Find the vector $\mathbf{v}$ such that
\[\operatorname{proj}_{\begin{pmatrix} 3 \\ 0 \end{pmatrix}} \mathbf{v} = \begin{pmatrix} -5 \\ 0 \end{pmatrix}\]and
\[\operatorname{proj}_{\begin{pmatrix} 1 \\ 3 \end{pmatrix}} \mathbf{v} = \begin{pmatrix} \frac{1}{10} \\ \frac{3}{10} \end{pmatrix}.\]
Let $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}.$

[asy]
usepackage("amsmath");

unitsize(1 cm);

pair P, Q, V;

V = (-5,2);
P = (-5,0);
Q = (1/10,3/10);

draw((-6,0)--(1,0));
draw((0,-1)--(0,3));
draw((0,0)--V,Arrow(6));
draw(V--P,dashed);
draw((0,0)--P,red,Arrow(6));
draw((-1/3,-1)--(1,3));
draw(V--Q,dashed);
draw((0,0)--Q,red,Arrow(6));

label("$\mathbf{v}$", V, W);
label("$\begin{pmatrix} -5 \\ 0 \end{pmatrix}$", P, S);
label("$\begin{pmatrix} \frac{1}{10} \\ \frac{3}{10} \end{pmatrix}$", Q, SE);
[/asy]

Since the projection of $\mathbf{v}$ onto $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ (or equivalently, the $x$-axis) is $\begin{pmatrix} -5 \\ 0 \end{pmatrix},$ we know that $x = -5.$  Then $\mathbf{v} = \begin{pmatrix} -5 \\ y \end{pmatrix}.$

Then by properties of projections,
\[\left( \begin{pmatrix} -5 \\ y \end{pmatrix} - \begin{pmatrix} \frac{1}{10} \\ \frac{3}{10} \end{pmatrix} \right) \cdot \begin{pmatrix} 1 \\ 3 \end{pmatrix} = 0.\]This leads to the equation
\[-\frac{51}{10} + \left( y - \frac{3}{10} \right) \cdot 3 = 0.\]Solving, we find $y = 2.$  Therefore, $\mathbf{v} = \boxed{\begin{pmatrix} -5 \\ 2 \end{pmatrix}}.$