12 g bullet travelling at 420 m/s is fired into the middle an unlatched 32.0 kg

Question

12 g bullet travelling at 420 m/s is fired into the middle an unlatched 32.0 kg door. The door in hinged on one end, is 1.20 m wide and has I = 1/3 Md2. The bullet hits the middle of the door. What is the angular speed of the door just after the bullet is imbedded in ti? Calculate whether the kinetic energy conserved.

Explanation / Answer

Moment of Inertia of door, I = 3.84 kg - m^2 Moment of Inertia of door+bullet system, (I_{net}) = 3.84108 kg - m^2 Angular momentum of bullet + door system, before the bullet hit the door, (L_{before}) = 3.024 J-s after the bullet hit the door, (L_{after}) = (I_{net} omega ) By Angular Momentum Conservation ==> (L_{after}) = (L_{before}) a) (omega =) 0.7873 rad/s Kinetic Energy before, (KE_{before}) = 529.2 J Kinetic Energy after, (KE_{after}) = 1.1904 J b) KE loss happens. (Delta KE) = - 528.0096 J

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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