B Conservation of Energy and Momentum in an Explosion

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The discussion centers on the conservation of kinetic energy and momentum during an explosion. A participant questions their calculations regarding the total kinetic energy (KE) before and after an explosion, initially calculating it as (26m)/4 and (18m)/4, respectively. They clarify that the KE of the bomb pieces remains consistent despite the explosion, illustrating this with specific values for KE before and after detonation. The conclusion emphasizes that the total KE remains the same, demonstrating that energy is conserved in the system. The conversation highlights the importance of accurately calculating and understanding energy transformations in explosive events.
JamesG23
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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.

IMG_2061.png
 
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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.
 
Doc Al said:
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
 
JamesG23 said:
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.
 
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!
 
Can someone here check my math? On an euc (electric unicycle), the motor has to exact the same amount of torque onto the wheel+tire that the rider exerts onto the euc, otherwise the frame would rotate (due to the motor exerting an equal in magnitude but opposite direction onto the frame than it does onto the wheel, a Newton third law like pair of torques). Choosing some arbitrary numbers: rider = 200 lb euc = 100 lb rider + euc = 300 lb tire radius = .83333 foot (10 inches) rider center...

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