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I have a problem with understanding the nature of collisions and their outcomes. From my understanding, I come to think that when a mass collides with another, both of them should always have equal velocities post-collision. For example, when a mass moving at v1, m1, collides with a mass at rest, m2, their velocity after collision should always be m1v1 / m1 + m2. My justification for this is that once they reach the said velocity, they are not colliding anymore, they are moving along together, they have zero kinetic energy, relative to each other. I don't understand how is it that there are cases in which one mass loses it momentum completely to the other mass and other cases in which m1 may even rebound: how can any change in momentum still occur after they have equal velocities?

I also have problems understanding what governs the magnitude of impulse at collision: is it possible to predict the magnitude of the force and the duration of which the applied force will last? It seems, from all the problems on impulse that I have seen, it's impossible because the problems always have to give some information about the momentum pre and post collision, never only pre collision.

What am I failing to understand?

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  • $\begingroup$ The formula given is valid only for a perfectly inelastic collision.Do you know about elastic collision? $\endgroup$ – Rajath Krishna R May 6 '16 at 13:30
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    $\begingroup$ Your second sentence is easily shown wrong by personal observation - bounce a ball off the ground. $\endgroup$ – Jon Custer May 6 '16 at 13:46
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Often when two object collide it is often represented as an instantaneous impulse exchange. However in reality this happens continuously. Namely both objects are not completely rigid and will deform during the collision, storing energy in the elastic deformation (like a spring) and dissipating energy with any inelastic deformation.

During such a collision there will indeed be an instant at which both masses will have relative velocity of zero, but any elastically stored energy will push the two masses away from each other. Only if all the kinetic energy relative to the center of mass is dissipated by inelastic deformation, then there will be no elastic energy to push the mass apart from each other and the two masses will have the same velocity.

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  • $\begingroup$ If they are completely rigid, will my hypothesis be correct? $\endgroup$ – Amanda May 10 '16 at 23:07
  • $\begingroup$ @Amanda Maybe, because there is no way to test this, since every thing in the real world has a finite stiffness. $\endgroup$ – fibonatic May 10 '16 at 23:14
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I come to think that when a mass collides with another, both of them should always have equal velocities post-collision.

To paraphrase Feynmann, no matter how beautiful you may believe your reasoning to be about this, if it doesn't agree with experiment, it is wrong.

If the two objects 'stick together' after the collision (the collision is totally inelastic), then both objects must, of course, have the same velocity.

how can any change in momentum still occur after they have equal velocities?

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What is conserved during a collision is not velocity, but momentum (mass times velocity).

If the collision is elastic, kinetic energy is also conserved: https://en.wikipedia.org/wiki/Elastic_collision.

If the collision is inleastic, only momentum will be conserved: https://en.wikipedia.org/wiki/Inelastic_collision

The resulting equations (which you can find on the Wikipedia pages above) are easy enough to solve.

Let's take your example: if a mass moving with momentum $m_1 u_1$ hits a mass $m_2$ at rest, in the case of an elastic collision, the final velocities will be

$$v_1 = \frac {u_1 (m_1 - m_2)} {m_1+m_2}$$

$$v_2 = \frac {2 m_1 u_1} {m_1+m_2}$$

Note that if $m_1 = m_2$, mass number 1 will stop completely and transfer all its momentum to mass number 2 ($v_1 = 0$ and $v_2=u_1$).

On the other hand, f the collision is inelastic, only momentum will be conserved, so that

$$m_1 u_1 = m_1 v_1 + m_2 v_2$$

In this case we can't really tell what the final velocities will be (we have one equation with two unknowns): it will depend on how much energy is lost during the collision.

The magnitude of the force acting during the collision cannot be trivially computed, and will depend on many things, including the material the masses are made of.

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Consider two objects, 1 and 2, colliding for some short time interval $\delta t$. During $\delta t$ let's ignore all forces except the contact force that 1 exerts on 2, $\vec{F}_{12}$ and the contact force that 2 exerts on 1, $\vec{F}_{21}$.

As long as the objects touch each other, both of these forces exist, and by the principle of Newton's 3rd Law we know that $\vec{F}_{12}= -\vec{F}_{21}.$ This is true even if the forces are not constant.

Now, a force acting on an object for a period of time will change the momentum of the object. We call that change in momentum the impulse, $\vec{J}$. Technically and mathematically, we would write $$\vec{J}_{12}=\int_0^{\delta t}\vec{F}_{12}\;dt$$ and we could write a similar statement for $\vec{J}_{21}$. Looking at those statements we can confidently say that the impulse that 1 produces on 2 is equal in magnitude and opposite in direction to the impulse that 2 produces on 1. Simplistically, whatever momentum is lost by 1 is gained by 2, so the net momentum change in the 1-2 system is zero during $\delta t$.

This true regardless of mass, stickiness, elasticity, etc. The size of the impulse can be a function of masses, materials, shapes, stickiness, elasticity, etc., but both objects will experience the same magnitude impulse.

The change in individual velocity will depend on the impulse and individual mass. A more massive object will experience a small velocity change.

If they don't stick together they cannot have a common velocity afterwards (by definition). It is possible they could have the same speed, but different directions.

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