Submarine Torpedo Momentum Question: Solving for Velocity Change

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Homework Help Overview

The problem involves a submarine firing a torpedo and seeks to determine the submarine's velocity after the event, focusing on momentum conservation principles. The subject area includes concepts from mechanics, specifically momentum and velocity changes in a system involving two bodies.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply momentum conservation but questions the sign of the resulting velocity and the setup of their equations. Some participants clarify the momentum equations and the treatment of relative velocities, while others discuss the implications of directionality in momentum calculations.

Discussion Status

Participants are actively engaging with the problem, offering different perspectives on the setup and calculations. There is a focus on clarifying the treatment of signs in momentum equations and the interpretation of relative velocities. No explicit consensus has been reached, but several productive lines of reasoning are being explored.

Contextual Notes

Participants are navigating potential confusion regarding the direction of momentum and the relative speed of the torpedo. The original poster's assumptions about the torpedo's speed and the submarine's mass after firing are also under discussion.

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Homework Statement


A 60 tonne submarine, intially traveling forward at 1.5 m/s, fires a 500 kg torpedo striaght ahead with a velocity of 21 m/s in relation to the submarine. What is the velocity of the submarine after it fires the torpedo?

Answer: 1.3 m/s [forward]


Homework Equations


1 ton = 1000 kg
p = p'


The Attempt at a Solution



p = p'
(60000)(1.5) = (500)(22.5) - (59500)(Vs')
Vs' = -1.32 m/s

I don't understand why I am getting that answer, as the answer is actually forward. Because doesn't the torpedo going forwards cause the submarine to move backwards.

Other than that, can you guys tell me whether I set it up right? Like the speed of the torpedo should be 22.5 m/s right, since the submarine is already traveling at 1.5 so it gets added? Also, the final mass of submarine shud be 59 500 kg, since the torpedo is no longer attached to it?

Thanks.
 
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Why do you have a minus sign?

p = mv
p_{sub_i} = p_{sub_f} + p_{torpedo}
60000 \times 1.5 = (59500 \times v_{sub_f}) + (500 \times 21)
v_{sub_f} = (60000 \times 1.5) - (500 \times 21) = 1.336

Forgive the lack of units.
 
Bhajun said:
Why do you have a minus sign?

p = mv
p_{sub_i} = p_{sub_f} + p_{torpedo}
60000 \times 1.5 = (59500 \times v_{sub_f}) + (500 \times 21)
v_{sub_f} = (60000 \times 1.5) - (500 \times 21) = 1.336

Forgive the lack of units.

Minus because the momentum of the torpedo being shot forward causes the sub to move back. Also, shouldn't the torpedo have a speed of 22.5 m/s, since its 21 m/s relative to the sub, so it has a speed of 22.5? Like say if you're running at 2m/s and you throw a tennis ball at 5 m/s, the speed of the tennis ball would be 5 relative to you, but 7 m/s in total.
 
?...
 
momentum of before = momentum of after

(60000)(1.5)=(59500)v+(500*22.5)

Yes, its 22.5. But there's no negative sign since their both doing in forward (Same direction.).
 
To avoid confusion with signs in the future, define a direction to be positive, then always use positive signs in the momentum equations (ex. m1v1=m2v2 + m3v3). Get the signs of the velocities correct and the sign of the final answer will be correct.
 
You are confusing change in momentum with final momentum. Yes indeed the CHANGE in the sub's momentum is in the negative dirrection. However, the final momentum for the ship is still in the positive direction.
 

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