A 0.004-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an
ID: 1476302 • Letter: A
Question
A 0.004-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 17.6-kg door, imbedding itself 12.0 cm from the side opposite the hinges. The 1.00-m-wide door is free to swing on its hinges. (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? (b) Is mechanical energy conserved in this collision? Answer without doing a calculation. (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. (Give the kinetic energy of the door-bullet system just after collision.)
Explanation / Answer
(a) Yes.
If an object has movement that is not parallel to a particular axis then that object has angular momentum about that axis.
(b) No.
Mechanical energy is conserved only in a perfectly elastic collision, such as the case where a bullet bounces off at the same relative speed before impact.
(c) Angular velocity is angular momentum divided by rotational moment of inertia.
= L / I
The angular momentum of the bullet is transferred to the door:
L = mv × r
L = ( 0.004 kg )( 10^3 m/s )( 1.000m - 0.12 m )
L = 3.52 kg·m²/s
Neglecting the mass of the bullet, we compute the rotational moment of inertia of the door:
I = ( 1/3 )Mw² + mr²
I = ( 1/3 )(17.6 kg )( 1.000m )² + ( 0.004 kg )( 1.000m - 0.12 m )²
I = 5.86 kg·m²
= 3.52 kg·m²/s / 5.87 kg·m²
= 0.599 rad/sec
(d)
The rotational kinetic energy of the system is:
RKE = ½I²
RKE = ½( 5.87 kg·m² )( 0.599 rad/sec )²
RKE = 1.05 J
The kinetic energy of the bullet before impact was:
TKE = ½mv²
TKE = ½( 0.004 kg )( 10^3 m/s )²
TKE = 2000 J
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