Say we have a system with two particles with mass $m_1=m$ and $m_2=m$ with positions described in cartesian coords. by $\mathbf r_1=(x_1=0, y_1=C-q_1)$ and $\mathbf r_2=(x_2=q_1+q_2, y_2=0)$. Its kinetic energy is then $$\mathcal T=\dfrac{1}{2}m_1(\dot x_1^2+\dot y_1^2)+\dfrac{1}{2}m_2(\dot x_2^2+\dot y_2^2)=\dfrac{1}{2}m(\dot x_2^2+\dot y_1^2)=\dfrac{1}{2m}(p_{x_2}^2+p_{y_1}^2),$$ where we used the fact that $$p_\square=\dfrac{\partial\mathcal T}{\partial \dot\square}$$
My question is whether there's a way to relate $p_{x_2}$ or $p_{y_1}$ with $p_{q_1}$ and $p_{q_2}$ in order to obtain the generalized kinetic energy.
I thought maybe you could apply chain rule like this $$p_{x_2}=\dfrac{\partial \mathcal T}{\partial \dot q_1}\dfrac{\partial \dot q_1}{\partial\dot x_2}+\dfrac{\partial \mathcal T}{\partial \dot q_2}\dfrac{\partial \dot q_2}{\partial\dot x_2}=p_{q_1}\partial_{\dot x_2}\dot q_1+p_{q_2}\partial_{\dot x_2}\dot q_2$$ but when checking whether it was the same as the kinetic energy obtained from substituing $\dot x_2=\dot q_1+\dot q_2$ and $\dot y_1=-\dot q_1$, I got different things.