In this model extension ever holds CH

Question

Let $\kappa$ be a measurable cardinal and consider the collapsing forcing $\mathbb{P}:=Coll(\omega_1,<\kappa)$. I know that for every generic extension $V^{\mathbb{P}}$ we get $\kappa=\aleph_2$. But, is it true that $V^{\mathbb{P}}\models CH$?

If the answer is negative. Which combinatorics properties we need to assume in $V$ to get a counter example?

Thanks for advance

Answer 1

Yes, it is true.

1. You are not adding any new reals since your forcing is $\sigma$-closed.
2. Whatever the continuum was in the ground model, it was well-below $\kappa$, so it must have collapsed to $\omega_1$ in the process.

$\sf 1+2 = CH$.

(Note that we don't really need that $\kappa$ is measurable, just larger than the continuum.)