Elastic collision

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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).

In the case of macroscopic bodies this will not be the case as some of the energy will become heat.

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.

1 Equations and calculation in the one-dimensional case (Low speed approximation)
2 Equations and calculation in the one-dimensional case (General Solution)
3 Two-dimensional collisions
4 See also
5 External links

[edit] Equations and calculation in the one-dimensional case (Low speed approximation)

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

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] Equations and calculation in the one-dimensional case (General Solution)

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 momentun 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 postulate in Special Relativity state that Laws of Physics should be invariant in all inertial frame of reference. That is, if total momentum is conserved in a particular inertial frame of reference, total momentum will also be conserved in any intertial 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,

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

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

$E^2 = m_0^2 c^4 + p^2 c^2$

$m_0 = \sqrt{E^2/c^4 - p^2/c^2}$

$E^2 = \frac{m_0^2 c^6}{c^2 - v^2}$

$E^2c^2-E^2v^2=m_0^2 c^6$

v =

± $\sqrt{c^2- \frac{m_0^2c^6}{E^2}}$

If total momentum is greater than zero, than 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 < < c

and

u 2 < < c

m 0

≈

m 1 + m 2

p

≈

m 1 u 1 + m 2 u 2

v

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

u 1'

≈

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'

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

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

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

v 2

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

Therefore, classical calculation only holds true when the speed of both colliding bodies is much lower than the speed of light.

[edit] Two-dimensional collisions

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 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

[edit] See also

[edit] External links

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

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Categories: Articles to be merged since March 2007 | Classical mechanics | Introductory physics