Hard tutorial question can anyone help

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Homework Help Overview

The problem involves a torpedo boat firing a torpedo and seeks to determine the momentary reduction in the boat's forward speed due to the firing. The context includes concepts from linear impulse and conservation of momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the torpedo's launch angle on momentum transfer. Questions about the launching method and its effects on the boat's reaction are raised. Some suggest converting units to SI for clarity, while others explore the conservation of momentum principles.

Discussion Status

There is an ongoing exploration of the problem with various interpretations being discussed. Some participants have offered calculations related to momentum loss, while others highlight the importance of the launch angle. No explicit consensus has been reached, but productive lines of reasoning are being explored.

Contextual Notes

Participants note the need to consider the angle of the torpedo's launch and the implications it has on the momentum calculations. There is also mention of unit conversion as a potential step to clarify the problem.

plug2002
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A torpedo boat with a mass of 50 tons is moving at 10 knots when it fires a 250-kg torpedo horizontally with the launch tube at the 30 degree angle. The torpedo has a relative velocity to the boat of 10m/s as it leaves the tube
1) Determine the momentary reduction v in forward speed of the boat?

it is meant as a linear impuse question, can anyone help me please!
 
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I can't help with the math at all, unfortunately. My first question would be regarding the launching method. Not all of them would impart the same reaction to the boat.
 
Have you done linear impulse questions before?

Start by converting all the units to SI and it might become a bit clearer.
 
The sub and torpedo as a whole has a combined linear momentum. Whe the two separate...
 
conservation of momentum
the mass of the sub + torpedo times its velocity before firing=
the mass of the sub*vsub +mass torpedo*vtorp.

or you could say that the sub is at rest altogether and fires the torpedo which then has momentum of 250*10=2500
so the sub must loose this amount of momentm
-2500=50,000*vsub
so vsub= -.05m/s
the sub looses .05m/s when it fires the the torp.

it aint rocket science, but wait, maybe it is
 
Unfortunately, phlegmy, you miseed an important portion of the problem: the torpedo is launched at a 30* angle. This will slightly decrease the momentum loss from the boat.
 
doh!
in that case loss of mom in x direction = 250*10*cos(30)= 2165
sub loosese .043 m/s in the forward direction! and starts to decend at
.025m/s
 

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