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Impulses of Squids

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Squids and octopuses propel themselves by expelling water. They do this by taking the water into a cavity and then suddenly contracting the cavity, forcing the water to shoot out of an opening. A 6.50 kg squid (including the water in its cavity) that is at rest suddenly sees a dangerous predator.

    If this squid has 1.75 kg of water in its cavity, at what speed must it expel the water to suddenly achieve a speed of 2.50 m/s to escape the predator? Neglect any drag effects of the surrounding water.

    2. Relevant equations

    m dv/dt=-Vex dm/dt
    v-vo- -vex loge (m1/m0)

    3. The attempt at a solution
    vex= V/ log (m1/M0)

    = -2.5 m/s / (log ((6.5-1.75)6.5)
    = 7.9704 m/s

    Now I figured out this problem based on the equations that i have and this is the only solution that i keep getting. Am I using the wrong equation or am I entering the numbers wrong? Thanks for the help!
  2. jcsd
  3. Nov 6, 2008 #2


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    Welcome to PF!

    Hi mdewdude! Welcome to PF! :smile:

    I suspect you're making this too complicated.

    When an exam question says "suddenly", it usually intends you to ignore calculus and assume, in this case, that the squid somehow expels all the water in one go.

    So try it with an ordinary "collision" equation. :smile:
  4. Nov 6, 2008 #3
    Mass of whale with water* desired velocity= Mass of whale escape * velocity of escape

    would this be the proper equation?
  5. Nov 6, 2008 #4
    Mmmmm... calamari.
  6. Nov 7, 2008 #5
    Sorry, I am horrible at this momentum stuff. Could somebody tell me if i'm on the track or could possibly steer me into it? Thatnk you
  7. Nov 7, 2008 #6
    Not to worry. I'm sorry myself, I am a software engineer with a mushy old brain, not one of these physics types, and I was only attracted into this thread by the discussion of squid.

    I feel bad for interrupting, so I'm going to take a crack at this.

    So from what tiny-tim told us there, we probably don't need to use any calculus, so I went off to find some algebra-based momentum equations. I think this one from here for inelastic collisions is what we need:

    [tex] m_1 \mathbf v_{1,i} + m_2 \mathbf v_{2,i} = \left( m_1 + m_2 \right) \mathbf v_f \,[/tex]

    I think this is going to be an inelastic collision except that events will play in reverse from what's described in the problem, just remember that. Wikipedia says "A common example of a perfectly inelastic collision is when two snowballs collide and then stick together afterwards."

    What we've got is basically as though the two snowball fighters, one of them took up a dry / empty squid in his hand, the other one took a blob of water in his hand, and they threw them at each other and they collided and stuck together. Do you see how that's the reverse? At the end of our model scenario, when they've collided and stuck together, that's the same as the start of the written problem when you've got the squid still full of water.

    I'm going to post this in case you come back to read it and work on the next steps in another post.
  8. Nov 7, 2008 #7
    Okay, let's start assigning variables.

    [tex] v_f \,[/tex] is literally "final velocity". But since this is backwards, this will refer to the starting velocity in the original problem. In that case the squid's at rest when the predator shows up and scares him, so
    [tex] v_f \, = 0 [/tex]​
    The masses are easy. Let's make [tex] m_1 [/tex] the water and [tex] m_2 [/tex] the squid. So based on the masses in the original problem we can say
    [tex] m_1 = 1.75 kg [/tex]​
    [tex] m_2 = 6.50 kg - 1.75 kg [/tex]​
    [tex] m_2 = 4.75 kg [/tex]​
    Conveniently, [tex] v_f \,[/tex] being zero gets rid of the entire left size of the equation, since anything multiplied by zero also equals zero.
    [tex] m_1 \mathbf v_{1,i} + m_2 \mathbf v_{2,i} = \left( m_1 + m_2 \right) \mathbf 0[/tex]​
    [tex] m_1 \mathbf v_{1,i} + m_2 \mathbf v_{2,i} = 0[/tex]​
    Now balance the equation...
    [tex] m_1 \mathbf v_{1,i} = - m_2 \mathbf v_{2,i} [/tex]​
    Now I'll plug the masses in.
    [tex] 1.75 kg \mathbf v_{1,i} = - 4.75 kg \mathbf v_{2,i} [/tex]​
    Now we've got an equation where we can either plug in the velocity of the water and solve for the velocity of the squid, or we can start off with the velocity of the squid and solve for the velocity of the water globule. The original equation gives us 2.50 m/s as the desired speed of the squid so we'll plug that in.
    [tex] 1.75 kg \mathbf v_{1,i} = - 4.75 kg \mathbf 2.50 m/s [/tex]​
    Now we want to isolate the remaining variable, so we'll divide both sides by 1.75 kg.
    [tex] v_{1,i} = - 2.714285 \mathbf 2.50 m/s [/tex]​
    And solve for [tex] v_{1,i} [/tex]
    [tex] v_{1,i} = - 6.7857142 m/s [/tex]​
  9. Nov 7, 2008 #8
    Like I said, it's been many years since I've done something like that, so if anyone wants to confirm or deny the approach...
  10. Nov 8, 2008 #9
    Thanks, for the help, sorry that i couldn't reply much sooner. you are right on the equation but having the numbers on which side of the equation was wrong. To find the kinetic energy from the escape, would it be just the simple KE=0.5 mv^2 the question would be, which mass would it be, the 6.5 or the 4.75 (the squids mass without water)
  11. Nov 8, 2008 #10


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    Warning! :bugeye: Warning, mdewdude! :bugeye:

    Energy is not conserved … this is an explosion, or reverse inelastic collision … as CaptainQuasar :smile: said:
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