A 0.004 00-kg bullet traveling horizontally with a speed of 1.00 X 10^3 m/s ente

Question

A 0.004 00-kg bullet traveling horizontally with a speed of 1.00 X 10^3 m/s enters an 17.9-kg door, imbedding itself 11.5 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

(b) If so, evaluate this angular momentum. (If not, enter zero.)
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?
.593 rad/s

(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.

KEf= ?
KEi = 2000 J

Explanation / Answer

b)
L=mvr
L=.004(1000)(1-.115)
L=3.54 kg-m2/s

e)

KE=(1/2)I2

KE=(1/2)[(1/3)(17.9)(1)2+(.004)(1-.115)2].5932

KE=1.049583 joules

Hope that helps

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