Elastic Collision is outer space

  • Thread starter RedBurns
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  • #1
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I think I'm getting lost in the numbers somewhere here.

Two astronauts, one of mass 60 kg and the other 84 kg, are initially at rest in outer space. They then push each other apart. How far apart are they when the lighter astronaut has moved 10 m?


m1= 60 kg
m2=84 kg
X initial = 0
X1 final= 10 m
V1 inital = 0
V2 initial= 0
V 1 final= 10m/s
T=1 S

.5*60*0^2 + .5*84*0^2 = .5*60*10^2 +.5*84*V2Final^2
0=3000 + 42* V2Final^2
-71=V2Final^2
V2Final = -8.5 m/s

8.5+10=18.5 m
 
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Answers and Replies

  • #2
Physics Monkey
Science Advisor
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Why did you say the final velocity for astronaut one is 10 m/s? This isn't right. You need to determine the velocity of each astronaut after they push off each other. Since the collision is elastic, you can use conservation of energy and momentum to find the two velocities. Notice that as it stands now, your astronauts have not conserved momentum!

Edit: Yikes, I knew there was some reason this question was bothering me. Redburns, I'm sorry, but I didn't read the question or your response very carefully. The square root of a negative number is imaginary, not another negative number. Clearly the "collision" or push can't be elastic since they start out with no kinetic energy. My apologies for posting too quickly without reading things carefully. You do need to use conservation of momentum.
 
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  • #3
andrevdh
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Their combined centre of mass will stay at the initial point and will not move after the interaction since they are experiencing only internal forces.
 
  • #4
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In the question you posted no information is given about the strength of the push given and so finding the *numerical* velocity of either person A or B is gonna be a bit tricky (did one push very gently with his little finger or give him a mighty big shove!)...knowing these facts are not necessary!

What is important is that the impulse that A exerts on B is the same impulse that B imparts on A

Whether the lighter dudes velocity was 5m/s or 0.0001m/s he is always going to have travelled a distance of 10 meters at precisely the same point when the heavy dude has travelled 'a' meters (where 'a' is constant) (assuming no other forces are acting)
 
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  • #5
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Thanks! I though I might be making this one harder than it should be.
 
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