Elastic_collision

As long as black-body radiation (not shown) doesn’t escape a system, atoms in thermal agitation undergo essentially elastic collisions. On average, two atoms rebound from each other with the same kinetic energy as before a collision. Here, room-temperature helium 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 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 black-body photons 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.

Contents 1 Equations 1.1 One-dimensional Newtonian 1.2 One-dimensional relativistic 1.3 Two- and three-dimensional 2 See also 3 External links

Equations

One-dimensional Newtonian

Consider two particles, denoted by subscripts 1 and 2. Let $m$ be the mass, $u$ be the velocity before collision and $v$ be the velocity after collision.

Total kinetic energy is the same before and after the collision, hence:

$\backslash frac\{m\_1u\_1^2\}2+\backslash frac\{m\_2u\_2^2\}2=\backslash frac\{m\_1v\_1^2\}2+\backslash frac\{m\_2v\_2^2\}2$

Total momentum remains constant throughout the collision:

$\backslash ,\backslash !\; m\_\{1\}v\_\{1\}+m\_\{2\}v\_\{2\}=m\_\{1\}u\_\{1\}\; +\; m\_\{2\}u\_\{2\}$

These equations may be solved directly to find $\backslash \; v\_\{1\}$ and $\backslash \; v\_\{2\}$. However, the algebra can get messy. A cleaner solution is to first change the frame of reference such that either $\backslash \; v\_\{1\}$ or $\backslash \; v\_\{2\}$ appears to be 0. The final velocities in the new frame of reference can then be determined followed by a conversion back to the original frame of reference to reach the same final result. Once either $\backslash \; v\_\{1\}$ or $\backslash \; v\_\{2\}$ is determined the other may be found by symmetry.

Note that these simultaneous equations have both a trivial and a non trivial solution. The reason for this is that their solution does not in fact describe velocities only after an elastic collision, but in fact the velocities with which two particles in an isolated system may be travelling after an arbitrary time interval, with the condition that the total kinetic energy at the end of the time interval is equal to that at the start. This interpretation highlights that the equations may also describe the case in which no interaction takes place.

$v\_\{1\}\; =\; \backslash frac\{u\_\{1\}(m\_\{1\}-m\_\{2\})+2m\_\{2\}u\_\{2\}\}\{m\_\{1\}+m\_\{2\}\}$ , $v\_\{2\}\; =\; \backslash frac\{u\_\{2\}(m\_\{2\}-m\_\{1\})+2m\_\{1\}u\_\{1\}\}\{m\_\{1\}+m\_\{2\}\}$

OR

$\backslash \; v\_\{1\}\; =\; u\_\{2\}$ , $\backslash \; v\_\{2\}\; =\; u\_\{1\}$, when we have same mass

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:

$\backslash \; v\_\{1\}-v\_\{2\}\; =\; u\_\{2\}-u\_\{1\}$

Derivation: Using the kinetic energy we can write

$\backslash \; m\_1(v\_1^2-u\_1^2)=m\_2(u\_2^2-v\_2^2)$

$\backslash Rightarrow\; m\_1(v\_1-u\_1)(v\_1+u\_1)=m\_2(u\_2-v\_2)(u\_2+v\_2)$

Rearrange momentum equation:

$\backslash \; m\_1(v\_1-u\_1)=m\_2(u\_2-v\_2)$

Dividing kinetic energy equation by the momentum equation we get:

$\backslash \; v\_1+u\_1=u\_2+v\_2$

$\backslash Rightarrow\; v\_1-v\_2\; =\; u\_2-u\_1$

• 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

Elastic collision of equal masses

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.

Elastic collision of masses in a system with a moving frame of reference

The velocity of the center of mass does not change by the collision:

The center of mass at time $\backslash \; t$ before the collision and at time $\backslash \; t\backslash \text{'}$ after the collision is given by two equations:

$\backslash bar\{x\}(t)\; =\; \backslash frac\{m\_\{1\}\; \backslash cdot\; x\_\{1\}(t)+m\_\{2\}\; \backslash cdot\; x\_\{2\}(t)\}\{m\_\{1\}+m\_\{2\}\}$, and $\backslash bar\{x\}(t\backslash \text{'})\; =\; \backslash frac\{m\_\{1\}\; \backslash cdot\; x\_\{1\}(t\backslash \text{'})+m\_\{2\}\; \backslash cdot\; x\_\{2\}(t\backslash \text{'})\}\{m\_\{1\}+m\_\{2\}\}$

Hence, the velocities of the center of mass before and after the collision are:

$\backslash \; v\_\{\; \backslash bar\{x\}\; \}\; =\; \backslash frac\{m\_\{1\}u\_\{1\}+m\_\{2\}u\_\{2\}\}\{m\_\{1\}+m\_\{2\}\}$, and $\backslash \; v\_\{\; \backslash bar\{x\}\; \}\backslash \text{'}\; =\; \backslash frac\{m\_\{1\}v\_\{1\}+m\_\{2\}v\_\{2\}\}\{m\_\{1\}+m\_\{2\}\}$

The numerator of $\backslash \; v\_\{\; \backslash bar\{x\}\; \}$ is the total momentum before the collsion, and numerator of $\backslash \; v\_\{\; \backslash bar\{x\}\; \}\backslash \text{'}$ is the total momentum after the collsion. Since momentum is conserved, we have $\backslash \; v\_\{\; \backslash bar\{x\}\; \}\; =\; \backslash \; v\_\{\; \backslash bar\{x\}\; \}\backslash \text{'}$.

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 $\backslash \; v\_\{1\}$ and $\backslash \; v\_\{2\}$ above we see that in the case of a large $\backslash \; u\_\{1\}$, the value of $\backslash \; v\_\{1\}$ 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.

Elastic collision of unequal masses

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,

$m\_\{1\}u\_\{1\}\; +\; m\_\{2\}u\_\{2\}\; =\; m\_\{1\}v\_\{1\}\; +\; m\_\{2\}v\_\{2\}\; =\; \{0\}\backslash ,\backslash !$

$m\_\{1\}u\_\{1\}^\{2\}\; +\; m\_\{2\}u\_\{2\}^\{2\}\; =\; m\_\{1\}v\_\{1\}^\{2\}\; +\; m\_\{2\}v\_\{2\}^\{2\}\backslash ,\backslash !$

$\backslash frac\{(m\_\{2\}u\_\{2\})^\{2\}\}\{2m\_1\}\; +\; \backslash frac\{(m\_\{2\}u\_\{2\})^\{2\}\}\{2m\_2\}\; =$

\frac{(m_{2}v_{2})^{2}}{2m_1} + \frac{(m_{2}v_{2})^{2}}{2m_2}\,\!

$(m\_\{1\}\; +\; m\_\{2\})(m\_\{2\}u\_\{2\})^\{2\}\; =\; (m\_\{1\}\; +\; m\_\{2\})(m\_\{2\}v\_\{2\})^\{2\}\backslash ,\backslash !$

$u\_\{2\}\; =\; -v\_\{2\}\backslash ,\backslash !$

$\backslash frac\{(m\_\{1\}u\_\{1\})^\{2\}\}\{2m\_1\}\; +\; \backslash frac\{(m\_\{1\}u\_\{1\})^\{2\}\}\{2m\_2\}\; =$

\frac{(m_{1}v_{1})^{2}}{2m_1} + \frac{(m_{1}v_{1})^{2}}{2m_2}\,\!

$(m\_\{1\}\; +\; m\_\{2\})(m\_\{1\}u\_\{1\})^\{2\}\; =\; (m\_\{1\}\; +\; m\_\{2\})(m\_\{1\}v\_\{1\})^\{2\}\backslash ,\backslash !$

$u\_\{1\}=-v\_\{1\}\backslash ,\backslash !$

According to Theory of Relativity, in a particular frame of reference, if the total momentum is equal to zero,

$\backslash frac\{m\_\{1\}\backslash ;u\_\{1\}\}\{\backslash sqrt\{1-u\_\{1\}^\{2\}/c^\{2\}\}\}\; +$

\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\,\!

$\backslash frac\{m\_\{1\}c^\{2\}\}\{\backslash sqrt\{1-u\_1^2/c^2\}\}\; +$

\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 are conserved. Rest mass of the two different colliding bodies 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 using the relativistic theory because it is 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,

$\backslash frac\{m\_\{1\}\backslash ;u\_\{1\}\}\{\backslash sqrt\{1-u\_\{1\}^\{2\}/c^\{2\}\}\}\; +$

\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

$\backslash frac\{m\_\{1\}c^\{2\}\}\{\backslash sqrt\{1-u\_1^2/c^2\}\}\; +$

\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 We can look at the two moving bodies as one system of which the total momentum is $p$, the total energy is $E$ and its velocity $v$ is the velocity of its center of mass. Relative to the center of mass the total momentum equals zero. It can be shown that $v$ is given by:

$v\; =\; \backslash frac\{p\; c^2\}\{E\}$

Now the velocities before the collision in the center of mass reference frame $u\_1\; \backslash \text{'}$ and $u\_2\; \backslash \text{'}$ are:

$u\_\{1\}\; \backslash \text{'}=\; \backslash frac\{u\_1\; -\; v\; \}\{1-\; \backslash frac\{u\_1\; v\}\{c^2\}\}$

$u\_\{2\}\; \backslash \text{'}=\; \backslash frac\{u\_2\; -\; v\; \}\{1-\; \backslash frac\{u\_2\; v\}\{c^2\}\}$

$v\_\{1\}\; \backslash \text{'}=-u\_\{1\}\; \backslash \text{'}$

$v\_\{2\}\; \backslash \text{'}=-u\_\{2\}\; \backslash \text{'}$

$v\_\{1\}\; =\; \backslash frac\{v\_1\; \backslash \text{'}\; +\; v\; \}\{1+\; \backslash frac\{v\_1\; \backslash \text{'}\; v\}\{c^2\}\}$

$v\_\{2\}\; =\; \backslash frac\{v\_2\; \backslash \text{'}\; +\; v\; \}\{1+\; \backslash frac\{v\_2\; \backslash \text{'}\; v\}\{c^2\}\}$

When and ,

$p$ ≈ $m\_1\; u\_1\; +\; m\_2\; u\_2$

$v$ ≈ $\backslash frac\{m\_1\; u\_1\; +\; m\_2\; u\_2\}\{m\_1\; +\; m\_2\}$

$u\_1\; \backslash \text{'}$ ≈ $u\_1\; -\; v$ ≈ $$

\frac {m_1 u_1 + m_2 u_1 - m_1 u_1 - m_2 u_2}{m_1 + m_2} = \frac {m_2 (u_1 - u_2)}{m_1 + m_2}

$u\_2\; \backslash \text{'}$ ≈ $\backslash frac\; \{m\_1\; (u\_2\; -\; u\_1)\}\{m\_1\; +\; m\_2\}$

$v\_1\; \backslash \text{'}$ ≈ $\backslash frac\; \{m\_2\; (u\_2\; -\; u\_1)\}\{m\_1\; +\; m\_2\}$

$v\_2\; \backslash \text{'}$ ≈ $\backslash frac\; \{m\_1\; (u\_1\; -\; u\_2)\}\{m\_1\; +\; m\_2\}$

$v\_1$ ≈ $v\_1\; \backslash \text{'}\; +\; v$ ≈ $$

\frac {m_2 u_2 - m_2 u_1 + m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{u_1 (m_1 - m_2) + 2m_2 u_2}{m_1 + m_2}

$v\_2$ ≈ $\backslash frac\{u\_2\; (m\_2\; -\; m\_1)\; +\; 2m\_1\; u\_1\}\{m\_1\; +\; m\_2\}$

Therefore, the classical calculation only holds true when the speed of both colliding bodies is much lower than the speed of light (about 300 million m/s).

Two- and three-dimensional

Newton\'s Rule (i.e. the conservation of momentum) applies to the components of velocity resolved along the common normal surfaces of the colliding bodies at the point of contact. In the case of the two spheres the velocity components involved are the components resolved along the line of centers during the contact. Consequently, the components of velocity perpendicular to the line of centers will be unchanged during the impact.

To solve an equation involving two colliding bodies in two-dimensions, the overall velocity of each body must be split into two perpendicular velocities: one tangent to the common normal surfaces of the colliding bodies at the point of contact, the other along the line of collision. Since the collision only imparts force along the line of collision, the velocities that are tangent to the point of collision do not change. The velocities along the line of collision can then be used in the same equations as a one-dimensional collision. The final velocities can then be calculated from the two new component velocities and will depend on the point of collision. Studies of two-dimensional collisions are conducted for many bodies in the framework of a two-dimensional gas.

Two-dimensional elastic collision

The momentum of two bodies depends upon their actual velocities and mass, so one cannot predict about the momentum of the two bodies if the kinetic energies of the two bodies are equal.

See also

• Inelastic collision
• Coefficient of restitution

External links

• Rigid Body Collision Resolution in three dimensions including a derivation using the conservation laws
• http://vam.anest.ufl.edu/physics/collisionphysics.html Free simulation of 2-particle collision with user-adjustable coefficient of restitution and particle velocities (Requires Adobe Shockwave)
• http://www.geocities.com/vobarian/2dcollisions/ Explanation of how to calculate 2-dimensional elastic collisions using vectors
• http://www.geocities.com/vobarian/bouncescope/ Free simulator of elastic collisions of dozens of user-configurable objects
• http://www.emanueleferonato.com/2007/08/19/managing-ball-vs-ball-collision-with-flash/ Flash script to manage elastic collisions among any number of spheres

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