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?
newtonian-mechanics energy momentum conservation-laws collision
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show 1 more comment
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?
newtonian-mechanics energy momentum conservation-laws collision
New contributor
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
|
show 1 more comment
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?
newtonian-mechanics energy momentum conservation-laws collision
New contributor
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
newtonian-mechanics energy momentum conservation-laws collision
New contributor
New contributor
edited 14 mins ago
Qmechanic♦
100k121821134
100k121821134
New contributor
asked 15 hours ago
Pires William
693
693
New contributor
New contributor
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
|
show 1 more comment
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
|
show 1 more comment
3 Answers
3
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.
add a comment |
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
add a comment |
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.
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
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
edited 8 hours ago
answered 15 hours ago
ggcg
78613
78613
add a comment |
add a comment |
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
add a comment |
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
add a comment |
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
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
answered 14 hours ago
Bill N
9,36712040
9,36712040
add a comment |
add a comment |
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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.
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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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