A 0.005-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 20.3-kg door, imbedding itself 10.6 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.
(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.
Use conservation of momentum (angular).
massbullet*velocitybullet*radius=massdoorbullet*w door
d) you do that, but energy was lost...
So (.005kg)(1000m/s)(1.06m)=(20.305kg)(w door)
w door=.26101?
I didn't check the math, but you need units ALWAYS. If your math is correct, w= .26 radians/sec
To solve parts (c) and (d) of the problem, we need to understand the conservation of angular momentum and kinetic energy.
(c) To find the angular speed at which the door swings open immediately after the collision, we can use the principle of conservation of angular momentum. Before the collision, the bullet has no angular momentum since it is traveling in a straight line. However, after the collision, the bullet becomes embedded in the door and starts moving in a circular path.
We can calculate the initial angular momentum of the bullet-door system by using the formula:
L_initial = m_bullet * v_bullet * r
where
m_bullet = mass of the bullet
v_bullet = velocity of the bullet
r = radius of circular motion (distance at which the bullet embeds in the door)
Substituting the given values:
m_bullet = 0.005 kg
v_bullet = 1.00 * 10^3 m/s
r = 0.106 m
L_initial = 0.005 kg * (1.00 * 10^3 m/s) * 0.106 m
Next, we need to find the final angular momentum of the system. After the collision, the bullet and door will move together as a single object. The door will rotate about its hinge, so its angular momentum will be given by:
L_final = I * ω
where
I = moment of inertia of the door (which is assumed to be a rod)
ω = angular speed of the door
Substituting the given values:
I = moment of inertia of a rod about one end = (1/3) * m * L^2
m = mass of the door = 20.3 kg
L = length of the door = 1.00 m
I = (1/3) * (20.3 kg) * (1.00 m)^2
We need to convert the length of the door from meters to centimeters to match the given units.
I = (1/3) * (20.3 kg) * (100 cm)^2
Finally, we can equate the initial and final angular momenta and solve for ω:
L_initial = L_final
0.005 kg * (1.00 * 10^3 m/s) * 0.106 m = (1/3) * (20.3 kg) * (100 cm)^2 * ω
Solving the equation above will give us the angular speed at which the door swings open immediately after the collision.
(d) To 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, we need to compare the kinetic energy before and after the collision.
The initial kinetic energy of the bullet is given by:
KE_initial = (1/2) * m_bullet * (v_bullet)^2
Substituting the given values:
m_bullet = 0.005 kg
v_bullet = 1.00 * 10^3 m/s
KE_initial = (1/2) * (0.005 kg) * (1.00 * 10^3 m/s)^2
Next, we need to calculate the final kinetic energy of the door-bullet system after the collision. To do this, we can use the fact that angular velocity affects kinetic energy and rotate the entire system to a reference frame in which the door is stationary.
The final kinetic energy of the door-bullet system will then be:
KE_final = (1/2) * (m_door + m_bullet) * ((v_bullet * r)^2 + ω^2 * I)
Substituting the given values:
m_door = 20.3 kg (mass of the door)
m_bullet = 0.005 kg
v_bullet = 1.00 * 10^3 m/s
r = 0.106 m (distance at which the bullet embeds in the door)
ω = angular speed of the door (calculated in part (c))
I = moment of inertia of the door (calculated in part (c))
After calculating the final kinetic energy, we can compare it with the initial kinetic energy to determine if it is less than or equal to the initial kinetic energy of the bullet.