A 0.006 00-kg bullet traveling horizontally with a speed of 1.00 x 10^3 m/s enters an 20.1-kg door, imbedding itself 10.9 cm from the side opposite the hinges as in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges.
At what angular speed does the door swing open immediately after the collision?
Calculate the 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 (what is the final and initial KE's)
I'll assume that 1.00 103 means 1x10^3
a)
Yes. Any object which has a velocity vector which does not pass directly through the chosen axis has angular momentum about that axis.
b)
No, mechanical energy is not conserved. Angular momentum is conserved, and any excess energy is dissipated within the door.
c)
W.before = W.after
Solve for both, then solve for final w. (note that W is angular momentum, w is angular velocity)
W.before = I.bullet * w.bullet + I.door * w.door
I.bullet = m.bullet * (.91 m)^2
w.bullet = V.bullet / (.91 m)
W.before = m.bullet * V.bullet * (.91 m) + 0
W.after = I.bullet * w.bullet + I.door + w.door
W.after = m.bullet * (.91 m)^2 * w.bullet + 1/3 * m.door * (1m)^2 * w.door
since w.bullet = w.door, we'll just call that w
W.after = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2)
Set W.before = W.after and solve for w
m.bullet * V.bullet * (.91 m) + 0 = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2)
.006 kg * 1000 m/s * .91 m = w * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1 m)^2)
5.46 kgm^2/s = w * 6.0716 kgm^2
w = .8992 radians/second
d)
Before:
Ek = 1/2 * m.bullet * V.bullet^2
Ek = 1/2 * .006 kg * 1000000 m^2/s^2
Ek = 3000 joules
After:
Ek = I.bullet * w^2 + I.door * w^2
Ek = (.8992 rad/sec)^2 * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1m)^2)
Ek = 4.909 joules
Note that kinetic energy after is significantly less than kinetic energy before.
Well, isn't this bullet quite the daredevil, embedding itself in a door? Let's get to the bottom of it!
To find the angular speed of the door after the collision, we can use the principle of conservation of angular momentum. The angular momentum before the collision is zero because the bullet is traveling horizontally, and the door is at rest. After the collision, the bullet gets embedded in the door, causing the door to start rotating.
We can use the equation:
L(initial) = L(final)
Since the initial angular momentum is zero, we have:
0 = (I_door + I_bullet) * ω
Here, I_door represents the moment of inertia of the door and I_bullet is the moment of inertia of the bullet. ω is the angular speed of the door after the collision.
Let's assume the door is a thin, rectangular plate rotating about an axis through one of its edges:
I_door = (1/3) * M * h^2
Where M is the mass of the door and h is the length of the door.
Since the bullet embeds itself 10.9 cm from the side opposite the hinges, we can calculate the moment of inertia of the bullet:
I_bullet = m * r^2
Here, m is the mass of the bullet, and r is the perpendicular distance of the bullet from the rotational axis.
With all the measurements, you can now calculate the angular speed, ω, to find out how fast that door swings open.
To calculate the angular speed at which the door swings open immediately after the collision, we need to apply the principle of conservation of angular momentum.
Angular momentum is conserved when there are no external torques acting on a system. In this case, assuming no external torques, the angular momentum before the collision should be equal to the angular momentum after the collision.
The angular momentum before the collision is given by the formula:
L1 = I * ω1
Where L1 is the angular momentum, I is the moment of inertia, and ω1 is the angular speed before the collision.
The angular momentum after the collision is given by:
L2 = I * ω2
Where L2 is the angular momentum, I is the moment of inertia, and ω2 is the angular speed after the collision.
Since the door is free to swing on its frictionless hinges, we can assume its moment of inertia remains constant. Therefore, we can write:
I * ω1 = I * ω2
Simplifying, we find:
ω2 = ω1
So, the angular speed at which the door swings open immediately after the collision is equal to the angular speed before the collision, which is the same as the bullet's speed.
To calculate the energy of the bullet-door system, we need to consider both kinetic energy and potential energy.
The initial kinetic energy of the bullet is given by:
KE_initial = 0.5 * m_bullet * v_bullet^2
where m_bullet is the mass of the bullet and v_bullet is its velocity.
Substituting the given values, we find:
KE_initial = 0.5 * 0.00600 kg * (1.00 x 10^3 m/s)^2
Calculating this, we get:
KE_initial = 3.00 J
The final kinetic energy of the system after the bullet has embedded itself in the door is given by:
KE_final = 0.5 * (m_bullet + m_door) * v_door^2
where m_door is the mass of the door and v_door is its velocity after the collision.
To find v_door, we can use the principle of conservation of linear momentum:
m_bullet * v_bullet = (m_bullet + m_door) * v_door
Rearranging the equation, we find:
v_door = (m_bullet * v_bullet) / (m_bullet + m_door)
Substituting the given values, we find:
v_door = (0.00600 kg * 1.00 x 10^3 m/s) / (0.00600 kg + 20.1 kg)
Calculating this, we get:
v_door = 9.00 m/s
Now, we can calculate the final kinetic energy:
KE_final = 0.5 * (0.00600 kg + 20.1 kg) * (9.00 m/s)^2
Calculating this, we find:
KE_final = 810 J
Comparing the initial and final kinetic energies, we have:
KE_final > KE_initial
So, the energy of the bullet-door system after the collision is greater than the initial kinetic energy of the bullet.