Question

A space vehicle travelling at a velocity of 1200m/s separates by a controlled explosion into two sections of mass 855 kg and 240 kg. The two parts carry on in the same direction with the heavier rear section moving 120 m/s slower than the lighter front section. Determine the velocity of each section after separation.

Answer 1

A space vehicle travelling at a velocity of $1200\mathrm{m/s}$ separates by a controlled explosion into two sections of mass $855\mathrm{kg}$ and $240\mathrm{kg}$. The two parts carry on in the same direction with the heavier rear section moving $120\mathrm{m/s}$ slower than the lighter front section. Determine the velocity of each section after separation.

Solution:

$V = 1200\frac{\mathrm{m}}{\mathrm{s}}$ – initial speed of the space vehicle;

$m_1 = 855\mathrm{kg}$ – mass of the first section;

$m_2 = 240\mathrm{kg}$ – mass of the second section;

$V_1$ – speed of the heavier section;

$V_2$ – speed of the lighter section;

$\Delta V = 120\frac{\mathrm{m}}{\mathrm{s}}$ – speed difference between the sections;

The law of conservation of momentum along the X-axis:

$$x: (m_1 + m_2)V = m_1V_1 + m_2(V_1 + \Delta V)$$ $$m_1V_1 + m_2V_1 = (m_1 + m_2)V - m_2\Delta V$$ $$V_1 = \frac{(m_1 + m_2)V - m_2\Delta V}{m_1 + m_2} = V - \frac{m_2\Delta V}{m_1 + m_2} = 1200\frac{\mathrm{m}}{\mathrm{s}} - \frac{240\mathrm{kg} \cdot 120\frac{\mathrm{m}}{\mathrm{s}}}{855\mathrm{kg} + 240\mathrm{kg}} = 1174\frac{\mathrm{m}}{\mathrm{s}}$$ $$V_2 = V_1 + \Delta V = 1174\frac{\mathrm{m}}{\mathrm{s}} + 120\frac{\mathrm{m}}{\mathrm{s}} = 1294\frac{\mathrm{m}}{\mathrm{s}}$$

Answer: speed of the heavier (855kg) section: $1174\frac{\mathrm{m}}{\mathrm{s}}$

Speed of the lighter (240kg) section: $1294\frac{\mathrm{m}}{\mathrm{s}}$