Curly arrow mechanism for reaction between sodium borohydride and water?

I was asked to compare the reaction between sodium hydride and water and sodium borohydride and water. I know that the former acts more violently than the latter, but apparently, I must use a curly arrow mechanism to show this.

I think that the HOMO for the latter reaction is the oxygen lone pairs, and the LUMO is the B-H bond, but I'm having difficulty moving on from there.

What attacks what first?

• I don’t see much point in a curly-arrow mechanism for the reaction of $\ce{NaBH4}$ with water. And I don’t see the point of doing it with $\ce{NaH}$, either. In one case, you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’ – Jan Oct 2 '15 at 22:17

Their mechanisms are easily searchable online.

This is from 2009 from art-xy.com:

Notice that the reaction would be different in the presence of metal catalysts:

$$\ce{NaBH4 + 2H2O ->[metal catalyst] NaBO2 + 4H2}$$

This is from 2007 from chembook.co.uk:

Which reaction is more violent is already explained by Jan:

In one case [$$\ce{NaH}$$], you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case [$$\ce{NaBH4}$$] you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’

Therefore the former is more violent.