A 0.006-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 21.1-kg door, imbedding itself 10.4 cm from the side opposite the hinges as in the figure below. The 1.00-m-wide door is free to swing on its hinges.
(c) At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.)
Answer: 0.749rad/s
(d) Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.
Answer:_____J
I can't figure out C
A 0.006-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 21.1-kg door, imbedding itself 10.4 cm from the side opposite the hinges as in the figure below. The 1.00-m-wide door is free to swing on its hinges.
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C) Angular momentum of the system is the same before and after. Assume rotation about the door hinge.
d) before (1/2)m v^2
after =(1/2) I w^2
it better be much less after.
Angular momentum of bullet alone about door hinge L = m v r = .006 * 10^3 * (1.000-.104) = 5.376 kg m^2/s
Angular momentum of system after collision (ignore mass of bullet)
L = I w where I = ( 1/3) m b^2 and w is angular velocity and b is door width.
L = (1/3)(21.1)(1)^2 w = 7.033 w
so
7.033 w = 5.376
w = .7644 rad/s
=
energy before = (1/2) m v^2
energy after = (1/2) I w^2
energy after better be much smaller than energy before
before (1/2) (.006)(10^3)^2 = 3000 Joules
after (1/2)(7.003)(.7644)^2 = 2.046 Joules
the bullet entering the wood turned most of the energy into heat
I'm sorry I meant I cant figure out part D.
By the way, we did not get the same answer for part c
Well, if I were a door, I'd be pretty dizzy after getting hit by a bullet too! But let's try to help you out with the calculations.
To solve part (c), we can start by using the principle of conservation of angular momentum. Before the collision, the bullet has no angular momentum because it's traveling horizontally. However, after the collision, the door-bullet system will start rotating.
We can calculate the initial angular momentum (L_i) of the door-bullet system using the equation:
L_i = I * ω_i
Where I is the moment of inertia of the door (which is the same as a rod with one end as the axis), and ω_i is the initial angular speed.
Given that the door has a moment of inertia of a rod, its moment of inertia is:
I = (1/3) * m * L^2
Where m is the mass of the door (21.1 kg) and L is the length of the door (1 m).
Substituting the values in, we get:
I = (1/3) * 21.1 kg * (1 m)^2
I = 7.033 kg·m^2
Now, we can calculate the initial angular speed (ω_i) using the equation:
L_i = I * ω_i
The initial angular momentum (L_i) is equal to the angular momentum (L_f) immediately after the collision:
L_i = L_f
Since the bullet embeds itself 10.4 cm from the side opposite the hinges, the length from the hinges to the point of embedment is 90.6 cm (1 m - 10.4 cm = 0.906 m).
The final angular momentum (L_f) can be calculated as:
L_f = (m_bullet * v_bullet * r)
Where m_bullet is the mass of the bullet (0.006 kg), v_bullet is the speed of the bullet (1.00 x 10^3 m/s), and r is the distance from the hinges to the point of embedment (0.906 m).
Substituting the values in, we get:
L_f = (0.006 kg * 1.00 x 10^3 m/s * 0.906 m)
L_f = 5.436 kg·m^2/s
Since L_i = L_f, we can solve for ω_i:
I * ω_i = L_f
(7.033 kg·m^2) * ω_i = 5.436 kg·m^2/s
Dividing both sides by (7.033 kg·m^2), we get:
ω_i = (5.436 kg·m^2/s) / (7.033 kg·m^2)
ω_i = 0.773 rad/s
So the angular speed immediately after the collision is approximately 0.773 rad/s. Oops, it looks like we have a different result from the given answer. So maybe I'm just a clown and got it wrong. Sorry about that!
To find the angular speed at which the door swings open immediately after the collision, you need to apply the principle of conservation of angular momentum. The initial angular momentum is zero since the door is stationary. After the collision, the bullet becomes embedded in the door, and together they will rotate.
The formula for angular momentum is:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Let's assume that the door rotates clockwise. The angular momentum of the bullet before the collision is given by the product of its linear momentum (mv) and the perpendicular distance from the rotation axis (which is the hinge of the door). So:
Lbullet = mv * d
Where m is the mass of the bullet, v is its velocity, and d is the distance from the hinge.
The angular momentum of the door is given by its moment of inertia (which is the same as a rod with an axis at one end) multiplied by the angular speed ω:
Ldoor = Idoor * ω
Now, according to the conservation of angular momentum, the total angular momentum before the collision should be equal to the total angular momentum after the collision:
Lbullet = Ldoor
Substituting the values, we have:
mv * d = Idoor * ω
Rearranging the equation for ω:
ω = (mv * d) / Idoor
The moment of inertia (Idoor) of a rod with an axis at one end is given by:
Idoor = (1/3) * mass * length^2
Substituting the known values:
Idoor = (1 / 3) * 21.1 kg * (1 m)^2
Now, you can plug in the values into the equation to calculate the angular speed (ω):
ω = (0.006 kg * 1.00 m/s * 0.104 m) / [(1/3) * 21.1 kg * (1 m)^2]
Simplifying the equation:
ω = 0.006 * 1.00 * 0.104 / [1/3 * 21.1]
ω = 0.624 / 7.03
ω = 0.0886 rad/s
Therefore, the angular speed at which the door swings open immediately after the collision is approximately 0.0886 rad/s.
I apologize for providing an incorrect answer earlier. The correct angular speed is 0.0886 rad/s, not 0.749 rad/s.