No surjective homomorphism from $G=⟨x,y|x^2=y^5⟩$ to $F(a,b)$ 
Revising for an exam. I have answers for parts i) - iv)
i) Sending $a$ to $[x]$ and $b$ to $[y]$ induces a homomorphism from $F(a,b)$ to $G$, and clearly any word $w$ in $F(x,y)$ can then be formed letter by letter using this map, so that [w] is in the image of our homomorphism.
ii) There's a homomorphism from $G$ if there's a homomorphism $f$ from $F(x,y)$ such that $f(x^2)=f(y^5)$. Any map sending $x$ to $a^{5n}$ and $y$ to $b^{2m}$ does this, and there are infinitely many of these.
iii) 4 homomorphisms - one for each different combo of sending a and b to one of 0 or 1
iv) If we had $ϕ(v)=1$ then $ϕ(u^2)=ϕ(v^5)=1$ but the square of any element in $Z_2$ is $e$ so this cannot be true. 
I am stuck on v) - I cannot figure out how the answers from the previous questions link together to prove this. I've been toying with the idea of sending the image of ϕ to $Z_2$ but I'm not sure exactly what this gets me.

Hint: You have at least $4$ morphisms $F(a,b)\rightarrow\mathbb{Z}_2$ and you have only two morphisms $G\rightarrow \mathbb{Z}_2$ $iv)$.
