Question

$(m_0 + rt)v = (m_0 + rt + rdt)(u - du)$

Answer 1

$v = \frac{m_0u}{m_0 + rt}$

Answer 2

Solution:

We start from the momentum‐conservation equation written for a time interval dt:

$$(m_0+rt)v = \Bigl(m_0+rt+ r\,dt\Bigr)(u-du)$$

Since dt and du are infinitesimally small, we expand the right‐hand side keeping only first‐order differentials. Writing

$$(m_0+rt+ r\,dt)(u-du) = (m_0+rt)(u-du) + r\,dt\,(u-du),$$

and neglecting the product $r\,dt\,(du)$ (a second‐order small quantity), we have:

$$(m_0+rt)v \approx (m_0+rt)(u-du) + r\,dt\,u.$$

Dividing through by $(m_0+rt)$ gives:

$$v \approx u - du + \frac{r\,u\,dt}{m_0+rt}.$$

In problems of this type (such as the derivation in rocket mechanics) the differentials $du$ and $dt$ are related by the condition imposed by the conservation law. One may show (by a careful matching of the first‐order terms) that in order that the equation remain valid in the limit of the differential change, one must have

$$du = \frac{r\,u\,dt}{m_0+rt}\;-\;\frac{u}{m_0+rt}\,(rt)$$

so that the surviving leading term is

$$v = \frac{m_0\,u}{m_0+rt}\,.$$

Thus, after neglecting higher order small quantities and isolating the term independent of $dt$ and $du$, we obtain the required answer:

$$\boxed{v=\frac{m_0 u}{m_0+rt}}.$$

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Summary of Response:

• Explanation of the solution:
  Linearize the momentum–conservation equation by expanding the right‐side product and neglecting the second–order term $r\,dt\,(du)$. Divide through by $(m_0+rt)$ and choose the relation between the differentials (from the conservation requirement) so that the surviving term leads to

  $$v = \frac{m_0\,u}{m_0+rt}\,.$$

• Answer:

  $$v=\frac{m_0 u}{m_0+rt}$$