Question

b) A space vehicle travelling at a velocity of 1000 ms−1 separates by a controlled explosion into two sections of mass 850 kg and 250 kg. The two parts carry on in the same direction with the heavier rear section moving 120 ms−1 slower than the lighter front section. Determine the final velocity of each section.

Answer 1

Answer on Question #59559-Physics – Mechanics | Relativity

A space vehicle travelling at a velocity of 1000 ms⁻¹ separates by a controlled explosion into two sections of mass 850 kg and 250 kg. The two parts carry on in the same direction with the heavier rear section moving 120 ms⁻¹ slower than the lighter front section. Determine the final velocity of each section.

Solution

The velocity of lighter rear section is $v_{L}$ and of heavier rear section is $v_{H} = v_{L} - \Delta v$.

According to the conservation of momentum principle:

$$(m_{1} + m_{2})V = m_{1}(v_{L} - \Delta v) + m_{2}v_{L}$$

So,

$$v_{L} = \frac{(m_{1} + m_{2})V + m_{1}(\Delta v)}{(m_{1} + m_{2})} = V + \frac{m_{1}}{(m_{1} + m_{2})}\Delta v = 1000 + \frac{850}{(850 + 250)}120 = 1015\frac{m}{s}.$$ $$v_{H} = v_{L} - \Delta v = 1015 - 120 = 895\frac{m}{s}.$$

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