# Angular Motion Of A Bullet Striking A Door

1. Dec 13, 2012

### Bashyboy

The problem is:

"A 0.00600-kg bullet traveling horizontally with a speed of 1.00*103 m/s enters an 19.2-kg door, imbedding itself 8.60 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? 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? Yes? No? (Answer without doing a calculation.)

(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."

I think I will be able to solve parts (d) and (e).

For (a), the answer is yes, but I can't quite figure out why the answer is yes.

For (c), the answer is no; but I am having trouble understanding how we can possibly know. Couldn't the kinetic energy of the bullet be completely absorbed as kinetic energy in the door? In this scenario, wouldn't mechanical energy be conserved?

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2. Dec 13, 2012

### Staff: Mentor

How is the angular momentum of a particle defined?

When the colliding objects end up merged and traveling together, that is a perfectly inelastic collision. You can show (using conservation of momentum) that mechanical energy cannot be conserved. Only when they bounce off each other can energy be conserved.

3. Dec 13, 2012

### Bashyboy

Angular momentum of a particle is found to be the product linear momentum of the particle and the particle's distance from the axis of rotation. Yes, I see now. I do have another question, however; for part (d), how do I calculate the momentum of inertia of the door?

4. Dec 13, 2012

### Staff: Mentor

You can treat it like a thin rod.

5. Dec 13, 2012

### Bashyboy

I don't have the entire length of the door, though.

6. Dec 13, 2012

### Staff: Mentor

Use the 'width' of the door, which is given.

7. Dec 13, 2012

### Bashyboy

Okay, I got parts (b), (d), and (e) wrong.

Part (b): $L=(0.00400~kg)(1.00\cdot 10^3~m/s)(8.10\cdot 10^{-2}~m)=0.324~kg\cdot m^2/s$

For (d), $L_i=L_f \rightarrow (0.324~kg\cdot m^2/s)+0=[(0.00400~kg)((8.10\cdot 10^{-2}~m/s)^2+1/3(15.8)(1.00~m)^2] \omega \rightarrow \omega = 0.0615~rad/s$

For (e), I solved for the initial kinetic energy, and got it right; but I apparently didn't properly solve the final kinetic energy: $K_f=1/2(0.00400~kg)((8.10\cdot 10^{-2}~m/s)^2+1/3(15.8)(1.00~m)(0.0615~rad/s)^2= 0.00997 J$

Last edited: Dec 13, 2012
8. Dec 13, 2012

### Staff: Mentor

I don't understand what are doing here. What's the momentum of the bullet? How far is it from the axis?

9. Dec 13, 2012

### Bashyboy

0.081 m, right? I accidentally put units of velocity with that number.

10. Dec 13, 2012

### Staff: Mentor

That's not right. How did you determine that distance?

11. Dec 13, 2012

### Bashyboy

Well, in the problem it states that the bullet is embedded into the door at a distance of 8.60 cm from the hinges from the door, so I took that to be the distance...Oh, I calculated 8.1 cm...Was that the error?

Last edited: Dec 13, 2012
12. Dec 13, 2012

### Staff: Mentor

The bullet hits 8.60 cm from the side opposite the hinges (see the figure). What's the distance to the axis?

13. Dec 13, 2012

### Bashyboy

Oh, 8.60 cm is the distance from where the bullet hits the door to the edge of the door not containing the hinges? If that's the case, I wouldn't know the other distance.

14. Dec 13, 2012

### Staff: Mentor

You have the width of the door.

15. Dec 13, 2012

### Bashyboy

But how is that the distance from the axis of rotation (hinges) to the other end of the door?

16. Dec 13, 2012

### SammyS

Staff Emeritus
How wide is the door ?

17. Dec 13, 2012

### Bashyboy

1.0 m, and then it's length is 8.60 m plus the distance beyond the bullet.

18. Dec 13, 2012

### SammyS

Staff Emeritus
The door is 1 meter wide. Therefore the far edge of the door is 1 meter from the edge with the hinges.

How far is the bullet from the hinges?

(The figure is from a perspective looking down on the door.)

Last edited: Dec 14, 2012
19. Dec 14, 2012

### Bashyboy

Oh, I see. I thought that the 1.0 m was how thick it was (width); it's the length of it.

20. Dec 14, 2012

### Bashyboy

I SOLVED IT!!!! Thank you both for your help!!!!