Answer:
[tex]v_{1f} = 90.12m/s[/tex]
Explanation:
As we know that Probe along with rocket system is an isolated system
So here net force on this system must be zero
and we can use momentum conservation for this system
So we will have
[tex](m_1 + m_2)v_{i} = m_1v_{1f} + m_2v_{2f}[/tex]
here we know that
[tex]7310 \times 88.3 = (7310- 54)v_{1f} + 54(v_{1f} - 246)[/tex]
[tex]645473 = 7310\times v_{1f} - 13284\\658757 = 7310 \times v_{1f}[/tex]
so we have,
[tex]v_{1f} = 90.12m/s[/tex]
Answer:
Final velocity of Probe; V_f = 656.314 m/s
Explanation:
We are given;
Mass of space probe; m_i = 7310 kg
Initial velocity of probe ; v_i = 88.3 m/s
Mass of exhaust fuel; m = 54 kg
V_rel = 246 m/s
Mass of fuel; m_f = m_i - m = 7310 kg - 54kg = 7256 kg
From Rocket equation;
V_f - V_i = V_rel[In(m_i/m_f)]
Where V_f is final velocity of probe.
Thus, plugging in the relevant values to get ;
V_f - 88.3 = 246[In(7310/7256)]
V_f - 88.3 = 246 x 2.309
V_f - 88.3 = 568.014
Thus, V_f = 568.014 + 88.3 = 656.314 m/s
