Consider the problem of projectile motion in 2 dimensions. Launch angle is constant. Range of projectile, $x$, then depends only on launch speed, $v$, and is given by
\begin{equation}
x=v^2, \quad v\in [0,1] \tag{1}
\end{equation}
Above equation has been non-dimensionalised (by taking maximum range as our length scale, and maximum launch speed as our velocity scale), so all quantities are dimensionless. Probability density function for launch speed is assumed uniform over the interval $[0,1]$:
\begin{equation}
f(v)=1, \quad \textrm{if}~v\in [0,1]\tag{2}
\end{equation}
and zero otherwise. I want to find p.d.f for range of projectile, $x$. An easy way of doing this 
\begin{equation}
f(x)=\left| \frac{dv}{dx}\right|f(v)=\frac{1}{2\sqrt{x}}, \quad x\in [0,1]\tag{3}
\end{equation}

However I wanted to solve the same problem using Dirac delta function:
\begin{align}
f(x) & =\int\_0^1 dv~f(x|v)~f(v) \\
& = \int\_0^1 dv~f(x|v) \\
& = \int\_0^1 dv~\delta(v^2-x)\tag{4}
\end{align}
Here $f(~|~)$ denotes conditional p.d.f.. Last line was arrived at because for given value of $v$, it is certain that we shall obtain that value of $x$ that satisfies the equation $v^2-x=0$. Now I make use of the identity for delta function
\begin{align}
\delta(g(x))=\sum\_i \frac{\delta(x-x\_i)}{|g'(x\_i)|}\tag{5}
\end{align}
Here $x\_i$ are roots of function $g(x)$, and $g'\equiv \dfrac{dg}{dx}$. Now $g(v)=v^2-x$, whose roots are $\pm \sqrt{x}$. We reject the negative root because $v\geq 0$. $g'=2v$. Hence
\begin{align}
f(x) & =\int\_0^1 dv~\delta(v^2-x) \\
& = \int\_0^1 dv~\frac{1}{2\sqrt{x}}\delta(v-\sqrt{x}) \\
& = \frac{1}{2\sqrt{x}}\tag{6}
\end{align}
which is correct.

However instead of $f(x|v)=\delta(v^2-x)$, we could equally well have begun with the equation $f(x|v)=\delta(v-\sqrt{x})$, because at least according to me, physical content of both equations is identical. However the last choice yields a completely different p.d.f.:
\begin{align}
f(x) & =\int\_0^1 dv~\delta(v-\sqrt{x})=1\tag{7}
\end{align}
I don't think I have done anything wrong mathematically (if I have, please point out). To a mathematician of course the two functions are different, and so the fact that they yielded different p.d.f.s is not surprising. But when the equations are put in their proper physical context, both have identical physical content (as far as I can see). This example makes me wonder if Dirac Delta function may be used unambiguously in solving physical problems. While this was a simple problem where a second method of solution was available and so we could compare, what does one do in more complicated situations where such a comparison is not possible?