A rocket with a structural mass m_s = 2 t and fuel mass m_z…

A rocket with a structural mass $m_s = 2\ \mathrm{t}$ and fuel mass $m_z = 1.5\ \mathrm{t}$ burns fuel at $k = 250\ \mathrm{kg/s}$; the exhaust speed relative to the rocket is $u = 1\ \mathrm{km/s}$. (a) After what time does it reach $v = 200\ \mathrm{m/s}$? (b) What is its final speed? (Neglect gravity and air resistance.)

Explanation of Method

Rocket speed while its fuel burns.

Step 1: (a) Time to 200 m/s

\[ v = u\ln\frac{m_0}{m_0 - kt}, \]

\[ t = \frac{m_0}{k}\left(1 - e^{-v/u}\right) = \frac{3500}{250}\left(1 - e^{-0.2}\right). \]

\[t \approx 2.54\ \mathrm{s}\]

Step 2: (b) Final speed

All the fuel burns in $1500/250 = 6\ \mathrm{s}$, leaving the 2 t structure: $v_f = u\ln\frac{3500}{2000} = 1000\ln 1.75$.

\[v_f \approx 560\ \mathrm{m/s}\]

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $2$ $4$ $6$ $200$ $400$ $560$ $2.54$ $\text{burnout}$ $v = u\ln\frac{m_0}{m_0 - kt}$

Rocket speed while the fuel burns (no gravity or drag). The curve steepens because the same thrust acts on an ever lighter rocket; all 1.5 t burn in 6 s.

Result

\[\boxed{\,t \approx 2.54\ \mathrm{s}\,}\]

\[\boxed{\,v_f \approx 560\ \mathrm{m/s}\,}\]

Remark

Check. The original results are correct (for motion without gravity, e.g. horizontal or in space). For a vertical launch, gravity would subtract $gt$: 25 m/s by 2.54 s and 59 m/s by burnout.