Number of planks required to stop the bullet [closed]

A bullet looses (1/n)th of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be?

Logically, to me the answer seems to be infinity, as always a fraction of velocity will get reduced. But in my book the answer is n^2/(2n-1) (that comes from energy balance). What is correct?

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closed as off-topic by John Rennie, Kyle Kanos, Brandon Enright, jinawee, Waffle's Crazy PeanutMay 1 '14 at 11:14

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The question is slightly misleading. It should say A bullet loses (1/n)th of its velocity passing through the first plank. The question is also assuming the energy loss per unit distance travelled through the planks is constant. – John Rennie Apr 28 '14 at 19:41
Actually this question appeared in a national level exam and the answer given in the key released by them is n^2/(2n-1). This question appeared as is, and I seriously think the answer should be infinite. I agree with your point but I don't think the question is telling that energy loss is constant. I have spent $16 and challenged this question. What do you think should be the correct answer with the language of question provided? – Ayush Apr 28 '14 at 19:56 Is this the exact wording of the question? If so, it's not entirely clear what the solution is supposed to even be. With John Rennie's additional wording "assuming the energy loss per unit distance travelled through the planks is constant" the answer comes out to be$\frac{n^2}{2 n-1}$, but without it, it's ambiguous. – DumpsterDoofus Apr 28 '14 at 20:27 @DumpsterDoofus Why is it ambiguous? Isn't the question telling that the bullet always loses 1/n th of its velocity no matter which plank? It's because of the answer provided we are thinking that the question is ambiguous, but the answer can also be wrong in making those assumptions, right? – Ayush Apr 28 '14 at 20:41 The question as described has little to do with the behavior of actual bullets or actual planks, both of which exhibit highly non-linear behaviors. Real planks are not going to absorb a uniform amount of energy independent of bullet velocity, so it would seem grossly unreasonable to require such an assumption without stating it. – supercat Apr 28 '14 at 23:05 3 Answers Ayush: Isn't the question telling that the bullet always loses 1/n th of its velocity no matter which plank? Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing$1/n^\text{th}$of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous). With this assumption, the energy loss becomes $$\Delta E=\frac{1}{2}mv^2-\frac{1}{2}m\left(v-\frac{v}{n}\right)^2$$ and the number of planks$N$becomes $$N=\frac{\frac{1}{2}mv^2}{\Delta E}=\frac{n^2}{2n-1}.$$ Otherwise, if you assume that the bullet loses$1/n^\text{th}$of its velocity per plank, then the answer is$N=\infty$. - The question is ambiguous because you know the answer. But, if you would have not known the answer what do you think is the answer you would have opted for? – Ayush Apr 28 '14 at 21:56 @Ayush: I don't really have an answer for that. Technically I agree that the question is ambiguous. It just isn't a good question. – DumpsterDoofus Apr 28 '14 at 22:03 Okay, thanks :) – Ayush Apr 28 '14 at 22:03 "Based on the answer provided" -> We do not need to base on the answer. The question clearly wants to test if a relationship between plank breaking prowess and bullet velocity change is known. The only reasonable assumption is that breaking planks takes energy. Bullets have lots of kinetic energy. Some relation must exist: (1/2)mv^2. – Phil Apr 29 '14 at 0:36 @Phil: Agreed, and I suspect that the test-writers had that idea in mind (energy loss due to material deformation from penetrating bullet is constant, and thus penetrating each plank takes the same amount of energy). So I guess it depends how literal you want to be: if you're taking the question literally, it has no answer (not enough information), but if you insert some semi-accurate-to-real-life assumptions, it can be answered. Unfortunately, that also makes this more of an opinion question, which are not well-suited to this site. – DumpsterDoofus Apr 29 '14 at 0:43 Even if you consider the same question, I think we will get the answer to be infinity. If we consider initial velocity of the bullet to be$v$, then its velocity after passing through first, second, ....$N$plank will be $$v(1-1/n) , v(1-1/n)^2, v(1-1/n)^3.....$$ respectively. You must notice that velocity of the bullet ceases iff$(1-1/n)^N=0$, then for any value of$N$, the value won't be equal to$0$. Thus,$n$must be equal to$1$in order to satisfy the above condition. If$n=1$, then the bullet losses$(1/1)$th of its velocity when passed through one plank, meaning its velocity remains constant inspite of passing through those planks. Thus, infinity number of planks is required to stop it. Read this extracted paragraph from Feynman's "Surely You're Joking, Mr. Feynman!": ....Then cdtnes the list of problems. It says, "John and his father go out to look at the stars. John sees two blue stars and a red star. His father sees a green star, a violet star, and two yellow stars. What is the total temperat ure of the stars seen by John and his father?"--and I would explode in horror. My wife would talk about the volcano downstairs. That's only an example: it was perpetually like that. Perpetual absurdity! There's no purpose whatsoever in adding the temperature of two stars. Nobody ever does that except, maybe, to then take the average temperature of the stars, but not to find out the total temperature of all the stars! It was awful! All it was was a game to get you to add, and they didn't understand what they were talking about. It was like reading sentences with a few typographical errors, and then suddenly a whole sentence is written backwards. The mathematics was like that. Just hopeless!...... Now you would understand why question is ambiguous (if the above calculation is right), no bullet will remain unaffected even if it passes though plank. - I think so that the answer is wrong and the answer should be infinite as no plank will take away full velocity of the bullet and it would never stop.Assuming that initial velocity is$u$, velocity after passing through first plank is$(u - u/n)$, then if we assume the length of the plank to be$d$, velocity after passing through first plank is$(u - u/n)$.We know that $$v^2 –u^2 = 2as$$ $$(u – u/n)^2 – u^2 = 2as$$ $$-2u^2/n + u^2/n^2 = 2as$$ Let the number of plank required to stop it be$N$: $$u^2 = 2Nas$$ $$N= - u^2/2as$$ $$N = - u^2/(2u^2/n + u^2/n^2)$$ $$N = n^2/(2n-1)$$ But here if we consider the case of second plank then the velocity will be$u – 2u/n – u^2/n^2\$

So the acceleration is not constant which will lead to no solution

So I think that the question has a wrong answer or has not sufficient information

And taking the logical approach the answer should be infinite.

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