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Momentum Help

  1. Oct 25, 2007 #1
    This is my first time to use this template by the way.....



    1. A hungry 12.5 kg predator fish is coasting from west to east at 70.0 cm/s} when it suddenly swallows a 1.15 kg} fish swimming from north to south at 3.60 m/s.

    Find the magnitude of the velocity of the large fish just after it snapped up this meal. Neglect any effects due to the drag of the water.

    Find the direction of the velocity of the large fish just after it snapped up this meal.




    2.conservation of momentum, p=m/v



    3. (12.5kg)(.7m/s)+(1.15)(3.6) = total momentum= 12.89, then i got the total mass by adding 12.5+1.15.... i use the equation p=m/v and solve for the veolicity.

    I get the wrong answer here and I am not sure how to find the direction..


    THANKS
     
  2. jcsd
  3. Oct 25, 2007 #2
    Because the two fish are travelling in different directions, you can't just add up the two momentums and get 12.89. Since one is travelling west to east (hint: along the 'x' axis) and the other north to south (in the direction of the 'y' axis), you have to calculate the two separately, then figure out the velocities after the 'meal' to get the direction and magnitude of the 'after' velocity.
     
  4. Oct 25, 2007 #3
    ok so i calculate the that the one moving on the x axis is 8.75, and the one moving on the y axis at 4.14, then do i square them, add them together, and then take the square root to get the "after" velocity?
     
  5. Oct 25, 2007 #4
    Not quite. (I'm a little shaky on this). You have to do the x and y separately, so you have 8.75 x and 4.14 y. So, on the other side of the equation, the x has to equal 8.75, but now the 'm' of the mv is 12.5 + 1.15, which is 13.65. So the 'v' in the x direction is

    [tex]\frac{8.75}{13.65}[/tex]

    Do the same for the 'y'.

    Then, you do the square root of the sum of the squares of the velocities to get 'the' velocity. Then you use trig to get the angle.

    Let me know if you run into problems.
     
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