A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 15.

Question

A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 15.1-kg door, embedding itself 10.9 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?

YesNo    

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

If not, explain why there is no angular momentum.

This answer has not been graded yet.

(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation.

YesNo    

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

Explanation / Answer

a)

Yes

b)

m = mass of bullet = 0.004 kg

v = speed = 1000 m/s

r = distance from axis = 10.9 cm = 0.109 m

angular momentum of bullet = mvr = 0.004 x 1000 x 0.109 = 0.436 kgm/s

c)

yes

d)

w = width = 1 m

M = mass of dorr = 15.1 kg

I = moment of inertia of door = M w2/3 = 15.1 (1)2 /3 = 5.03 kgm2

I' = moment of inertia of bullet = mr2 = 0.004 (0.109)2 = 0.0000475 kgm2

W = angular speed

using conservation of angular momentum

mvr = (I + I') W

0.436 = (5.03 + 0.0000475) W

W = 0.087 rad/s

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