Question

A 0.00600 kg bullet traveling horizontally with speed 1.00 10³ m/s strikes a 21.4 kg door, embedding itself 11.3 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 door shown from above such that its hinge is on the top side of the figure with the door going down. A bullet is traveling horizontally to the right towards the door on the opposite end as the hinge.

(a)

Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation?

Yes No    

(b)

If so, evaluate this angular momentum (in kg · m²/s). (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.

Yes No    

(d)

At what angular speed (in rad/s) 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. (Enter your answers in J.)

KEf	=	J
KEi	=	J

(f)

What If? Imagine now that the door is hanging vertically downward, hinged at the top, so that the figure is a side view of the door and bullet during the collision. What is the maximum height (in cm) that the bottom of the door will reach after the collision?

cm

Answer 1

a.

Yes (because it has some speed and a perpendicular distance from the axis of rotation.

b.

Angular momentum of bullet = mvr

= 0.006×103×0.113 = 0.678 kgm2/s

c.

No mechanical energy is not conserved. (as the system has external forces, they do work)

d.

By angular momentum conservation

Li = Lf

mvr = Iω

Momentum of Inertia of door+bullet(I) = ML2/3 + ml2​​​​​​

= 21.4×12/3 + 0.06×(0.113)2 = 7.13 kgm2​​​​​​

So putting the values in above equation

0.678 = 7.13×ω

ω = 0.095 rad/s

e.

K. E. i = 1/2mv2 = 1/2×0.06×(103)2 = 3000 J

K. E.f = 1/2Iω2 = 1/2×7.31×(0.095)2 = 0.033 J

So final Kinetic Energy is less then initial.