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Cat and momentum

  1. Feb 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Two 21.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 3.41 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 4.04 m/s relative to the ice. What is the final speed of sled 2? (Assume the ice is frictionless.)


    2. Relevant equations



    3. The attempt at a solution
    (mass of cat + mass of sled)*(horizontal speed of cat) = (mass of sled)*(final speed of sled 2) - (mass of cat)*(speed of cat with respect to ground)

    -- how do i find the speed of cat with respect to ground
     
  2. jcsd
  3. Feb 23, 2008 #2
    Your formula doesn't make sense. The speed of the cat with respect to ground is just the speed of the cat with respect to the ice.
    You can use conservation of momentum for the cat and sled_2 The cat starts out with a momentum of (jumping speed of cat) * (mass of cat) and sled_2 starts out with a momentum of 0. After the cat jumps on the sled they will now have the same speed.
    so [(mass of cat)+(mass of sled_2)] * (speed sled_2 after first jump) should be equal to this.
    once you've found the speed of sled_2 after the first jump you can use conservation of momentum to see what happens if the cat jumps off
     
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