There have been questions posted about this before, but those were well over 12 years ago, and I was wondering if someone could elucidate some of the formulas used.
"A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 17.1-kg door, embedding itself 11.2 cm from the side opposite the hinges as shown in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges."
a) Before it hits the door, does the bullet have angular momentum relative to the door's axis of rotation
b) If so, evaluate this angular momentum
c) Is mechanical energy of the bullet-door system constant in this collision? Answer without a calculation
d) At what angular speed does the door swing open immediately after the collision
e) Calculate the total energy of the bullet-door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.
I have answered A by knowing that the bullet is travelling with constant velocity perpendicular to the hinge, B, by utilizing the formula L = mvrsin(phi), C, by knowing it's a perfect inelastic collision, D by utilizing L = Iw, where I = 1/3Mr^2. My question lies with part e.
My approach was to utilize the conservation of energies formula, and break that down into knowing that the initial state of kinetic energy, and the final state of rotational energy should appear as:
Ki = Tf
1/2mvi^2 = Iw^2
1/2mvi^2 = 1/3(M+m)r^2*w^2
plugging in the numbers yields a correct result for the initial kinetic energy, but not the final. Why?
3 answers