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Why is it that object of more mass attains less velocity

  1. Mar 6, 2012 #1
    Why is that an object of more mass attains less velocity as compared to another object of less mass on application of equal force?
    I know f=ma governs this but, is it related to "weight" of the body or is it just "natural"?
    Also another question:
    Suppose there are 2 objects of same mass one on earth and another one on moon. If equal force is applied on both objects, which one(moon/earth) will have much acceleration?
    Thanks.
     
  2. jcsd
  3. Mar 6, 2012 #2

    PhysicoRaj

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    Of course this is natural. it is governed by F=ma. But u hav to see that friction also plays a role> when it comes to the moon, the gravity is less & hence friction also. hence there, the object gains more accelaration and covers more distance.
     
  4. Mar 6, 2012 #3
    If equal NET force is applied on the moon or earth or in space or inside the sun, the acceleration will be equal. F=ma mentions nothing about gravity. Friction and such might deduct from your net force, but that's speculation. There are plenty of cases where friction ADDS to the net force. What we should be interested in is the NET force, and F=ma does not care where or what from the F or m arises.

    As to "why" a more massive particle is harder to accelerate...that's not so easy to answer. I will regurgitate some basic Wikipedia reading and say that it's because of the Higgs boson (otherwise known as the "God" particle, which the Large Hadron Collider is currently trying to find), but of course this is very unsatisfactory at our level of understanding. Even if you do manage to understand what this means, it will inevitably lead to another "why" question....which can of course go on forever. "This is natural" is perhaps what we should settle on for now.
     
  5. Mar 6, 2012 #4

    sophiecentaur

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    The answer to this question depends entirely on the level you want to discuss it. It's one of those "why" questions that Feynman says have no complete answer and may or may not be answered to one's satisfaction. A 'satisfactory answer' will include enough already familiar concepts to make you happy with 'another step' in your understanding.

    You can talk in terms of classical Energy and Momentum, both of which are 'conserved' in a Classical world of Physics:
    If you think of this in terms of Momentum. Momentum (in Classical terms) is Mass times Velocity. The momentum change is equal to the Force applied times the time, for which it is applied. So, for a certain Force times Time (Impulse) you have a trade-off between Mass and velocity change - small mass/big velocity change or large mass/small velocity change. That's a perfectly reasonable argument - if you are prepared to accept it.
    In terms of Energy - if you apply a force over a certain distance (doing work) by pushing an object then the Kinetic Energy given to the object will be Force times distance. KE is mv2/2 and, again, increasing the M will decrease the v if you started off with the same amount of Energy. (Another argument which includes already familiar concepts).

    You may want to have your answer in terms of General Relativity - or, as mentioned earlier, in terms of a Particle.
    They are just other levels in the 'Why Heirarchy'. Enjoy your reading round - that's the way into it all.
     
  6. Mar 6, 2012 #5
    First of all thanks for replying. Theres one thing that id like to ask (it might be stupid. Forgive me if it is):
    If earth pulls that object with much force, shouldnt it be harder to produce same acceleration as to that on moon. Because it is already occupied with a force towars the centre of earth.
    I think it is something related to forces being perpendicular.
    But im not sure. So, anyone on this?
     
  7. Mar 6, 2012 #6

    sophiecentaur

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    OK - so we're in the Classical World. (Don't let's be distracted further 'till we sort this out)

    Are you talking about the difference in gravitational forces - or is there some other force involved here?
     
  8. Mar 6, 2012 #7
    Again it's all about net force. The earth is constanly pulling you down, but if you're on the ground then the net force on you is zero, since the ground is pushing you up just as hard. Accelerating sideways, then, should make no difference (minus any frictional losses). The perpendicular force down has no effect on the sideways force.

    If the net downward force is 100000 N then you will accelerate downwards by 100000 N=ma. If the net sideways force is 5 N then you will accelerate sideways by 5 N=ma. If you have both forces acting on you simultaneously, then you will accelerate down by 100000 N=ma AND sideways by 5 N=ma. When the forces are perpendicular, one doesn't care about the other.

    If you want to accelerate up, however, then clearly it will be harder on the earth since you have to overcome the force of your weight first, which is greater on earth than on the moon. If you're on top of a mountain, though, and want to accelerate down, then obviously it's easier on earth.
     
  9. Mar 7, 2012 #8
    Lsos thanks! You really did understand my question.

    Ok so consider
    Case 1:
    I throw a ball parallel to the ground while standing(from some height). Ball has 2 forces - gravity and one that i applied. Forces are perpendicular.
    Case 2:
    Same Ball is dropped just from the same height as in case 1. Ball has only one force - gravity.

    According to your answer, in both cases acceleration due to gravity is the same(obvious).
    But imagine it practically, in case 1 the ball covers more distance and time before touching the ground than the distance and time in case 2.
    What i think: Because dist and time both increase, the acceleration is same in both cases!
    Right?

    By the way, Thank you all guy for replying. You Rock!
     
  10. Mar 7, 2012 #9
    The first ball does cover more distance but in the same amount of time then the second ball. They fall on the ground the same time because there is the same down force on them. The horizontal force is not added to the vertical force.

    In other words the horizontal force does not magically add to the time of flight of the ball. Only vertical force (gravity) determines time of flight.

    Think of it this way. You throw the 1st ball with acceleration equal to gravity and let go of the second one. 1st ball falls 1m and moves 1m in a direction, second ball just falls 1m, repeat that till both fall to the ground. Which one fell first? The only difference is the first one moved in x,y and z and the second one only in z.
     
    Last edited: Mar 7, 2012
  11. Mar 8, 2012 #10
    Really? They fall at the same time?
     
  12. Mar 8, 2012 #11
    Hi,

    Yes they do fall at the same time, given that there is no force acting on the bodies. There is this famous experiment by Galileo, where he dropped a 10 pd wt and a 1 pd wt from the leaning tower of pisa, only to conclude that both reached the ground at the same time! here's an animated version of the experiment:

    www.youtube.com/watch?v=3YZ4xkMNB1s

    Two bodies will move at the same speed, and take the same time to reach the surface of the earth only if there was no force of gravity acting on the bodies. According to Newton's 2nd law, the expression can also be written as F=mg, where g is the accn due to gravity.
    Now if g=0, then F=0. and comparing with F=ma, a =0. Therefore, two bodies dropped from a height h, will fall down, in the same time, irrespective of their masses, according to the equation, h=vt, or t=h/v, where time taken is independent of mass!

    Any doubts, feel free to ask me,
    regards,
    math_way
     
  13. Mar 8, 2012 #12
    If I dropped a bullet without the casing, an identical bullet shot out of a pistol hits the ground at the exact same time. A similar simple example, if I horizontally shot an arrow from a bow, it would hit the ground at the same time a dropped arrow would.

    The fired bullet's displacement along x means nothing regarding gravity's pull.

    Think about this: if you were to somehow travel with the shot bullet along the ground, at the exact same speed as the bullet so that you were always next to it, as it fell, it would look EXACTLY like a bullet just being dropped next to you while still, ignoring air resistance.
     
  14. Mar 8, 2012 #13
    Yes really.

    Imagine a net x,y,z. and lets say you have 3 rocks. You re at a point (0,0,10)
    1st one you drop. Drops at (0,0,0) in some amount of time.
    2nd you throw parallel to the x,y grid but only in x direction. So it drops somewhere (x,0,0) in a different time (by your understanding).
    So with your logic if you throw a rock parallel to the x,y but at an angle so it lands (x,y,0) it would travel longer. Because it traveled in z, x and y direction.

    And that's just not the case. Each of the rocks travel exactly the same amount of time, since x,y direction does not affect the time of flight. They're all falling at same acceleration thus the same amount of time.
     
  15. Mar 8, 2012 #14
    You mean to say when they're just dropped?
    But what if one out of the two objects was thrown parallel to the ground...
    For example, in leaning tower of pisa case
    What if the second object was dropped by applying force perpendicular to gravity but from the same height?
    Even then the time will remain same? (should as per other replies.)
     
  16. Mar 8, 2012 #15
    Yes the time would remain the same. If the force is perpendicular to the gravity then the time is the same.
     
  17. Mar 8, 2012 #16

    sophiecentaur

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    This is just one example of how Vectors work in every day life. Force, Velocity, Acceleration and Displacement are well known Vector quantities. They have magnitude (size) and direction.

    A displacement can be broken down into three independent values - the xyz co ordinates used on a graph are examples. (We're all familiar with plotting points on xy graph paper). Move parallel to the y axis and the x co ordinate (x displacement) doesn't change. Your y displacement changes independently of the x displacement. (The secret is to choose your two (or three) axes to be at Right Angles to each other)

    With motion, it's the same. You can consider vertical and horizontal motion quite independently. g only acts along the y axis so it will not affect any motion along the x axis. This allows you to break down what may look like a difficult problem into two simpler problems.

    It works the same for forces. The (vertical) weight of a 20m narrow boat makes it impossible to lift against gravity but, with only a bit of effort, you can get it to glide through the water in a horizontal direction. Two vectors at right angles again.
     
  18. Mar 8, 2012 #17

    cmb

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    Mass may be equally defined as its resistance to acceleration when subject to a force - thus your question can be turned into a tautology.
     
  19. Mar 8, 2012 #18

    sophiecentaur

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    There are many more of these 'threesome' relationships in Science. That "tautology" comment applies in a lot of places and just shows that nothing, on its own, is 'fundamental'. It's relationships that count. We tend to choose the variable that are easiest to measure / specify and call them the basics but it's often only by chance. For instance, you could argue that Momentum is a far more 'fundamental' quantity than Mass (being one of those quantities that's conserved) but it wouldn't be a practical quantity to use as a base.
     
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