Elastic collision

From Wikipedia, the free encyclopedia

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.

1 Equations
2 See also
3 References
4 External links

[edit] Equations

[edit] 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:

$\frac{m_1u_1^2}2+\frac{m_2u_2^2}2=\frac{m_1v_1^2}2+\frac{m_2v_2^2}2.$

Total momentum remains constant throughout the collision:

$\,\! m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1} + m_{2}u_{2}.$

These equations may be solved directly to find $\ v_{1}$ and $\ v_{2}$ . However, the algebra can get messy. A cleaner solution is to first change the frame of reference such that either $\ v_{1}$ or $\ v_{2}$ 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 $\ v_{1}$ or $\ v_{2}$ 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.

$v_{1} = \frac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2}}{m_{1}+m_{2}}$ , $v_{2} = \frac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}$

$\ v_{1} = u_{2}$ , $\ v_{2} = u_{1}$ , 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:

$\ v_{1}-v_{2} = u_{2}-u_{1}$

Derivation: Using the kinetic energy we can write

$\ m_1(v_1^2-u_1^2)=m_2(u_2^2-v_2^2)$

$\Rightarrow m_1(v_1-u_1)(v_1+u_1)=m_2(u_2-v_2)(u_2+v_2)$

Rearrange momentum equation:

$\ m_1(v_1-u_1)=m_2(u_2-v_2)$

Dividing kinetic energy equation by the momentum equation we get:

$\ v_1+u_1=u_2+v_2$

$\Rightarrow v_1-v_2 = u_2-u_1$

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 $\ t$ 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.

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.

[edit] One-dimensional relativistic

According to Special Relativity,

$p = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$

Where p denotes momentum of any massive particle, v denotes velocity, c denotes the speed of light.

in the center of momentum frame where the total momentum equals zero,

p 1 = - p 2

$p_1^2 = p_2^2$

$\sqrt {m_1^2c^4 + p_1^2c^2} + \sqrt {m_2^2c^4 + p_2^2c^2} = E$

$p_1 = \pm \frac{\sqrt{E^4 - 2E^2m_1^2c^4 - 2E^2m_2^2c^4 + m_1^4c^8 - 2m_1^2m_2^2c^8 + m_2^4c^8}}{cE}$

u 1 = - v 1

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, $p 1$ denotes the momentum of the first colliding body, $p 2$ denotes the momentum of the second colliding body and $c$ denotes the speed of light in vacuum, $E$ denotes the total energy of the system (i.e. the sum of rest masses and kinetic energies of the colliding bodies).

Since the total energy and momentum of the system are conserved and the rest mass of the colliding body do not change, it is shown that the momentum of the colliding body is decided by the rest masses of the colliding bodies, total energy and the total momentum. The magnitude of the momentum of the colliding body does not change after collision but the direction of movement is opposite relative to the center of momentum frame.

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 the center of momentum frame, according to Classical Mechanics,

$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} = {0}\,\!$

$m_{1}u_{1}^{2} + m_{2}u_{2}^{2} = m_{1}v_{1}^{2} + m_{2}v_{2}^{2}\,\!$

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

$(m_{1} + m_{2})(m_{2}u_{2})^{2} = (m_{1} + m_{2})(m_{2}v_{2})^{2}\,\!$

$u_{2} = -v_{2}\,\!$

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

$(m_{1} + m_{2})(m_{1}u_{1})^{2} = (m_{1} + m_{2})(m_{1}v_{1})^{2}\,\!$

$u_{1}=-v_{1}\,\!$

It is shown that $u 1 = - v 1$ remains true in relativistic calculation despite other differences. 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 any,

$\frac{m_{1}\;u_{1}}{\sqrt{1-u_{1}^{2}/c^{2}}} + \frac{m_{2}\;u_{2}}{\sqrt{1-u_{2}^{2}/c^{2}}} = \frac{m_{1}\;v_{1}}{\sqrt{1-v_{1}^{2}/c^{2}}} + \frac{m_{2}\;v_{2}}{\sqrt{1-v_{2}^{2}/c^{2}}}=p_T$

$\frac{m_{1}c^{2}}{\sqrt{1-u_1^2/c^2}} + \frac{m_{2}c^{2}}{\sqrt{1-u_2^2/c^2}} = \frac{m_{1}c^{2}}{\sqrt{1-v_1^2/c^2}} + \frac{m_{2}c^{2}}{\sqrt{1-v_2^2/c^2}}=E$

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

$v_c = \frac{p_T c^2}{E}$

Now the velocities before the collision in the center of momentum frame $u 1'$ and $u 2'$ are:

$u_{1} '= \frac{u_1 - v_c }{1- \frac{u_1 v_c}{c^2}}$

$u_{2} '= \frac{u_2 - v_c }{1- \frac{u_2 v_c}{c^2}}$

v 1' = - u 1'

v 2' = - u 2'

$v_{1} = \frac{v_1 ' + v_c }{1+ \frac{v_1 ' v_c}{c^2}}$

$v_{2} = \frac{v_2 ' + v_c }{1+ \frac{v_2 ' v_c}{c^2}}$

When $u 1 < < c$ and $u 2 < < c$ ,

p T

≈

m 1 u 1 + m 2 u 2

v c

≈ $\frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

u 1'

≈

u 1 - v c

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

u 2'

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

v 1'

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

v 2'

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

v 1

≈

v 1' + v c

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

v 2

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

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).

[edit] 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.

[edit] See also

[edit] References

Elastic collision in one dimension in special relativity

[edit] External links

Rigid Body Collision Resolution in three dimensions including a derivation using the conservation laws
VNE Rigid Body Collision Simulation Small Open Source 3D engine with easy-to-understand implementation of elastic collisions in C
Visualize 2-D Collision Free simulation of 2-particle collision with user-adjustable coefficient of restitution and particle velocities (Requires Adobe Shockwave)
2-Dimensional Elastic Collisions Without Trigonometry Explanation of how to calculate 2-dimensional elastic collisions using vectors
Bouncescope Free simulator of elastic collisions of dozens of user-configurable objects
Managing ball vs ball collision with Flash Flash script to manage elastic collisions among any number of spheres

Categories: Classical mechanics | Introductory physics

Elastic collision

From Wikipedia, the free encyclopedia

Contents

[edit] Equations

[edit] One-dimensional Newtonian

[edit] One-dimensional relativistic

[edit] Two- and three-dimensional

[edit] See also

[edit] References

[edit] External links

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