just for the record: I did my (necessary) correction job before you unveiled anything. So I don't understand your post of today 8:08 a.m. nor the one right now. Well, my posting yesterday telling you about the impossibility of your result (and giving my own result) was definitely long before you've made any corrections to your calculation - just look at the times of my posting and your correction! And about today - you don't really expect that I re-read every posting every day just to see if there may have been made any changes/corrections, do you? Quote: The IMHO aggressive attitude shown seems having causes which might be interesting for specialists in other fields - but, alas, that's an old topic. Sorry, but I dont see any aggressiveness in my posting at all - I just wondered a bit about you, first asking for my calcualtions and then saying I shouldn't waste my time with it. But your antipathy against me (and Austrians in general) is well known here ... Franz ▼ Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-20-2013, 10:21 AM I don't remember having read any other Austrian here so I won't blame a whole nation for one ... but I'll better be careful before the Austrian navy will kill me. d;-) Bunuel66 Member Posts: 59 Threads: 5 Joined: Jul 2011 01-20-2013, 06:43 PM The 'inaccuracy' of my result is may be due to the fact that all the bounces are symmetric (up and down) except the first (down only), that the last bounce occurs with less than 1 J, and the value of g taken as 9.81 ms-2. With the same hypothesis, I'll be glad to see where the difference lies. That said, if part of the energy is dissipated in the horizontal direction, each bounce will dissipate less in the vertical one and it will take longer. ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 01-20-2013, 07:01 PM Also, you've forgotten to extract the square root in Quote: sqrt(2*h0/g)~2.04s=t0 The difference is only 0.61 seconds, however. Regards, Gerson. fhub Posting Freak Posts: 1,216 Threads: 75 Joined: Jun 2011 01-21-2013, 04:51 AM Quote: I'll be glad to see where the difference lies. Well, there's nothing wrong at all in your solution of this problem, and you've explained it indeed very well. The 'inaccuracy' is just in 2 points: First, in t0=2.04 you forgot the 'sqrt()' (as Gerson already mentioned), and then also for your total formula T=2.04+(2/9.81)sqrt(2/5)[sqrt(489.5)+sqrt(488.5)+..+sqrt(0.5)] the result would be 934.4 instead of 931.3. But else everything is ok ... :-) Franz ▼ Bunuel66 Member Posts: 59 Threads: 5 Joined: Jul 2011 01-21-2013, 03:30 PM Thank's for the comment. Actually, I made two mistakes, the first one is the square root as mentioned, the second one is in the sum: for some funny reason rather to use a HP calculator I wrote a quick and dirty program in python and I made a mistake in the range of the loop. With those corrections, I got 933.82 which seems to be consistent with 934.4 where the square root correction is missing. Conclusion, don't try to run faster than your legs...;-) Csaba Tizedes (Hungary) Member Posts: 59 Threads: 9 Joined: Apr 2008 01-23-2013, 01:56 PM My solution is 2836.9s, the steps find below: For running program set variables as 6.) point, clear stat variables, clear PRGM, press R/S then make a coffee and enjoy... After 11min 58sec continous RUNNING my calc stopped with Error0 message, this means the total energy of ball not enough to more bouncing. You can read out number of bounces in stat reg R1: RCL1: 1490 (that is correct!!), and the total bouncing time in R2: RCL2: 2836.2s This is the total time without t0. t0 is simply calculated from free-fall equations: t0=1/2*SQRT(2*10m/(9.81N/kg))=0.71s, the total time of moving is 2836.9sec. Csaba ▼ Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 01-23-2013, 02:43 PM Fine - but to = 1.429s. d;-) ▼ Csaba Tizedes (Hungary) Member Posts: 59 Threads: 9 Joined: Apr 2008 01-23-2013, 02:51 PM :-D yes, you're totally right, I was fallen into my trap: the 'half'-bouncing duration confused me..., so the right solution is 2836.2+1.4=2837.6s Thanks! Csaba David Hayden Senior Member Posts: 528 Threads: 40 Joined: Dec 2008 01-23-2013, 11:02 PM I might be missing something, but it appears to me that your first bounce hasn't accounted for the potential energy converted to kinetic energy when the ball falls. Isn't the energy at the first bounce 1/2 m v0^2 + m*g*h0? Dave ▼ Csaba Tizedes (Hungary) Member Posts: 59 Threads: 9 Joined: Apr 2008 01-24-2013, 03:33 AM Hi! When we set initial values for variables the velocity calculated from total energy: ```v0=SQRT(2*Etot/m)=SQRT(2*1490.5J/5kg)=24.42m/s STO .3 vy0=SQRT(2*g*H0)=SQRT(2*9.81N/kg*10m)=14.01m/s STO .4 ``` To set initial values R.0, R.1, R.2: ```RCL .4 RCL .3 / STO .0 2 ENTER 5 / STO .1 2 ENTER 9.81 / STO .2 ``` Then CLEAR SUM, CLEAR PGM and R/S When you get Error0, press RCL 1 for number of bounces (1490), and the total bouncing time RCL 2 (2836.2s), then add 1.4s for the 0th half bouncing and you'll get the right answer: 2837.6s, approx. 47-48 minutes.

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