If an object the mass of the moon was to hit earth's surface at an angle that would be in the opposite direction of earth's spin, how can you find how much momentum the object needs to cause earth to spin in the opposite direction?
A simplified answer is obtained by ignoring all possible effects and dependencies other that the mass of the earth (M), its angular velocity ($w$), the mass of the impacting object($m$,) and its velocity ($v$).
The rotational energy of the earth is $I w^2/2$, the kinetic energy of the object is $mv^2/2$. To stop the rotation and then make it go in reverse, kinetic energy must be twice the rotational energy, therefore :$$mv^2/2 = Iw^2$$ Since the moment of inertia (I) for a solid sphere is $2Mr^2/5$ and $w^2 =v^2/r^2 $, substituting in the above equation, one gets: $$ mv^2/2 = (2Mr^2/5) (v^2/r^2)$$simplifying gives:$$m=.8M$$ The speed $v$ (at the equator) is obtained from the known earth parameters = 464m/s. So, an object 80% of the mass of the earth, with a speed of 464 m/s, hitting tangentially at the equator (and "sticking" to it), should do the job.