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Blocks, a spring, and a collision

  1. Feb 23, 2008 #1

    ~christina~

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    1. The problem statement, all variables and given/known data

    Block 1 of mass 0.200 kg is sliding to the right over a frictionless elevated surface at a speed of 8.00 m / s as shown in the figure below. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5 N / m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.140 s, and block 1 slides off the opposite end of the elevated surface landing a distance d from the base of that surface after falling height h = 4.90 m.

    (a) Write an expression that gives the displacement of block 2 as a function of time. This expression must include the values of the amplitude of vibration and the angular frequency.

    (b) Use differential calculus to obtain expressions for the velocity and acceleration of block 2 as functions of time. What are the displacement, velocity, and acceleration of block 2 at t = 0.520 s?

    (d) What is the value of d?

    [​IMG]

    2. Relevant equations
    hm...
    [tex]T= 2\pi / \omega [/tex]

    3. The attempt at a solution

    I have

    m1= 0.200kg
    v1= 8.00m/s

    T= 0.140s
    h= 4.90m
    d= ?

    I was thinking that it's the conservation of momentum...but not sure exactly...

    a) find a exprssion that gives the displacement of block 2 as a function of time.
    Well I can find [tex]\omega [/tex] from the equation
    [tex] T= 2 \pi / \omega [/tex]

    [tex] \omega= 2 \pi / T [/tex]

    [tex] \omega= 2 \pi / 0.140s [/tex]

    [tex] \omega = 44.880rad/s [/tex]

    then I can find the mass of the block m2 I think from this equation:

    [tex] \omega = \sqrt{ k/ m_2} [/tex]

    [tex] m_2= k/ \omega^2 [/tex]

    [tex] m_2= 1208.5N/m / (44.880 rad/s)^2 [/tex]

    [tex] m_2= 0.5999kg [/tex]

    for the displacement vs time equation would it be this=>
    [tex]x(t)= A cos(\omega t + \phi )[/tex]

    A= ?
    [tex] \omega= 44.880rad/s [/tex]
    [tex] \phi = 0? [/tex] => I think phi is = 0 since when the block collides with the other block it's at equillibrium...thus :
    x= 0
    v= vi of block 1
    [tex]\phi= \pm \pi/2[/tex] (here I THINK it's [tex]+ \pi /2[/tex] since it's moving to the right initially)
    A= ? not sure how to find it...do I just plug in 0 for the time distance equation and find it
    that way?

    [tex] x(t)= A cos( 44.880rad/s t + \pi /2) [/tex] => is this correct for the displacement equation?

    But how do I find A?? (I'm stumped as to how to find that)


    b) Use differential calculus to obtain expressions for the velocity and acceleration of block 2 as functions of time. What are the displacement, velocity, and acceleration of block 2 at t = 0.520 s?

    well still don't know A but...

    [tex]v(t)= dx/dt= A d/dt cos (44.880rad/s t + \pi/2) = -\omega A sin (44.880rad/s*t + \pi/2) [/tex]

    and

    [tex]a(t)= dv/dt = -\omega A d/dt sin(44.880rad/s t + \pi/2) = -\omega^2 cos(\omega*t + \pi/2)[/tex]

    at t= 0.520s

    I need A and I'm not sure how to find that....so I need help on that.
    d) value of d

    I think I'd find v that the block has after colliding and than use that and put that into the UAM equations to find d since I have y= 4.90 or height.

    would I use the period given then plug that into the equation for velocity as a function of time then go and use that velocity as the velocity that block 1 leaves with and thus the velocity that it has as it goes off the edge?
    ( I would use the distance to the edge but it's not given and the surface is frictionless so that would be the same velocity as it has after being propelled by the spring back to the edge) right?


    THANKS ALOT
     
    Last edited: Feb 23, 2008
  2. jcsd
  3. Feb 24, 2008 #2

    siddharth

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    Right.
    To find these, use the initial conditions. at t=0, what's the displacement of the block? How about it's velocity?

    Yeah, looks right.

    You don't need to do all that. Remember that since the collision is elastic, energy is conserved. The total energy of oscillation is 1/2kx^2, when the velocity of the oscillating block is zero. You only need to find what the maximum displacement is.

    Then, since energy is conserved, you can find the velocity of block 1 after collision.

    Yes. After that, it's a simple projectile problem.
     
  4. Feb 24, 2008 #3
    Be careful with your phase angles...Think about what your displacement is if phi and t are zero, and you are using a cosine function (as opposed to a sin function).
     
  5. Feb 24, 2008 #4

    ~christina~

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    Hm...so would it be [tex]- \pi/2 [/tex] ??

    thanks
     
  6. Feb 24, 2008 #5

    ~christina~

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    Um..at t= 0
    x= 0
    v= um 0? (not quite sure about that one)

    I still don't know..... I don't think I'd plug into the equation with t=0 since that would give me 0 for the displacement at 0 and that is Not the amplitude :cry:
    goshes this is getting me stuck otherwise I could do this problem I think... For the velocity it's at max velocity but would it be the same velocity that the other block hits it with? thus 8.0m/s ???


    [tex]x(t)= A cos(\omega t + \phi )[/tex]

    [tex] x(t)= A cos( 44.880rad/s t - \pi /2) [/tex] => is this correct for the displacement equation?

    [tex]v(t)= dx/dt= A d/dt cos (44.880rad/s t - \pi/2) = -\omega A sin (44.880rad/s*t + \pi/2) [/tex]

    and

    [tex]a(t)= dv/dt = -\omega A d/dt sin(44.880rad/s t - \pi/2) = -\omega^2 cos(\omega*t + \pi/2)[/tex]

    at t= 0.520s

    I changed the phi sign to [tex]-\pi /2 [/tex] because matthewpowers said it was incorrect and I suspect the sign is incorrect....Not sure if that's wrong though

    Max displacement is A isn't it???

    Not sure if I understand but would I go and equate that with

    KE + PE= KE + PE

    when the block after one collision I'm not sure just where the block is...is it back at equillibrium ?

    I'm thinking initially at x= 0
    KE= 1/2kA^2
    vmax= [tex]- \omega A [/tex]
    PE= 0


    1/2kA^2= 1/2 kx^2

    I have this equation in my book anhow that says

    [tex] v= \pm \sqrt{k/m(A^2 - x^2) = \pm \omega \sqrt{A^2-x^2} [/tex]
    (where v is afunction of position for a simple harmonic oscilator)

    but do I use this for finding v that I need? (I don't have x though)


    Oh good grief...sort of lost here

    Please help Thanks
     
    Last edited: Feb 24, 2008
  7. Feb 24, 2008 #6
    It is with this last statement to which I was initially referring...you talked here about using a phi of zero when t=0 to make x=0. This is what prompted me to make the post..

    However, you seemed to realize that phi=0 was not right for the displacement here.

    But you do still want to think about what it means to use +pi/2 versus -pi/2. If you use +pi/2, then at t=0 you willl have cos(pi/2), but as time goes up by a small mount, the argument of the cosine will change from pi/2 toward pi (if I add a small amount to pi/2 I get something between pi/2 and pi). Think about a cosine graph between pi/2 and pi and ask yourself if this is what the displacement actually does (this really depends on what you defined to be positive and negative...did you define to the right of x=0 to be positive, or to the left?).

    Go through the same thought process assuming phi is -pi/2. If time increases by a small amount, does the displacement do what it should. Thats how I decide on the sign of the phase...

    However, I was hoping you might realize something else that might make your life easier. You want the displacement of the mass to start at zero (at t=0), go to max positive (if you have defined positive to the right), then back through zero, to negative max, and back to zero. You want the velocity to be max positive (if you defined positive to be to the right), to to zero, go negative, etc. If you think about sin and cosine, there is an obvious choice for the function that you might use for the displacement, and not have to rack your brain for the phase angle (and since the velocity is the derivative of the displacement, that makes things easier too)...

     
  8. Feb 24, 2008 #7

    siddharth

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    [tex]x(t)= A cos(\omega t + \phi )[/tex]
    [/quote]

    If the displacement of the block 2 as a function of time is [tex]x(t)= A cos(\omega t + \phi )[/tex], then at t=0, the position is
    [tex]x(0) = A cos(\phi) [/tex]

    In this problem, block 2 is stationary and initially not extended, which means x(0) = 0.
    The velocity is

    [tex]v(t)= -A\omega sin(\omega t + \phi )[/tex]

    and at t=0, this is
    [tex] v(0) = -A \omega sin(\phi) [/tex]

    What is v(0)? It's NOT simply v1, because block 1 is still moving after the collision. You have to use the conservation of momentum and energy to calculate the final velocity.

    Note that you can also find what the velocity of block 1 is after collision this way (which will be negative).

    Once you find v(0), you can solve the 2 equations for A and phi to find consistent solutions.

    Yes, the max displacement is A, because the cos function takes values between -1 and 1. For a Simple harmonic motion, the max displacement is always A.

    First, you should have already calculated the velocity of block 1 by the conservation of momentum and energy. So you don't need to do this, as I'd initially suggested.

    What I actually meant was this. Initially, block 2 isn't moving, so its energy is 0, while block 1 has only kinetic energy. After the collision, but before block 1 falls of the height, it is moving with some velocity v and corresponding KE.

    But now, block 2 is still oscillating with some KE + PE. How would you calculate this? Note the fact that, the total energy of oscillation of block 2 is constant. So, when the velocity of block 2 is 0 at the maximum displacement, all the energy is the potential energy. But this is just the total energy at all times, because energy is conserved.From this, you can find the velocity of
    block 1 after collision.

    Just remember that you can directly find v1 from the conservation of momentum and energy after collision.
     
    Last edited: Feb 24, 2008
  9. Feb 24, 2008 #8
    Although, I do believe you have to find m2 by some other means (which the original poster did just fine I think) before you can use conservation of energy and momentum to solve for the velocity...otherwise you have 3 unknowns and 2 equations
     
  10. Feb 26, 2008 #9
    Before I go any further, I want to make sure that we are thinking of the situation in the same way. When I first read this problem I was confused because I thought that what happened was block one hit block 2, and both block one and block two together compressed the spring, and then block one was released when the spring decompressed back to it's original (equilibrium) position. However, after thinking it over more, and realizing that it was key that the collision was elastic, this could not be the case. If the two blocks stayed together and compressed the spring, then they would have to have hit and stuck together. When two objects hit and stick, this is not an elastic collision (and energy is not conserved, even though momentum is). To be elastic, they must hit and bounce off each other (hence, elastic, or springy, and energy is conserved as well as momentum).

    Realizing that, then it was clear what was happening. The two blocks hit and bounce off each other. Block one bounced back in the opposite direction, and block 2 goes to the right (just as if the spring weren't there). After the collision, block two begins to compress the spring, and block one eventually falls off the edge.

    OK, now on to your post.

    Since the blocks don't stick together, they probably have different V's (certainly different signs, since we know one goes to the left, and one goes to the right after the collision).

    Looks good to me.

    I suspect that most of this was based on the assumption that the blocks hit and stuck together while compressing the spring. Now we know that they hit and bouce off before the spring even enters the situation (the collision is assumed to happen instantaneously, so fast that there was no time for the spring to compress until after the collision was finished, and each block has it's new velocity due to the collision). So you can treat this problem in 2 parts: one part is the collision, and you can find the final velocities of both bolcks after the collision as if the spring was not there (using conservation of energy and momentum during the collision, and since you already know m2 from the angular frequency and spring constant of the spring, you have 2 equations and 2 unknowns, so you're good to go). And then you can do the rest, starting from the situation just after the collision finished (block 2 has it's initial velocity from the collision, and now begins to compress the spring).



    Carefull....What is cos(pi/2)? What is cos(pi)? What is sin(pi/2)? What is sin(pi)?
     
  11. Feb 26, 2008 #10

    ~christina~

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    um..well I deleted the post since I came up with something different.
    here it is.

    T= 0.140s
    v1i= 8.00m/s
    v2i= 0m/s
    v1f= ?
    v2f?

    M1= 0.200kg
    k= 1208.5N/m

    a)[tex] x(t)= Acos(\omega*t + \pi/2 [/tex]

    A=?
    omega=?
    m2= ?

    [tex]T= 2\pi / \omega = 2 \pi \sqrt{m/k} [/tex]

    [tex] \omega = 2\pi / T= 2\pi / 0.140 = 44.879rad/s [/tex]

    [tex]m_2= (T/(2 \pi))^2k = (0.140/ (2\pi))^2(1208.5N/m) = 0.5999kg [/tex]

    m1v1i + m2v2i = m1v1f + m2v2f

    (0.200kg)(8m/s) + (0.599kg)(0) = (0.200kg)v1f + (0.599kg)v2f

    v1i-v2i = -(v1f-v2f)
    v1i-v2i= -v1f + v2f

    8m/s= -v1f+ v2f

    0.200kg(8m/s)= -0.200kg*v1f + 0.200kg*v2f

    1.6kgm/s= -0.200kg*v1f+ 0.2*v2f

    9.0m/s= v2f


    8m/s= -v1f + 9.

    v1f= 1m/s

    Is this right?

    (I'm working on it now and ..well I'll post as soon as I try to figure out the rest of it)
     
  12. Feb 26, 2008 #11

    ~christina~

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    I thougth that was happening as well..then I reread the question. I get it but technically it should be elastic not inelastic in reality but the question says differently.

    I think I found them.
    Yes it was. Then I reread the question...




    well I have issues with this...
    (the equation is cos in my text)
    okay lets see.

    cos pi/2 = max displacement thus 0
    cos pi= -1
    sin (pi/2)= 0
    sin pi= -1

    but the problem is I'm confused when it shifts and which side would be negative vs positive in sign when I say for example the block's right is possitive as opposed to the left side. I'm not sure how that would change.

    what I think is:
    block moving right compressing spring (possitive direction):
    t= 0 = > if cos graph then it would be cos pi/2
    t= _ => reach A => it would be cos pi
     
  13. Feb 28, 2008 #12
    Everything looked good up until this very last equation. How did you get 9=v2f? Did you add the equation before it to the conservation of momentum equation earlier so that the v1f term went to zero? If so, you just did your addition wrong...You should have:

    1.6 = .2*v1f + .6*v2f -> from conservation of momentum earler
    1.6 = -.2*v1f + .2*v2f -> the equation right before the last one

    Just add these 2 equations and solve for v2f, and you've got it. Then, as you do below, plug this in to one of the above equations.

     
  14. Feb 28, 2008 #13
    Well, the question des say it's elastic, and thats why you can use both conservation of momentum and conservation of energy like you do to solve for the two final velocities, and then use v2f as the initial velocity for it compressing the spring.

    Right procedure, looks like you just messed up on the algebra somewhere...

    Hmmm...Here's the correct answers for these:

    cos(pi/2) = 0
    cos(pi) = -1
    sin(pi/2) = 1
    sin(pi) = 0

    You'll want to know the sin and cos graphs (and unit circle) like the back of your hand.

    And can you explain more what you mean by "max displacement thus 0"?

    Also, when you say that the equation in your text is cos, do you mean they gave the answer as a cos function?

    Well, what you know is that at t=0 the block hasn't yet moved from the equilibrium position, so it's position is 0. Since it's position is zero, the equation needs to give you zero when you plug zero in for t. Hence, there are two choices:

    cos(pi/2) = 0
    cos(-pi/2) = 0

    But, which one is right? Well, for that we need to know what happens next. If we add a small amount to pi/2 we get something between pi/2 and pi, and cos(pi) = -1, so the displacement is somewhere between cos(pi/2) =0 and cos(pi) = -1...the displacement will be between 0 and -1 after a small amount of time goes by.

    Is this what the displacement really does? Not if we define to the right as positive...The block has moved a small amount to the right after a small amount of time goes by, so it's position is a small positive value.

    Thus, if we use +pi/2 as the phase, the equation won't be right, since it says the block has a negative position after a small amount of time goes by.

    How about -pi/2? Well if a small amount of time goes by, we get something between -pi/2 and 0. Cos(-pi/2) = 0, cos(0) = 1, so the displacement will be between 0 and positive 1 after a small amount of time goes by. Is that what the block really does? Yes, so we're good. If we use a cos function, and to the right is positive, then the phase should be -pi/2, because then the function actually describes the displacement correctly.

    Basically, with this method for finding the correct phase, make sure the sign of the function matches the sign of the displacement at different times in the motion.

    Hopefully you can now answer this one for yourself. This would be incorrect (unless the block has gone all the way back through the equilibrium position to reach A on the left side...). Why?

    Now, here's the other thing...if we were to choose to use Asin(wt +phi) as our equation to describe the motion, what would our phase be?

    Well, at t=0 we need the equation to equal zero, so the possible phases are:

    sin(0) = 0
    sin(pi) = 0

    Which one is right? Well, we need the equation to become a small positive value after a small amount of time goes by. So, if phi=0, we have something between 0 and pi/2, so displacement is between sin(0) = 0 and sin(pi/2) = 1. Sounds right, the equation goes positive after a small time goes by.

    Lets look at if phi = pi. After a small time, we are between pi and 3pi/2, so we are between sin(pi) = 0 and sin (3pi/2) = -1. Nope, this makes the block go negative after a small time.

    So if we were using sin, our equation would just be:

    x = Asin(wt)

    I'd say thats simpler than Acos(wt - pi/2)...

    Figuring this out all hinges on knowing sin and cos like the back of your hand, so you can visualize their graphs and understand what they mean.

    We need a sinusoidal function that starts at zero, goes to max positive, then back through zero, and so on. We know this because we understand the physical situation. If you know the graphs of sin and cos, you know that this is exactly what what the sin function does with no phase adjustment (or a phase of zero...), so it's the obvious choice.

    Hope this helps. :)
     
  15. Mar 5, 2008 #14

    ~christina~

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    T= 0.140s
    v1i= 8.00m/s
    v2i= 0m/s
    v1f= ?
    v2f?

    M1= 0.200kg
    k= 1208.5N/m

    a)[tex] x(t)= Acos(\omega*t + \pi/2) [/tex]

    A=?
    omega=?
    m2= ?

    [tex]T= 2\pi / \omega = 2 \pi \sqrt{m/k} [/tex]

    [tex] \omega = 2\pi / T= 2\pi / 0.140 = 44.879rad/s [/tex]

    [tex]m_2= (T/(2 \pi))^2k = (0.140/ (2\pi))^2(1208.5N/m) = 0.5999kg [/tex]

    m1v1i + m2v2i = m1v1f + m2v2f

    (0.200kg)(8m/s) + (0.599kg)(0) = (0.200kg)v1f + (0.599kg)v2f

    v1i-v2i = -(v1f-v2f)
    v1i-v2i= -v1f + v2f

    8m/s= -v1f+ v2f

    0.200kg(8m/s)= -0.200kg*v1f + 0.200kg*v2f

    1.6kgm/s= -0.200kg*v1f+ 0.2*v2f

    Oh..yep I actually did this again and got something else.

    1.6kg m/s= -0.200kg v1f + 0.200kg v2f
    + 1.6kg m/s = 0.200kg v1f + 0.599kg v2f
    3.2kgm/s= 0.799kg v2f

    v2f= 4.00m/s

    8.00m/s= -v1f + 4.00m/s

    v1f= -4m/s => not sure if this is correct since it's just 4.00m/s as well just in opposite direction but the equation is right so I guess it's fine.

    Thank you matthewpowers
     
    Last edited: Mar 5, 2008
  16. Mar 5, 2008 #15

    ~christina~

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    aw... I was actually looking at a cos and sine graph online and I thought the cos was the sin line.

    I made a error..Thinking about it again I would say that at max equillibrium it would be 0 displacement.

    No, I meant that the book shows these equations in cosine rather than sine

    I get it.
    yep, I see that (looking at the sin/cos graph)

    It goes toward possititve max displacement, yes I get this too.

    alright :smile:

    I said this before: t= _ => reach A => it would be cos pi

    I see that the block would have to go through almost a full cycle (is a full cycle from where it starts to going back to that position again? [I'm basing my thoughts on this]) thus to reach A it would be cos 0 there.

    It would be sin 0.
    It is simpler. I see what your saying.

    I guess I have to work on that. I actually memorized this in precal class and It's been awhile since that class.

    so it would be sin is the best choice.

    sure does. Thank you so much matthewpowers (you explain it better than my book or anywhere else I tried to find this info)
     
  17. Jan 25, 2009 #16
    The methods used above were quite unclear. What is the best method to find (d)?
     
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