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I am a tutor, and one of my students is practicing for the BPHO. One of the problems he wanted to go over is below.

screenshot of the BPHO problem

I decided to use the fact that

$$F_{net}=\frac{dp}{dt}=M\frac{dv}{dt}+\frac{dM}{dt}v$$

for $M$ the total mass. From there, you can set up an ODE for $v$ and solve. However, in the official solutions for that problem, they instead set up their ODE as $$\frac{dv}{dt}=\frac{F_{net}}{M}$$

These ODEs have different (though similar) solutions, but I can't think of why my approach would be considered incorrect.

EDIT: To be clear, my issue here is why taking $F_{net}=Ma$, as the examiners did, is considered to be correct, while $F_{net}=\frac{dp}{dt}$ is not. Am I missing or forgetting some reason the latter would just simplify to the former?

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  • $\begingroup$ @WillO Yes, I am aware. Hence why I have a dM/dt term in that momentum derivative. Similarly, the official solution also has M as a function of time. $\endgroup$ Nov 5, 2022 at 17:31
  • $\begingroup$ I misread what you were saying, so my comment is irrelevant. I am deleting it, with apologies. $\endgroup$
    – WillO
    Nov 6, 2022 at 2:05
  • $\begingroup$ I have voted to reopen. It seems to me there is an interesting conceptual question here. $\endgroup$
    – WillO
    Nov 6, 2022 at 9:17
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    $\begingroup$ I think your method is right, don't really see why there wouldn't be a $\dot m v$ term. I have heard from others that BPhO can sometimes be incorrect and this may be one of those problems (do take that with a grain of salt as I have never done any BPhO problem before). $\endgroup$ Nov 7, 2022 at 23:24
  • $\begingroup$ Unless in the BPHO solution they didn't somehow put the $\dot{M}v$ term into their $F_{net}$ I would agree with you and Ashmit Dutta. $\endgroup$
    – kricheli
    Nov 8, 2022 at 11:59

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