# Solving the Bomb Explosion Problem

• strugis
In summary, the conversation involved solving for the magnitude and direction of a third piece of a bomb that exploded into three pieces, with two equal pieces flying off at 80° with a speed of 100m/s. The third piece has three times the mass of the other pieces. The homework equations used were potential energy equations for the x and y axis, and the attempt at a solution involved setting up equations with an arbitrary zero direction and solving for theta and velocity. However, there was confusion about the direction and the final solution was unclear.
strugis
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)

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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