How deep was the Scorpion when its torpedo exploded?

  • Thread starter ero
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In summary, the conversation revolves around a student struggling with solving a physics problem involving the sinking of the U.S.S. Scorpion. They are trying to use a formula for acceleration through a fluid to find the depth of the submarine at the time of the torpedo explosion. However, they are unsure of how to solve for the resistance of water and are attempting to use previously learned equations for velocity and position. The conversation also touches on the constant nature of the drag force and the use of calculus in solving the problem. Overall, the student expresses frustration and confusion with the problem and seeks help from others.
  • #1
ero
Yeah, yet another nimrod with absolutely no clue on how to solve problems. I seriously think I need to stick with literature... something I actually get. >.<;

Anyhow, I have this problem I need to solve, but have no idea how. It's AP Physics B, so we're still based on Algebra, not Calculus. :)

Here's my problem:

Q: How deep was the Scorpion when its torpedo exploded?

Given/Background: In 1968, the attack submarine U.S.S. Scorpion mysteriously sank in the North Atlantic Ocean. On the night of October 26, 1993, CBS news aired a report on the sinking of the Scorpion with the loss of over 90 seamen. The report interviewed a United State's Navy Admiral who said the Scorpion sank itself when one of its torpedoes accidentally armed itself and exploded. The Admiral also stated that the wreck of the Scorpion was found in 11,500 of water. The evidence from the wreckage indicated that the sub. hit the ocean floor at 30 mph and took only 90 seconds to reach the ocean floor after the explosion. The U.S.S. Scorpion was 30 meters in length, 9 meters in width, and weighed (mass) 3.6 million kg. Water weighs about 1 kg per liter.


So, yeah... all I know is this: *sigh*

  • The formula for finding acceleration of a body through a fluid:
    a= W-R/M
  • a= acce. of sub. through water
  • W= weight of sub.
  • R= resistance of water
  • m= mass of falling object through water

So far, for W, I have 352800000 Newtons (w= m x g -> w= 3.6 x [tex]10^6[/tex] x 9.8) and 3,600,000 kg for m.

... And that's all. -_-;
To tell the truth, I don't see how solving for a helps much. That and I have no idea how to solve for R (resistance of water). I know I have to take in count the time when the sub. reaches terminal velocity (so that within 90s, it's going at constant velocity - 30mph.)I have to get the answers in feet, so I know I have to go from meters to feet and that something has to be cubed.

So, please help. I was able to get the equation and decipher what each letter represents, but I have no idea what to do, or how to proceed next!

I would really, really appreciate the help! :)
 
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  • #2
Where did you get this formula a = W-R/M? The drag force is actually dependent upon the square of the velocity.

- Warren
 
  • #3
Hm? But isn't the velocity changing? How can it be squared when a number is constantly changing? Wouldn't it have to be at constant velocity?

Do I know anything? No, not really. >.<;

Oh, and that was our only given equation. We're to use that for finding the acceleration of the sub. though the water (keeping in mind water resistance, I guess.).
 
  • #4
Originally posted by ero
Hm? But isn't the velocity changing? How can it be squared when a number is constantly changing? Wouldn't it have to be at constant velocity?
If the velocity is changing, and the drag depends upon the velocity, then the drag is also changing.
Oh, and that was our only given equation.
What do you plug in for this 'R' when you solve the problem? Your teacher is probably just approximating the drag force as a constant to make the problem easier to solve. Truth be told, however, the drag force is not at all constant, but depends upon the square of the velocity.

In any case, you need to solve for the acceleration of the sub. If R is constant, so is a. You could then use the equation

[tex]v(t) = v_0 + a t[/tex]

to find the velocity as a function of time. That's why your teacher is approximating the drag force as constant -- if it weren't constant, you'd have to use calculus to solve the problem. You're more interested in the position, though, so you can use the equation

[tex]x(t) = x_0 + v_0 t + \frac{1}{2} a t^2[/tex]

to find the position as a function of time.

Do these equations make sense to you? Let me know if they don't, and we can talk about where they come from.

- Warren
 
  • #5
Well, one of the biggest reasons for my confusion is the fact that we have to solve for R, and I've no idea how. No one (in class) does, actually. Somehow, R, water resistance, has to be solved and left in Newtons.

Does the this: [tex]H_2O[/tex] = 1g = 1[tex]cm^3[/tex] play into the problem at all?

I believe it does, since something has to be cubed, and the only cube I see is in that "formula". But then, would dimensions come into play, to get the cube, I mean? So far, there's only two... but what about the third?

The first equation does make sense, but that's only because I've seen that before. But yet, how to get the velocity to plug in?
The second, I'm afraid, I've never seen before. I guess I'm taking dummy-physics or something... and even then... Oh, jeez.

Oh, and thanks for the help thus far. I'm so grateful! :)
 

1. How deep was the Scorpion when its torpedo exploded?

The USS Scorpion was approximately 400 miles southwest of the Azores at a depth of 1,350 feet when its torpedo exploded on May 22, 1968.

2. What caused the Scorpion's torpedo to explode?

The exact cause of the explosion is still unknown, but it is believed that a faulty battery in the torpedo caused a chemical reaction, resulting in the explosion.

3. Were there any survivors from the Scorpion?

No, unfortunately all 99 crew members on board the Scorpion perished in the explosion.

4. How long did it take for the Scorpion to sink after the explosion?

It is estimated that it took only a matter of minutes for the Scorpion to sink to the ocean floor after the torpedo explosion.

5. Has the wreckage of the Scorpion ever been found?

Yes, the wreckage of the Scorpion was eventually found in October 1968 by a team from the USNS Mizar using a towed sled equipped with sonar.

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