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atoms are slowed down two trillion fold. Five atoms are colored red to facilitate following their motions.An
elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only if there is no conversion of kinetic energy into other forms. The collisions of atoms are elastic collisions (Rutherford backscattering is one example).
The
molecules — as distinct from
atoms — of a gas or liquid rarely experience perfectly elastic collisions because kinetic energy is exchanged between the molecules’ translational motion and their internal
Degrees of freedom (physics and chemistry) with each collision. At any one instant, half the collisions are, to a varying extent,
inelastic collisions (the pair possesses less kinetic energy in their translational motions after the collision than before), and half could be described as “super-elastic” (possessing
more kinetic energy after the collision than before). Averaged across the entire sample, molecular collisions can be regarded as essentially elastic as long as
Planck's law of black body radiation are not permitted to carry away energy from the system.
In the case of macroscopic bodies, elastic collisions are an ideal never fully realized, but approximated by the interactions of objects such as billiard balls.
Equations
One-dimensional Newtonian
Total kinetic energy is the same before and after the collision, hence:
\frac{m_1u_1^2}2+\frac{m_2u_2^2}2=\frac{m_1v_1^2}2+\frac{m_2v_2^2}2
Total
momentum remains constant throughout the collision:
\,\! m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1} + m_{2}v_{2}
To solve for \ v_{1}, we use the following steps. First, solve the momentum equation for \ v_{2} . Then, substitute the result into the energy equation, and use the quadratic formula to solve for \ v_{1}. The equation for \ v_{2} is symmetrical.
v_{1} = \frac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2-->{m_{1}+m_{2-->
and
v_{2} = \frac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1-->{m_{1}+m_{2-->
For example:
Ball 1: mass = 3 kg,
v = 4 m/s
Ball 2: mass = 5 kg,
v = −6 m/s
After collision:
Ball 1:
v = −8.5 m/s
Ball 2:
v = 1.5 m/s
Property:
\ v_{1}-v_{2} = u_{2}-u_{1}
- the relative velocity of one particle with respect to the other is reversed by the collision
- the average of the momenta before and after the collision is the same for both particles
As can be expected, the solution is invariant under adding a constant to all velocities, which is like using a frame of reference with constant translational velocity.
The velocity of the
center of mass does not change by the collision:
The center of mass at time \ t before the collision and at time \ t' after the collision is given by two equations:
\bar{x}(t) = \frac{m_{1} \cdot x_{1}(t)+m_{2} \cdot x_{2}(t)}{m_{1}+m_{2-->, and \bar{x}(t') = \frac{m_{1} \cdot x_{1}(t')+m_{2} \cdot x_{2}(t')}{m_{1}+m_{2-->
Hence, the velocities of the center of mass before and after the collision are:
\ v_{ \bar{x} } = \frac{m_{1}u_{1}+m_{2}u_{2-->{m_{1}+m_{2-->, and \ v_{ \bar{x} }' = \frac{m_{1}v_{1}+m_{2}v_{2-->{m_{1}+m_{2-->
The numerator of \ v_{ \bar{x} } is the total momentum before the collsion, and numerator of \ v_{ \bar{x} }' is the total momentum after the collsion. Since momentum is conserved, we have \ v_{ \bar{x} } = \ v_{ \bar{x} }' .
With respect to the center of mass both velocities are reversed by the collision: in the case of particles of different mass, a heavy particle moves slowly toward the center of mass, and bounces back with the same low speed, and a light particle moves fast toward the center of mass, and bounces back with the same high speed.
From the equations for \ v_{1} and \ v_{2} above we see that in the case of a large \ u_{1}, the value of \ v_{1} is small if the masses are approximately the same: hitting a much lighter particle does not change the velocity much, hitting a much heavier particle causes the fast particle to bounce back with high speed.
Therefore a neutron moderator (a medium which slows down fast neutrons, thereby turning them into
thermal neutrons capable of sustaining a
chain reaction) is a material full of atoms with light nuclei (with the additional property that they do not easily absorb neutrons): the lightest nuclei have about the same mass as a neutron.
One-dimensional relativistic
Classical Mechanics is only a good approximation. It will give accurate results when it deals with the object which is macroscopic and running with much lower speed than the speed of light. Beyond the classical limits, it will give a wrong result. Total momentum of the two colliding bodies is frame-dependent. In a particular frame of reference, if the total momentum equals zero, according to Classical Mechanics,
m_{1}u_{1}+m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2} = 0
m_{1}u_{1}^{2} + m_{2}u_{2}^{2} = m_{1}v_{1}^{2} + m_{2}v_{2}^{2}
\frac{(m_{2}u_{2})^{2-->{2m_1} + \frac{(m_{2}u_{2})^{2-->{2m_2} =
\frac{(m_{2}v_{2})^{2-->{2m_1} + \frac{(m_{2}v_{2})^{2-->{2m_2}
(m_{1} + m_{2})(m_{2}u_{2})^{2} = (m_{1} + m_{2})(m_{2}v_{2})^{2}
u_{2} = -v_{2}
\frac{(m_{1}u_{1})^{2-->{2m_1} + \frac{(m_{1}u_{1})^{2-->{2m_2} =
\frac{(m_{1}v_{1})^{2-->{2m_1} + \frac{(m_{1}v_{1})^{2-->{2m_2}
(m_{1} + m_{2})(m_{1}u_{1})^{2} = (m_{1} + m_{2})(m_{1}v_{1})^{2}
u_{1}=-v_{1}
According to Theory of Relativity, in a particular frame of reference, if the total momentum is equal to zero,
\frac{m_{1}\;u_{1-->{\sqrt{1-u_{1}^{2}/c^{2-->} +
\frac{m_{2}\;u_{2-->{\sqrt{1-u_{2}^{2}/c^{2-->} =\frac{m_{1}\;v_{1-->{\sqrt{1-v_{1}^{2}/c^{2-->} +\frac{m_{2}\;v_{2-->{\sqrt{1-v_{2}^{2}/c^{2-->} = 0\frac{m_{1}c^{2-->{\sqrt{1-u_1^2/c^2--> +\frac{m_{2}c^{2-->{\sqrt{1-u_2^2/c^2--> =\frac{m_{1}c^{2-->{\sqrt{1-v_1^2/c^2--> +\frac{m_{2}c^{2-->{\sqrt{1-v_2^2/c^2-->
Where m_1 represents the rest mass of the first colliding body, m_2 represents the rest mass of the second colliding body, u_1 represents the initial velocity of the first collidng body, u_2 represents the initial velocity of the second colliding body, v_1 represents the velocity after collision of the first colliding body, v_2 represents the velocity after collision of the second colliding body.
When u_1=-v_1, u_2=-v_2, both kinetic energy and momentum is conserved. Rest mass of the two different colliding body do not change after the collision. It is shown that classical calculation is correct in this frame of reference where the total momentum is equal to zero. However, classical calculation will differ greatly with relativistic calculation when the total momentum is not equal to zero. Here, we do not solve directly by using the relativistic theory because it is quite impossible since the power of the equation is too high. One of the postulates in Special Relativity states that the Laws of Physics should be invariant in all inertial frames of reference. That is, if total momentum is conserved in a particular inertial frame of reference, total momentum will also be conserved in any inertial frame of reference, although the amount of total momentum is frame-dependent. Therefore, by transforming from an inertial frame of reference to another, we will be able to get the desired results. In a particular frame of reference where the total momentum could be anything,
\frac{m_{1}\;u_{1-->{\sqrt{1-u_{1}^{2}/c^{2-->} +
\frac{m_{2}\;u_{2-->{\sqrt{1-u_{2}^{2}/c^{2-->} =\frac{m_{1}\;v_{1-->{\sqrt{1-v_{1}^{2}/c^{2-->} +\frac{m_{2}\;v_{2-->{\sqrt{1-v_{2}^{2}/c^{2-->}=p
\frac{m_{1}c^{2-->{\sqrt{1-u_1^2/c^2--> +
\frac{m_{2}c^{2-->{\sqrt{1-u_2^2/c^2--> =\frac{m_{1}c^{2-->{\sqrt{1-v_1^2/c^2--> +\frac{m_{2}c^{2-->{\sqrt{1-v_2^2/c^2-->=E
E^2 = m_0^2 c^4 + p^2 c^2
m_0 = \sqrt{E^2/c^4 - p^2/c^2}
E^2 = \frac{m_0^2 c^6}{c^2 - v^2}
E^2c^2-E^2v^2=m_0^2 c^6
v=±\sqrt{c^2- \frac{m_0^2c^6}{E^2-->
If total momentum is greater than zero, then v will be greater than zero. If total momentum smaller than zero, v will be smaller than zero.
u_{1} '= \frac{u_1 - v }{1- \frac{u_1 ' v}{c^2-->
u_{2} '= \frac{u_2 - v }{1- \frac{u_2 ' v}{c^2-->
v_{1} '=-u_{1} '
v_{2} '=-u_{2} '
v_{1} = \frac{v_1 ' + v }{1+ \frac{v_1 ' v}{c^2-->
v_{2} = \frac{v_2 ' + v }{1+ \frac{v_2 ' v}{c^2-->
When u_1
atoms are slowed down two trillion fold. Five atoms are colored red to facilitate following their motions.An
elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only if there is no conversion of kinetic energy into other forms. The collisions of atoms are elastic collisions (
Rutherford backscattering is one example).
The
molecules — as distinct from atoms — of a gas or liquid rarely experience perfectly elastic collisions because kinetic energy is exchanged between the molecules’ translational motion and their internal Degrees of freedom (physics and chemistry) with each collision. At any one instant, half the collisions are, to a varying extent,
inelastic collisions (the pair possesses less kinetic energy in their translational motions after the collision than before), and half could be described as “super-elastic” (possessing
more kinetic energy after the collision than before). Averaged across the entire sample, molecular collisions can be regarded as essentially elastic as long as Planck's law of black body radiation are not permitted to carry away energy from the system.
In the case of macroscopic bodies, elastic collisions are an ideal never fully realized, but approximated by the interactions of objects such as billiard balls.
Equations
One-dimensional Newtonian
Total kinetic energy is the same before and after the collision, hence:
\frac{m_1u_1^2}2+\frac{m_2u_2^2}2=\frac{m_1v_1^2}2+\frac{m_2v_2^2}2
Total momentum remains constant throughout the collision:
\,\! m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1} + m_{2}v_{2}
To solve for \ v_{1}, we use the following steps. First, solve the momentum equation for \ v_{2} . Then, substitute the result into the energy equation, and use the quadratic formula to solve for \ v_{1}. The equation for \ v_{2} is symmetrical.
v_{1} = \frac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2-->{m_{1}+m_{2-->
and
v_{2} = \frac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1-->{m_{1}+m_{2-->
For example:
Ball 1: mass = 3 kg,
v = 4 m/s
Ball 2: mass = 5 kg,
v = −6 m/s
After collision:
Ball 1:
v = −8.5 m/s
Ball 2:
v = 1.5 m/s
Property:
\ v_{1}-v_{2} = u_{2}-u_{1}
- the relative velocity of one particle with respect to the other is reversed by the collision
- the average of the momenta before and after the collision is the same for both particles
As can be expected, the solution is invariant under adding a constant to all velocities, which is like using a frame of reference with constant translational velocity.
The velocity of the
center of mass does not change by the collision:
The center of mass at time \ t before the collision and at time \ t' after the collision is given by two equations:
\bar{x}(t) = \frac{m_{1} \cdot x_{1}(t)+m_{2} \cdot x_{2}(t)}{m_{1}+m_{2-->, and \bar{x}(t') = \frac{m_{1} \cdot x_{1}(t')+m_{2} \cdot x_{2}(t')}{m_{1}+m_{2-->
Hence, the velocities of the center of mass before and after the collision are:
\ v_{ \bar{x} } = \frac{m_{1}u_{1}+m_{2}u_{2-->{m_{1}+m_{2-->, and \ v_{ \bar{x} }' = \frac{m_{1}v_{1}+m_{2}v_{2-->{m_{1}+m_{2-->
The numerator of \ v_{ \bar{x} } is the total momentum before the collsion, and numerator of \ v_{ \bar{x} }' is the total momentum after the collsion. Since momentum is conserved, we have \ v_{ \bar{x} } = \ v_{ \bar{x} }' .
With respect to the center of mass both velocities are reversed by the collision: in the case of particles of different mass, a heavy particle moves slowly toward the center of mass, and bounces back with the same low speed, and a light particle moves fast toward the center of mass, and bounces back with the same high speed.
From the equations for \ v_{1} and \ v_{2} above we see that in the case of a large \ u_{1}, the value of \ v_{1} is small if the masses are approximately the same: hitting a much lighter particle does not change the velocity much, hitting a much heavier particle causes the fast particle to bounce back with high speed.
Therefore a
neutron moderator (a medium which slows down fast neutrons, thereby turning them into
thermal neutrons capable of sustaining a chain reaction) is a material full of atoms with light nuclei (with the additional property that they do not easily absorb neutrons): the lightest nuclei have about the same mass as a
neutron.
One-dimensional relativistic
Classical Mechanics is only a good approximation. It will give accurate results when it deals with the object which is macroscopic and running with much lower speed than the speed of light. Beyond the classical limits, it will give a wrong result. Total momentum of the two colliding bodies is frame-dependent. In a particular frame of reference, if the total momentum equals zero, according to Classical Mechanics,
m_{1}u_{1}+m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2} = 0
m_{1}u_{1}^{2} + m_{2}u_{2}^{2} = m_{1}v_{1}^{2} + m_{2}v_{2}^{2}
\frac{(m_{2}u_{2})^{2-->{2m_1} + \frac{(m_{2}u_{2})^{2-->{2m_2} =
\frac{(m_{2}v_{2})^{2-->{2m_1} + \frac{(m_{2}v_{2})^{2-->{2m_2}
(m_{1} + m_{2})(m_{2}u_{2})^{2} = (m_{1} + m_{2})(m_{2}v_{2})^{2}
u_{2} = -v_{2}
\frac{(m_{1}u_{1})^{2-->{2m_1} + \frac{(m_{1}u_{1})^{2-->{2m_2} =
\frac{(m_{1}v_{1})^{2-->{2m_1} + \frac{(m_{1}v_{1})^{2-->{2m_2}
(m_{1} + m_{2})(m_{1}u_{1})^{2} = (m_{1} + m_{2})(m_{1}v_{1})^{2}
u_{1}=-v_{1}
According to Theory of Relativity, in a particular frame of reference, if the total momentum is equal to zero,
\frac{m_{1}\;u_{1-->{\sqrt{1-u_{1}^{2}/c^{2-->} +
\frac{m_{2}\;u_{2-->{\sqrt{1-u_{2}^{2}/c^{2-->} =\frac{m_{1}\;v_{1-->{\sqrt{1-v_{1}^{2}/c^{2-->} +\frac{m_{2}\;v_{2-->{\sqrt{1-v_{2}^{2}/c^{2-->} = 0\frac{m_{1}c^{2-->{\sqrt{1-u_1^2/c^2--> +\frac{m_{2}c^{2-->{\sqrt{1-u_2^2/c^2--> =\frac{m_{1}c^{2-->{\sqrt{1-v_1^2/c^2--> +\frac{m_{2}c^{2-->{\sqrt{1-v_2^2/c^2-->
Where m_1 represents the rest mass of the first colliding body, m_2 represents the rest mass of the second colliding body, u_1 represents the initial velocity of the first collidng body, u_2 represents the initial velocity of the second colliding body, v_1 represents the velocity after collision of the first colliding body, v_2 represents the velocity after collision of the second colliding body.
When u_1=-v_1, u_2=-v_2, both kinetic energy and momentum is conserved. Rest mass of the two different colliding body do not change after the collision. It is shown that classical calculation is correct in this frame of reference where the total momentum is equal to zero. However, classical calculation will differ greatly with relativistic calculation when the total momentum is not equal to zero. Here, we do not solve directly by using the relativistic theory because it is quite impossible since the power of the equation is too high. One of the postulates in Special Relativity states that the Laws of Physics should be invariant in all inertial frames of reference. That is, if total momentum is conserved in a particular inertial frame of reference, total momentum will also be conserved in any inertial frame of reference, although the amount of total momentum is frame-dependent. Therefore, by transforming from an inertial frame of reference to another, we will be able to get the desired results. In a particular frame of reference where the total momentum could be anything,
\frac{m_{1}\;u_{1-->{\sqrt{1-u_{1}^{2}/c^{2-->} +
\frac{m_{2}\;u_{2-->{\sqrt{1-u_{2}^{2}/c^{2-->} =\frac{m_{1}\;v_{1-->{\sqrt{1-v_{1}^{2}/c^{2-->} +\frac{m_{2}\;v_{2-->{\sqrt{1-v_{2}^{2}/c^{2-->}=p
\frac{m_{1}c^{2-->{\sqrt{1-u_1^2/c^2--> +
\frac{m_{2}c^{2-->{\sqrt{1-u_2^2/c^2--> =\frac{m_{1}c^{2-->{\sqrt{1-v_1^2/c^2--> +\frac{m_{2}c^{2-->{\sqrt{1-v_2^2/c^2-->=E
E^2 = m_0^2 c^4 + p^2 c^2
m_0 = \sqrt{E^2/c^4 - p^2/c^2}
E^2 = \frac{m_0^2 c^6}{c^2 - v^2}
E^2c^2-E^2v^2=m_0^2 c^6
v=±\sqrt{c^2- \frac{m_0^2c^6}{E^2-->
If total momentum is greater than zero, then v will be greater than zero. If total momentum smaller than zero, v will be smaller than zero.
u_{1} '= \frac{u_1 - v }{1- \frac{u_1 ' v}{c^2-->
u_{2} '= \frac{u_2 - v }{1- \frac{u_2 ' v}{c^2-->
v_{1} '=-u_{1} '
v_{2} '=-u_{2} '
v_{1} = \frac{v_1 ' + v }{1+ \frac{v_1 ' v}{c^2-->
v_{2} = \frac{v_2 ' + v }{1+ \frac{v_2 ' v}{c^2-->
When u_1
Elastic Collision : Leveraging the Power of Virtual Worlds : What We ...
Elastic Collision is a full-service agency that develops content and provides strategic advice for organizations establishing a virtual world presence.
Elastic Collision : Leveraging the Power of Virtual Worlds : Our ...
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Elastic collision - Wikipedia, the free encyclopedia
An elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision.
Elastic Collision in Two Dimensions
Exact and general solution ... Firstly a note in order to avoid any misunderstandings: the exact kinematics of a particle collision is rarely of interest in plasma physics as it is ...
Elastic Collision in Three Dimensions
Exact and general solution for 3D collision ... Firstly a note in order to avoid any misunderstandings: the exact kinematics of a particle collision is rarely of interest in plasma ...
Elastic and Inelastic Collisions
Elastic and Inelastic Collisions. A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in ...
Air Track Collisions
Elastic Collision (1-D) Calculator - computes the final velocities for ...
Elastic Collision (1-D) Calculator - computes the final velocities for an elastic collision of two masses in one dimension
Elastic and Inelastic Collision
Java Applet: Collision Processes ... This Java applet deals with the extreme cases of a collision process illustrated by two wagons:
elastic collision - Hutchinson encyclopedia article about elastic ...
In physics, a collision between two or more bodies in which the total kinetic energy of the bodies is conserved (remains constant); none is converted into any other form of energy.
