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?
To solve part (e) of the problem, let's first breakdown the energy conservation equation correctly.
The initial kinetic energy (Ki) of the system is given by the bullet's kinetic energy before the collision. It can be calculated using the formula: Ki = 1/2 * m * v^2.
The final energy (Tf) of the system is the sum of the bullet's kinetic energy after the collision and the rotational energy of the door. The bullet will be embedded in the door, and the door will start rotating around its hinge.
The bullet's kinetic energy after the collision is given by 1/2 * m * vf^2, where vf is the final velocity of the bullet after embedding itself in the door. Since the bullet is embedded 11.2 cm from the side opposite the hinges, its velocity will have two components: linear velocity due to the bullet's motion and tangential velocity due to the rotation of the door.
To determine the final angular velocity (w) of the door, we can use the conservation of angular momentum. The initial angular momentum of the system (Li) is zero because the bullet does not have angular momentum relative to the door's axis of rotation before the collision. The final angular momentum (Lf) can be calculated by considering the bullet's angular momentum and the door's initial angular momentum (which is zero).
The bullet's angular momentum after the collision is given by L_bullet = m * vf * r, where r is the distance at which the bullet is embedded in the door. The door's initial angular momentum (zero) is equal to the final angular momentum after the collision, which is L_door = I * w, where I is the moment of inertia of the door.
Now, let's proceed to solve part (e) using these equations.
1. Calculate the initial kinetic energy (Ki):
Ki = 1/2 * m * v^2
= 1/2 * 0.00400 kg * (1.00 * 10^3 m/s)^2
2. Calculate the final angular velocity (w) of the door:
L_bullet = L_door
m * vf * r = I * w
0.00400 kg * vf * 0.112 m = (1/3 * (17.1 kg + 0.00400 kg) * (0.112 m)^2) * w
3. Substitute the value of vf obtained from step 2 into the expression for bullet's kinetic energy after the collision:
K_bullet = 1/2 * m * vf^2
= 1/2 * 0.00400 kg * (vf)^2
4. Substitute all the values calculated above into the final energy equation (Tf = K_bullet + I * w^2):
Tf = K_bullet + I * w^2
= 1/2 * 0.00400 kg * (vf)^2 + (1/3 * (17.1 kg + 0.00400 kg) * (0.112 m)^2) * w^2
By substituting the values correctly and solving this equation, you should be able to find the correct final energy (Tf) of the bullet-door system. Make sure to double-check your substitutions and calculations to ensure accuracy.