Solution

Newton's second law of motion:

$$OX: \;T - T \sin\alpha = m_0 a_{0x}$$

$$T \sin\alpha = m_{a1x}$$

$$OY: \;N - mg - T \cos\alpha = m_0 a_{0y} = 0$$

$$T \cos\alpha - mg = m_{a1y}$$

$$\tan\alpha = \frac{x_0 - x_1}{y_0 - y_1}$$

Since $\alpha = \text{const}$

$$\frac{d}{dt} \left( \tan\alpha\right) = \frac{\left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) - \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1)}{(y_0 - y_1)^2} = 0$$

$$\frac{d}{dt} \left( \left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) \right) = \frac{d}{dt} \left( \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1) \right)$$

$$\left(\frac{d^{2}x_{0}}{dt^{2}}-\frac{d^{2}x_{1}}{dt^{2}}\right)(y_{0}-y_{1})+\left(\frac{dx_{0}}{dt}-\frac{dx_{1}}{dt}\right)\left(\frac{dy_{0}}{dt}-\frac{dy_{1}}{dt}\right)=$$

$$\left(\frac{d^{2}y_{0}}{dt^{2}}-\frac{d^{2}y_{1}}{dt^{2}}\right)(x_{0}-x_{1})+\left(\frac{dy_{0}}{dt}-\frac{dy_{1}}{dt}\right)\cdot\left(\frac{dx_{0}}{dt}-\frac{dx_{1}}{dt}\right)$$

$$(a_{0x} - a_{1x})(y_0 - y_1) = (a_{0y} - a_{1y})(x_0 - x_1)$$

$$\tan\alpha (a_{0y} - a_{1y}) = a_{0x} - a_{1x}$$

From (1)

$$a_{0y} = 0$$

$$-\tan\alpha\, a_{1y} = a_{0x} - a_{1x}$$

Length of the rope:

$$L = x_w - x_0 + \frac{y_0 - y_1}{\cos\alpha}$$

Since the rope is inextensible

$$\frac{d^2 L}{dt^2} = -\frac{d^2 x_0}{dt^2} + \frac{1}{\cos\alpha} \left( \frac{d^2 y_0}{dt^2} - \frac{d^2 y_1}{dt^2} \right) = 0$$

$$\left( \frac{d^2 x_0}{dt^2} = a_{0x}, \frac{d^2 y_0}{dt^2} = a_{0y}, \frac{d^2 y_1}{dt^2} = a_{1y} \right)$$

$$a_{0x} \cdot \cos\alpha = a_{0y} - a_{1y}$$

From $(1)$: $a_{0y} = 0$

$$a_{0x} \cos\alpha = -a_{1y}$$

By dividing the equations from (1):

$$\frac{T \sin\alpha}{T \cos\alpha} = \frac{m a_{1x}}{m (g + a_{1y})} \to \tan\alpha (g + a_{1y}) = a_{1x}$$

$$\tan\alpha a_{1y} = a_{1x} - g \tan\alpha$$

and using (4)

$$\boxed{a_{0x} = g \tan\alpha}$$

By dividing the equations from (1):

$$\frac{T(1 - \sin\alpha )}{T \cos\alpha} = \frac{M}{m} \cdot \frac{a_{0x}}{g + a_{1y}}$$

Note that:

$$a_{1y} = -g \sin\alpha$$

$$\frac{1 - \sin\alpha}{\cos\alpha} \frac{M}{m} = \frac{g \tan\alpha}{g - g \sin\alpha} \to m = \frac{M \sin\alpha}{(1 - \sin\alpha )^2}$$

$$\boxed{m = \frac{M \sin\alpha}{(1 - \sin\alpha )^2}}$$