Submarine Torpedo Momentum Question: Solving for Velocity Change

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The discussion revolves around calculating the final velocity of a submarine after it fires a torpedo. Initially, the submarine weighs 60 tonnes and moves at 1.5 m/s, while the torpedo, weighing 500 kg, is fired at 21 m/s relative to the submarine. The correct final velocity of the submarine is determined to be approximately 1.3 m/s forward, despite some confusion regarding the negative sign in momentum calculations. Participants clarify that the torpedo's speed should indeed be 22.5 m/s when considering the submarine's initial velocity. The key takeaway is to maintain consistent directional signs in momentum equations to avoid confusion in calculations.
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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.
 
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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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