Conservation of Momentum in two dimensions

In summary, the two spacecrafts were originally moving at different speeds, but they were linked together and now their speeds are the same. The original speed of the two spacecrafts when they were linked together is 125 m/s.
  • #1
Hey Everyone,

I've just signed up for this forum because I'm currently taking a correspondence course and there's one problem in particular that I'm having problems with and there isn't really anybody I can ask. Can anybody here help me with this?

Two spacecraft s from different nations have linked in space and are coasting with their engines off, heading directly towards 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 crafts when they were linked together.

I arrived at an answer of 125 m/s. But I know that's way off! Again, can anybody help me out please?
 
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  • #2
Mitchtwitchita said:
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.
I presume that the spacecraft move on opposite sides of the original direction. (For example: if one angles north of the original direction, the other angles south.)

I arrived at an answer of 125 m/s.
Show how you arrived at your answer.
 
  • #3
yes you are correct that they move on opposite sides of the origin.

This is how I arrived at that answer:

Given: m1=1.9 x 10^4 Kg
m2=1.7 x10^4 Kg
v1f=3.5 x 10^3 Km/h = 972 m/s
v2f=3.4 x 10^3 Km/h = 944 m/s

Also: v1 = v2 = Pti

Initial Momentum:

P1i =m1v1
=unknown

P2 =m2v2i
=unknown

Final Momentum:

P1f =m1v1f
=1.9 x 10^4 Kg * 972 m/s [5.1 degrees Up]
=1.8 x 10^7 Kg*m/s

P2f =m2v2f
=1.7 x 10^4 * 944 m/s [5.9 degrees down]
=1.6 x 10^7 Kg*m/s

Since P1i + P2i =Pti and Pti = Ptf

Therefore, Pti = P1f + P2f

(Pti)^2 = (P2f)^2 + (P1f)^2 - 2(P1f)*(P2f)Cos11degrees
=(1.6 x 10^7 Kg*m/s)^2 + (1.8 x 10^7 Kg*m/s)^2 - 2(1.8 x 10^7 Kg*m/s) * (1.6 x 10^7 Kg*m/s)Cos11degrees
=2.6 x 10^14 + 3.4 x 10^14 - 5.9 x 10^14 Cos11degrees
=2.6 x 10^14 + 3.4 x 10^14 - 5.8 x 10^14
=2 x 10^13 (Kg*m/s)^2

Pti = the square root of 2 x 10^13 Kg*m/s
= 4.5 x 10^6 Kg*m/s

Therefore, the magnitude of the total initial momentum of the rocket was 4.5 x 10^6 Kg*m/s

P=mv
Therefore, 4.5 x 10^6 Kg*m/s = 36000 Kg(v)
v=4.5 x 10^6 Kg*m/s / 36000 Kg
= 125 m/s

Somehow I don't think I had the right approach.
 
  • #4
vector addition

Nothing wrong with using the law of cosines to find the resultant, but you have the wrong triangle and thus the wrong angle.

Draw a diagram of the vector addition you are trying to model: Draw an arrow representing P1i (draw it making an angle of 5.1 degrees below the x-axis), then add to it an arrow representing P2i (draw it with an angle of 5.9 degrees above the x-axis). Make sure to put the tail of P2i at the head of P1i to correctly represent vector addition.

Once you've drawn the triangle properly, then you can apply the law of cosines. What's the angle between those two sides of the new triangle?
 
  • #5
Perhaps its worth pointing out that a solution can also be found by finding the components of P1f and P2f parallel to the original direction of motion. Add these components and you have the initial momentum Pti.

The perpendicular components should be equal and opposite so do not contribute to Pti.
It gets the same answer of course but is slightly quicker I think.
 
  • #6
apelling said:
Perhaps its worth pointing out that a solution can also be found by finding the components of P1f and P2f parallel to the original direction of motion. Add these components and you have the initial momentum Pti.
That's exactly how I would solve this problem. But there are several ways to skin this cat. To the OP: Do it both ways and compare! :wink:
 
  • #7
Thanks Doc The new vector diagram worked! I was always crappy with Trigonometry! I appreciate it Doc.
 

1. What is conservation of momentum in two dimensions?

Conservation of momentum in two dimensions is a fundamental principle in physics that states that in a closed system, the total momentum remains constant in all directions. This means that the total momentum of all objects before a collision or interaction is equal to the total momentum after the collision or interaction.

2. How is momentum conserved in two dimensions?

Momentum is conserved in two dimensions through the conservation of both the x and y components of momentum. This means that the total momentum in the x-direction before a collision is equal to the total momentum in the x-direction after the collision. The same applies for the y-direction.

3. What is the formula for calculating momentum in two dimensions?

The formula for calculating momentum in two dimensions is p = m*v, where p is the momentum, m is the mass of the object, and v is the velocity of the object. In two dimensions, this formula can be applied separately for the x and y components of momentum.

4. Can momentum be transferred between objects in two dimensions?

Yes, momentum can be transferred between objects in two dimensions. This transfer of momentum occurs during collisions or interactions between objects. The total momentum before and after the collision must be equal in both the x and y directions.

5. How does conservation of momentum in two dimensions relate to real-world situations?

Conservation of momentum in two dimensions is a fundamental principle in physics that applies to many real-world situations. For example, it explains the motion of objects in collisions, such as billiard balls on a pool table, and can also be seen in the movement of objects in nature, such as planets orbiting around the sun.

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