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can someone please check my ans

Discussion in 'MCAT Study Question Q&A' started by inaccensa, Aug 11, 2011.

  1. inaccensa

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    This is a qualitative q from kaplan, but if i had to solve for it, this is probably how i'd do it

    Two wooden balls of equal volume, but different density are held beneath the surface of water. Ball A --density =.5g/cm3 Ball B-.7g/cm3, once released, they accelerate upwards. identify their acceleration relation.

    Ball A , Fb-W=m*a , Fb=m*a+W, similarly Fb =m'*b +W' (,a'w',m'- ball b)
    Since the Fb is identical, m*a+W =m*a+W' , m(a+g) =m'(a'+g)
    m/m' =(a'+g)/(a+g)
    d/d' = a'/a (g doesnt cancel, but they add the same value, so ignored it for simplicity)
    a' = some fraction of a

    Is this right

    I do know that the mass of A should be less than mass of B(d=m/v ,equal v, higher density = larger mass) and F=ma dictates that the a of mass A will be higher, but just wanted to check my understanding.

    Is there a better way to solve a problem like this one, if it actually needs to be solved

    thanks
     
    #1 inaccensa, Aug 11, 2011
    Last edited: Aug 11, 2011
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  3. MD Odyssey

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    Here is how I would approach the problem. Draw a free-body diagram for both balls. In general, there are two forces on the ball, the buoyant force and the gravitational force. Newton's 2nd law gives you the net acceleration for each ball. Regardless of which has the larger mass, you get the following set of equations:

    [​IMG]

    [​IMG]

    Notice that the density of the balls is irrelevant here. All it tells you is which of the masses is larger and the buoyant force is independent upon the density (This is what the problem is really trying to test).

    Now, if you want to know the relationship between the two accelerations, you just divide the two, solve for the ratio and you get

    [​IMG]

    Notice that neither the density or the acceleration due to gravity are present in the final equation, although you would need to use the density to calculate the masses if you wanted to actually solve for the relationship.

    Of course, qualitatively, it should be pretty obvious that the acceleration on the less massive ball is going to be larger than the acceleration on the more massive ball, since the gravitational force is less, but the buoyant force is the same. One way to show this is to take the equations for the accelerations of the ball, solve for the buoyant force, and set them equal to each other. With a bit of algebra, this gives you the following

    [​IMG]

    From this, we can deduce which of the accelerations is the larger, based upon the ratio of the masses or the densities (since the volumes are the same).
     
    #2 MD Odyssey, Aug 11, 2011
    Last edited: Aug 11, 2011
  4. inaccensa

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    your work is much neater than mine:) Well, my only q is that Fb is equal in both instances, right?
     
  5. MD Odyssey

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    Correct, because the buoyant force is only a function of the volume of the mass. This is why denser fluids sink in water and less dense fluids float on top (intermolecular forces notwithstanding).
     
  6. indianjatt

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    Quick question. We can assume that the bouyant forces are equal because the objects both have the same volume correct?
     
  7. MD Odyssey

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    It's not an assumption. It's true, since the buoyant force is equal to the mass of water that is displaced by the object, not the objects mass. And, if the volumes are the same, then the mass of water displaced by each is the same.

    When an object is submerged, the buoyant force is always the mass of water displaced. This means that a plastic Whiffle ball and a ball made of lead that's the same size have precisely the same buoyant force when submerged in water. The reason the lead ball sinks is because the gravitational force on the lead ball is much larger and the net acceleration is downwards.

    This sort of thing becomes a lot simpler if you draw a picture and label the two forces. Sum the forces and solve for the net acceleration using Newton's law and all should become clear.
     

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