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Submarine Fires Torpedo, Submarine's Recoil? |
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| Apr9-09, 09:24 PM | #1 |
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Submarine Fires Torpedo, Submarine's Recoil?
1. The problem statement, all variables and given/known data
A submarine of mass 2.5 x 10^6 kg and initially at rest fires a torpedo of mass 260 kg. The torpedo has an initial speed of 100.4 m/s. What is the initial recoil speed of the submarine? Neglect the drag force of the water. 2. Relevant equations m1v1=-m2v2 3. The attempt at a solution mass one = 2.5x10^6 (submarine) velocity one = 0 (sub isnt moving) mass two = 260 (torpedo) velocity two = 100.4 (torpedo) How can I find the recoil? I thought maybe it'd just be velocity one (instead of zero solve for x) v1 = (260*100.4)/2.5x10^6 = -0.01 m/s But that's wrong... what am i missing? |
| Apr9-09, 09:34 PM | #2 |
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I think you have the right idea, but you need to do it like this:
m1v1 (of sub) + m1v1 (of torpedo) = m2v2 (of sub) + m2v2 (of torpedo) and since initial velocity of the sub is 0 m1v1 (of torpedo) = m2v2 (of sub) + m2v2 (of torpedo) I think this is right, hope this helps! |
| Apr9-09, 09:45 PM | #3 |
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Recognitions:
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| Apr9-09, 11:57 PM | #4 |
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Submarine Fires Torpedo, Submarine's Recoil?
Yeah, but if we have:
m1v1 (of torpedo) = m2v2 (of sub) + m2v2 (of torpedo) doesnt m1v1 (torpedo) = m2v2 (torpedo) ? thus you'd just subtract one from the other, giving zero, resulting in: 0 = m2v2 (of sub) Result would be undefined. What am I missing? |
| Apr10-09, 12:22 AM | #5 |
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Initially in the problem, before the torpedo is fired, what is the momentum of the system? The total momentum after the torpedo is fired must be the same.
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| Apr10-09, 12:24 AM | #6 |
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so for m1v1 (torpedo) is the initial mass the sub plus the torpedo?
Not sure I follow, can you be more specific? |
| Apr10-09, 12:26 AM | #7 |
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Yes it would be. However, I think you're missing the point. The sub initially has no velocity, and the torpedo initially has no velocity. Therefore, what is the initial momentum of the system?
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| Apr10-09, 12:29 AM | #8 |
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zero...
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| Apr10-09, 12:31 AM | #9 |
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That's correct. If the initial momentum of the system is zero, then because of the conservation of momentum, the final momentum of the system must be zero. Does that help you at all?
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| Apr10-09, 12:35 AM | #10 |
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I still don't get it...
Initial Momentum of System (aka zero) = m_sub*v_sub + m_torpedo*v_torpedo Rearrange: - m_sub*v_sub = m_torpedo*v_torpedo Resulting in: - (2500000)(recoil) = (260)(100.4) -Recoil = ((260)(100.4))/2500000 = -0.0104416, which isn't accepted answer... |
| Apr10-09, 12:39 AM | #11 |
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Do you happen to know what the correct answer is? How do you know it's not the accepted answer?
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| Apr10-09, 12:41 AM | #12 |
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Our assignments are administered online via webassign.com.
It says -0.01 is wrong... |
| Apr10-09, 12:44 AM | #13 |
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Have you tried it without the negative sign? And/or with more or less significant figures? I don't know why that answer would be wrong. However, if you're required to have the correct number of significant figures based on the problem, your answer should be 0.010.
I hope one of these solutions works for you |
| Apr10-09, 12:46 AM | #14 |
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Odd...
The program allows for a 2% error and I've tried with any number of significant figures/variations (this is just practice, so I can try as many times as I want with no penalty) Just frustrating because I understand all other recoil problems except this one... |
| Apr10-09, 12:47 AM | #15 |
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I have a similar problem (somewhat different concept...vertical height acheived).
http://www.physicsforums.com/showthr...=1#post2154921 We can't seem to figure it out...care to take a look for me? |
| Apr10-09, 12:53 AM | #16 |
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Recognitions:
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| Apr10-09, 12:56 AM | #17 |
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Yes,
These are wrong: -0.01 0.01 0.010, etc 0.01044etcetc Very odd |
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