A 10 gram bullet traveling horizontally with speed 1.00 times 103 m/s strikes a
ID: 1497571 • Letter: A
Question
A 10 gram bullet traveling horizontally with speed 1.00 times 103 m/s strikes a 18-kg door, embedding itself 70 cm from the side opposite the hinges as shown in the figure below. The 80-cm wide door is free to swing on its frictionless hinges. Hinge Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? If so, evaluate this angular momentum. II not, explain why there is no angular momentum. Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. At what angular speed does the door swing open immediately after the collision? 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)
m = mass of bullet = 0.01 kg
v = speed = 1000 m/s
r = distance from axis of rotation = 0.70 m
angular momentum of bullet = mvr = 0.01 x 1000 x 0.7 = 7 kgm/s
c)
No since the collision in inelastic
d)
M = mass of door = 18 kg
L = length = 80 cm = 0.80 m
I = moment of inertia of door = ML2/3 = 18 (0.8)2/3 = 3.84 kgm2
using conservation of momentum
mvr = (I + mr2) w
7 = (3.84 + 0.01 (0.7)2) w
w = 1.82 rad/s
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