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F = MA 2011 #'s 19, 20 (Force, Kinematics)

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Questions 19 and 20 refer to the following information
    A particle of mass 2.00 kg moves under a force given by
    F~ = −(8.00 N/m)(i + j)
    where i and j are unit vectors in the x and y directions. The particle is placed at the origin with an initial velocity
    v = (3.00 m/s)i + (4.00 m/s)j

    19. After how much time will the particle first return to the origin?
    (A) 0.785 s
    (B) 1.26 s
    (C) 1.57 s
    (D) 2.00 s
    (E) 3.14 s
    The correct answer is C
    20. What is the maximum distance between the particle and the origin?
    (A) 2.00 m
    (B) 2.50 m
    (C) 3.50 m
    (D) 5.00 m
    (E) 7.00 m
    The correct answer is B


    2. Relevant equations
    F = ma


    3. The attempt at a solution
    I keep messing up somewhere.
    If f = -8i - 8j
    a = -4i - 4j
    Antidifferentiate:
    v = (-4t + 3)i + (-4t+4)j
    r = (-2t^2 + 3t)i +(-2t^2 + 4t)j
    I can't find an answer that is one of the above choices.
     
    Last edited: Jan 26, 2013
  2. jcsd
  3. Jan 26, 2013 #2

    Simon Bridge

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    Please show the rest of your calculation.
    It will help if you are consistent with your variable names - you have x and y directions, then you use x as the name for the displacement vector as well. Try using ##\vec{r}## instead so you can use x and y for the components.
     
  4. Jan 26, 2013 #3
    Ok, so we find when the particle returns to the origin.
    This occurs when the position vector is 0.
    (-2t^2+3t)^2 + (-2t^2+4t)^2 = 0
    4t^4 - 12t^3 + 9t^2 + 4t^4 - 16t^3 + 16t^2 = 0
    t^2(8t^2 - 28t + 25) = 0
    no real solutions beside t= 0, which doesn't make sense?
    I'm stuck here.
     
  5. Jan 27, 2013 #4

    fgb

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    Weird, I agree with your equations but they don't give an answer for Question 19 - unless I'm making the same mistake you are. It seems to me that it is a simple question, you want x = 0i + 0j, which gives
    -2t² + 3t = 0
    -2t² + 4t = 0
    and there is no time that satisfies both equations.

    If you think about the distance of the particle to the origin, ie, [itex]d=\sqrt{(-2t^2 + 3t)^2 + (-2t^2 + 4t)^2}[/itex], and you plug in the answer (1.57), you get a non zero distance, which verifies that the particle is not at the origin. Also, the distance goes to infinity as t gets bigger and bigger, so Question 20 does not even make sense.

    Maybe i'm being picky, but is it really 8 N/m? (emphazys in the unit) That would suggest there's something missing, like the force being F = -8x*i -8y*j or something like that, in order to produce the correct units for F. That would change the equations and so on.

    Or we are both making the same stupid mistake and will feel stupid later :P
     
  6. Jan 27, 2013 #5

    tms

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    The way I read the problem, the force is of magnitude of 8/m in the direction of [itex]-\hat i - \hat j[/itex].
     
  7. Jan 27, 2013 #6
    Ya woops, it's 8xi and 8yi, I omitted those on accident.
    How would I incorporate that?
     
  8. Jan 27, 2013 #7

    fgb

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    Oh, I had not paid attention to that.

    UPDATE: Oh, the formula for F had something missing, too.
     
  9. Jan 27, 2013 #8
  10. Jan 27, 2013 #9

    tms

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  11. Jan 27, 2013 #10
    Hmm then how would I approach it?
    Would I just continue to anti-differentiate assuming x,y are constants?
    In that case I would have:
    (-2xt^2+3t)^2 + (-2yt^2+4t)^2 = 0
    4x^2t^4 - 12xt^3 + 9t^2 +4y^2t^4-16yt^3 + 16t^2 = 0
    Divide by t^2, since the solution t = 0 is physically meaningless anyways
    (4x^2+4y^2)t^2 + (12x - 16y)t + 25 = 0
    Is there a simpler way to do this? This seems unnecessarily complex.
     
  12. Jan 27, 2013 #11

    tms

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    You have a position-dependent force, so you can't simply integrate with respect to time to find the velocity and position. What do you get if you integrate the force with respect to position? In other words, the derivative of what gives you the force?
     
  13. Jan 27, 2013 #12
    The integral of force wrt position gives Work:
    W = integral(-8xdx - 8ydy)
    W = -4x^2 - 4y^2 + C
    I don't see how this helps.

    Work = ΔK + ΔU, maybe, but here ΔU = 0
    so Work = ΔK
    taking x,y separately
    -4x^2 = 1/2mv^2 - 1/2m(3^2)
    Dividing by m (given as 2 kg), multiplying by two
    -4x^2 = v^2 - 9
    v_x^2 = 4x^2 - 9

    Similarly with y:
    v_y^2 = 4x^2 - 16

    How do I proceed?
     
  14. Jan 27, 2013 #13

    tms

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    Actually, I was thinking of this:
    [tex]F(x) = - \frac{dU(x)}{dx,}[/tex]
    looking at the 1-dimensional case.
     
  15. Jan 27, 2013 #14
    How could I make use of that?
     
  16. Jan 27, 2013 #15
    Integrating, we find:
    -U = -4x^2i - 4y^2j
    U = 4x^2i + 4y^2j
    How can we relate U to position at origin?
     
  17. Jan 27, 2013 #16

    fgb

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    I would take a slightly different approach. You have therefore that
    [itex]a_x = -4x \\
    a_y = -4y[/itex]
    where the x subscript is for the x component, and so on. Therefore
    [itex]\dfrac{d^2x}{dt^2}=-4x \\
    \dfrac{d^2y}{dt^2}=-4y
    [/itex]

    So for x(t), ie, the x position of the particle with respect to time, you need a function whose second derivative is -4 times the original function, and that vanishes when t = 0 (that's one of the initial conditions). A sine function fits perfectly. So, x(t) = Asin(2t) (you need the 2 inside to make the factor of 4 when you differentiate twice).

    You could use standard differential equations techniques to get this result, but I believe it is not hard to "guess" what the correct function is.

    Also, realize that dx/dt = Vx = 3 at t=0, which means that A = 3/2. You can procede in a similar way to realize that y(t) = 2sin(2t). In summary, we have

    [itex]x(t) = \frac{3}{2}\sin{(2t)}\\
    y(t) = 2\sin{(2t)}[/itex]

    It is easy to see, from these functons, that x = y = 0 again at [itex]t=\dfrac{\pi}{2}[/itex], which is 1.57s (C).

    Also, the distance from the origin is [itex]\sqrt{x^2 + y^2} = \sqrt{\dfrac{9}{4}\sin^2{2t} + 4\sin^2{2t}} = \sqrt{\dfrac{25}{4}\sin^2{2t}}=\dfrac{5}{2}\sin{2t},[/itex], which one can easily see has a maximum value of 2.50m :)
     
  18. Jan 27, 2013 #17
    Could you explain in more detail or give me a link to how you derived those expressions?
     
  19. Jan 27, 2013 #18

    fgb

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    The ones I called x(t) and y(t)? :)
     
  20. Jan 27, 2013 #19
    Yes - how did you guess/derive those, I don't really understand your explanation.
     
  21. Jan 27, 2013 #20

    fgb

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    So, you have the following differential equations
    [itex]\dfrac{d^2x}{dt^2}=-4x \\
    \dfrac{d^2y}{dt^2}=-4y
    [/itex]

    Notice that they are very similar, so I will describe my solution to x(t), but that can be extended to y(t). The formal solution requires a little background on differential equations. If you want to, you could look up "2nd order homogeneous differential equation with constant coefficients" on Google - that's the kind of equation you have for x(t) and y(t). I will try to explain how I ended up "guessing" the function. Look at

    [itex]\dfrac{d^2x}{dt^2}=-4x [/itex]

    that means that the second derivative of x(t) is -4 times x(t). Ignore the 4 for a while and think about that: you want for x(t) a function that, when you derive twice, you get the function again, but with a minus sign in front of it. If you take a sine function, you have
    [itex] \sin'{t} = \cos{t} [/itex]
    if you derive again
    [itex] \sin''{t} = \cos'{t} = -\sin{t} [/itex]

    So sin(t) is a suitable function for x(t). However, I want to also produce a 4 when I differentiate twice, you can check that it works for sin(2t). This will actually also work if you have a number in front of the sine, so x(t) = Asin(2t) is actually a more general solution.

    And how am I supposed to find out how much is A worth? Well, you know that the initial velocity in the x direction, that is, dx/dt at t=0, is 3.

    x'(t) = 2*A*cos(2t), at t=0 you get
    x'(t) = 2*A*cos(0) = 3
    therefore A = 3/2.

    So, x(t) = (3/2)sin(2t).


    You may wonder how the hell did I know that a sine would fit. Well, that comes from experience. If you do not remember that a sine and a cosine have that property that, if you differentiate them twice, you get them back with a minus sign in front of it, all you can do is solve the differential equation by a formal methor - which is somewhat like following a cake recipe you have to memorize.

    As a tip, anytime you have x''(t) + k²x(t) = 0 as an equation, the general solution is x(t) = Acos(kt) + Bsin(kt).

    It is somewhat hard to explain the guessing process, is that a little more clear now? :/
     
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