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A 7.18 x10³ kg space vehicle and its empty 6.32 x10² kg booster unit are moving together through space at a speed of 369.66 m/s. An explosion lasting 2.81 s is used to separate the two parts. If the speed of the space vehicle after the separation increased to 444.57 m/s what is the magnitude force on the booster unit?

User Jovany
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The magnitude of the force on the booster unit can be calculated using the principle of conservation of momentum. According to this principle, the momentum of a system is conserved before and after an interaction, as long as no external forces act on the system.

Before the separation, the total momentum of the system is:

m_total * v_initial = (7.18 x 10^3 kg) * (369.66 m/s)

After the separation, the momentum of the space vehicle is:

m_vehicle * v_final = (7.18 x 10^3 kg) * (444.57 m/s)

And the momentum of the booster unit is:

m_booster * v_final = (6.32 x 10^2 kg) * v_final

Since the total momentum is conserved, the initial momentum must equal the final momentum:

m_total * v_initial = m_vehicle * v_final + m_booster * v_final

Solving for v_final, we find:

v_final = (m_total * v_initial - m_vehicle * v_final) / m_booster

Now we can find the magnitude of the force on the booster unit using Newton's Second Law, which states that the force acting on an object is equal to its mass times its acceleration:

F = m_booster * a = m_booster * (dv/dt)

where dv/dt is the change in velocity over time, which can be approximated as (v_final - v_initial) / time.

Substituting the values we have found into this equation, we find:

F = (6.32 x 10^2 kg) * ((v_final - v_initial) / (2.81 s))

This equation can be evaluated to find the magnitude of the force on the booster unit.

User Chaudhary Amar
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