I guess I can't leave comments just yet so this will have to be in the form of an answer:
Assuming that the distributions have the same number of degrees of freedom ($n$), I think the answer will look something like (with some abuse of notation)
$\mathcal{H}(k_aA + t_a \mid k_bB + t_b) = \log(\frac{k_b}{k_a}) + \frac{n+1}{2} \left( E[ \log(1 + (\frac{Yk_a + t_a - t_b}{k_b})^2 \frac{1}{n}) ] - E[ \log (1 + \frac{Y^2}{n}) ]\right) $,
where the r.v. $Y$ has the Student's t distribution with $n$ degrees of freedom. If the number of degrees of freedom are assumed to be different between the two distributions you will get additional terms that correspond to the logarithm of the "$\Gamma$-parts" of the density functions and in the above formula the $n$'s will be changed accordingly. Expectation will still be w.r.t. the Student's t with $n$ deg. of freedom (or whatever degree that is associated with the r.v. $A$).
OK, let me fully address the question since there is no easy way out. The
normal approach is to maximize the "likelihood" of the data under the
parameter. The key question here is how to define likelihood for a
mixed distribution. Let's use the standard approach as our guide.
Parameter estimation is usually based on the idea that we want to
choose parameters that make our data "the most likely." For a discrete
probability distribution, we interpret this to mean that our data is
the most probable. But this breaks down in the case of continuous
probability distributions, where, no matter our choice of parameters,
our data has probability zero.
Statisticians thus replace the probability with the probability
density for continuous distributions. Here is the justification for
this. Instead of actually having a set of numbers drawn from the
probability distribution, you have a highly accurate
measurement---say, your sequence $\{x_i\}$ for $i = 1,\dots,n$ tells
you that the true value of the (still unknown) sequence $\{g_i\}$ satisfies $|x_i -
g_i| < \varepsilon$ for all $i$. When $\varepsilon$ is sufficiently
small, the replacement
$$
\mathbb{P}(|x_i - g_i|) < \varepsilon )\approx \varepsilon p_{g}(x_i)
$$
is very accurate, where $p_g$ is the pdf of $g_i$. Assuming that
your sequence is iid, we are led to the approximation
$$
\mathbb{P}(|x_i - g_i| < \varepsilon \text{ for all } i)
\approx \varepsilon^n \prod_{i=1}^n p_g(x_i).
$$
We thus choose the pdf from our family which maximizes the right
hand side of the above equation, reproducing the standard maximum
likelihood method.
Now the question is, what do we do with mixed distributions? When
there is a mass at a point $x_i$, that is $\mathbb{P}(x_i=g_i) > 0$,
our first approximation is incorrect; for very small $\varepsilon$, we
have the approximation
$$ \mathbb{P}(|x_i - g_i| < \varepsilon) \approx \mathbb{P}(x_i = g_i)
$$
If we let $\mathcal{N}$ be the index set of the "massless" samples, we can approximate
the probability of our data as
$$ \mathbb{P}(|x_i - g_i| < \varepsilon) \approx \varepsilon^n
\prod_{i \in \mathcal{N}} p_g(x_i) \prod_{i \notin \mathcal{N}} \mathbb{P}(x_i = g_i).
$$
where $n$ is the number of elements in $\mathcal{N}$. That is, we can reasonably define our maximum likelihood estimate
for a parameter $m$ as the value of the parameter that maximizes
$$
\prod_{i \in \mathcal{N}} p_g(x_i) \prod_{i \notin \mathcal{N}} \mathbb{P}(x_i = g_i).
$$
In your case, it is fairly simple to write down the value of the likelihood function above. First, note that
$$\mathbb{P}(x=0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-m}
e^{-x^2/2}dx.$$
For $x>0$, you have the standard Gaussian pdf
$p_g(x) = \tfrac{1}{\sqrt{2\pi}} e^{-(x-m)^2/2}$.
I won't do any more here; suffice it to say that the standard approach
to maximizing the likelihood involves taking the logarithm of the likelihood function
and setting its derivative to zero. You will probably get a
transcendental equation that you will need to solve numerically.
Best Answer
I think you want the law of total variance:
The variance of $\operatorname{Bin}(n,X)$ where $X$ is uniform on $[a,b]$ is $$\begin{eqnarray} \operatorname{Var}(\operatorname{Bin}(n,X))&=&E_X[\operatorname{Var}(\operatorname{Bin}(n,X)|X)] + \operatorname{Var}_X(E[\operatorname{Bin}(n,X)|X])\newline &=& E[nX(1-X)] + \operatorname{Var}(nX) \newline &=& n(E[X]-E[X^2]) + n^2 \operatorname{Var}(X) \newline & =& n \left( \frac{(a+b)}{2} - \frac{(a^2+ab+b^2)}{3}\right) + n^2 \frac{(b-a)^2}{12} \end{eqnarray}$$
For example, if $[a,b] =[1/20,1/10]$, this is $\frac{83n}{1200} + \frac{n^2}{4800} = 0.0691 n + 0.000208n^2$.
If $X$ takes the values $a$ and $b$ with probabilities equal to $1/2$, then
$$\begin{eqnarray} \operatorname{Var}(\operatorname{Bin}(n,X))&=&E_X[\operatorname{Var}(\operatorname{Bin}(n,X)|X)] + \operatorname{Var}_X(E[\operatorname{Bin}(n,X)|X])\newline &=& n\left(\frac{a(1-a)+b(1-b)}{2}\right) + n^2 \frac{(b-a)^2}{4}\end{eqnarray}$$
If $a=1/20$ and $b=1/10$, this is $\frac{11n}{160} + \frac{n^2}{1600} =0.0688 n + 0.000625n^2$.