Gr. 12 - Conservation of Momentum in Two Dimensions

[Deceit]

Homework Statement

This question is from a correspondence course I'm taking:

Two spacecraft from different nations have linked in space and are coasting with their engines off, heading directly toward Mars. The spacecraft s are thrust apart by large springs. Spacecraft 1, with a mass of 1.9 x 10^4 kg, then has a velocity of 3.5 x 10^3 km/h at 5.1 degrees to its original direction. Spacecraft 2, whose mass is 1.7 x 10^4 kg, has a velocity of 3.4 x 10^3 km/h at 5.9 degrees to its original direction. Determine the original speed of the two craft when they were linked together.

Now, my first thought was that I could find the original speed by finding the component of the final speed of one of the crafts that is parallel to the original direction of motion. Because the spring seems to be accelerating the crafts perpendicular to their motion, their components of speed parallel to the direction of motion should be unchanged, correct? But this seems too simplistic and doesn't really deal with the momentum of the system (which is the topic on hand)... and the numbers don't quite work out.

Also, p(f) of the system would not equal p(i) of the system, is that correct? The spring is applying a force, thus adding kinetic energy to the system. So would conservation of momentum not apply here?

Any help is appreciated!

 

[Denken]

i don't really feel like number crunching to check if my idea is correct but for the most part i agree with how you attempted this and i would have started the same way. the difference i believe would be that if you factor in a ratio of their mass's. ... If you think of it as two separate objects with different mass's that are traveling with the same velocity they will each have a different momentum so when put together their momentum should change but since the mass is only added together when you divide mass from momentum you should get their initial velocity.

 

[Deceit]

But even though their momentums are different, they should still have the same velocity along the x component, right?

 

[cepheid]

Now, my first thought was that I could find the original speed by finding the component of the final speed of one of the crafts that is parallel to the original direction of motion. Because the spring seems to be accelerating the crafts perpendicular to their motion, their components of speed parallel to the direction of motion should be unchanged, correct?

You don't know this for sure. The spring force could be directed along almost any axis (except one that is perfectly aligned with the original direction of motion). You should use conservation of momentum in order to treat this situation as generally as possible.[STRIKE] EDIT: Another way to understand why what you were saying can't be right is to realize that the final total momentum in the original (let's call it x) direction has to be less than the original total momentum in that direction, since some of that x-momentum has now been turned into y-momentum.[/STRIKE] EDIT 2: Yeah that was really stupid. What is actually correct is that the final total x-momentum equals the initial total momentum (which was all x). The final total y-momentum is zero, just as it was originally, which means that the two final y-momenta have to cancel each other out.

Also, p(f) of the system would not equal p(i) of the system, is that correct? The spring is applying a force, thus adding kinetic energy to the system. So would conservation of momentum not apply here?

No, this is not correct. The idea here is that there are no net external forces acting on the system. Therefore, the total momentum of the system is conserved. There are only contact forces between the spacecraft , which are forces that are internal to the system. Think about this this way: Newton's 3rd Law of Motion says that if spacecraft 1 exerts a force on spacecraft 2 for some duration before they separate, then spacecraft 2 also exerts a force on spacecraft 1 for that duration that is equal in magnitude and opposite in direction. Therefore, the change in momentum experienced by spacecraft 1 is exactly opposite to the change in momentum experienced by spacecraft 2. So, although the momenta of the individual components in the system change, these changes cancel each other out, keeping the total momentum of the system constant.

 

[Deceit]

Thanks cepheid! That really helped me understand the problem better.

Now here's my attempt at a solution:
p(i) = v(m1 + m2) (the two craft have the same velocity at this point)
p(f) = m1v1 + m2v2
p(i) = p(f)

p(f) = (19000kg)(972 m/s) + (17000kg)(944 m/s)
p(f) = 3.45 x 10^7 kg m/s

p(i) = p(f) = v(m1 + m2)
v = p(f)/(m1 + m2)
v = (3.45 x 10^7 kg m/s) / (19000kg + 17000kg)
v = 959 m/s

Therefore the two crafts were moving with a speed of 9.6 x 10^2 m/s when they were linked together.

This solution almost seems too simple, I was expecting some trigonometry. Does my solution make sense?

EDIT: No, it doesn't make sense. I didn't take direction into account. As it turns out, I do need some trigonometry.

Let's try this again:
p(f) = p1(f) + p2(f)
p(i) = p(f)
p(i) = p1(f) + p2(f)

So, to find p(i), I can draw a vector diagram adding p1(f) and p2(f) together.
The angles within the triangle are 5.1, 5.9, and 169 degrees.
Knowing all of the angles and two of the sides, I use sin law to find p(i):

a/sinA = b/sinB
p1(f) / (sin5.9) = p(i) / (sin169)
(1.85 x 10^7) / (sin5.9) = p(i) / (sin169)
p(i) = 3.43 x 10^7

Now I can find the velocity:
p(i) = v(m1 + m2)
v = (3.43 x 10^7 kg m/s) / (19000kg + 17000kg)
v = 954 m/s

Therefore the two crafts were moving with a speed of 9.5 x 10^2 m/s when they were linked together.

Very close to the first answer I got, but a tad more accurate I suppose.

I feel much more confident with this answer; does it make sense?