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The Clam and Its Forces

  1. Sep 16, 2008 #1
    SOLVED


    1. The problem statement, all variables and given/known data
    Sea clams move by pushing water in one direction so that the water pushes them in the opposite direction. In doing so a 0.5 kg clam can accelerate 0.6 kg of water from rest to a speed of 1.9 m/s in 0.43 seconds. Calculate the magnitude of the clam's acceleration


    2. Relevant equations
    F= ma
    Vf^2= V0^2 + 2ax
    Vf= V0 + at
    x= V0*t + .5a(t^2)

    3. The attempt at a solution

    I'm not sure why my answer isn't checking out. The initial velocity is 0. The final velocity is 1.9. The time is .43. These all fit into the third equation: 1.9= 0 + .43a. a then equals 4.42. This, however, is not working as my answer. What am I missing?

    Thanks in advance!

    ~Phoenix
     
    Last edited: Sep 17, 2008
  2. jcsd
  3. Sep 16, 2008 #2
    Apply conservation of momentum
    0.6*1.9=0.5*v
    v=2.28
    [tex]a=\frac{\Delta v}{\Delta t}[/tex]
    a=2.28/0.43
    a=5.3
    Does this help u?
     
  4. Sep 17, 2008 #3

    Ygggdrasil

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    Science Advisor
    2015 Award

    The clam accelerates the water from 0 to 1.9 m/s in 0.43s. You need to calculate the acceleration of the clam, not the water. (hint: remember Newton's third law)
     
  5. Sep 17, 2008 #4
    OH! Alrighty, gotcha. That makes sense to me. Thank you so much. I realize now that I was mixing variables that shouldn't be used to solve for this.

    Thank you so very much!
     
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