Number of planks required to stop the bullet 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?
 A: 
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 $N=\infty$.
A: 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.
A: 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.
