What Happens When a Soviet and US Satellite Collide Elastically?

  • Thread starter hobo
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In summary, the Soviet satellite has a higher velocity and will thus travel further than the American satellite.
  • #1
hobo
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I have no clue on where to start on this. I have spent time playing with that cart thing and the momentum balls and I have ideas floating around, but nothing concrete.

There is a 248.2kg Soviet satellite traveling due east at a velocity of 1025km/hr and a 149.9kg US satellite traveling due west at a velocity of 1181.8km/hr. The colission is perfectly elastic. What happens afterwards?
 
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  • #2
Use conservation of momentum (no external forces) and conservation of energy (fully elastic collision)
 
  • #3
From what you told me, I used

m1v1 + m2v2 = m1v1' + m2v2'
(mv1^2)/2 + (mv2^2)/2 = (mv1'^2)/2 + (mv2'^2)/2

I solved both for V1' then I plugged in the numbers.After that, it came out to

5364792208 - 35930354.61v2' +59675.19v2'^2

I then plugged it into the quadratic equation, getting 328.08333m/s which is v2, and 274.0153711m/s, which is not the answer my teacher gave me
 
  • #4
Did you take into account that the velocities are in different directions?
 
  • #5
Now, let's solve this problem in full:
I will assume that you gave me the right information, in particular that the collision was fully ELASTIC.

The equations for two objects, 1 and 2, to be solved are:
[tex]m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}[/tex]

This is conservation of momentum, as you had; b and a refer to initial and final velocities (before and after collision).

Energy conservation:
[tex]\frac{1}{2}m_{1}v_{1,a}^{2}+\frac{1}{2}m_{2}v_{2,a}^{2}=\frac{1}{2}m_{1}v_{1,b}^{2}+\frac{1}{2}m_{2}v_{2,b}^{2}[/tex]

This is also, I believe, what you meant.
Note that the energy equation can be simplified in the following manner:
[tex]m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a})+m_{2}(v_{2,b}-v_{2,a})(v_{2,b}+v_{2,a})=0[/tex]

Rewrite the momentum equation as:
[tex]m_{2}(v_{2,b}-v_{2,a})=-m_{1}(v_{1,b}-v_{1,a})[/tex]

Hence, the energy equation may be written as:
[tex]m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a}-(v_{2,b}+v_{2,a}))=0[/tex]

The root [tex]v_{1,b}=v_{1,a}[/tex] corresponds to no collision, so we have the following system to solve:
[tex]v_{1,b}+v_{1,a}=v_{2,b}+v_{2,a}[/tex]
[tex]m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}[/tex]

Solving for [tex]v_{1,a},v_{2,a}[/tex] yields:
[tex]v_{1,a}=\frac{(m_{1}-m_{2})v_{1,b}+2m_{2}v_{2,b}}{m_{1}+m_{2}}[/tex]
[tex]v_{2,a}=\frac{(m_{2}-m_{1})v_{2,b}+2m_{1}v_{1,b}}{m_{1}+m_{2}}[/tex]

Now, we have simplified the symbolic expressions maximally; it is time to enter in the values!
We set:
[tex]m_{1}=149.9,v_{1,b}=1181.8[/tex]
[tex]m_{2}=248.2,v_{2,b}=-1025[/tex]

The answers should be quite different from the ones you gave
 

1. What is the difference between impulse and momentum?

Impulse is the change in momentum of an object over a period of time, while momentum is the quantity of motion an object has. In other words, impulse is a force applied to an object for a certain amount of time, while momentum is a measure of the object's mass and velocity.

2. How is impulse calculated?

Impulse is calculated by multiplying the force applied to an object by the time duration of the force. Mathematically, it can be represented as J = F x Δt, where J is impulse, F is force, and Δt is the change in time.

3. Can impulse and momentum be negative?

Yes, both impulse and momentum can be negative. A negative impulse would indicate a decrease in the object's momentum, while a negative momentum would indicate that the object is moving in the opposite direction of the chosen reference point.

4. How does impulse affect an object's motion?

Impulse causes a change in an object's momentum, which in turn affects its motion. If the impulse is applied in the same direction as the object's motion, it will increase its speed. If the impulse is applied in the opposite direction, it will decrease the object's speed or even bring it to a complete stop.

5. What is the principle of conservation of momentum?

The principle of conservation of momentum states that in a closed system, the total amount of momentum remains constant. This means that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction, as long as there are no external forces acting on the system.

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