Question

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 at 973 ms−1. Determine the final velocity of the lighter section.

Answer 1

The velocity of lighter rear section is $v_L$ and of heavier rear section is $v_H$

According to the conservation of momentum principle:

$$(m_1+m_2)\text{v}=m_1v_H+m_2v_L$$

$$v_L=\dfrac{(m_1+m_2)\text{v}-m_1v_H}{m_2}$$

$$v_L=\dfrac{(850kg+250kg)1000ms^{-1}-850kg(973ms^{-1})}{250kg}$$

$$=1091.8ms^{-1}$$