(a) Before it hits the door, does the bullet have angular momentum relative the
ID: 1476583 • Letter: #
Question
(a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation?
Yes /No
(b) If so, evaluate this angular momentum. (If not, enter zero.)
kg · m2/s
If not, explain why there is no angular momentum
(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation.
Yes/ No
(d) At what angular speed does the door swing open immediately after the collision?
rad/s
(e) Calculate the total 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.
Explanation / Answer
a)) Yes
b) bold is vectorial quantity
L = r x p p = m v
L = r m v without (Ang) = m r v
The direction between speed and RADIUS are perpendicular Ang = 90 °
L = 0.006 1 1 103 = 6 Kg m2/s
c) Yes.
Mechanical energy is constant since is composed of two parts the kinetic energy and potential energy. The kinetic energy changes already that the bullet is stuck in the door, by what part of the initial kinetic energy is transformed into potential energy specifically in heat, this is a demonstration of the conservation of energy.The system is composed of the bullet and the door by which all forces are internal.
The kinetic energy is not conserved
d) We use conservation of angular momentum
Li = mb v r
Lf = (b I + Ip) w
B i = m r2
Ip = Icm = 1/3 Mp (r)2
Lf = (m r2 + 1/3 Mp r2) w
Li = Lf m v r = (m + 1/3 M) r2 w
. (w = m /(m+1/3M) (v) w = 17.3 0.006/(0.006+1/3) (1031) = 6/5.772 = 1.04 rad/s
e) We calculate kinetic energy
Ki = ½ v2 m
KF = ½ (b I + Ip) w2 rotational movement
Ki = ½ 0.006 (10-3)2 = 6000 J
KF = ½ (m r2 + 1/3 M r2) w2 Kf = ½ (0.006 12 + 1/3 17.3 12) 1.042 = 3.12 J
The rest of the energy is transformed into internal energy of the system
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