The general term of the expansion of $(1+ax)^5$ is $T_{r+1} = \binom{5}{r}a^rx^r$.

Therefore, the term containing $x$ is $\binom{5}{1}ax = 5ax$.

According to the problem, we have $5a = 10$, which implies $a = 2$.

Then, $b = \binom{5}{2}a^2 = 10 \times 4 = 40$.

Hence, the answer is $\boxed{40}$.