Conservation of energy, A bomb explodes into three pieces

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SUMMARY

The discussion focuses on a physics problem involving the conservation of momentum during an explosion. Two pieces of equal mass fly off at 100 m/s at an angle of 60 degrees to each other, while the third piece has three times the mass of each of the other pieces. The calculated direction of the third piece is 150 degrees, and the momentum equation used is 2*100 m/s cos(180-150) - 3mv = 0, leading to a solution that appears correct based on the responses from other forum members.

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ErikiaMeno
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Homework Statement


A bomb explodes into three pieces, two have an equal mass and fly off at 60 degress to one another with the speed of 100 m/s. The third piece has three times the mass of each of the other pieces. Find the direction and magnitude of the third piece immediatly after the explosion



Homework Equations


Is my solution correct?
If not, how should I re-approach the problem


The Attempt at a Solution


2A + 60 = 360
2A = 300
A = 150 degres, A is the direction of the third angle

2*100m/s cos(180-150)-3mv = 0
200/3cos(30)
57.7
 
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Welcome to PF!

Hi ErikiaMeno! Welcome to PF! :smile:
ErikiaMeno said:
2A + 60 = 360
2A = 300
A = 150 degres, A is the direction of the third angle

2*100m/s cos(180-150)-3mv = 0
200/3 cos(30)
57.7

Yes, that looks ok …

what is worrying you about that? :confused:
 

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