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Jumping cat & momentum

  1. Oct 27, 2005 #1
    Two 18.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 4.72 kg cat, initially standing on sled 1, jumps across to sled 2 and then jumps back to sled 1. Both jumps are made at a horizontal speed of 3.79 m/s relative to the ice. What is the final speed of sled 2? (Assume the ice is frictionless.)

    I tried using conservation of momentum.
    (18.0 + 4.72)(3.79)= (18.0)v
    solving for v gave me 4.78 m/s which wasn't right.
    Can someone help?
    thanks
     
  2. jcsd
  3. Oct 28, 2005 #2

    Tide

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    You didn't write momentum conservation correctly.

    If v is the speed of the cat (relative to the ground) and m it's mass then the cat's momentum upon leaving sled 1 is mv. After landing on the sled (mass = M) the momentum is (m + M) V1 so mv = (m + M)V1. When the cat jumps off that sled, momentum is again conserved so (m+M)V1 = MV2 - mv. (Notice that the first sled doesn't really enter into the picture for this particular problem.)

    You should be able to take it from there! :)
     
  4. Oct 28, 2005 #3
    Thank you. I got that part, the answer was 1.99 m/s.
    The second part asks for the final speed of sled 1.
    I tried using what you did before, only this time the initial sled was sled 2 and final was sled 1.
    MV2-mv= (m+M)V1
    (18)(1.99)-(4.72)(3.79)= (18 + 4.72)V1
    Solving for v gave me .789 m/s, which wasn't right.
    Can someone help?
     
  5. Oct 28, 2005 #4

    Doc Al

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    You're solving for V2, not V1. V1 = 1.99 m/s, not V2.
     
  6. Oct 28, 2005 #5
    Ok I used (m+M)V1= MV2-mv
    (4.72+18)(1.99)= (18)V2- (4.72)(3.79)
    Solving for V2 gave me 3.51 m/s.. which wasn't right.
    Is my equation even right?
     
  7. Oct 28, 2005 #6

    Tide

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    Punch,

    From what I wrote previously you can take a shortcut:

    [tex]mv = (M+m)V_2 - mv[/tex]

    and the solution should jump right out! :)
     
  8. Oct 29, 2005 #7

    Doc Al

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    Staff: Mentor

    Your equation is correct, but the speed of sled #2 after the first jump (V1) is not 1.99 m/s! (Sorry for not checking before.) You can redo your first calculation or...

    Even better is to use Tide's last suggestion, which recognizes that the total momentum of "cat + sled #2" after the first jump must equal the momentum of the jumping cat: (m + M)V1 = mv. Thus your equation:
    [tex](m + M) V_1 = M V_2 - mv[/tex]
    becomes [tex]mv = M V_2 - mv[/tex]
     
  9. Oct 29, 2005 #8
    Ok. I got it. Thanks a bunch!
     
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