Imagine a bullet fired at a series of partitions stacked one after the other. Given that the bullet looses half its velocity in crossing each partition, velocity of the same is a geometric progression(GP) with $a= v_0$ and $r=0.5$.

The question is, how many partitions is it supposed to successfully cross, or at what partition would the bullet be stopped?

  • $\begingroup$ There's something missing from the question. What's the minimum velocity needed to break through a single partition? $\endgroup$
    – 410 gone
    Apr 2, 2012 at 12:03
  • 2
    $\begingroup$ Are you trying to recreate Zeno's paradox? $\endgroup$
    – Alexander
    Jun 1, 2012 at 11:07

1 Answer 1


Looks like the answer is $\infty$/"never", since the velocity after the $i^{\rm{th}}$ partition is $v_i=ar^{i-1}$ by normal GP formulae; and an exponential can only approach zero. But you probably knew that.

Maybe there's a mistake in the question?

  • $\begingroup$ Yup, that's why I asked it. I distinctly remember it having a specific answer, and probably had something to do with losing Kinetic energy at each collision. $\endgroup$ Apr 2, 2012 at 7:27
  • $\begingroup$ Does this make any sense? the amount of kinetic energy lost at each collision is 3/4 of the energy before the collision, and the total energy to be lost is one unit, and hence the bullet with stop at the second partition. I'd have to recheck the calculations however. $\endgroup$ Apr 2, 2012 at 7:50
  • $\begingroup$ @Vaibhav No, you'll get an infinite series there as well. The common ratio will be $\frac34$ instead. Are you sure that the fraction $\frac12$ was of the velocity before hitting the plate or of the initial velocity? In the latter case the problem is trivial, answer is 2 (you get an AP, not a GP) $\endgroup$ Apr 2, 2012 at 7:53
  • 1
    $\begingroup$ It is an infinite series, but it will take a finite time. $\endgroup$
    – Bernhard
    Apr 2, 2012 at 8:11
  • $\begingroup$ @Bernhard: Uhh, how? Its getting slower, not faster. If it gets slower, it will take more and more time to reach the next barrier. So the time taken will be $\sum\limits_{i=0}^\infty \frac{d}{v}=\sum \frac{d\times \Large{2^i}}{v_0}=\infty$ $\endgroup$ Apr 2, 2012 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.