Solving the Bomb Explosion Problem

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Homework Help Overview

The problem involves a bomb explosion resulting in three pieces, with two pieces having equal mass and moving at an angle of 80° to each other. The third piece has three times the mass of the others, and the goal is to determine its magnitude and direction. The discussion centers around the application of momentum conservation in both the x and y axes.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to establish a reference direction for the momentum equations. There is confusion regarding the angles for the two equal mass pieces and how to express the equations accordingly. Some participants explore the implications of using different angles and the resulting equations for momentum conservation.

Discussion Status

The discussion is ongoing, with participants providing insights into the setup of the equations. Some have suggested corrections to the equations while others are questioning the assumptions made about the directions of the pieces. There is no clear consensus yet, but various approaches are being explored.

Contextual Notes

Participants note the absence of specified directions for the pieces, leading to uncertainty in the calculations. The problem's constraints and the need for clarity in the setup of the equations are acknowledged.

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1. The Problem: A bomb at rest explodes, breaking into three pieces. Two of the pieces hace equal mass and fly off at 80° of each other with a speed of 100m/s. The tird piece has three times the mass of the other pieces. What is the magnitude and the direction of the third piece?

2. Homework Equations
potential energy equations for the x and y axis
is have
px: m0v0 = m1v1 +m2v2 +(3)m3v3
py: m0v0 = m1v1 +m2v2 +(3)m3v3




3. The Attempt at a Solution
This is what i have so far, my confusion is that i was not given any direction and am not sure what i am suppose to put in for theta for m1 and m2 so that i can find v3 and theta3? thank you for any help in advance!

px: 0 = m1v1cos(theta) +m2v2cos(theta) +(3)m3v3cos(theta)
py: 0 = m1v1sin(theta) +m2v2sin(theta) +(3)m3v3sin(theta)

px: 0 = m1(100)cos(theta) + m2(100)cos(theta) +(3)m3v3cos(theta)
py: 0 = m1(100)sin(theta) + m2(100)sin(theta) +(3)m3v3sin(theta)


 
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Clearly you need to specify a 'zero' direction then express your answer in relation to it. An obvious choice would be the line bisecting the trajectories of the two equal masses.
 


ok so iif i split it each angle would be 40 degrees in opposite directions giving me

px: 0 = m1(100)cos(40) + m2(100)cos(40) +(3)m3v3cos(theta)
py: 0 = m1(100)sin(40) + m2(100)sin(40) + (3)m3v3sin(theta)

all the m1 and m2 would then cancel when i divide py by px giving me:
.8355+ tan(theta) = 0
so theta would be -39.8 degrees? i still think i am doing something wrong
 


strugis said:
px: 0 = m1(100)cos(40) + m2(100)cos(40) +(3)m3v3cos(theta)
py: 0 = m1(100)sin(40) + m2(100)sin(40) + (3)m3v3sin(theta)
Check the signs in the second equation. m1 and m2 are not moving in the same direction.
 


if m2 is negative then they would end up being zero giving me 0 for theta and velocity, which doesn't make sense.
px: 0 = m1(100)cos(40) - m2(100)cos(40) +(3)m3v3cos(theta)
py: 0 = m1(100)sin(40) - m2(100)sin(40) + (3)m3v3sin(theta)

0 + tan(theta) = 0
 


strugis said:
0 + tan(theta) = 0
What are all the solutions to that equation?
 


(inverse) tan 0 = 0 and therefore so does velocity
 


i suppose tan(180) = 0 also, but then velocity would still be 0
 


Just noticed you corrected one equation but corrupted the other. You should have
px: 0 = m1(100)cos(40) + m2(100)cos(40) +(3)m3v3cos(theta)
py: 0 = m1(100)sin(40) - m2(100)sin(40) + (3)m3v3sin(theta)
With theta = pi, you should get a nonzero solution for v3.
 

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