How much of the momentum is being transfer in case of collision relative to elastically and momentum difference? if we take a look at two objects that has been collision ,we may use the the third law of newton to calculate the velocity of each one if we know the force they use on each other. for example if we take newton balls we may say that 100% of the momentum his being transfer to the other ball because one ball is stopped moving so we can assume it pushed the other ball with force that equal to it momentum ,so the other direction force  would be equal to the  force it momentum . but this is not always the case for example if 1000kg object moving in velocity of 100m/s would hit a another object with 1kg in rest it not like the 1 kg velocity would be 10000m/s an the 1000kg object would stop moving , it doesn't make any sense. So my question is whether there is any formula to calculate how much of the momentum is being transfer relative to the mass difference and speed difference?. and also if is it depends on how the objects are elasticity
 A: First off, please check your spelling and grammar, it's really hard to follow the question when it is poorly phrased.
Second, momentum is $\textbf{always}$ conserved. $\textbf{Always}$. This a fundamental fact about collisions and everything in Classical Mechanics really, the Conservation of Momentum alone is equivalent to Newton's Laws. Now that we got that out of the way; why are some collisions different than others? Energy.
Consider the case with two masses $m_1$ and $m_2$ travelling at $v_1$ and $v_2$ respectively. After their collision, we can't just "solve" for the resulting velocities since we have 2 variables and 1 equation. You can have all the momentum go to $m_1$ or to $m_2$ or have it split equally, ... So we need an extra imposed constraint which comes in the form of energy. Notice energy is also always "conserved" but not always in the form of Kinetic Energy, some of it might be the sound of the collision, some of it might be heat, but Kinetic Energy isn't in general conserved.
In introductory physics they normally give you the added constraint, either in the form of one of the velocities, like "after the collision $m_1$ moves at $v_1'$; now calculate $v_2'$". Or the constraint is given in the form of a "collision type", something you have probably heard as Inelastic vs. Elastic Collisions. Inelastic collisions are in fact just a constraint on the velocity; can be formalized as "after the collision $m_1$ and $m_2$ moves at the same $v$, calculate $v$", this is because the objects are literally moving at the same speed together. Elastic collisions have their constraint in a bit more of a nuanced way, and that is; Kinetic Energy is conserved. This gives you another equation, so now you have 2 equations 2 variables and a solution exists.
In reality collisions are never perfectly elastic or perfectly inelastic, nonetheless moment is still conserved.
A: We can say that momentum is conserved in the collision. So, in your example: $m_1 = 1000$, $v_1 = 100 \implies p_1 = 10000$. The total momentum after the collision is also $10000$. But there are infinite combinations of velocities for object 1 and 2 that fulfill that requirement. There is no formula to find the final velocity of each object.
If we are talking about rigid bodies, part of the kinetic energy of object 1 is transformed in sound and/or heat. And part is transformed in rotational kinetic energy. One of them, or both objects will rotate after the collision, even if the object 1 is not initially rotating.
Their final spin depends on the exact point of collision and its form.
