Firing machine question 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 $a_n-a_{n-1}= T \cdot \frac{{p-1+N}}{p}$.
 A: 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.) 
