Conservation of Energy and Momentum in an Explosion

[JamesG23]

Hey, I have a question about explosions and how kinetic energy works during them. I have outlined my question on the attached image. Please let me know if something is wrong or needs clarifying. Thank you.

 

[Doc Al]

It would be easier to comment if you had typed up your work. But anyway:

Viewed from the lab frame, you calculated the total KE after the explosion = (26m)/4. Sounds good. Not sure why you set that equal to 4m.

The KE before explosion = (1/2)m(9) = (18m)/4. Subtract that from the KE after the explosion and see what you get.

 

[JamesG23]

It would be easier to comment if you had typed up your work. But anyway:

Viewed from the lab frame, you calculated the total KE after the explosion = (26m)/4. Sounds good. Not sure why you set that equal to 4m.

The KE before explosion = (1/2)m(9) = (18m)/4. Subtract that from the KE after the explosion and see what you get.

Oh shoot I don't know why I simplified like that. Maybe I thought it was 24/6. Thank you

 

[Baluncore]

I have a question about explosions and how kinetic energy works during them.

In an atmosphere, the explosion of a flying bomb produces a sphere of hot combustion gas that has a very low density compared to the original explosive charge.
That sphere is effectively stopped immediately by it's low mass and the area of it's greater cross-section.
The original KE is not lost, it is just insignificant when applied to the huge mass of atmosphere that encloses the explosion.

 

[Drakkith]

Let's say the bomb pieces are 10kg each.
KE of each piece at ±2 m/s: 1/2mv²=20J
Total KE of bomb pieces: 40 J

KE of both bomb pieces at +3 m/s before detonation: 90 J
KE of bomb piece at +5m/s: 125 J
KE of bomb piece at +1 m/s: 5 J
Total KE of bomb pieces after explosion: 130 J

But look. 125+5-90 = 40 J
The same as when the bomb is stationary!