Calculating Momentum in Explosions

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Homework Help Overview

The problem involves calculating the momentum of three pieces resulting from an explosion, with specific velocities given for two pieces. The context is centered around the conservation of momentum in a system where the initial momentum is zero.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of momentum conservation, questioning the relationship between speed and momentum. There are attempts to verify calculated speeds and directions based on vector components.

Discussion Status

There is an ongoing exploration of the problem with various interpretations of the speed and direction of the third piece. Some participants offer calculations and seek verification, while others clarify the principles of momentum conservation.

Contextual Notes

Participants are working under the assumption that the masses of the pieces are equal and that the initial momentum of the system is zero. There is a focus on using trigonometric methods to resolve vector components.

mindhater
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Need help on this problem - exactly as stated

Homer the human powder keg, initially at rest, suddenly explodes into three pieces each with equal mass. One piece moves east at 30 m/s and a second piece moves at 30 m/s southeast.

Caculate the speed of the third piece and calculate the direction in which it travels

Thanks for the help
 
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I'm going to go with 55.4m/s at 22.5degrees North of East.

Hopefully someone can verify that answer.
 
i agree with the direction but the speed wud be equal would it not
 
Originally posted by FUNKER
i agree with the direction but the speed wud be equal would it not

Why? Momentum is conserved, not 'speed'. Before the explosion momentum was zero. It must be afterwards too.
 
Since the three pieces had equal mass, you can factor the mass out of momentum and just use speed, if you want. The point is that the vector sum of the three vectors has to be 0, and so you have to do some trig.
 
yeah, add the x components up = 8.8 m/s East
add the y components up = 21.2 m/s South

so 8.8 m/s West and 21.2 m/s North will give you the opposite vector, thereby conserving momentum

answer = 22.95 m/s @ 22.5 deg West of North
 

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