A bullet (m = 0.005 kg) travelling horizontally with speed v = 1103 m/s strikes
ID: 1774206 • Letter: A
Question
A bullet (m = 0.005 kg) travelling horizontally with speed v = 1103 m/s strikes a door (M = 18 kg, L = 1 m), imbedding itself 10cm from the side opposite the hinge as shown in the figure below. The 1cm-wide door is free to swing on its frictionless hinge. (a) Before the bullet hits the door, does it have angular momentum relative to the door’s hinge? (b) If so, calculate this angular momentum. If not, explain why there is no angular momentum. (c) Is the mechanical energy of the bullet-door system constant during the collision? Answer and explain without calculation. (d) At what angular speed does the door swing open immediately after the collision? (e) Calculate the total energy of the bullet-door system and determine how it compares with the kinetic energy of the bullet before the collision.
2. A bullet (m 0.005 kg) travelling horizontally with speed v 1x10 m/s strikes a door (M 18 kg, L m), imbedding itself 10cm from the side opposite the hinge as shown in the figure below. The 1cm-wide door is free to swing on its (a) Before the bullet hits the door, does it have angular momentum relative to the door's hinge? (b) If so, calculate this angular momentum. If not, explain why there is no angular momentum (c) Is the mechanical energy of the bullet-door system constant during the collision? Answer and explain without calculation. (d) At what angular speed does the door swing open immediately after the collision? (e) Calculate the total energy of the bullet-door system and determine how it compares with the kinetic energy of the bullet before the collision. frictionless hinge. Hinge 18.0 kg 0.005 00 kgExplanation / Answer
a)
YES
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(b)
angular momentum about the hing Li = m*v*r*sintheta
theta = 90 degrees
r = 100cm - 10 cm = 90
Li = 0.005*10^3*0.9*sin90 = 0.9 kgm^2/s
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(c)
No
(d)
after the bullet imbedds in door
final angular momentum Lf = ((1/3)*M*L^2 + m*r^2)*wf
Lf = Li
((1/3)*M*L^2 + m*r^2)*wf = m*v*r*sin90
((1/3)*18*1 + (0.005*0.9^2))*wf = 0.005*10^3*0.9*sin90
wf = 0.75 rad/s
(e)
total kinetic energy of bullet door Ktot = (1/2)*I*w^2
Ktot = (1/2)*((1/3)*18*1 + (0.005*0.9^2))*0.75^2 = 1.68 J
kinetic energy of bullet Ki = (1/2)*m*v^2 = (1/2)*0.005*(10^3)^2 = 2500 J
(Ktot)/Ki *100 = 0.0672%
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