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A wagon of initial mass $M$ is moving with an initial velocity $V_o$. Now sand starts accumulating on this wagon, increasing the total mass from $M$ to $M+s$. We have to find the final velocity of the wagon. Provided is the fact that the accumulating sand does not have any initial velocity.

Let p(t) be the momentum of the wagon and the sand on it.

$$p(t)=(M+m)v \\$$ where m = m(t) is the sand accumulated on the wagon $$p(t + dt) = (M+m+dm)(v+dv)$$ Since momentum is conserved, hence: $$P_{initial} = P_{final}$$

$$p(t)=p(t + dt)$$

$$(M+m)v = (M+m+dm)(v+dv)$$

$$(M+m)v = (M+m)v+(M+m)dv+vdm+dmdv$$

$$0 = (M+m)dv+vdm+dmdv$$ Since dmdv is very small, we get $$(M+m)dv+vdm = 0$$ $$\frac {dv}{v} = -\frac{dm}{(M+m)}$$

We observe that dv is negative, making v+dv lesser than v.

Integrating, we get

$$\int_{V_o}^{V} \frac {dv}{v} = -\int_{o}^{s}\frac{dm}{(M+m)}$$ $$V = \frac{V_o M}{M+s}$$

But now if we write -dv instead of dv, and redo the calculation, then: $$p(t)=(M+m)v \\ p(t + dt) = (M+m+dm)(v-dv)\\ P_{initial} = P_{final}$$

$$p(t)=p(t + dt)$$

$$(M+m)v = (M+m+dm)(v-dv)$$

$$(M+m)v = (M+m)v-(M+m)dv+vdm-dmdv$$

$$0 = -(M+m)dv+vdm-dmdv$$ Since dmdv is very small, we get $$-(M+m)dv+vdm = 0$$ $$\frac {dv}{v} = \frac{dm}{(M+m)}$$

We see that dv is positive, making v-dv less than v.

Integrating, we get

$$\int_{V_o}^{V} \frac {dv}{v} = \int_{o}^{s}\frac{dm}{(M+m)}$$ $$V = \frac{V_o (M+s)}{M}$$

The answer in this case is inverted. What exactly is wrong in the second version of the calculation. In most calculations, changing the sign of dv doesn't affect the final answer, as it readjusts to give out the same answer in either case.

So does this mean that we should always write a change in quantity $x$ as $x+dx$, or is there anything that I specifically did wrong in the second calculation?

Qmechanic
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Sidd
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    This reminds me of the classic rocket-losing-fuel problem. This is related. – HDE 226868 Nov 30 '14 at 22:13
  • Ya, it is one of the same kind. I know how to do it, but I don't understand why one way works and another doesn't. – Sidd Nov 30 '14 at 22:15
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    I suspect it's something along the lines of: since you replaced $v+dv$ with $v-dv$, you no longer integrate from $v_{i}$ to $v_{f}$, but from $v_{f}$ to $v_{i}$. – G. Paily Nov 30 '14 at 22:42
  • That occurred to me, but I couldn't find a reason for doing that. – Sidd Nov 30 '14 at 22:43
  • Okay, I found the reason why and how. I was assuming that velocity goes from $V_o$ to $V$, but that was not the case, since in integration, you vary the variable which occurs as the differential element and not f(x). Just like in $\int f(x) dx$, it's $x$ that varies, in this one it's $v$ that varies based on the sign on $dv$. I on the other hand was varying the $v(t)$ part of it. In actuality, a negative $dv$ means that $V_o - \int dv = V$ and a positive $dv$ means that $V_o + \int dv = V$, consequentially giving the change of velocity, i.e., $dv$ as $V_o - V$ and $V - V_o$ respectively. – Sidd Nov 30 '14 at 22:58

2 Answers2

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You don't decide yourself to make $dv$ negative or positive, you only find out what it is! For example, take the equation $x < 0$. Well that means $x$ is negative, so (by your logic) I should replace the equation with $-x < 0$, but now it says $x$ is positive?? This is in essence what you're doing, so I hope you can see the flaw.
The way derivatives work, is that you add the perturbations $dx$ to your variables, and then let the math do the work and it will tell you how the quantities behave.

Chris Gerig
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edit - issue with the second method - the way I see it that if you change the sign of $dv$ you also need to change the sign of $v$.

Yes normally, in a problem like this we use integration with $dx$, but in the middle I would normaly use $\delta x$ when we talk about small quantities. The change to $dx$ when integrating with infinitessimally small step size.

I don't understand why you went to $-dv$ from $dv$ in the second calculation - this is the problem. As momentum is conserved it makes sense that if you increase mass the velocity will decrease, but that doesn't mean that we can change to $v-dv$ it messes up the integration...... the way I see it that if you change the sign of $dv$ you also need to change the sign of $v$.

In Fact the final answer for the first part you can get without integration if you just look at $mv$ before and after and consider conservation of momentum. Let us say

$m_iv_i = m_fv_f$

where $m_i$ and $v_i$ are initial mass and velocity, and $m_f$ and $v_f$ are the final mass and velocity.

tom
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