Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Suppose we have a firing machine on a frictionless surface at point $x=0$. It fires a bullet of mass $m$ every $T$ seconds. Each bullet has the same constant velocity $v_0$. There's a body of mass $pm$ ($p$ times larger than the bullet's mass) at point $x=x_0$. I know, according to the law of momentum conservation (inelastic collision) that the velocity of the body as a function of bullets inside of it will be: $v(n)=\frac{v_0 n}{n+p}$.

How can I know the elapsed time $\Delta t$ from the $n-1$ bullet hit to $n$ bullet hit? Is there any way to solve this without a recursion?

What I mean by recursion is this (the answer to the question): $ \left\{ \begin{array}{l} a_1=\frac{x_0}{v_0}\\ a_n=a_{n-1}+ T \cdot \frac{{p-1+N}}{p} \end{array} \right. $

This series represent the time when nth bullet hits the object. I do not quite understand why $a_n-a_{n-1}= T \cdot \frac{{p-1+N}}{p}$.

share|cite|improve this question

1 Answer 1

up vote 3 down vote accepted

This problem has a recursive flavor that we'll not try to avoid.

Conservation of momentum tells us that $$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$ Imposing the boundary condition $v(0)=0$ we find $$v(n) = \frac{n}{n+p}v_0$$ as claimed.

Let $a_n$ be the time at which the $n$th bullet strike occurs. We have $a_1=x_0/v_0$ and $$\begin{equation*} v_0(a_n-(n-1)T) = v_0(a_{n-1}-(n-2)T) + v(n-1)(a_{n}-a_{n-1}) \end{equation*}$$ In words, the distance between the block and the gun at the $n$th strike is the distance between the block and the gun at the $(n-1)$th strike plus the distance the block travels between the strikes. Rearranging we find $$a_n = a_{n-1} + \frac{p+n-1}{p} T$$ as claimed. This recursion can be solved by standard techniques. We find $$a_n = \frac{x_0}{v_0} + \frac{(n-1)(n+2p)}{2p} T.$$ As a consistency check we take the limit where $p$ is large. Then $$a_n \sim \frac{x_0}{v_0} + (n-1)T.$$ This is the result we should expect for bullets fired at an immovable wall.

Addendum: Let $x_n^B$ and $x_n^b$ be the location of the block and the bullet at the $n$th strike, respectively. Note that $x_n^B = x_n^b$ for any $n$. We have $$\begin{eqnarray*} x_1^B &=& x_0 \\ x_1^b &=& v_0 a_1 \\ x_2^B &=& x_1^B + v(1)(a_2-a_1) \\ x_2^b &=& v_0(a_2-T) \\ x_3^B &=& x_2^B + v(2)(a_3-a_2) \\ x_3^b &=& v_0(a_3-2T) \\ &\vdots& \\ x_{n}^B &=& x_{n-1}^B + v(n-1)(a_{n}-a_{n-1}) \\ x_{n}^b &=& v_0(a_{n}-(n-1) T). \end{eqnarray*}$$ Intuitively, the block is where it was before the strike plus the distance it travelled at the new speed before being struck again. The bullets are shot every $T$ seconds so the $n$th bullet is only in flight for $a_n-(n-1)T$ seconds. Thus, we have $$\begin{equation*} v_0(a_n-(n-1)T) = v_0(a_{n-1}-(n-2)T) + v(n-1)(a_{n}-a_{n-1}) \tag{1} \end{equation*}$$ and so $$\begin{equation*} v_0 (a_n - T) = v_0 a_{n-1} + v(n-1)(a_n - a_{n-1})\tag{2} \end{equation*}$$ as claimed. (I'll replace the equation in the original argument above with (1) for clarity.)

share|cite|improve this answer
Thank you very much, sir. It was much easier than I thought. Your solution is quite elegant and neat. Thank you again. – grjj3 May 19 '13 at 7:31
@grjj3: Glad to help. Thank you for the interesting question. – user26872 May 19 '13 at 15:38
@oen - Why the distance between the block and the gun is necessarily $v_0(a_n-T)$? $v_0$ is the velocity of the bullet, and not of the block. Also, the block changes its velocity after each strike, so the distance it moved is supposed to be a far more complicated, isn't it? – www Jun 1 '13 at 22:27
@WalterWhite: Thanks for the question - I was wondering if someone would ask about that relation. I've added some further explanation above. – user26872 Jun 1 '13 at 23:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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