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Two different masses in an elastic collision

  1. Aug 9, 2014 #1
    Before you go through all the trouble of reading all this back story the real information is towards the bottom starting at KEM1=KEM2=KEt. Hello, i was currently playing with a two separate masses ( a yoga ball, and a small bouncy ball, yes i know juvenile), while i bounced the two objects with the lighter mass on the top of the larger mass, the object with smaller mass accelerated from the ground at a rate that was perceptibly higher than the object with the higher mass. I started to work out an equation that was based off of the algebraic formula for Kinetic Energy (KE=1/2*m*V^2). I assumed that the collision is 100% elastic. I also assumed that the total Kinetic Energy created by the acceleration of the two masses by gravity was transferred to both masses equally to create the following formula:
    1/2*(M1+M2)*Vt^2=(1/2*M1*VM1^2)+(1/2*M2*VM2^2)
    Where,
    M1= Mass one.
    M2= Mass two.
    Vt= Velocity of both masses at the instant before impact with surface of earth.
    VM1= Velocity of M1 at the instant it breaks contact with M2.
    VM2= Velocity of M2 at the instant it breaks contact with M1.
    This formula is correct but it complicates the process by needing the velocity of the more massive object in order to find the velocity of the less massive object. After discovering this i decided to go about this with a new thought process. I assumed, as i did above, that the Kinetic Energy of both masses at the instant before impact was equal to the Kinetic energy transferred to the larger mass AND the lesser mass. Therefore:
    KEM1=KEM2=KEt
    Where:
    KEM1= The kinetic energy transferred to mass one.
    KEM2= The kinetic energy transferred to mass two.
    KEt= The total kinetic energy created by the displacement of the two masses in earths gravitational field.
    From this i determined that:
    KEt=1/2*(M1+M2)*Vt^2
    Because of the above statement i will set this equal to the kinetic energy of the lesser mass:
    1/2*(M1+M2)*Vt^2=1/2*M1*VM1^2
    In order to find Vt i used a simplification of:
    Vf^2=Vi^2+2*acceleration*change in distance
    Which is:
    Vf=square root(2*acceleration*change in distance)

    Vf=Vt
    Therefore:
    1/2*(M1+M2)*(2*acceleration*change in distance)=1/2*M1*VM1^2
    I then solved for VM1:
    VM1=square root(((M1+M2)*(19.6*x))/M1)
    Where:
    2*acceleration=19.6 meters/second^2
    x= change in the distance in the direction of the acceleration
    Is this correct? I have never taken calculus so the math that is used may be beyond my current comprehension.
     
  2. jcsd
  3. Aug 9, 2014 #2

    mfb

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    Transferred where?
    In the initial falling process? That is not true, the larger mass will gain more kinetic energy.
    They also won't have the same kinetic energy after the collision process.
    That would mean a very small ball would shoot through your ceiling like a bullet - something that just does not happen.

    The total kinetic energy is not very useful in this problem. A better approach is to split the process in two collisions: let the bottom ball collide with the ground, and then let the two balls collide with each other. That gives a good approximation for the total process.
     
  4. Aug 9, 2014 #3
    Thank you for responding. To answer your question about where the kinetic energy was transferred, for KEM2, which i assumed to be the larger mass which actually comes in contact with the ground, the ground supplied a force opposite the masses velocity, i also assumed that this force can be calculated using the kinetic energy formula. For KEM1, i assumed that the total kinetic energy created by both masses accelerating toward earth had been transferred to M2 due to my assumption of a 100% elastic collision. Because of this i was picturing both masses to act as a single object until the more massive object collided with the ground where there combined kinetic energy was applied to both objects causing a larger acceleration of the less massive object. I understand that due to its larger mass, M2 will have a higher kinetic energy than M1, but i was viewing the two objects masses as one until after the collision.

    About your suggestion that the total kinetic energy is not useful. Would this be how you would go about solving the problem:
    The kinetic energy of the larger mass object before collision is:
    KEM2=1/2*M2*(19.6*x)
    The kinetic energy of the lesser mass object before collision is:
    KEM1=1/2*M1*(19.6*(x-(radius of M2+radius of M1))
    At the instant before collision between M2 and M1, M2 is traveling upwards having collided with the ground at that instant so no kinetic energy is lost to deceleration due to gravity, while M1 is traveling downwards with KEM1.
    Kinetic energy of the two objects is:
    KE=KEM2-KEM1
    The reasoning behind this is that M1 is traveling in the opposite direction to M2.
    To solve for velocity of M1:
    V=square root(2*M1*KE)
    Using this process where,
    M1=1 kilograms
    M2=2 kilograms
    x=5 meters
    V=9.9 meters/second
    At this initial velocity the smaller mass would travel 5.0 meters.
    If i used my original equation:
    VM1=square root(((M1+M2)*(19.6*x))/M1)
    using the same masses and distance i get,
    VM1=17.1 meters/second

    Are my equations based off your suggestions correct? Or where you suggesting i completely abandon kinetic energy and use simple kinematic formulas?
     
  5. Aug 10, 2014 #4

    mfb

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    That is not possible, in a simplified model you do not have access to forces at all - they depend on details of the collision process (in particular, the timescale) you cannot control.
    No. Why should it?
    That is not true in general.
    Where does the factor 2 compared to the gravitational acceleration come from? And please don't forget units.

    Why do you think it falls down a different distance?
    Right.
    That does not make sense. Kinetic energy does not have a direction. Momentum has, but even then the momentum itself cares about the sign and total momentum is the sum, not the difference.

    I don't understand what you do afterwards. Do you know formulas for elastic collisions of two objects?
     
  6. Aug 10, 2014 #5

    sophiecentaur

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    It strikes me that confusion arises if the problem is not solved by considering the Momentum rather than the KE.
    The heavy ball will bounce back with -v (Earth has 'infinite mass') and the small ball will bounce off it according to the Momentum situation.
    In the limit as M/m increases , the maximum velocity of the lighter ball will be twice its impact velocity so it should reach √2 of the height the two balls were released.
    Imagine the height reached by the top one of a stack of falling balls, with progressively smaller masses.
     
  7. Aug 10, 2014 #6

    mfb

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    I agree that the whole kinetic energy thing does not help.
    Twice the velocity gives 4 times the height.
     
  8. Aug 10, 2014 #7

    sophiecentaur

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    Oh boy. I can be so sloppy. So it's even better then.
     
  9. Aug 10, 2014 #8

    vanhees71

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    Let's consider central collisions and assume that particle 2 is at rest in the initial state. Then you just need the conservation of energy and momentum to calculate the velocities in the final state. From energy conservation you get (I multiply the whole equation by 2):
    [tex]m_1 v_1^2=m_1 v_1'^2+m_2 v_2'^2.[/tex]
    The momentum conservation reads
    [tex]m_1 v_1=m_1 v_1' + m_2 v_2'.[/tex]
    Now we can rewrite the first equation as
    [tex]m_1 (\vec{v}_1^2-\vec{v}_1'^2)=m_1 (v_1-v_1')(v_1+v_1')=m_2 v_2'^2.[/tex]
    Now using the momentum-conservation equation on the last equation gives
    [tex]m_2 v_2' (v_1+v_1')=m_2 v_2'^2 \; \Rightarrow \; (v_1+v_1')=v_2'. \qquad (*)[/tex]
    Plugging this into the energy-conservation equation again, we find
    [tex]m_1 v_1^2 = m_1 v_1'^2+m_2(v_1+v_1')^2.[/tex]
    This is a quadratic equation for [itex]v_1'[/itex] with the two solutions
    [tex]v_{1,1}'=v_1, \quad v_{1,2}'=\frac{m_1-m_2}{m_1+m_2} v_1.[/tex]
    From (*) we get
    [tex]v_{2,1}'=0, \quad v_{2,2}'=\frac{2 m_1}{m_1+m_2} v_1.[/tex]
    The first solution is obviously the trivial case that no collision has happened. The 2nd solution is thus the one we want, i.e., a collision happened, i.e., the velocities after a collision has taken place.

    If you make [itex]m_2=m_1[/itex] you have the case of the famous toy named "Newton's craddle": After the collision, the first particle comes to rest and the second particle runs with the original velocity of particle 1.

    For [itex]m_2 \gg m_1[/itex] you get [itex]v_1' \simeq -v_1[/itex] and [itex]v_2' \simeq 0[/itex], i.e., the light particle is simply reflected by the heavy one, which (nearly) doesn't move at all.

    For [itex]m_2 \ll m_1[/itex] you get [itex]v_1' \simeq v_1[/itex] and [itex]v_2'=2 v_1[/itex], i.e., the heavy particle's velocity is nearly unaffected by the collision, while the light one, originally at rest, runs with twice the speed.
     
  10. Aug 10, 2014 #9

    A.T.

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    That's still wrong. For M>>m the limit is 9 times the original height.
     
  11. Aug 10, 2014 #10

    mfb

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    Note that we have to change the coordinate system to get that here.

    @A.T.: Ah right, didn't think about that number.
     
  12. Aug 11, 2014 #11

    vanhees71

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    Ok, then I didn't understand your setup.
     
  13. Aug 11, 2014 #12

    sophiecentaur

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    Another instance where a diagram was called for in the OP but we have all assumed our own mental diagrams.
     
  14. Aug 11, 2014 #13

    A.T.

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    I assumed this arrangement:

    http://www.luiseduardo.com.br/mechanics/conservationlaws/conservationlawsproblems.htm [Broken]
    http://www.luiseduardo.com.br/mechanics/conservationlaws/conservationlawsproblems_arquivos/image021.jpg [Broken]

    If m2 >> m1, collisions perfectly elastic, no air resistance, then m2 bounces back to ~9h (plus lower ball size).
     
    Last edited by a moderator: May 6, 2017
  15. Aug 11, 2014 #14

    sophiecentaur

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    That has to be right. The relative (approach) velocity between the two balls is 2 but the big one is going up at 1, already, giving a velocity for the small ball of -3. I didn't spot that. Those damned reference frames again.
    (Useful picture.)
     
  16. Aug 11, 2014 #15

    vanhees71

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    Ah, that's a funny problem :-)).

    Consider the time [itex]t_1[/itex] right before the big ball hits the floor. Then the total kinetic energy is given by [itex]E_{\text{tot}}=(m_1+m_2) g h[/itex]. The common velocity of the two balls thus is given by [itex]\frac{m_1+m_2}{2} V^2=E_{\text{tot}}[/itex], i.e.,
    [tex]V=\sqrt{2 g h}.[/tex]
    Now the big ball bounces elastically off the floor, which we can consider as of being of infinite mass. This means right after the collision of the big ball with the floor is finished, it has a velocity [itex]v_1=V[/itex] (pointing upwards, counted positive) and the little ball still has a velocity [itex]v_2=-V[/itex].

    This we take as initial conditions for the elastic scattering between the big and the small ball. Momentum and energy conservation then leads to
    [tex](m_1-m_2) V=m_1 v_1'+m_2 v_2', \quad \frac{m_1+m_2}{2} V^2=m_1 v_1^2 + m_2 v_2^2.[/tex]
    The nontrivial solution for the velocities right after the collision between the big and the little ball is finished are thus
    [tex]v_1=\frac{m_1-3m_2}{m_1+m_2}V, \quad v_2=\frac{3m_2-m_1}{m_1+m_2} V.[/tex]
    The maximal heights then follow from energy conservation again. Putting [tex]V=\sqrt{2 g h}[/tex] into the final solution gives
    [tex]\frac{m_1}{2} v_1'^2=m_1 g h_1' \; \Rightarrow \; h_1'=\left (\frac{m_1-3m_2}{m_1+m_2} \right)^2 h \simeq h[/tex]
    and
    [tex]\frac{m_1}{2} v_2'^2=m_1 g (h_2'-d) \; \Rightarrow \; h_2'=d+\left (\frac{3m_1-m_2}{m_1+m_2} \right)^2 h \simeq d+9 h.[/tex]
    For the approximate formulae I used the limit [itex]m_1 \rightarrow \infty[/itex] because of the assumption [itex]m_1 \gg m_2[/itex].
     
  17. Aug 11, 2014 #16

    A.T.

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    Yes, it is also a nice practical demonstration.

    And in that link:
    http://www.luiseduardo.com.br/mechanics/conservationlaws/conservationlawsproblems.htm [Broken]
    there is also the generalized version with N balls stacked in this way (nor very practical any more).
     
    Last edited by a moderator: May 6, 2017
  18. Aug 11, 2014 #17

    vanhees71

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    Then you've just to iterate the calculation [itex]n[/itex] times. Have a lot of fun!
     
  19. Aug 11, 2014 #18

    A.T.

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    The point is of course to have a formula for the height as a function of ball count, so you don't have to iterate. It's given in the link.
     
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