*Collision between two particles is the short term mutual interaction between those particles. *Collision results in the change of the energy and the momentum of the particles. Collision between two stones is an example of collision. Physical contact is not necessary for the collision to occur. In Rutherford’s alpha ray scattering experiment, collision between alpha particles and nucleus takes place but without physical contact. Thus, in physics a collision is said to have occurred, if the two particles physically collide against each other or even when the path of the motion of one particles is affected by the other.

There are two types of collision;

1. Elastic Collision

2. Inelastic Collision

## Elastic Collision

The collision in which momentum as well as kinetic energy of the system gets conserved is known as elastic collision.

Consider two bodies having mass $m_1$ and $m_2$ are moving with velocities $u_1$ and $u_2$ such that $u_1>u_2$. Suppose they collide with each other and gets separated from each other with velocities $v_1$ and $v_2$.

Then, for elastic collision,

I. Momentum gets conserved. \[m_1u_1+m_2u_2=m_1v_1+m_2v_2\]

II. Kinetic energy gets conserved. \[\frac{1}{2}m_1{u_1}²+\frac{1}{2}m_2{u_2}²=\frac{1}{2}m_1{v_1}²+\frac{1}{2}m_2{v_2}²\]

**Characteristics of elastic collision;**

- Momentum is conserved.
- Total energy is conserved.
- Kinetic energy is conserved. Forces involved during the collision are conservative.
- Mechanical energy is not converted into other form (sound, light, heat, etc.) of energy.

The collisions between atomic and subatomic particles are elastic in nature.

## Inelastic Collision

*The collision in which momentum is conserved but kinetic energy is not conserved is called inelastic collision. *

In our practical life, most of the collisions are elastic. **Characteristics of inelastic collision;**

- Momentum is conserved.
- Total energy is conserved.
- Kinetic energy is not conserved.
- Some or all of the forces involved during the collision are non conservative.
- A part or whole of the mechanical energy may be converted into other form (sound, light, heat, etc.) of energy.

## One Dimensional Elastic Collision

Consider two bodies having mass $m_1$ and $m_2$ are moving with velocities $u_1$ and $u_2$ such that $u_1>u_2$. Suppose they collide with each other and gets separated from each other with velocities $v_1$ and $v_2$.

Then, for elastic collision,

I. Momentum gets conserved. \[m_1u_1+m_2u_2=m_1v_1+m_2v_2\] \[m_1(u_1-v_1)=m_2(v_2-u_2)\text{__(1)}\]

II. Kinetic energy gets conserved. \[\frac{1}{2}m_1{u_1}²+\frac{1}{2}m_2{u_2}²=\frac{1}{2}m_1{v_1}²+\frac{1}{2}m_2{v_2}²\] \[m_1\left({u_1}²-{v_1}²\right)=m_2\left({v_2}²-{u_2}²\right)\text{__(2)}\] Dividing equation (2) by (1), \[\frac{m_1\left({u_1}²-{v_1}²\right)}{m_1(u_1-v_1)}=\frac{m_2\left({v_2}²-{u_2}²\right)}{m_2(v_2-u_2)}\] \[u_1+v_1=v_2+u_2\] \[u_1-u_2=v_2-v_1\text{___(3)}\] Relative velocity of approach $=$ Relative velocity of separation

Therefore, in one dimensional elastic collision, the relative velocity of approach of first body before collision is equal to the relative velocity of separation of second body after collision.

**Velocity of bodies after elastic collision**From $(3)$, \[u_1-u_2=v_2-v_1\] \[v_2=u_1-u_2+v_1\] Putting the value of $v_2$ in equation $(1)$, \[m_1(u_1-v_1)=m_2(u_1-u_2+v_1-u_2)\] \[m_1u_1-m_1v_1=m_2u_1-2m_2u_2+m_2v_1\] \[m_1u_1-m_2u_1+2m_2u_2=m_2v_1+m_1v_1\] \[(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1\] \[v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}\text{__(4)}\] This is the velocity of the first body after collision. Similarly, velocity of another body can be determined which is found to be, \[v_2=\frac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}\text{__(5)}\]

## Special Cases

**Case I:** If the two bodies have equal masses i.e. $m_1=m_2=m$ then,

From $(4)$, \[v_1=\frac{(m-m)u_1+2mu_2}{m+m}\] \[v_1=u_2\] \[\text{From (5)},\] \[v_2=\frac{(m-m)u_2+2mu_1}{m+m}\] \[v_2=u_1\] Thus, if the bodies have equal masses, they will exchange their velocities after elastic collision.

**Case II:** If the second body is initially at rest i.e. $u_2=0$ then, From $(4)$, \[v_1=\frac{(m_1-m_2)u_1}{m_1+m_2}\text{__(6)}\] From $(5)$, \[v_2=\frac{2m_1u_1}{m_1+m_2}\text{__(7)}\]

For the case II, the following sub-cases may arise;

**Sub-case I:** If the two bodies have equal masses i.e. $m_1=m_2=m$ then, From $(6)$, \[v_1=0\] From $(7)$, \[v_2=u_1\] Thus, when the two bodies of equal masses collide elastically such that the second body is at rest, the first body comes to rest while the second body starts to move with the initial velocity of the first body.

**Sub-case II:** If the mass of second body is negligible as compared to the first body i.e. $m_1>>m_2$ then, neglecting $m_2$ in equation $(6)$ and $(7)$, we get, \[v_1=\frac{m_1u_1}{m_1}=u_1\] \[v_2=\frac{2m_1u_1}{m_1}=2u_1\] Thus, when a very heavy body collides elastically with a light body at rest, the heavy body moves with its same initial velocity and the light body also starts to move with the double velocity of the heavy body.

**Sub-case III:** If the mass of the first body is negligible as compared to the first body i.e. $m_2>>m_1$ then, neglecting $m_1$ in equation $(6)$ and $(7)$, \[v_1=\frac{-m_2u_1}{m_2}=-u_1\] \[v_2=\frac{2m_1u_1}{m_2}=0\;\;\;[m_2>>m_1]\] Thus, when a light body collides elastically with a heavy body at rest, the heavy body stays at rest while the light body starts to move with its same initial velocity but in opposite direction.