Question

How do I find total momentum before, and final momentum after collision? I have elastic and inelastic

Answer 1

Can you provide an example of what you mean? Or tell us what we know and can use to find the momentum?

Elastic collisions have no mechanical energy lost. We'll use the variables:
\(T\equiv\quad\) Kinetic energy before
\(U\equiv\quad\) Potential energy before
\(T'\equiv\quad\) Kinetic energy after
\(U'\equiv\quad\) Potential energy after
We can write the total mechanical energy at any time (total mechanical energy is kinetic plus potential energies, remember) is equal to the total mechanical energy at any other time.
That is,
\(T+U\qquad=\qquad T'+U'\)
:) Take a moment to take that in! For multiple objects, the kinetic energy is the sum of the kinetic energies of the objects. Same for potential energy; the potential energy is the sum of the potential energy of the objects.

For an inelastic collision, that is not true. So, we know more concepts for elastic equations, and we can use them as tools to solve those.

In both cases, we use the conservation of momentum, which is always true (at least classically) [in a closed system].
The conservation of momentum states that the sum of the momentums at one time will equal the sum of the momentums at any other time.
We'll use the variables:
\(m\equiv\quad\) mass
\(v\equiv\quad\) velocity before
\(v'\equiv\quad\) velocity after
For two objects,
\(\large m_1v_1+m_2v_2\qquad=\qquad m-1v_1'+m_2v_2'\)
For three,
\(\large m_1v_1+m_2v_2+m_3v_3=m-1v_1'+m_2v_2'+m_3v_3'\)
And so on...

We can also use summation notation to show addition of the momentum of any number of particles. Using \(i\) to indicate the specific object, having \(n\) objects total.
Since the momentums before and after are equal,
\[\Large \sum_i^n m_iv_i=\sum_i^n m_i v_i'\]

I hope this helped! Feel free to ask any questions on that, too.