A 0.005 kg bullet traveling horizontally with speed 1 x 103 m/s strikes an 18.0

Question

A 0.005 kg bullet traveling horizontally with speed 1 x 103 m/s strikes an 18.0 kg door, embedding itself 10 cm from the side opposite the hinges as shown in the figure below. The 1 m wide door is free to swing Hinge 1 meter Bullet 10cm (a) Before it hits the door, does the bullet have angular momentum relative to the door's axis of rotation? (b) If so, evaluate this angular momentum, both magnitude and direction (into or out of the page). If not, explain why there is no angular momentum. (c) Is the kinetic energy of the bullet-door system constant during this collision? Answer without doing any calculation (d) At what angular speed does the door swing open immediately after the collision? (e) Calculate the total kinetic 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. Explain what you find.

Explanation / Answer

a)

yes the bullet has angular momentum relative to door's axis of rotation

b)

r = distance of the path of bullet from hinge = 100 cm - 10 cm = 90 cm = 0.90 m

v = velocity = 1000 m/s

m = mass of bullet = 0.005 kg

Lbi = angular momentum of the bullet = m v r = 0.005 x 1000 x 0.90 = 4.5 kgm2/s

c)

no the kinetic energy is not conserved

d)

Idoor = ML2/3 = 18(1)2/3 = 6 kgm2

using conservation of angular momentum

mvr = (m r2 + Idoor) w

4.5 = (0.005 (0.9)2 + 6) w

w = 0.75 rad/s

kinetic energy before collision = KEi = (0.5) m v2 = (0.5) (0.005) (1000)2 = 2500 J

kinetic energy before collision = KEf = (0.5) (m r2 + Idoor) w2 = (0.5) (0.005 (0.9)2 + 6) (0.75)2 = 1.7 J

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