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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 $a_n-a_{n-1}= T \cdot \frac{{p-1+N}}{p}$.

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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.)

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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

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