The collision of solid objects is a common event in the day to day life. This collision is guided by the laws of motions and momentum conservation laws. In such a collision in which kinetic energy of the system is not conserved but the momentum is conserved. An inelastic collision is any collision between objects in which some energy is lost due to it. A special case of this is also called the perfectly inelastic collision. In this topic, we will discuss the concept of inelastic collision and inelastic collision formula with some examples. Let us begin the concept!
Source: en.wikipedia.org
Inelastic Collision Formula
Concept of inelastic collision:
An inelastic collision is such a type of collision which takes place between two objects. Also, there will be some loss of energy. In these cases of inelastic collision, momentum is always conserved but the kinetic energy is not conserved. Most of the collisions are inelastic in nature.
Perfectly Inelastic Collision:
The special case of inelastic collision is referred to as a perfectly inelastic collision. Also after the collision, two objects stick together. For example, when wet mud ball is thrown against a wall mud ball stick to the wall.
In twodimensional inelastic collision conservation of momentum is separately applied separately along each axis. As momentum is a vector equation and there is one conservation of momentum equation. Similarly, there will be only one conservation of energy equation.
The formula for Inelastic collision:
Mass of object 1 × initial velocity 1 + Mass of object 1 × initial velocity 1 = (Mass of 1 + mass of 2) × final velocity of combined objects)
In the form of the equation:
\( m_1u_1 + m_2u_2 = (m_1+m_2) v \)
Where,
\(m_1\)  mass of a first object (kg)

\(m_2\)  mass of a second object (kg)

\(u_1 \)  the initial velocity of the first object (m/s)

\(u_2\)  the initial velocity of the second object (m/s)

v  the final velocity of the combined objects (m/s) 
Solved Examples
Q1: A man shoots a ball at an old can on a fencepost. The ball pellet has a mass of 0.200 g, and they can have a mass of 15.0 g. The ball hits the can at a velocity of 90.0 \(ms^{1}\). The full mass of the ball sticks to the can and knocks it off the post. Then, what will be the final velocity of the combined paintball and can?
Solution:
The given parameters are:
\( m_1 \) = 0.200 g = 0.0002 kg
\(m_2 \) = 15.0 g = 0.015 kg
\( u_1 \) = 90.0 \(ms^{1}\)
\(u_2\) = 0.0 \(ms^{1}\) ( due to at rest position )
Thus, the final velocity can be found for the combined ball and can together, by using the formula:
\(m_1u_1 + m_2u_2 = (m_1+m_2) v \)
Rearranging it, we get,
v = \(\frac { m_1 u_1 + m_2 u_2 }{ ( m_1 + m_2 ) } \)
i.e. v = 1.18 \(ms^{1}\)
v = \(\frac { m_1 u_1 + m_2 u_2 }{ (m_1 + m_2 ) } \)
= \(\frac { 0.0002 \times 90 + 0.015 \times 0.0 }{ (0.0002 + 0.015 ) } \)
= \(\frac { 0.0180 } { 0.0152 }\)
= \(1.18 \;ms^{1} \)
Therefore, velocity will be 1.18 \(ms^{1}\)
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
M=f/g
Interesting studies
It is already correct f= ma by second newton formula…