Does the total kinetic energy change during an elastic collision?











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If two balls of same mass with speed $v$ and $-v$ undergo an elastic collision, the kinetic energy will be the same after the collision as before.



However, during the collision, does it also remain the same? Isn't there a moment where both balls have zero velocity and hence zero kinetic energy?










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  • 2




    It's called an _______ collision because you need to imagine the material of the balls as _______
    – Joshua Ronis
    13 hours ago












  • Note that for most elastic collisions, there will not be a moment where both the balls have zero velocity. For example, if a ball moving at $v$ hits another ball at rest, there is never an instant where both of them are at rest.
    – Michael Seifert
    12 hours ago






  • 1




    @MichaelSeifert but two balls of the same mass colliding with equal and opposite velocity...
    – leftaroundabout
    12 hours ago










  • Some things I wrote about the treatment of collisions in an earlier answer might be helpful in thinking about this kind of question.
    – dmckee
    10 hours ago










  • One could consider a ball bouncing off the ground to eliminate the push back about a collision not ever reaching zero velocity. It still retains the essence of the question of turning kinetic energy into potential energy and then back again.
    – CramerTV
    7 hours ago















up vote
10
down vote

favorite












If two balls of same mass with speed $v$ and $-v$ undergo an elastic collision, the kinetic energy will be the same after the collision as before.



However, during the collision, does it also remain the same? Isn't there a moment where both balls have zero velocity and hence zero kinetic energy?










share|cite|improve this question









New contributor




Pires William is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    It's called an _______ collision because you need to imagine the material of the balls as _______
    – Joshua Ronis
    13 hours ago












  • Note that for most elastic collisions, there will not be a moment where both the balls have zero velocity. For example, if a ball moving at $v$ hits another ball at rest, there is never an instant where both of them are at rest.
    – Michael Seifert
    12 hours ago






  • 1




    @MichaelSeifert but two balls of the same mass colliding with equal and opposite velocity...
    – leftaroundabout
    12 hours ago










  • Some things I wrote about the treatment of collisions in an earlier answer might be helpful in thinking about this kind of question.
    – dmckee
    10 hours ago










  • One could consider a ball bouncing off the ground to eliminate the push back about a collision not ever reaching zero velocity. It still retains the essence of the question of turning kinetic energy into potential energy and then back again.
    – CramerTV
    7 hours ago













up vote
10
down vote

favorite









up vote
10
down vote

favorite











If two balls of same mass with speed $v$ and $-v$ undergo an elastic collision, the kinetic energy will be the same after the collision as before.



However, during the collision, does it also remain the same? Isn't there a moment where both balls have zero velocity and hence zero kinetic energy?










share|cite|improve this question









New contributor




Pires William is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











If two balls of same mass with speed $v$ and $-v$ undergo an elastic collision, the kinetic energy will be the same after the collision as before.



However, during the collision, does it also remain the same? Isn't there a moment where both balls have zero velocity and hence zero kinetic energy?







newtonian-mechanics energy momentum conservation-laws collision






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share|cite|improve this question




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edited 14 mins ago









Qmechanic

100k121821134




100k121821134






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asked 15 hours ago









Pires William

693




693




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New contributor





Pires William is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 2




    It's called an _______ collision because you need to imagine the material of the balls as _______
    – Joshua Ronis
    13 hours ago












  • Note that for most elastic collisions, there will not be a moment where both the balls have zero velocity. For example, if a ball moving at $v$ hits another ball at rest, there is never an instant where both of them are at rest.
    – Michael Seifert
    12 hours ago






  • 1




    @MichaelSeifert but two balls of the same mass colliding with equal and opposite velocity...
    – leftaroundabout
    12 hours ago










  • Some things I wrote about the treatment of collisions in an earlier answer might be helpful in thinking about this kind of question.
    – dmckee
    10 hours ago










  • One could consider a ball bouncing off the ground to eliminate the push back about a collision not ever reaching zero velocity. It still retains the essence of the question of turning kinetic energy into potential energy and then back again.
    – CramerTV
    7 hours ago














  • 2




    It's called an _______ collision because you need to imagine the material of the balls as _______
    – Joshua Ronis
    13 hours ago












  • Note that for most elastic collisions, there will not be a moment where both the balls have zero velocity. For example, if a ball moving at $v$ hits another ball at rest, there is never an instant where both of them are at rest.
    – Michael Seifert
    12 hours ago






  • 1




    @MichaelSeifert but two balls of the same mass colliding with equal and opposite velocity...
    – leftaroundabout
    12 hours ago










  • Some things I wrote about the treatment of collisions in an earlier answer might be helpful in thinking about this kind of question.
    – dmckee
    10 hours ago










  • One could consider a ball bouncing off the ground to eliminate the push back about a collision not ever reaching zero velocity. It still retains the essence of the question of turning kinetic energy into potential energy and then back again.
    – CramerTV
    7 hours ago








2




2




It's called an _______ collision because you need to imagine the material of the balls as _______
– Joshua Ronis
13 hours ago






It's called an _______ collision because you need to imagine the material of the balls as _______
– Joshua Ronis
13 hours ago














Note that for most elastic collisions, there will not be a moment where both the balls have zero velocity. For example, if a ball moving at $v$ hits another ball at rest, there is never an instant where both of them are at rest.
– Michael Seifert
12 hours ago




Note that for most elastic collisions, there will not be a moment where both the balls have zero velocity. For example, if a ball moving at $v$ hits another ball at rest, there is never an instant where both of them are at rest.
– Michael Seifert
12 hours ago




1




1




@MichaelSeifert but two balls of the same mass colliding with equal and opposite velocity...
– leftaroundabout
12 hours ago




@MichaelSeifert but two balls of the same mass colliding with equal and opposite velocity...
– leftaroundabout
12 hours ago












Some things I wrote about the treatment of collisions in an earlier answer might be helpful in thinking about this kind of question.
– dmckee
10 hours ago




Some things I wrote about the treatment of collisions in an earlier answer might be helpful in thinking about this kind of question.
– dmckee
10 hours ago












One could consider a ball bouncing off the ground to eliminate the push back about a collision not ever reaching zero velocity. It still retains the essence of the question of turning kinetic energy into potential energy and then back again.
– CramerTV
7 hours ago




One could consider a ball bouncing off the ground to eliminate the push back about a collision not ever reaching zero velocity. It still retains the essence of the question of turning kinetic energy into potential energy and then back again.
– CramerTV
7 hours ago










3 Answers
3






active

oldest

votes

















up vote
16
down vote













Good point. The comparison of initial and final energies is done before and after contact. During contact there must be some work done to bring them to rest and turn around. But for an elastic collision these internal forces are conservative, like the elastic force. If you watch slow motion photography of a collision you will see the balls deform slightly then come back to their original shape. This is due to the elasticity of the materials in each ball. In real life there is no such material that is perfectly conservative (at least as far as I know) but it's a good approximation for many materials. So in short, while they are at rest for a moment the kinetic energy is stored as potential in the balls.






share|cite|improve this answer






























    up vote
    13
    down vote













    You are correct. While there is a universal principle that momentum is conserved for ALL interactions, and momentum of isolated systems will remain constant, there is no universal conservation of kinetic energy.



    In the perfectly elastic collision system, the interaction forces are modeled as conservative spring-like forces. The interactions of the objects result in kinetic energy being efficiently transformed into potential energy of "springy" surfaces, then 100% transformed back to kinetic energy.



    For partially-elastic collisions (real-world collisions), the transformation to and from elastic potential energy is not 100%. Some KE goes into sound waves, deformation/stress of material, and internal ("thermal") energy






    share|cite|improve this answer




























      up vote
      4
      down vote













      During this collision process, kinetic energy is converted to internal energy. More specifically, elastic potential energy! While it may surprise you, each ball can actually be modeled as compressible, like a spring, under the study of Hertzian Contact Mechanics. This is due to the compressibility and deformation of the balls during collision.



      In fact, length of compression between the 2 balls can be defined as
      $$d^3=frac{9F^2}{16E*^22/R},$$ where Poisson's ratio and the elastic moduli of the ball can affect $E*$.



      Of course however, we are assuming no friction or energy loss to the surroundings, a key basis for Hertzian Contact Mechanics.



      Consequently, this elastic potential energy will be converting back to kinetic energy.



      You can read up on 2 research articles in 1975 and 1981 by N. Maw, J. R. Barber and J. N. Fawcett titled "The Oblique Impact of Elastic Spheres" and "The Role of Elastic Tangential Compliance in Oblique Impact" respectively.






      share|cite|improve this answer























      • Do those articles have a derivation of the equation you cited? If not, can you please provide a reference.
        – Chester Miller
        8 hours ago










      • Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two.
        – QuIcKmAtHs
        4 hours ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

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      up vote
      16
      down vote













      Good point. The comparison of initial and final energies is done before and after contact. During contact there must be some work done to bring them to rest and turn around. But for an elastic collision these internal forces are conservative, like the elastic force. If you watch slow motion photography of a collision you will see the balls deform slightly then come back to their original shape. This is due to the elasticity of the materials in each ball. In real life there is no such material that is perfectly conservative (at least as far as I know) but it's a good approximation for many materials. So in short, while they are at rest for a moment the kinetic energy is stored as potential in the balls.






      share|cite|improve this answer



























        up vote
        16
        down vote













        Good point. The comparison of initial and final energies is done before and after contact. During contact there must be some work done to bring them to rest and turn around. But for an elastic collision these internal forces are conservative, like the elastic force. If you watch slow motion photography of a collision you will see the balls deform slightly then come back to their original shape. This is due to the elasticity of the materials in each ball. In real life there is no such material that is perfectly conservative (at least as far as I know) but it's a good approximation for many materials. So in short, while they are at rest for a moment the kinetic energy is stored as potential in the balls.






        share|cite|improve this answer

























          up vote
          16
          down vote










          up vote
          16
          down vote









          Good point. The comparison of initial and final energies is done before and after contact. During contact there must be some work done to bring them to rest and turn around. But for an elastic collision these internal forces are conservative, like the elastic force. If you watch slow motion photography of a collision you will see the balls deform slightly then come back to their original shape. This is due to the elasticity of the materials in each ball. In real life there is no such material that is perfectly conservative (at least as far as I know) but it's a good approximation for many materials. So in short, while they are at rest for a moment the kinetic energy is stored as potential in the balls.






          share|cite|improve this answer














          Good point. The comparison of initial and final energies is done before and after contact. During contact there must be some work done to bring them to rest and turn around. But for an elastic collision these internal forces are conservative, like the elastic force. If you watch slow motion photography of a collision you will see the balls deform slightly then come back to their original shape. This is due to the elasticity of the materials in each ball. In real life there is no such material that is perfectly conservative (at least as far as I know) but it's a good approximation for many materials. So in short, while they are at rest for a moment the kinetic energy is stored as potential in the balls.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 8 hours ago

























          answered 15 hours ago









          ggcg

          78613




          78613






















              up vote
              13
              down vote













              You are correct. While there is a universal principle that momentum is conserved for ALL interactions, and momentum of isolated systems will remain constant, there is no universal conservation of kinetic energy.



              In the perfectly elastic collision system, the interaction forces are modeled as conservative spring-like forces. The interactions of the objects result in kinetic energy being efficiently transformed into potential energy of "springy" surfaces, then 100% transformed back to kinetic energy.



              For partially-elastic collisions (real-world collisions), the transformation to and from elastic potential energy is not 100%. Some KE goes into sound waves, deformation/stress of material, and internal ("thermal") energy






              share|cite|improve this answer

























                up vote
                13
                down vote













                You are correct. While there is a universal principle that momentum is conserved for ALL interactions, and momentum of isolated systems will remain constant, there is no universal conservation of kinetic energy.



                In the perfectly elastic collision system, the interaction forces are modeled as conservative spring-like forces. The interactions of the objects result in kinetic energy being efficiently transformed into potential energy of "springy" surfaces, then 100% transformed back to kinetic energy.



                For partially-elastic collisions (real-world collisions), the transformation to and from elastic potential energy is not 100%. Some KE goes into sound waves, deformation/stress of material, and internal ("thermal") energy






                share|cite|improve this answer























                  up vote
                  13
                  down vote










                  up vote
                  13
                  down vote









                  You are correct. While there is a universal principle that momentum is conserved for ALL interactions, and momentum of isolated systems will remain constant, there is no universal conservation of kinetic energy.



                  In the perfectly elastic collision system, the interaction forces are modeled as conservative spring-like forces. The interactions of the objects result in kinetic energy being efficiently transformed into potential energy of "springy" surfaces, then 100% transformed back to kinetic energy.



                  For partially-elastic collisions (real-world collisions), the transformation to and from elastic potential energy is not 100%. Some KE goes into sound waves, deformation/stress of material, and internal ("thermal") energy






                  share|cite|improve this answer












                  You are correct. While there is a universal principle that momentum is conserved for ALL interactions, and momentum of isolated systems will remain constant, there is no universal conservation of kinetic energy.



                  In the perfectly elastic collision system, the interaction forces are modeled as conservative spring-like forces. The interactions of the objects result in kinetic energy being efficiently transformed into potential energy of "springy" surfaces, then 100% transformed back to kinetic energy.



                  For partially-elastic collisions (real-world collisions), the transformation to and from elastic potential energy is not 100%. Some KE goes into sound waves, deformation/stress of material, and internal ("thermal") energy







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 14 hours ago









                  Bill N

                  9,36712040




                  9,36712040






















                      up vote
                      4
                      down vote













                      During this collision process, kinetic energy is converted to internal energy. More specifically, elastic potential energy! While it may surprise you, each ball can actually be modeled as compressible, like a spring, under the study of Hertzian Contact Mechanics. This is due to the compressibility and deformation of the balls during collision.



                      In fact, length of compression between the 2 balls can be defined as
                      $$d^3=frac{9F^2}{16E*^22/R},$$ where Poisson's ratio and the elastic moduli of the ball can affect $E*$.



                      Of course however, we are assuming no friction or energy loss to the surroundings, a key basis for Hertzian Contact Mechanics.



                      Consequently, this elastic potential energy will be converting back to kinetic energy.



                      You can read up on 2 research articles in 1975 and 1981 by N. Maw, J. R. Barber and J. N. Fawcett titled "The Oblique Impact of Elastic Spheres" and "The Role of Elastic Tangential Compliance in Oblique Impact" respectively.






                      share|cite|improve this answer























                      • Do those articles have a derivation of the equation you cited? If not, can you please provide a reference.
                        – Chester Miller
                        8 hours ago










                      • Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two.
                        – QuIcKmAtHs
                        4 hours ago















                      up vote
                      4
                      down vote













                      During this collision process, kinetic energy is converted to internal energy. More specifically, elastic potential energy! While it may surprise you, each ball can actually be modeled as compressible, like a spring, under the study of Hertzian Contact Mechanics. This is due to the compressibility and deformation of the balls during collision.



                      In fact, length of compression between the 2 balls can be defined as
                      $$d^3=frac{9F^2}{16E*^22/R},$$ where Poisson's ratio and the elastic moduli of the ball can affect $E*$.



                      Of course however, we are assuming no friction or energy loss to the surroundings, a key basis for Hertzian Contact Mechanics.



                      Consequently, this elastic potential energy will be converting back to kinetic energy.



                      You can read up on 2 research articles in 1975 and 1981 by N. Maw, J. R. Barber and J. N. Fawcett titled "The Oblique Impact of Elastic Spheres" and "The Role of Elastic Tangential Compliance in Oblique Impact" respectively.






                      share|cite|improve this answer























                      • Do those articles have a derivation of the equation you cited? If not, can you please provide a reference.
                        – Chester Miller
                        8 hours ago










                      • Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two.
                        – QuIcKmAtHs
                        4 hours ago













                      up vote
                      4
                      down vote










                      up vote
                      4
                      down vote









                      During this collision process, kinetic energy is converted to internal energy. More specifically, elastic potential energy! While it may surprise you, each ball can actually be modeled as compressible, like a spring, under the study of Hertzian Contact Mechanics. This is due to the compressibility and deformation of the balls during collision.



                      In fact, length of compression between the 2 balls can be defined as
                      $$d^3=frac{9F^2}{16E*^22/R},$$ where Poisson's ratio and the elastic moduli of the ball can affect $E*$.



                      Of course however, we are assuming no friction or energy loss to the surroundings, a key basis for Hertzian Contact Mechanics.



                      Consequently, this elastic potential energy will be converting back to kinetic energy.



                      You can read up on 2 research articles in 1975 and 1981 by N. Maw, J. R. Barber and J. N. Fawcett titled "The Oblique Impact of Elastic Spheres" and "The Role of Elastic Tangential Compliance in Oblique Impact" respectively.






                      share|cite|improve this answer














                      During this collision process, kinetic energy is converted to internal energy. More specifically, elastic potential energy! While it may surprise you, each ball can actually be modeled as compressible, like a spring, under the study of Hertzian Contact Mechanics. This is due to the compressibility and deformation of the balls during collision.



                      In fact, length of compression between the 2 balls can be defined as
                      $$d^3=frac{9F^2}{16E*^22/R},$$ where Poisson's ratio and the elastic moduli of the ball can affect $E*$.



                      Of course however, we are assuming no friction or energy loss to the surroundings, a key basis for Hertzian Contact Mechanics.



                      Consequently, this elastic potential energy will be converting back to kinetic energy.



                      You can read up on 2 research articles in 1975 and 1981 by N. Maw, J. R. Barber and J. N. Fawcett titled "The Oblique Impact of Elastic Spheres" and "The Role of Elastic Tangential Compliance in Oblique Impact" respectively.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 4 hours ago

























                      answered 15 hours ago









                      QuIcKmAtHs

                      2,4154828




                      2,4154828












                      • Do those articles have a derivation of the equation you cited? If not, can you please provide a reference.
                        – Chester Miller
                        8 hours ago










                      • Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two.
                        – QuIcKmAtHs
                        4 hours ago


















                      • Do those articles have a derivation of the equation you cited? If not, can you please provide a reference.
                        – Chester Miller
                        8 hours ago










                      • Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two.
                        – QuIcKmAtHs
                        4 hours ago
















                      Do those articles have a derivation of the equation you cited? If not, can you please provide a reference.
                      – Chester Miller
                      8 hours ago




                      Do those articles have a derivation of the equation you cited? If not, can you please provide a reference.
                      – Chester Miller
                      8 hours ago












                      Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two.
                      – QuIcKmAtHs
                      4 hours ago




                      Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two.
                      – QuIcKmAtHs
                      4 hours ago










                      Pires William is a new contributor. Be nice, and check out our Code of Conduct.










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                      Pires William is a new contributor. Be nice, and check out our Code of Conduct.












                      Pires William is a new contributor. Be nice, and check out our Code of Conduct.
















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