Solving Exchange of Momentum by Conserving Energy & Momentum

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Discussion Overview

The discussion revolves around a physics problem involving the exchange of momentum and energy conservation when a bullet is fired into a block of wood resting on supports. Participants explore the effects of firing the bullet at different points (center vs. off-center) and the resulting motion of the block, including the center of mass and rotational dynamics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes a scenario where a bullet is fired into a block of wood and asks which case results in a higher rise of the center of mass when fired at the center versus near the edge.
  • Another participant suggests that the energy conservation equation may not apply in the second case due to the introduction of rotational motion.
  • A participant notes that firing the bullet off-center will cause rotation, complicating the energy calculations and requiring consideration of both translational and rotational kinetic energy.
  • A video referenced by a participant indicates that both blocks achieve the same height despite differing energy distributions, raising questions about conservation of energy and momentum.
  • One participant proposes that the conservation of linear momentum necessitates both blocks reaching the same height, while the difference in energy is attributed to work done by the bullet in the wood, translating into angular energy.
  • Another participant acknowledges a misunderstanding regarding the energy equation and clarifies that the height will be the same in both cases, as the equations account for both translational and rotational energies.

Areas of Agreement / Disagreement

Participants express differing views on the application of energy conservation in the context of rotational dynamics, with some arguing that the height reached by the center of mass should be the same in both scenarios, while others question the implications of rotational motion on energy distribution.

Contextual Notes

The discussion includes assumptions about the system's behavior under different conditions and the dependence on definitions of energy forms. The implications of rotational dynamics on energy conservation remain unresolved.

pukb
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Imagine a block of wood resting on two supports at each end. You point a gun upward under the block and shoot the block upwards.

One scenario is you shoot the block upwards in the exact center. Another scenario is you shoot upwards near one of the support ends. The block is thick enough that in both cases the bullet comes to rest inside the block.

Question: in which case does the centre of mass of the block rise the highest?

I could quantify the first case (both for velocity and distance rise of cog) using conservation of momentum and conservation of energy as follows:
mu = (m+M)V
1/2(mu^2) = (m+M)gh

Can somebody help me in the second case?
 
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hi pukb! :smile:

hint: in the second case, why does 1/2 mu2 = (m+M)gh not work? :wink:
 
yes, it does apply - what i meant was the heights to which cog is lifted are not same when force is applied at center and at corner because the latter case will cause some rotation as well. The problem is to quantify that
 
hi pukb! :smile:

(btw, it's better to say centre of mass, com, than centre of gravity, cog :wink:)
pukb said:
…the latter case will cause some rotation as well.

yes! … so the kinetic energy will be the "translational" kinetic energy 1/2 mv2 plus the rotational kinetic energy 1/2 Iω2 :wink:
 
There's a video of this very problem on YouTube. The results are not exactly what you'd expect. The video does compare two wooden blocks, one shot thru the CG and the other shot off-center. Both went the same height into the air. Of course, this raises the question of what happened to the conservation of energy. The block that's spinning should not go as high into the air due to energy being transferred as angular momentum. But it does. What gives?

The video proposed an answer that both blocks must go the same height in order to conserve linear momentum. The difference in energy is due to the bullet not traveling as far into the wood (doing work in the process) and that difference in work is getting translated into angular energy.

Yes, it's not a comforting explanation. But it does seem to be correct.
 
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Hi Ryoko! :smile:

Yes, I didn't think this through fully :redface:: the energy equation which I said didn't work …
tiny-tim said:
hint: in the second case, why does 1/2 mu2 = (m+M)gh not work? :wink:

should be 1/2 mu2 + 1/2 Iω2 = (m+M)gh + 1/2 Iω2,

which of course is the same equation, with the same u (since u is fixed by conservation of momentum, irrespective of ω and of the slight difference in I).

So h (the height) will be the same in both cases.

Thanks, Ryoko. :smile:​
 

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