Question

As part of a carnival game, a ball is thrown at a stack of tall, objects and hits with a perfectly horizontal velocity of mb = 0.628 kg 17.3 cm mo = 0.403 kg b = 10.8 m/s. ,i suppose that the ball strikes the topmost object. immediately after the collision, the ball has a horizontal velocity of in the same direction, the topmost object has an angular velocity of 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. b,f = 3.60 m/s o = 5.23 rad/s if the object's center of mass is located below the point where the ball hits, what is the moment of inertia of the object about its center of mass? = 12.1 cm o o = kg⋅m 2 what is the center of mass velocity of the tall object immediately after it is struck? o,

Answer 1

The moment of inertia of the object about its center of mass is 0.585 kg·m^2. The center of mass velocity of the tall object immediately after it is struck is 7.682 m/s.he center of mass velocity of the tall object immediately after it is struck is 7.682 m/s.

Step-by-step explanation:

To find the moment of inertia of the object about its center of mass, we can use the conservation of angular momentum. Since the ball is not rotating initially, the angular momentum before the collision is zero. After the collision, the ball has a horizontal velocity of 3.60 m/s. The angular momentum of the topmost object about its center of mass can be calculated using the equation:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Rearranging the equation, we have:

I = L/ω

Substituting the given values, we have:

I = (0.403 kg)(5.23 rad/s) / 3.60 m/s

I = 0.585 kg·m2

The moment of inertia of the object about its center of mass is 0.585 kg·m2.

To find the center of mass velocity of the tall object immediately after it is struck, we can use the law of conservation of linear momentum. Before the collision, the total linear momentum is zero since the tall objects are undisturbed. After the collision, the ball and the topmost object move in the same direction with velocities of 3.60 m/s and 5.23 rad/s respectively. The total linear momentum after the collision is:

p = (mb + mo)vb + moωb

where p is the total linear momentum, mb is the mass of the ball, mo is the mass of the topmost object, vb is the velocity of the ball, and ωb is the angular velocity of the topmost object. Substituting the given values, we have:

p = (0.628 kg + 0.403 kg)(3.60 m/s) + (0.403 kg)(5.23 rad/s)

p = 7.682 kg·m/s

The center of mass velocity of the tall object immediately after it is struck is 7.682 m/s.