As part of a carnival game, a 0.578-kg ball is thrown at a stack of 24.3-cm tall

Question

As part of a carnival game, a 0.578-kg ball is thrown at a stack of 24.3-cm tall, 0.343-kg objects and hits with a perfectly horizontal velocity of 11.6 m/s. Suppose the ball strikes the very top of the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 3.85 m/s in the same direction, the topmost object now has an angular velocity of 4.03 rad/s and all the objects below are undisturbed. If the object's center of mass is located 17.0 cm below the point where the ball hits, what is the moment of inertia of the object?

What is the center of mass velocity of the tall object immediately after it is struck?

Explanation / Answer

at the time of the hit the angular momentum about the base would be conserved then before hitting angular momentum = 0.578 * 11.6 * 0.243= 1.629 Kg m2 s-1 angular momentum of objects = 0 after hitting angular momentum of ball = 0.578 * 0.243 * 3.85 = 0.5407 Kg m2 s-1 angular momentum of objects = I ? 0.5407 + I * 4.03 = 1.629 therefore moment of inertia (I) = 0.27 kg m2 let velocity of center of mass = v v/(0.243- 0.17) = 4.03 v= 0.294 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.