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roane
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I'm not very confident with my final answer for this question, wondering where I might have gone off track...
2 spacecraft are thrust apart, what was the original speed of the 2 craft when they were linked together?
Mass spacecraft 1 = 1.9 x 10^4kg
Velocity spacecraft 1 after separation is 3.5 x 10^3 km/h @ 5.1º [NE]
Mass spacecraft 2 = 1.7 x 10^4kg
Velocity spacecraft 2 after separation is 3.4 x 10^3 km/h @ 5.9º [SE]
For spacecraft 1
p = m*v
p = 1.8 x 10^7 kg.m/s
For spacecraft 2
p = m*v
p = 1.6 x 10^7 kg.m/s
The first vector is 1.8 x 10^7 kg.m/s @ 5.1 [NE]
The second vector is 1.6 x 10^7 kg.m/s @ 5.9 [SE]
The angle between these 2 vectors is 169º.
Using law of cosines...
c^2 = (1.8 x 10^7)^2+(1.6 x 10^7)^2 -[2(1.8 x 10^7)(1.6 x 10^7)cos(169)]
c^2 = 3.24 x 10^14 + 2.56 x 10^14 -[2(2.88 x 10^14)cos(169)]
c^2 = 5.8 x 10^14 + 5.7 x 10^14
c^2 = 11.5 x 10^14
c = 3.4 x 10^7kg.m/s
Final momentum = Initial momentum
p = m.v
and combined mass before separation is 36000kg
3.4 x 10^7kg.m/s = 36000kg.v
v = 944m/s
2 spacecraft are thrust apart, what was the original speed of the 2 craft when they were linked together?
Mass spacecraft 1 = 1.9 x 10^4kg
Velocity spacecraft 1 after separation is 3.5 x 10^3 km/h @ 5.1º [NE]
Mass spacecraft 2 = 1.7 x 10^4kg
Velocity spacecraft 2 after separation is 3.4 x 10^3 km/h @ 5.9º [SE]
For spacecraft 1
p = m*v
p = 1.8 x 10^7 kg.m/s
For spacecraft 2
p = m*v
p = 1.6 x 10^7 kg.m/s
The first vector is 1.8 x 10^7 kg.m/s @ 5.1 [NE]
The second vector is 1.6 x 10^7 kg.m/s @ 5.9 [SE]
The angle between these 2 vectors is 169º.
Using law of cosines...
c^2 = (1.8 x 10^7)^2+(1.6 x 10^7)^2 -[2(1.8 x 10^7)(1.6 x 10^7)cos(169)]
c^2 = 3.24 x 10^14 + 2.56 x 10^14 -[2(2.88 x 10^14)cos(169)]
c^2 = 5.8 x 10^14 + 5.7 x 10^14
c^2 = 11.5 x 10^14
c = 3.4 x 10^7kg.m/s
Final momentum = Initial momentum
p = m.v
and combined mass before separation is 36000kg
3.4 x 10^7kg.m/s = 36000kg.v
v = 944m/s