In physics, an elastic collision is a collision process between two physical objects in which the total kinetic energy of the two bodies remains the same before and after the collision. In an ideal, perfectly elastic collision, there is no net conversion of kinetic energy into other forms of energy such as heat, sound, 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 the other 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 radiation is negligible or doesn't escape. 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 or after a collision may also play a role.
Equations
One-dimensional Newtonian
In any collision without an external force, momentum is conserved; but in an elastic collision, kinetic energy is also conserved. Consider particles A and B with masses mA, mB, and velocities vA1, vB1 before collision, vA2, vB2 after collision. The conservation of momentum before and after the collision is expressed by:
m A v A 1 + m B v B 1 = m A v A 2 + m B v B 2 . {\displaystyle m_{A}v_{A1}+m_{B}v_{B1}\ =\ m_{A}v_{A2}+m_{B}v_{B2}.}
In an elastic collision, kinetic energy is conserved and can be expressed by:
1 2 m A v A 1 2 + 1 2 m B v B 1 2 = 1 2 m A v A 2 2 + 1 2 m B v B 2 2 . {\displaystyle {\tfrac {1}{2}}m_{A}v_{A1}^{2}+{\tfrac {1}{2}}m_{B}v_{B1}^{2}\ =\ {\tfrac {1}{2}}m_{A}v_{A2}^{2}+{\tfrac {1}{2}}m_{B}v_{B2}^{2}.}
These equations may be solved directly to find v A 2 , v B 2 {\displaystyle v_{A2},v_{B2}} when v A 1 , v B 1 {\displaystyle v_{A1},v_{B1}} are known:
v A 2 = m A − m B m A + m B v A 1 + 2 m B m A + m B v B 1 v B 2 = 2 m A m A + m B v A 1 + m B − m A m A + m B v B 1 . {\displaystyle {\begin{array}{ccc}v_{A2}&=&{\dfrac {m_{A}-m_{B}}{m_{A}+m_{B}}}v_{A1}+{\dfrac {2m_{B}}{m_{A}+m_{B}}}v_{B1}\\[.5em]v_{B2}&=&{\dfrac {2m_{A}}{m_{A}+m_{B}}}v_{A1}+{\dfrac {m_{B}-m_{A}}{m_{A}+m_{B}}}v_{B1}.\end{array}}}
If both masses are the same, we have a trivial solution:
v A 2 = v B 1 v B 2 = v A 1 . {\displaystyle {\begin{aligned}v_{A2}&=v_{B1}\\v_{B2}&=v_{A1}.\end{aligned}}}
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 = -4 m/s After collision
Ball A: velocity = −6 m/s Ball B: velocity = 2 m/s Another situation:
The following illustrate the case of equal mass, m A = m B {\displaystyle m_{A}=m_{B}} .
In the limiting case where m A {\displaystyle m_{A}} is much larger than m B {\displaystyle 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 A 1 {\displaystyle v_{A1}} , the value of v A 2 {\displaystyle 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 A 2 , v B 2 , {\displaystyle v_{A2},v_{B2},} rearrange the kinetic energy and momentum equations:
m A ( v A 2 2 − v A 1 2 ) = m B ( v B 1 2 − v B 2 2 ) m A ( v A 2 − v A 1 ) = m B ( v B 1 − v B 2 ) {\displaystyle {\begin{aligned}m_{A}(v_{A2}^{2}-v_{A1}^{2})&=m_{B}(v_{B1}^{2}-v_{B2}^{2})\\m_{A}(v_{A2}-v_{A1})&=m_{B}(v_{B1}-v_{B2})\end{aligned}}}
Dividing each side of the top equation by each side of the bottom equation, and using
a 2 − b 2 ( a − b ) = a + b , {\displaystyle {\tfrac {a^{2}-b^{2}}{(a-b)}}=a+b,} gives:
v A 2 + v A 1 = v B 1 + v B 2 ⇒ v A 2 − v B 2 = v B 1 − v A 1 {\displaystyle v_{A2}+v_{A1}=v_{B1}+v_{B2}\quad \Rightarrow \quad v_{A2}-v_{B2}=v_{B1}-v_{A1}}
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 A 2 , v B 2 , {\displaystyle v_{A2},v_{B2},} ; this is done by regarding
m A , m B , v A 1 , v B 1 {\displaystyle m_{A},m_{B},v_{A1},v_{B1}} as constants:
{ v A 2 − v B 2 = v B 1 − v A 1 m A v A 1 + m B v B 1 = m A v A 2 + m B v B 2 . {\displaystyle \left\{{\begin{array}{rcrcc}v_{A2}&-&v_{B2}&=&v_{B1}-v_{A1}\\m_{A}v_{A1}&+&m_{B}v_{B1}&=&m_{A}v_{A2}+m_{B}v_{B2}.\end{array}}\right.}
Once v A 2 {\displaystyle v_{A2}} is determined, v B 2 {\displaystyle v_{B2}} can be found by symmetry.
Center of mass frame With respect to the center of mass, both velocities are reversed by the collision: 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. The velocity of the center of mass does not change by the collision. To see this, consider the center of mass at time t {\displaystyle t} before collision and time t ′ {\displaystyle t'} after collision:
x ¯ ( t ) = m A x A ( t ) + m B x B ( t ) m A + m B x ¯ ( t ′ ) = m A x A ( t ′ ) + m B x B ( t ′ ) m A + m B . {\displaystyle {\begin{aligned}{\bar {x}}(t)&={\frac {m_{A}x_{A}(t)+m_{B}x_{B}(t)}{m_{A}+m_{B}}}\\{\bar {x}}(t')&={\frac {m_{A}x_{A}(t')+m_{B}x_{B}(t')}{m_{A}+m_{B}}}.\end{aligned}}}
Hence, the velocities of the center of mass before and after collision are:
v x ¯ = m A v A 1 + m B v B 1 m A + m B v x ¯ ′ = m A v A 2 + m B v B 2 m A + m B . {\displaystyle {\begin{aligned}v_{\bar {x}}&={\frac {m_{A}v_{A1}+m_{B}v_{B1}}{m_{A}+m_{B}}}\\v_{\bar {x}}'&={\frac {m_{A}v_{A2}+m_{B}v_{B2}}{m_{A}+m_{B}}}.\end{aligned}}}
The numerators of v x ¯ {\displaystyle v_{\bar {x}}} and v x ¯ ′ {\displaystyle v_{\bar {x}}'} are the total momenta before and after collision. Since momentum is conserved, we have v x ¯ = v x ¯ ′ . {\displaystyle v_{\bar {x}}=v_{\bar {x}}'\,.}
One-dimensional relativistic According to special relativity,
p = m v 1 − v 2 c 2 {\displaystyle p={\frac {mv}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}
where p denotes momentum of any particle with mass m, v denotes velocity, and c is 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 E = m 1 2 c 4 + p 1 2 c 2 + m 2 2 c 4 + p 2 2 c 2 p 1 = ± E 4 − 2 E 2 m 1 2 c 4 − 2 E 2 m 2 2 c 4 + m 1 4 c 8 − 2 m 1 2 m 2 2 c 8 + m 2 4 c 8 2 c E u 1 = − v 1 . {\displaystyle {\begin{aligned}p_{1}&=-p_{2}\\p_{1}^{2}&=p_{2}^{2}\\E&={\sqrt {m_{1}^{2}c^{4}+p_{1}^{2}c^{2}}}+{\sqrt {m_{2}^{2}c^{4}+p_{2}^{2}c^{2}}}\\p_{1}&=\pm {\frac {\sqrt {E^{4}-2E^{2}m_{1}^{2}c^{4}-2E^{2}m_{2}^{2}c^{4}+m_{1}^{4}c^{8}-2m_{1}^{2}m_{2}^{2}c^{8}+m_{2}^{4}c^{8}}}{2cE}}\\u_{1}&=-v_{1}.\end{aligned}}}
Here m 1 , m 2 {\displaystyle m_{1},m_{2}} represent the rest masses of the two colliding bodies, u 1 , u 2 {\displaystyle u_{1},u_{2}} represent their velocities before collision, v 1 , v 2 {\displaystyle v_{1},v_{2}} their velocities after collision, p 1 , p 2 {\displaystyle p_{1},p_{2}} their momenta, c {\displaystyle c} is the speed of light in vacuum, and E {\displaystyle E} denotes the total energy, the sum of rest masses and kinetic energies of the two bodies. Since the total energy and momentum of the system are conserved and their rest masses 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. Relative to the center of momentum frame, the momentum of each colliding body does not change magnitude after collision, but reverses its direction of movement. Comparing with classical mechanics, which gives accurate results dealing with macroscopic objects moving much slower than the speed of light, 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