At the launch of a rocket of mass m = 200 kg , a quarter of…
At the launch of a rocket of mass $m = 200\ \mathrm{kg}$, a quarter of its mass burns instantly and is ejected backward as gas, at $v_1 = 1800\ \mathrm{m/s}$ relative to the rocket. Find (1) the rocket's speed after the ejection, and (2) how far it flies if it is launched at $\alpha = 30^\circ$ to the horizontal (no air resistance, level ground).
Explanation of Method
A rocket ejecting a quarter of its mass.
Step 1: Speed after the ejection
The rocket (150 kg) moves forward at $V$; the gas (50 kg) leaves 1800 m/s slower than the rocket, i.e. with ground velocity $V - 1800$. The total momentum stays zero: $150\,V + 50\,(V - 1800) = 0 \;\Longrightarrow\; 200\,V = 90\,000$.
\[V = 450\ \mathrm{m/s}\]
(The gas then moves backward at 1350 m/s relative to the ground.)
Step 2: Range
\[ d = \frac{V^2\sin 2\alpha}{g} = \frac{450^2 \cdot 0.866}{9.81}. \]
\[d \approx 17\,900\ \mathrm{m} \approx 17.9\ \mathrm{km}\]
Check
Momentum: $ 150 \cdot 450 = 67\,500 $, gas $ 50 \cdot 1350 = 67\,500\ \checkmark $
Flight time $ 2 \cdot 225/9.81 = 45.9\ \mathrm{s} $; $ d = 389.7 \cdot 45.9 \approx 17.9\ \mathrm{km}\ \checkmark $
Result
\[\boxed{\,V = 450\ \mathrm{m/s}\,}\]
\[\boxed{\,d \approx 17\,900\ \mathrm{m} \approx 17.9\ \mathrm{km}\,}\]
Remark
About the original solution. It treated 1800 m/s as the gas speed relative to the ground and got 600 m/s. The problem gives it relative to the rocket, which is itself moving forward after the ejection, so the correct speed is 450 m/s and the range about 17.9 km (not 31.75 km). (Even with 600 m/s, the range would be 31.78 km; 31.75 is an arithmetic slip.)