As part of a carnival game, a ?b=0.503 kgmb=0.503 kg ball is thrown at a stack of 22.8 cm22.8 cm tall, ?o=0.363 kgmo=0.363 kg objects and hits with a perfectly horizontal velocity of ?b,i=11.6 m/s.vb,i=11.6 m/s. Suppose that the ball strikes the topmost object. Immediately after the collision, the ball has a horizontal velocity of ?b,f=3.35 m/svb,f=3.35 m/s in the same direction, the topmost object has an angular velocity of ?o=2.03 rad/sωo=2.03 rad/s about its center of mass, and all the remaining objects are undisturbed. Assume that the ball is not rotating and that the effect of the torque due to gravity during the collision is negligible.If the object's center of mass is located ?=16.0 cmr=16.0 cm below the point where the ball hits, what is the moment of inertia ?oIo of the object about its center of mass? ## Problem DescriptionAs part of a carnival game, a ball with mass \( m_b = 0.503 \, \text{kg} \) is thrown at a stack of objects each with a height of 22.8 cm and a mass \( m_o = 0.363 \, \text{kg} \). The ball is thrown with a perfectly horizontal velocity of \( v_{b,i} = 11.6 \, \text{m/s} \). It strikes the topmost object in the stack. Immediately after the collision, the ball has a reduced horizontal velocity of \( v_{b,f} = 3.35 \, \text{m/s} \) in the same direction. The topmost object receives an angular velocity \( \omega_o = 2.03 \, \text{rad/s} \) about its center of mass, while the other objects remain undisturbed. Assume the ball is not rotating and that gravitational torque effects during the collision are negligible.### Moment of Inertia CalculationThe center of mass of the object is located a distance \( r = 16.0 \, \text{cm} \) below the point where the ball impacts. Determine the moment of inertia \( I_o \) of the object about its center of mass.\[ I_o = \underline{\hspace{2cm}} \, \text{kg}\cdot\text{m}^2 \]### Center of Mass Velocity CalculationDetermine the velocity \( v_{o,cm} \) of the center of mass of the topmost object immediately after being struck by the ball.\[ v_{o,cm} = \underline{\hspace{2cm}} \, \text{m/s} \]## Diagrams1. **First Diagram**: - Illustrates the setup before collision, showing the ball approaching the top of a stack of three objects, aligned vertically. Arrow indicates the initial horizontal velocity \( v_{b,i} \).2. **Second Diagram**: - Depicts the situation immediately after the collision. The ball's reduced velocity \( v_{b,f} \) is indicated. The top object is shown with angular velocity \( \omega_o \), rotating about its center of mass, and the new velocity direction is shown with an arrow.

Given that, mb=0.503 kgmo=0.363 kg, ω=2.03 rad/svb,i=11.6 m/s, L=22.8 cmvb,f=3.35 m/s, r=16 cm Moment of inertia of any object is proportional to the product of mass and square of the position from the axes.a) By using law of conservation of momentum, mbvb,i+movo.i=mbvb,f+movo,f where v(o, i) is the initial velocity of object and v(o, f) if the final velocity of the object let it is u, therefore 0.503×11.6+0=0.503×3.35+0.363×uu=11.43 m/s L=Iω⇒mou=Iω0.363×11.43=I×2.03 I=2.04 kg·m2 b). Since, v=rω where r is distance from the axes of rotation. Therefore, v=0.16×2.03v=0.32 m/s