Sphere strikes hinged rod, e = 0.890

Take moments about the hinge $A$: during the impact the hinge reaction passes through $A$ and the weights are non-impulsive, so the angular momentum about $A$ of the sphere and rod together is conserved. Let $L = 1.2$ m be the distance from $A$ to the point of impact (the lower end $B$), $v_s'$ the sphere's velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

(The figure label shows 5 m/s but the text gives 7 m/s; 7 m/s is used.)

Data

$m_s = 2.5$ kg, $v_s = 7$ m/s, $m_r = 10$ kg, $e = 0.89$

$$I_A = \frac13 m_rL^2 = \frac13(10)(1.2)^2 = 4.800\ \text{kg m}^2$$

Conservation of angular momentum about A

$$m_sv_sL = I_A\omega + m_sv_s'L \quad (1)$$

Restitution

The velocity of the rod's lower end is $\omega L$:

$$e = \frac{\omega L - v_s'}{v_s} \;\Rightarrow\; v_s' = \omega L - e\,v_s \quad (2)$$

Solving

Substituting (2) in (1):

$$m_sv_sL(1+e) = (I_A + m_sL^2)\,\omega$$ $$\omega = \frac{2.5(7)(1.2)(1+0.89)}{4.800 + 2.5(1.2)^2} = 4.73\ \text{rad/s}$$ $$v_s' = \omega L - e\,v_s = 4.73(1.2) - 0.89(7) = -0.56\ \text{m/s}$$

Answer: $\omega = 4.73$ rad/s; just after impact the sphere moves at $0.56$ m/s to the left (its direction is reversed, $v_s' = -0.56$ m/s).