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


2 Answers 2


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.

  • $\begingroup$ Conservation of momentum results from Newton's laws for an isolated system. But they are not equivalent. Newton's laws work even for situations when the momentum is not conserved. Actaully Newton's second law tells you how the momentum changes. $\endgroup$
    – nasu
    Commented Jan 19, 2021 at 17:57
  • $\begingroup$ That isn't correct, every non-isolated system is coupled to another non-isolated system with the exact opposite momentum change, so the total momentum is always conserved. This is Newton's third law. The conservation of momentum is equivalent to newton's laws and Hamilton's principle all of which are theoretical frameworks of Classical mechanics. $\endgroup$ Commented Jan 19, 2021 at 18:05
  • $\begingroup$ To illustrate it even clearler, consider an inelastic collision, ofcourse in each individual colliding piece momentum is not conserved (its velocity clearly changes), but overall, momentum is conserved; the same can be said about literally any force...again because of Newtons Third Law $\endgroup$ Commented Jan 19, 2021 at 18:06
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    $\begingroup$ Well, it depends on what system you are looking at. "Always" is a dangerous word. The students will take without qualification and will expect to apply it even when the system is not isolated. Same as they expect entropy to "always" increase. I am not saying that there is something wrong with your answer, just that it will be useful to specify the meaning of "always". :) $\endgroup$
    – nasu
    Commented Jan 19, 2021 at 18:20
  • $\begingroup$ Yes thats definitely true haha, however in my own experience I remember only truly grasping where Newton's Laws come from once we derived them from momentum, but you are correct that this may misleading in the same way as the entropy statement is. $\endgroup$ Commented Jan 19, 2021 at 18:23

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.

  • $\begingroup$ I know the total momentum is the same but do there is any formula for how much momentum is being transfer I think It have to be related to the speed difference because when the bodies are at the same speed there isn't any force between them but the problem is that one body is slowing down and the other speed up so the question is how much would one object speed up and the other would slow down is it relative to it mass? $\endgroup$
    – daniel
    Commented Jan 20, 2021 at 8:35

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