Elastic collision explained

In physics, an elastic collision is an encounter (collision) between two bodies in which the total kinetic energy of the two bodies remains the same. In an ideal, perfectly elastic collision, there is no net conversion of kinetic energy into other forms such as heat, noise, or potential energy.

During the collision of small objects, kinetic energy is first converted to potential energy associated with a repulsive or attractive 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).

Collisions of atoms are elastic, for example Rutherford backscattering.

A useful special case of elastic collision is when the two bodies have equal mass, in which case they will simply exchange their momenta.

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 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 forbids energy from being carried away by black-body photons.

In the case of macroscopic bodies, perfectly 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

In any collision, momentum is conserved; but in an elastic collision, kinetic energy is also conserved. Consider particles A and B with masses m_A, m_B, and velocities v_A1, v_B1 before collision, v_A2, v_B2 after collision. The conservation of momentum before and after the collision is expressed by:$$m\_v\_+m\_v\_\; \backslash \; =\backslash \; m\_v\_\; +\; m\_v\_.$$

Likewise, the conservation of the total kinetic energy is expressed by:$$\backslash tfrac12\; m\_v\_^2+\backslash tfrac12\; m\_v\_^2\; \backslash \; =\backslash \; \backslash tfrac12\; m\_v\_^2\; +\backslash tfrac12\; m\_v\_^2.$$

These equations may be solved directly to find

v _{A2} ,v _{B2} when v _{A1} ,v _{B1} are known:

Alternatively the final velocity of a particle, v₂ (v_A2 or v_B2) is expressed by:

v = ( 1+ e)v _{CoM} - eu, v _{CoM} = \dfrac{m _{Av} _{A1} + m _{Bv} _{B1} }

Where:

• e is the coefficient of restitution.
• v_CoM is the velocity of the center of mass of the system of two particles.
• v₁ (v_A1 or v_B1) is the initial velocity of the particle.

If both masses are the same, we have a trivial solution:This simply corresponds to the bodies exchanging their initial velocities with each other.

As can be expected, the solution is invariant under adding a constant to all velocities (Galilean relativity), which is like using a frame of reference with constant translational velocity. Indeed, to derive the equations, one may first change the frame of reference so that one of the known velocities is zero, determine the unknown velocities in the new frame of reference, and convert back to the original frame of reference.

Examples

Before collision:

Ball A: mass = 3 kg, velocity = 4 m/s

Ball B: mass = 5 kg, velocity = 0 m/s

After collision:

Ball A: velocity = −1 m/s

Ball B: velocity = 3 m/s

Another situation:

The following illustrate the case of equal mass,

m _{A=m} _{B} .

In the limiting case where

m _{A} is much larger than m _{B} , such as a ping-pong paddle hitting a ping-pong ball or an SUV hitting a trash can, the heavier mass hardly changes velocity, while the lighter mass bounces off, reversing its velocity plus approximately twice that of the heavy one.

In the case of a large

v _{A1} , the value of v _{A2} 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. This is why 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 which do not easily absorb neutrons: the lightest nuclei have about the same mass as a neutron.

Derivation of solution

To derive the above equations for

v _{A2} ,v _{B2} , rearrange the kinetic energy and momentum equations:

Dividing each side of the top equation by each side of the bottom equation, and using

\tfrac{a ^{2-b} ^{2}{(a-b)}} = a + b, gives:

$$v\_+v\_=v\_+v\_\; \backslash quad\backslash Rightarrow\backslash quad\; v\_-v\_\; =\; v\_-v\_$$

That is, the relative velocity of one particle with respect to the other is reversed by the collision.

Now the above formulas follow from solving a system of linear equations for

v _{A2} ,v _{B2} , regarding m _{A,m} _{B,v} _{A1} ,v _{B1} as constants:$$\backslash left$$

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