Let $p$ be a prime and $K$ a finite extension of $\mathbb Q_p$ with ramification index $e$. Let $\mathcal O_K$ be the ring of integers of $K$ and $k$ its residue field and the unique maximal ideal. Let $0 \to A \to B \to C \to 0$ be an exact sequence of abelian varieties over $K$. Let $A', B', C'$ be the Neron models of $A,B$ and $C$ over $\mathcal O_K$. Then we get an induced sequence $A' \to B' \to C'$. We know the following:
Corollary 1.1 of [Maz78]: Assume that $B$ has semistable reduction and that $e < p-1$, then we have the following exact sequences of free $\mathcal O_K$ modules: $$0 \to Tan_0 A' \to Tan_0 B' \to Tan_0 C' \to 0$$ and $$0 \to Cot_0 C' \to Cot_0 B' \to Cot_0 A' \to 0.$$
Now I was wondering what happens in general? Is there some nice description of $Ker(Cot_0 C' \to Cot_0 B)$$Ker(Cot_0 C' \to Cot_0 B')$ so that is clear that $Ker(Cot_0 C' \to Cot_0 B) = 0$$Ker(Cot_0 C' \to Cot_0 B') = 0$ if $e < p-1$?
A consequence of modules in the above sequences are free is that the are split. In particular they stay exact after tensoring with $k$.
Now I was wondering what happens when $p-1>e$ after tensoring with $k$? Is there some nice description of $Ker(Cot_0 C' \otimes k \to Cot_0 B \otimes k)$$Ker(Cot_0 C' \otimes k \to Cot_0 B' \otimes k)$ so that is clear that $Ker(Cot_0 C' \otimes k \to Cot_0 B \otimes k) = 0$$Ker(Cot_0 C' \otimes k \to Cot_0 B' \otimes k) = 0$ if $e < p-1$?
I would already be happy with an answer in the case $K = \mathbb Q_2$ and $B$ has good reduction at $2$.
[Maz78]: Mazur, B., Rational isogenies of prime degree. (With an appendix by D. Goldfeld), Invent. Math. 44, 129-162 (1978). ZBL0386.14009.