A rocket of mass m = 8400 kg is launched from the ground at…

A rocket of mass $m = 8400\ \mathrm{kg}$ is launched from the ground at $v_0 = 350\ \mathrm{m/s}$ and $\alpha = 80^\circ$ to the horizontal. At the highest point of its path a part of mass $m_1 = 40\ \mathrm{kg}$ is ejected backward with speed $v = 200\ \mathrm{m/s}$ relative to the rocket. What is the rocket's speed immediately after the ejection?

Explanation of Method

Speed of a rocket after ejecting a part at the top.

Step 1: Speed at the top

At the highest point only the horizontal part of the velocity is left: $v_A = v_0\cos 80^\circ = 350 \cdot 0.17365 \approx 60.78\ \mathrm{m/s}$.

Step 2: Conservation of momentum

The ejection is sudden, so horizontal momentum is conserved. Let $u$ be the rocket's speed after; the part then moves at $u - v$ (backward relative to the rocket): $m\,v_A = (m - m_1)\,u + m_1\,(u - v) = m\,u - m_1 v$.

\[ u = v_A + \frac{m_1}{m}\,v = 60.78 + \frac{40}{8400} \cdot 200 = 60.78 + 0.95. \]

\[u \approx 61.7\ \mathrm{m/s} \ \text{(horizontal, forward)}\]

Check

After: $ 8360 \cdot 61.73 + 40 \cdot (61.73 - 200) = 516\,063 - 5\,531 = 510\,532 $

Before: $ 8400 \cdot 60.78 \approx 510\,527\ \checkmark $ (equal up to rounding)

$\text{just before (top of the path)}$ $m$ $60.78\ \mathrm{m/s}$ $\text{just after}$ $m - m_1$ $u \approx 61.73\ \mathrm{m/s}$ $m_1$ $u - 200 \approx -138\ \mathrm{m/s}$

At the top of its path the rocket moves horizontally. Throwing a small part backward gives the rest a small extra forward push; total momentum stays the same.

Result

\[\boxed{\,u \approx 61.7\ \mathrm{m/s} \ \text{(horizontal, forward)}\,}\]

Remark

Note. The original result, 61.63 m/s, used $350\cos 80^\circ = 60.68$ instead of 60.78. (If the 200 m/s is taken relative to the rocket's speed before ejection, the answer is essentially the same, 61.73 m/s.) This last problem uses momentum conservation, which goes slightly beyond pure kinematics.