A 0.006 00-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters

Question

A 0.006 00-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 18.2-kg door, imbedding itself 10.3 cm from the side opposite the hinges as in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges. a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? (yes or no) b) If so, evaluate this angular momentum. (If not, enter zero.) c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. (yes or no) d) At what angular speed does the door swing open immediately after the collision? e) Calculate the 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. -KEfinal=? -KEinitial=?

Explanation / Answer

a) Yes

b) Linear momentum p = mv = 0.006*1*10^3 = 6 kg-m/s

Perpendicular distance from hinge = 1 - 10.2*10^-2 = 0.898 m

Angular momentum = 6*0.898 = 5.388 kg-m^2/s

c)

No

d) Inertia of door+bullet = m_door*L^2/3 + m_bullet*r^2 = 15.4*1^2/3 + 0.006*0.898^2 = 5.1381 kg-m^2

Angular momentum conservation: 5.388 = 5.1381*

= 1.0486 rad/s

e) KEf = 1/2*Iw^2 = 1/2*5.1381*1.0486^2 = 2.82495 J

KEi = 1/2*mv^2 = 1/2*0.006*(1*10^3)^2 = 3000 J

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