Angular Momentum: There is no external torque about the vertical axis. \[ L_i = L_f \] Initially, the disc is at rest and the ball is at the center ($r=0$). So, $L_i = 0$. Finally, the ball exits at radius $R$ with tangential velocity $v$. The disc rotates with angular velocity $\omega$. \[ 0 = I\omega + m v R \]

Energy: There is no friction, so mechanical energy is conserved. \[ \frac{1}{2}mu^2 = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] (Assuming radial velocity is zero at exit due to semicircular path tangent to edge).

Given: $M_{disc} = 1.0$ kg, $m_{ball} = 1.5$ kg, $u = 2.0$ m/s.
Moment of Inertia $I = \frac{1}{2}MR^2 = 0.5 R^2$.

From Angular Momentum: \[ 0.5 R^2 \omega + 1.5 v R = 0 \implies R\omega = -3v \]

Substitute $\omega = -3v/R$ into Energy equation: \[ \frac{1}{2}(1.5)(2)^2 = \frac{1}{2}(1.5)v^2 + \frac{1}{2}(0.5 R^2) \left( \frac{-3v}{R} \right)^2 \] \[ 3 = 0.75v^2 + 0.25 (9v^2) \] \[ 3 = 0.75v^2 + 2.25v^2 \] \[ 3 = 3v^2 \implies v = 1.0 \text{ m/s} \]