I Doubt about variable mass systems

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The discussion centers around the application of Newton's second law to variable mass systems, specifically questioning Halliday's derivation. Participants express confusion over the treatment of mass changes and the implications of using terms like M - ΔM, which can lead to misunderstandings about mass loss. There is debate about whether the velocity of the mass being ejected (u) should be considered constant or variable as time approaches zero. Some contributors suggest that breaking the system into multiple components could clarify the derivation, while others find existing explanations insufficiently clear. Overall, the conversation highlights the complexities and ongoing confusion surrounding the treatment of variable mass in physics.
  • #31
Hak said:
OK, thanks. Were you able to see the posts in the thread I referred you to in post #26?
The OP in that thread is guilty of setting up the reader for confusion.
First the OP says "The rocket now has velocity v+dv and mass M+dM, where the change in mass dM is a negative quantity.". That's fine.
Then the OP says "If I'm allowed to change M+dM with M-dM and -dM with dM then Eq. 9-38 becomes
M * v = dM * U + (M - dM) * (v + dv)"
Equation 9.38 is
M * v = -dM * U + (M + dM) * (v + dv) ... (9-38)

Well, having defined dM as a negative quantity, one is not allowed to change dM to -dM but to -|dM| to avoid conflict with the definition plus confusion about the variables. Then Eq. 9-38 should be
M * v = |dM| * U + (M - |dM|) * (v + dv) ... (9-38a)
OP writes the equation
M * v = dM * U + (M - dM) * (v + dv) ... (9.38b)

Do you see the difference between equations (9.38a) and (9.38b)? They are the same but only if dM in (9.39b) is defined as a positive quantity which is in direct contradiction to its original definition. Is that what confused you?

The equation $$m~\frac{dv}{dt}=(u-v)\frac{dm}{dt}$$ gets around the issue whether ##dm## is positive or negative. The mass transfer to or from the system of interest 1 (e.g the rocket) and the secondary system 2 (e.g. the fuel) is such that ##dm_1+dm_2=0.## Study the derivation in the insight. Which one is positive and which one is negative is irrelevant. In the equation ##m## and ##dm## are ##m_1## and ##dm_1## with the subscripts dropped as explained in the derivation.
 

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