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Given these two identical objects, if one is stationary, and the centre of mass of the other object collides head on with the centre of mass of the object that is stationary, i.e it does not come into contact at an angle, will the object that was initially moving come to a stand still, or continue to move in the direction it was going in?

You see, I wonder, given Newton's third law, would the equal and opposite force completely counteract the initial force of the moving object?

Furthermore, would the stationary object somewhat 'take', or 'absorb' all the force of the moving object upon contact?

Remember, I am presuming these objects are identical to simplify matters, which means they have equal mass.

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  • $\begingroup$ Are the objects rigid, that they are "absolutely hard" and doesn't change shape upon contact? Note that even diamond changes shape under external force, though so slightly that nobody can see. $\endgroup$
    – busukxuan
    Dec 20, 2015 at 3:49
  • $\begingroup$ Sure, why not? Whatever it takes to run this mental simulation :) $\endgroup$
    – user108262
    Dec 20, 2015 at 3:51

3 Answers 3

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The fact that the objects are in vacuum has very little to do with Newton's Law.

Instead, as always there will be an exchange of momentum; exactly how much momentum is exchanged depends on whether the collision is elastic or inelastic (most collisions are somewhere in between the two...)

Newton's law can be restated as "the change in momentum of one object is equal and opposite to the change in momentum of the other object". But to get the final velocities, you need to know the mass (and the energy after the collision).

It is usually helpful to analyze collisions in the center of mass frame. Since the two objects in your example have the same mass (you said they are identical), the center of mass moves at half the velocity of the incoming particle. In that frame of reference, one particle appears to come from the left at $v/2$, and the other comes from the right at the same velocity.

In a perfectly inelastic collision, they will hit each other and stick: all relative velocity is gone. So if they have $v'=0$ in a frame moving at $v/2$ then their final velocity is $v/2$ for both.

In a perfectly elastic collision, they will bounce off each other and continue with the speed they started with: so the particle that was initially stationary, and therefore was moving at $v/2$ to the right in the c.o.m. frame, ends up moving with $v/2$ to the left; transforming back to the lab frame, it will have velocity $v$. The other particle started out with $v$, and by a similar argument ends up stationary.

Now if some energy is lost during the collision, then the final velocities in the c.o.m. frame will not be $±v/2$ - they will be slightly less (but still equal and opposite). That means in turn that the velocity of the first particle will not be reduced completely to zero, and the second particle will not get all the energy/momentum of the first.

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Assuming that these two objects will not connect with each other. According to the theorem of momentum:

m1v0=m1v1+m2v2

For the momentum of an object is a vector, the equation should be placed in 2 directions(x-axis and y-axis)

The theorem of momentum is applied on condition that there's no external force acting on the objects(or the time is short that the impulse of the external force can be neglected)

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Elastic Collisions:

Conservation of momentum of object 1 and object 2: \begin{equation} (m_1v_1)_{before} = (m_1v_1 + m_2v_2)_{after} \end{equation} Conservation of kinetic energy (elastic collision assumption): \begin{equation} (\frac{1}{2}m_1v_1^2)_{before} = (\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2)_{after} \end{equation} Since equal mass, momentum equation becomes: \begin{equation} (v_1)_{before} = (v_1 + v_2)_{after} \end{equation} and kinetic energy equation becomes: \begin{equation} (v_1^2)_{before} = (v_1^2 + v_2^2)_{after} \end{equation} We know $v_1$ before the collision. So we have 2 equations and 2 unknown velocities after the collision.

Solve these equations to get the velocities after the collision. \begin{equation} (v_1)_{after} = 0 \end{equation} \begin{equation} (v_2)_{after} = (v_1)_{before} \end{equation} For instance if the moving object was moving at $1$ $m/s$ before the collision, then it will completely stop and the second object will start moving at $1$ $m/s$.

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