Question: Conservation of Momentum

[roane]

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

 

[Redbelly98]

Welcome to Physics Forums

I really don't follow your derivation, in particular:

• Are the angles 5.1° and 5.9° taken from the x axis or the y axis?
  
• I don't see the significance of c here. I get that it's supposed to be the difference between the momenta of the two spacecraft , but that isn't useful in any way I'm aware of.

That being said -- the usual way to solve these problems is to set up two equations, expressing the conservation of momentum in the x and y directions. This will give you the x and y components of the initial velocity.

 

[roane]

Hi, thanks for responding!

My thinking was...first I'd calculate the total final momentum of each craft after separation.

Then create a diagram of the vector addition and use Law of Cosines where the resultant would be the "c" variable.

The arrow representing P after separation of the first craft would be 5.1° below the x-axis.
The arrow representing P after separation of the second craft would be 5.9° above the x-axis.
The angle between the 2 vectors is 169°

Since initial momentum = final momentum, the resultant here will equal total initial momentum which I arrived at 3.4 x 10^7kg.m/s

And two masses combined before separation is 36000kg

3.4 x 10^7kg.m/s = 36000kg.v

v = 944m/s

 

[Redbelly98]

Ah, now I understand. Your answer looks pretty good, though there may be some roundoff error. You might try expressing your intermediate calculations in 3 significant figures, then round off to the correct sig figs at the very end.

 

[roane]

OK, Thanks so much!