Question

A space vehicle travelling at a velocity of 1206ms−1 separates by a controlled explosion
into two sections of mass 952 kg and 332 kg. The two parts carry on in the same direction
with the lighter front section moving 145 ms−1 faster than the heavier rear section. Determine the
final velocity of each section. Ensure you include a labelled diagram.

Answer 1

Question # 74440

A space vehicle travelling at a velocity of 1206ms⁻¹ separates by a controlled explosion into two sections of mass 952 kg and 332 kg. The two parts carry on in the same direction with the lighter front section moving 145 ms⁻¹ faster than the heavier rear section. Determine the final velocity of each section. Ensure you include a labelled diagram.

Answer:

Here we have an inelastic collision problem (see the figure below).

The momentum conservation equation before and after the explosion $s$ given by:

$$m _ {0} v _ {0} = m _ {1} v _ {1} + m _ {2} v _ {2},$$

where $m_{1} = 952 \, \mathrm{kg}$ and $m_{2} = 332 \, \mathrm{kg}$ – masses of two parts of the vehicle after the explosion;

$v_{1}$ and $v_{2}$ – velocities of two parts of the vehicle after the explosion;

$m_0 = m_1 + m_2 - \text{mass of the vehicle before the explosion};$

$v_{0} = 1206 \, \text{m.s}^{-1}$ – velocity of the vehicle before the explosion;

Since

$$v _ {2} = v _ {1} + 145 \, \text{m.s}^{-1},$$

equation (1) gives us the following:

$$(m _ {1} + m _ {2}) v _ {0} = m _ {1} v _ {1} + m _ {2} (v _ {1} + 145),$$ $$v _ {1} = \frac {(m _ {1} + m _ {2}) v _ {0} - 145 m _ {2}}{m _ {1} + m _ {2}} = \frac {(952 + 332) \cdot 1206 - 145 \cdot 332}{952 + 332} = 1168.5 \, \text{m.s}^{-1},$$ $$v _ {2} = 1168.5 + 145 = 1313.5 \, \text{m.s}^{-1}.$$

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