Often, when constraint forces are very difficult to calculate, it's easier to use a Lagrangian approach with generalized coordinates. Consider a parameter ##\theta## which is the angle of the ball relative to the middle of the bottom face of the pipe block while the ball is in the pipe are in contact. It goes from 0 to pi.
Energy, momentum, and angular momentum are conserved. Assuming there is no deformation, all the energy is kinetic energy, so you have:
##T=(1/2)*m_b*v_b^2 + (1/2)*m_p*v_{cp}^2 + (1/2)*I_p*w_p^2##
The second and third terms are the translational and angular kinetic energy of the pipe block. We neglect the angular kinetic energy of the ball.
##v_{cp}## is the velocity of the center of mass of the pipe block.
The system momentum is ##p=m_b v_b + m_p v_p##
Let ##\alpha## be the orientation of the pipe block (which varies in time). Let ##(x_p, y_p)## be the position of the center of the bottom face of the pipe block.
Let ##(x_1,y_1)## be the position of the center of mass of the pipe block relative to ##(x_p, y_p)##. So the center of mass of the pipe block is
##(x_{cp},y_{cp}) = (x_p + x_1 \cos(\alpha) + y_1 \sin(\alpha), y_p + y_1 \cos(\alpha) - x_1 \sin(\alpha))##
The center of mass of the system is
##pt/(m_b+m_p) \equiv C_m##
(moving slightly upward)
so we can write the coordinates of the ball in terms of the generalized coordinates using the center of mass.
##m_b x_b + m_p (x_p + x_1 \cos(\alpha) + y_1 \sin(\alpha)) = C_{mx}##
##m_b y_b + m_p (y_p + y_1 \cos(\alpha) - x_1 \sin(\alpha)) = C_{my}##
We need to write the angular velocity of the block
##\omega_p = \dot{\alpha}##
Now, we need to write the Lagrangian in terms of ##\theta, \alpha, x_p, y_p##
This part is messy, but I think the equations are all above.
Using the Euler-Lagrange equation, you should be able to calculate the equations of motion for ##\alpha, \theta, x_1, y_1##. Then you know how to get ##x_b, y_b##.