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Collisions in 2 dimensions

  • #1
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how do you if the collision is in 2 dimensions?
 

Answers and Replies

  • #3
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so if there is an angle present the collison is 2d. So collision of ball or pucks are 1d.
i have a difficulty in this question A 3000 kg space capsule travelling in outer space with a velocity of 200 m/s. In an effort to alter its course, it fires a 25.0 kg projectile perpendicular to its original direction of motion at a speed of 2000 m/s. What is the new speed of the space capsule and by what angle has its direction changed? I just dont understand the question. could you explain it to me?
 
  • #4
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ok so this has got to do something with impulse, right? I can find P1 of the capsule and p1 of the projectile...and i think i can find F1 by F=mg? and find out P2 by using P2=P1+Ft
But i dont have t..
 
  • #5
Doc Al
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what's conserved?

Hint: In an "explosion" (which this can be treated as) what quantity is conserved?
 
  • #6
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momentum is conserved
 
  • #7
Doc Al
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momentum is conserved
Yes. Apply momentum conservation. That's all you need. You have the initial total momentum; and you have the final momentum of one piece. Remember that momentum is a vector, so direction is important.
 
  • #8
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So inititial momentum of the capsule+ im of the projectile= fm of capsule+fm of projectile
im of projectile is 0
then we find an equation for final v of the casule
so now we apply this to the x direction right?
 
  • #9
Doc Al
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So inititial momentum of the capsule+ im of the projectile= fm of capsule+fm of projectile
OK
im of projectile is 0
No. Initially the projectile was part of the capsule, so treat them as one piece at that point.
then we find an equation for final v of the casule
so now we apply this to the x direction right?
I suggest that you call the initial direction of the capsule to be in the +x direction. And call the direction of the projectile to be the -y direction. (Draw yourself a picture.)
 
  • #10
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ok so that means, the initial momentu of the capsule includes the projectile as well,
So it will be 3000v= 25*v of projectile+3000*Vf of capsule
Now we isolate Vf.
Vf= (3000v-25*V of projectile)/3000
k so i tried to draw the diagram, but i still dont understand
 
  • #11
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k so i got, Vfx=(3000*Vx-25*Vy of projectile)/3000

here Vx will be 200, Vy will be 2000, right? So now how do i got about finding the angle?
 
  • #12
Doc Al
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44,874
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ok so that means, the initial momentu of the capsule includes the projectile as well,
So it will be 3000v= 25*v of projectile+3000*Vf of capsule
Now we isolate Vf.
Vf= (3000v-25*V of projectile)/3000
k so i tried to draw the diagram, but i still dont understand
Remember to treat momentum as a vector. First find the x and y components of the initial momentum of capsule + projectile:

Px = (M + m) v = (3000 + 25)*200
Py = 0

(Although its not clear from the problem statement, I'm treating the capsule without the projectile as having mass 3000 kg and the projectile as an additional 25 kg.)

We know the final momentum of the projectile:
P_Proj_x = 0
P_Proj_y = mv = 25*(-2000) = -50,000

Apply momentum conservation to figure out the x and y components of the capsule's final momentum. That will tell you its speed and direction.
 
  • #13
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Ok so,
mv+Mv= mV1+MV2
So now we apply this to the x direction,
(3000+25)200= 0+3000V2
And we can find V2 (x)
For the y direction,
0=-50000+3000V2
we can find V2(y)
So now using the phythagorean theorem, we can find V2 and the angle..sounds k..?
 
  • #14
Doc Al
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44,874
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Sounds good.
 

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