ALL ANSWERS BESIDE PART E ARE CORRECT. A 0.00400-kg bullet traveling horizontall

Question

ALL ANSWERS BESIDE PART E ARE CORRECT.

A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 19.9-kg door, embedding itself 11.9 cm from the side opposite the hinges as shown 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

(b) If so, evaluate this angular momentum. (If not, enter zero.) 3.52 kg · m2/s

(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. No

(d) At what angular speed does the door swing open immediately after the collision? 0.53rad/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.

KEf =_______ J

KEi =________J

Explanation / Answer

(A) Yes.

(B) Li = m v r = 0.004 x 1000 x (1 - 0.119 )

= 3.524 kg m^2 /s  

(C) No

(D) Applying angular momentum conservation,

3.524 = ((0.004 (1-0.119)^2) + (19.9 x 1^2 / 3) ) w

3.524 = ( 0.0031 + 6.63333 ) w

w = 0.531 rad/s

(E) KEf = I w^2 / 2 = (0.0031 + 6.6333) (0.531^2)/2

= 0.9356 J

KEi = m v^2 / 2 = (0.004) (1000^2)/2

= 2000 J

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