Consider the category of torsion abelian groups. This category doesn't have enough projectives by the following argument.
Suppose $C_2$ (cyclic group of order 2) is the homomorphic image of a projective $P$, where $x\in P$ maps to the generator of $C_2$. There is a surjective map $C_{2^k}\to C_2$ by sending one generator to the other. Since $P$ is projective, it lifts to a map $P\to C_{2^k}$ sending $x$ to some generator. Hence $x$ has order at least $2^k$. Since $k$ is arbitrary, $x$ has unbounded order, a contradiction since $P$ is torsion.
Question: does this category have any (non-trivial) projectives? If so, what is an example?
Let $P$ be a nonzero projective in the category. Consider its injective envelope $D$ in the category of abelian groups, which is a torsion divisible module. Thus $D$ is a direct sum of Prüfer groups and so there is a nonzero morphism $f\colon P\to G$ where $G$ is a Prüfer group, say $G=\mathbb{Z}(p^\infty)$, for some $p$.
Let $G_0=\{0\}\subset G_1\subset G_2\subset\dotsb$ be the chain of subgroups of $G$; then, since $f\ne0$, there is $x\in P$ such that $f(x)\in G_1$, $f(x)\ne0$.
Now, for each $n>0$, the multiplication by $p^{n-1}$ defines a surjective morphism $\mu_n\colon\mathbb{Z}(p^\infty)\to\mathbb{Z}(p^\infty)$ and the morphism $f$ lifts to a morphism $g_n\colon P\to\mathbb{Z}(p^\infty)$ such that $\mu_n\circ g_n=f$.
In particular $g_n(x)\in G_n$ (otherwise $\mu_n(g_n(x))=0$), so the order of $x$ must be greater than $p^n$.
This is a contradiction, so $P$ doesn't exist.