A 0.005kg bullet travelling horizontally with a speed of 1000m/s enters an 18kg door, embedding itself 10cm from the side opposite the hinges as in figure. The 1m wide door is free to swing on the hinges.

a) Before it hits the door, does the bullet have angular momentum relative the door’s axis of rotation? Explain.

b) Is mechanical energy conserved in this collision?

c) If the door opens immediately after collision, calculate the angular speed the door swing. The door has the same moment of inertia as a rod with axis at one end.

d) Calculate the energy of the door-bullet system and determine whether it is less or equal to the kinetic energy of the bullet before collision.

d) Calculate the energy of the door-bullet system and determine whether it is less or equal to the kinetic energy of the bullet before collision.

1. Of course m * R x V = .005 * .9 * 1000
2. No reason it should be, heat generated when bullet strikes. However Angular momentum does not change.
3. I am going to ignore the I of the bullet after collision which is .005*.81
so
.005 * .9 * 1000 = I w = (1/3)18(1)^2 w^2
So w = 0.866 radians/second

a) Well, let's think about it. Imagine a bullet doing ballet. How glamourous! Now, if the bullet is traveling horizontally, and the door is also horizontal, they must be dancing in sync. So, relative to the door's axis of rotation, the bullet has no angular momentum, because they are both moving in the same direction.

b) Ah, conservation of energy, the superhero of physics. Unfortunately, in this collision, mechanical energy is not conserved. It's like trying to keep all your money while playing Monopoly – it just doesn't work. Some energy is lost as the bullet embeds itself in the door and causes deformation.

c) Now, let's calculate the door's exuberant swing! Since the door opens immediately after the collision, it's like a dramatic exit stage left. The angular speed of the door swing can be determined using the principle of conservation of angular momentum. Since the door has the same moment of inertia as a rod with one end as the axis, we can use the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

d) Lastly, let's talk about the energy of the door-bullet system. Is it less or equal to the kinetic energy of the bullet before the collision? Well, since we know that mechanical energy is not conserved in this collision, it's safe to say that the energy of the door-bullet system is less than the kinetic energy of the bullet before the collision. It's like giving a comedian a bad joke – the energy just goes down, down, down.

a) Before it hits the door, the bullet does not have angular momentum relative to the door's axis of rotation. This is because the bullet is traveling horizontally, and therefore its motion is parallel to the axis of rotation. Angular momentum is typically associated with rotational motion around an axis, and since the bullet is not rotating around the door's axis, it does not possess angular momentum in this context.

b) Mechanical energy is not conserved in this collision. When the bullet enters the door, some of its kinetic energy is transferred to the door as it embeds itself into the door. Additionally, there may be energy lost due to any friction between the bullet and the door. Therefore, the total mechanical energy before and after the collision is not the same.

c) To calculate the angular speed of the door swing, you need to first calculate the initial angular momentum of the bullet-door system. The initial angular momentum is given by the product of the mass, velocity, and perpendicular distance of the bullet from the axis of rotation.

Angular momentum (initial) = mass × velocity × distance
= 0.005 kg × 1000 m/s × 0.1 m
= 0.5 kg·m^2/s

Since angular momentum is conserved in absence of external torques, the final angular momentum of the bullet-door system will also be 0.5 kg·m^2/s.

The moment of inertia of a rod with an axis at one end is given by the formula:

I = (1/3) × mass × length^2

Since the door is 1m wide, its length is 1m. Given that the door's mass is 18 kg, the moment of inertia is:

I = (1/3) × 18 kg × (1m)^2
= 6 kg·m^2

Using the conservation of angular momentum, we can equate the initial and final angular momenta:

Initial angular momentum = Final angular momentum
(0.5 kg·m^2/s) = (6 kg·m^2) × (angular speed of the door swing)

Solving for the angular speed of the door swing, we get:

Angular speed = (0.5 kg·m^2/s) / (6 kg·m^2)
≈ 0.0833 1/s

Therefore, the angular speed of the door swing is approximately 0.0833 radians per second.

d) To determine the energy of the door-bullet system, you need to calculate the kinetic energy of the bullet before the collision and add it to the rotational kinetic energy of the door after the collision.

The kinetic energy of the bullet before the collision is given by:

KE (bullet) = (1/2) × mass × velocity^2
= (1/2) × 0.005 kg × (1000 m/s)^2
= 2500 J

The rotational kinetic energy of a rotating object is given by the formula:

KE (rotational) = (1/2) × moment of inertia × angular speed^2

Substituting the given moment of inertia and angular speed:

KE (rotational) = (1/2) × 6 kg·m^2 × (0.0833 1/s)^2
≈ 0.0208 J

Therefore, the energy of the door-bullet system is:

Energy of the system = KE (bullet) + KE (rotational)
= 2500 J + 0.0208 J
≈ 2500.02 J

Comparing this to the kinetic energy of the bullet before the collision, we can see that the energy of the door-bullet system is greater than the kinetic energy of the bullet before the collision.

a) Before the bullet hits the door, it is traveling horizontally, which means it has linear momentum. Angular momentum, on the other hand, is a measure of rotation. Since the bullet is not rotating before it hits the door, it does not have any angular momentum relative to the door's axis of rotation.

b) To determine if mechanical energy is conserved in this collision, we need to examine whether there are any external forces doing work on the system. In this case, the only external force is gravity, which is acting vertically downward and is not doing work on the horizontal motion of the bullet and the door. Therefore, mechanical energy is conserved in this collision.

c) To calculate the angular speed of the door swing after the collision, we need to apply the principle of conservation of angular momentum. The initial angular momentum of the system, considering only the bullet and the door, is zero since the bullet does not have any initial angular momentum. After the bullet embeds itself into the door, the system will begin to rotate due to the conservation of angular momentum.

Considering the moment of inertia of the door is the same as a rod with an axis at one end, we can use the formula for the moment of inertia of a rod rotating about one end, which is given by:

I = (1/3) * m * L^2

where I is the moment of inertia, m is the mass of the door, and L is the length of the door (which is 1m in this case).

To find the angular speed (ω) of the door swing, we can use the following equation:

ω = L * v / I

where v is the velocity of the bullet embedded in the door. Given that v = 1000 m/s and L = 1m, we can substitute these values into the equation to find the angular speed.

d) To calculate the energy of the door-bullet system, we need to consider the kinetic energy of both the door and the bullet before the collision. The kinetic energy of the door is given by:

KE_door = (1/2) * I * ω^2

Substituting the values for I and ω, we can calculate the kinetic energy of the door. The kinetic energy of the bullet before the collision is given by:

KE_bullet = (1/2) * m_bullet * v^2

Substituting the values for m_bullet and v, we can calculate the kinetic energy of the bullet. By adding the kinetic energy of the door and the bullet, we can determine whether it is less than or equal to the kinetic energy of the bullet before the collision.

I think you can do (1/2) m v^2 and (1/2) I w^2