# Elastic collision: Wikis

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# 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. 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 net conversion of kinetic energy into other forms. During the collision kinetic energy is first converted to potential energy associated with a repulsive force between the particles (when the particles move against this force, i.e. the angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. the angle between the force and the relative velocity is acute).

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.

When considering energies, possible rotational energy before and/or after a collision may also play a role.

## Equations

### One-dimensional Newtonian

Consider two particles, denoted by subscripts 1 and 2. Let mi be the masses, ui the velocities before collision and vi the velocities after collision.

The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision, and is expressed by the equation

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

Likewise, the conservation of the total kinetic energy is expressed by the equation

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

These equations may be solved directly to find vi when ui are known or vice versa. However, the algebra[1] can get messy. A cleaner solution is to first change the frame of reference such that one of the known velocities is zero. The unknown velocities in the new frame of reference can then be determined and followed by a conversion back to the original frame of reference to reach the same result. Once one of the unknown velocities is determined, the other can be found by symmetry.

Solving these simultaneous equations for vi we get:

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

OR

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

The latter is the trivial solution, corresponding to the case that no collision has taken place (yet).

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.

### 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,

p1 = − p2
$p_1^2 = p_2^2$
$\sqrt {m_1^2c^4 + p_1^2c^2} + \sqrt {m_2^2c^4 + p_2^2c^2} = E$
$p_1 = \pm \frac{\sqrt{E^4 - 2E^2m_1^2c^4 - 2E^2m_2^2c^4 + m_1^4c^8 - 2m_1^2m_2^2c^8 + m_2^4c^8}}{cE}$
u1 = − v1

Where m1 represents the rest mass of the first colliding body, m2 represents the rest mass of the second colliding body, u1 represents the initial velocity of the first collidng body, u2 represents the initial velocity of the second colliding body, v1 represents the velocity after collision of the first colliding body, v2 represents the velocity after collision of the second colliding body, p1 denotes the momentum of the first colliding body, p2 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 u1 = − v1 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 pT, the total energy is E and its velocity vc 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 vc is given by:

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

Now the velocities before the collision in the center of momentum frame u1' and u2' are:

$u_{1} '= \frac{u_1 - v_c }{1- \frac{u_1 v_c}{c^2}}$
$u_{2} '= \frac{u_2 - v_c }{1- \frac{u_2 v_c}{c^2}}$
v1' = − u1'
v2' = − u2'
$v_{1} = \frac{v_1 ' + v_c }{1+ \frac{v_1 ' v_c}{c^2}}$
$v_{2} = \frac{v_2 ' + v_c }{1+ \frac{v_2 ' v_c}{c^2}}$

When u1 < < c and u2 < < c,

pTm1u1 + m2u2
vc$\frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$
u1'u1vc$\frac {m_1 u_1 + m_2 u_1 - m_1 u_1 - m_2 u_2}{m_1 + m_2} = \frac {m_2 (u_1 - u_2)}{m_1 + m_2}$
u2'$\frac {m_1 (u_2 - u_1)}{m_1 + m_2}$
v1'$\frac {m_2 (u_2 - u_1)}{m_1 + m_2}$
v2'$\frac {m_1 (u_1 - u_2)}{m_1 + m_2}$
v1v1' + vc$\frac {m_2 u_2 - m_2 u_1 + m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{u_1 (m_1 - m_2) + 2m_2 u_2}{m_1 + m_2}$
v2$\frac{u_2 (m_2 - m_1) + 2m_1 u_1}{m_1 + m_2}$

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

For the case of 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

In a center of momentum frame at any time the velocities of the two bodies are in opposite directions, with magnitudes inversely proportional to the masses. In an elastic collision these magnitudes do not change. The directions may change depending on the shapes of the bodies and the point of impact. For example, in the case of spheres the angle depends on the distance between the (parallel) paths of the centers of the two bodies. Any non-zero change of direction is possible: if this distance is zero the velocities are reversed in the collision; if it is close to the sum of the radii of the spheres the two bodies are only slightly deflected.

Assuming that the second particle is at rest before the collision, the angles of deflection of the two particles, $\vartheta_1$ and $\vartheta_2$, are related to the angle of deflection θ in the system of the center of mass by [2]

$\tan \vartheta_1=\frac{m_2 \sin \theta}{m_1+m_2 \cos \theta},\qquad \vartheta_2=\frac{\pi}{2}-{\theta}.$

The velocities of the particles after the collision are:

$v'_1=\frac{\sqrt{m_1^2+m_2^2+2m_1m_2\cos \theta}}{m_1+m_2},\qquad v'_2=\frac{2m_1}{m_1+m_2}\sin \frac{\theta}{2}.$

## References

1. ^ Algebraic Derivation of Post-Collision Velocities (1D)
2. ^ Landau LD and Lifshitz EM (1976) Mechanics, 3rd. ed., Pergamon Press. ISBN 0-08-021022-8 (hardcover) and ISBN 0-08-029141-4 (softcover).

Elastic collision in one dimension in special relativity

# Simple English

An elastic collision is when two objects collide and bounce back with little or none deformation. For example, two rubber balls bouncing together would be elastic. Two cars hitting each other would be inelastic, as the cars crumple, and do not bounce back. In a perfectly elastic collision (the simplest case), no kinetic energy is lost, and so the kinetic energy of the two objects after the collision is equal to their total kinetic energy before the collision. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms (heat, sound). The other rule to remember when working with elastic collisions is that momentum is conserved.

## One-dimensional Newtonian

Consider two particles, indicated by subscripts 1 and 2. Let m1 and m2 be the masses, u1 and u2 be the velocities before the collision and v1 and v2 be the velocities after collision.

### Using Conservation of Momentum to write one formula

Since it is an elastic collusion, the total momentum before the collision is the same as the total momentum after the collision. Given that momentum (p) is calculated as

$\,\! p=mv$

We can calculate the momentum before the collision to be:

$\,\! m_\left\{1\right\}u_\left\{1\right\}+m_\left\{2\right\}u_\left\{2\right\}$

and the momentum after the collision to be:

$\,\! m_\left\{1\right\}v_\left\{1\right\}+m_\left\{2\right\}v_\left\{2\right\}$

Setting the two equal gives us our first equation:

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

### Using Conservation of Energy to write a second formula

The second rule we use is that the total kinetic energy remains the same, meaning that the initial kinetic energy is equal to the final kinetic energy.

The formula for kinetic energy is:

$\frac\left\{mv^2\right\}2$

So, using the same variables as before: The initial kinetic energy is:

$\frac\left\{m_1u_1^2\right\}2+\frac\left\{m_2u_2^2\right\}2$

The final kinetic energy is:

$\frac\left\{m_1v_1^2\right\}2+\frac\left\{m_2v_2^2\right\}2.$

Setting the two to be equal ( since the total kinetic energy remains the same):

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

### Putting those two equations together

These equations may be solved directly to find vi when ui are known or vice versa. Here is a sample problem, which can be solved using either conservation of momentum, or conservation of energy:

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 = unknown ( We'll represent it with v )

Using Conservation of Momentum:

$\,\! m_\left\{1\right\}u_\left\{1\right\}+m_\left\{2\right\}u_\left\{2\right\}=m_\left\{1\right\}v_\left\{1\right\} + m_\left\{2\right\}v_\left\{2\right\}.$
$\ 3*4 + 5*\left(-6\right) = 3*\left(-8.5\right) + 5*v$

After doing multiplication, and then subtracting $3*\left(-8.5\right)$ from both sides, we get:

$\ 12 - 30 + 25.5 = 5*v$

Summing the left side, then dividing by $5$ gives us:

$\frac\left\{7.5\right\}5 = v$, and doing the final division gives us: $\ 1.5 = v$

We could have also solved this problem using Conservation of Energy:

$\frac\left\{m_1u_1^2\right\}2+\frac\left\{m_2u_2^2\right\}2=\frac\left\{m_1v_1^2\right\}2+\frac\left\{m_2v_2^2\right\}2$
$\frac\left\{3*4^2\right\}2+\frac\left\{5*\left(-6\right)^2\right\}2=\frac\left\{3\left(-8.5\right)^2\right\}2+\frac\left\{5v^2\right\}2$

Multiplying both sides by $2$, and then do all the required multiplications gives us:

$\ 48+180=216.75+5v^2$

Adding the numbers on the left, subtracting $216.75$ from both sides, and dividing by $5$ gives us:

$\ 2.25 = v^2$

Taking the square root of both sides gives us an answer of $v = \pm1.5$.

Unfortunately, we'd still need to use conservation of momentum to figure out whether $v$ is positive or negative.