Let $\mathcal{M}$ be a category, let $\mathcal{M}^\circ$ be a reflective (full) subcategory of $\mathcal{M}$, and let $L : \mathcal{M} \to \mathcal{M}^\circ$ be the reflector.
Question. Does there exist a model structure on $\mathcal{M}$ with the following properties?
The weak equivalences in $\mathcal{M}$ are precisely the morphisms inverted by $L$.
Every morphism in $\mathcal{M}$ is a cofibration.
More specifically, does every morphism in $\mathcal{M}$ factor as a weak equivalence (= trivial cofibration) followed by a fibration (= morphism with the rlp wrt the weak equivalences)? This is related to the existence of reflectors for each slice category $\mathcal{M}_{/ S}$:
Proposition. Let $S$ be an object in $\mathcal{M}$. (Assuming $\mathcal{M}$ has cokernel pairs and products with $S$.) The following are equivalent:
Every morphism in $\mathcal{M}$ with codomain $S$ factors as a weak equivalence followed by a fibration.
The full subcategory $(\mathcal{M}_{/ S})^\circ$ of fibrations over $S$ is reflective in $\mathcal{M}_{/ S}$.
Note also that given a morphism $f : X \to Y$ in $\mathcal{M}$ where $Y$ is fibrant, we can factor $f$ as a weak equivalence $\eta_X : X \to L X$ followed by a fibration $(\eta_Y)^{-1} \circ L f : L X \to Y$, where $\eta$ is the unit of the reflector. ($L \eta_X : L X \to L L X$ is an isomorphism, $\eta_Y : Y \to L Y$ is an isomorphism, and every morphism in $\mathcal{M}^\circ$ is a fibration in $\mathcal{M}$.) So the crux of the matter is the factorisation of morphisms with non-fibrant codomain, or equivalently, the reflectivity of $(\mathcal{M}_{/ S})^\circ$ in $\mathcal{M}_{/ S}$ when $S$ is not fibrant.
Such a model structure would be a left Bousfield localisation of the trivial model structure on $\mathcal{M}$ (where the weak equivalences are the isomorphisms). Conversely, every left Bousfield localisation of the trivial model structure on $\mathcal{M}$ can be characterised in this way. I believe it should be possible to apply Smith's theorem when $\mathcal{M}$ is locally presentable and $L$ is accessible as an endofunctor on $\mathcal{M}$. But surely this is true in greater generality?
For example, suppose the class of weak equivalences is closed under pullbacks. (This happens if and only if $L$ preserves pullback squares where either the bottom or right edge is a weak equivalence.) Then, given a morphism $f : X \to Y$ in $\mathcal{M}$, the desired factorisation is $\langle f, \eta_X \rangle : X \to Y \times_{L Y} L X$ followed by the projection $Y \times_{L Y} L X \to Y$: after all, $L f : L X \to L Y$ is a fibration and the class of fibrations is closed under pullbacks, and $\langle f, \eta_X \rangle$ is a weak equivalence because $\eta_X$ is a weak equivalence and the projection $Y \times_{L Y} L X \to L X$ is a pullback of a weak equivalence.
Proof of proposition. Since $\mathcal{M}$ has cokernel pairs and $L : \mathcal{M} \to \mathcal{M}^\circ$ preserves them, a morphism in $\mathcal{M}$ has the rlp wrt the weak equivalences if and only if it is right orthogonal to the weak equivalences.
Suppose given a weak equivalence $e : X \to \hat{X}$, a morphism $f : X \to Y$, and fibrations $p : \hat{X} \to S$ and $q : Y \to S$ in $\mathcal{M}$ such that $q \circ f = p \circ e$. That is, we have the following commutative diagram: $$\require{AMScd} \begin{CD} X @>{f}>> Y \\ @V{e}VV @VV{q}V \\ \hat{X} @>>{p}> S \end{CD}$$ Since $q$ is right orthogonal to $e$, we have a unique lift $\hat{f} : \hat{X} \to Y$ fitting into this commutative square. Hence, for all fibrations $q : Y \to S$, $$e^* : \mathcal{M}_{/ S} ( (\hat{X}, p), (Y, q) ) \to \mathcal{M}_{/ S} ( (X, p \circ e), (Y, q) )$$ is a bijection. Therefore, if every morphism in $\mathcal{M}$ with codomain $S$ factors as a weak equivalence followed by a fibration, then $(\mathcal{M}_{/ S})^\circ$ is reflective in $\mathcal{M}_{/ S}$.
Conversely, suppose $(\mathcal{M}_{/ S})^\circ$ is reflective in $\mathcal{M}_{/ S}$. Consider the unit of the reflector. By definition, it is left orthogonal in $\mathcal{M}_{/ S}$ to every fibrant object in $\mathcal{M}_{/ S}$ (= fibration over $S$). So suppose we have a morphism $f : X \to Y$ and fibration $q : Y \to S$ such that $f$ is left orthogonal to every fibration over $S$. Consider the following commutative diagram: $$\begin{CD} X @>{\langle \eta_X, q \circ f \rangle}>> L X \times S @>>> L X \\ @V{f}VV @VV{L f \times \textrm{id}_S}V @VV{L f}V \\ Y @>>{\langle \eta_Y, q \rangle}> L Y \times S @>>> L Y \\ \end{CD}$$ The projections $L X \times S \to S$ and $L Y \times S \to S$ are fibrations, so by orthogonality we can find a lift $Y \to L X \times S$ filling in the left commutative square. This yields a lift $Y \to L X$ filling in the outer rectangle. But the horizontal composites are weak equivalences, so by 2-out-of-6, $f : X \to Y$ is also a weak equivalence. Hence, the unit of the reflector $\mathcal{M}_{/ S} \to (\mathcal{M}_{/ S})^\circ$ is a natural weak equivalence, so every morphism in $\mathcal{M}$ with codomain $S$ factors as a weak equivalence followed by a fibration. ◼
