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I am having a tough time grasping the concept that how is the work done during different types of collisions. I will not take the collision between point particles because, in reality, the bodies deform during collisions.

1.) Perfectly Elastic Collision:- Let's take two balls $A$ and $B$ having different masses, $m$ and $m'$ respectively. Ball $A$ is moving with a constant velocity, $v$ and $B$ is moving with a constant velocity, $v'$ towards each other. The two balls collide and bounce back and start moving with some velocities (Let it be $v''$ for ball A and $v'''$ for Ball B) but in opposite directions.

What is the work done by Ball A on Ball B and by Ball B on Ball A?

I think that Ball B does a work of $\frac{1}{2}mv''^2 - \frac{1}{2}mv^2$ on Ball A.

Its magnitude would be equal to the transfer of kinetic energy from Ball A to Ball B, wouldn't it?

Similarly, Ball A does a work of $\frac{1}{2}mv'''^2 - \frac{1}{2}mv'^2$ on Ball B.

What is the net work done?

As the Total Kinetic Energy of a system remains conserved, there will be no net work.

2.) Partial Inelastic Collision:- I can calculate the work done by Ball $A$ on Ball $B$ and by Ball $B$ on Ball $A$ by using the same principle as above (calculate the change of $KE$ for each ball). Suppose after the collision both the balls started moving in the same direction (Let it be the direction of the initial velocity of Ball $ A$).

Then the transfer of $KE$ from Ball $A$ to Ball $B$ would be equal to the magnitude of work done by Ball $B$ on Ball $A$, wouldn't it?

What is the net work done?

Again, the Net Work, in this case, would be the Total change in KE of the system.

3.) Perfectly Inelastic Collision:- This time the balls $A$ and $B$, after the collision, stick together and start moving with the same velocity, $V$ in the direction of the initial velocity of Ball $A$.

What would be the Work done on each ball and the Net Work done?

Again the work is done on each ball would be the change in KE of each ball and the Net Work would be the Total change in $KE$ of the system. Am I right?

What if the balls stick together and don't move after the collision (Is this even possible?).

It would mean that there is no transfer of $KE$ from one ball to another and thus there would be no work done by one ball on the other ball.

Would there still be some Net Work done?

I think the Net Work would still be equal to Total Change in KE of the system.

Let's suppose that Ball $B$ is at rest initially and Ball $A$ comes with a velocity, $v$ towards Ball $B$.

After the collision both the balls stick together and don't move at all (Is it possible?).

Again, there is no transfer of $KE$ from Ball $A$ to Ball $B$ so there would be no work done on either ball by the other balls. Still, there would be some Net Work that would be equal to the Total change in $KE$ of the system. Correct me if I am wrong.

What would happen in all the above cases if the Balls now move with some acceleration, $a$ and $a'$ respectively. Would there be any effect on the "Work done"?

In the above cases, I saw Work done as a change in $KE$. But in reality, the Work is equal to $F.dx$. So, where are the displacement and the force in the above cases?

Is the displacement, deformation undergone by the balls for a brief amount of time when they collide?

How can we find the force? In the above cases, no external force was applied (That's why the balls are moving with constant velocities.) How does force come into the picture?

I would want you to tell me how the work would be done in all the above cases and correct my concepts. It would be highly appreciated.

Note:- The parts written in blockquotes are my main Queries. Please try to answer them. Sorry for my long Question.

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Two things might be helpful here.

First - collisions are always easier to look at in the center of mass frame (that is, from the point of view of an observer who is traveling at the velocity of the center of mass). If you have two objects with mass $m_1, m_2$ and velocities $v_1, v_2$ then the center of mass moves at velocity

$$v_{com}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$

Second - during an elastic collision, you can consider there being a small spring between the masses: in the center of mass frame all kinetic energy will convert to potential energy of the spring (since the two masses will be stationary at one instant) after which it is "given back" to the masses (and each gets the same amount of energy as before, in this frame of reference).

Now if we have net motion (we look in a different frame of reference) then the mass that is moving in the same direction as the center of mass is doing more work during compression (it is pushing on a spring that is moving away from it). This means that that object will end up losing energy (doing work on the other one), and vice versa. The magnitude of the work done by A on B can be found by looking at the kinetic energy before and after the collision.

When you include losses during the collision the same kind of analysis can be used - but now part of the work done "during compression of the spring " is lost. This means you first need to figure out what fraction of the "after" energy is due to the mass itself (by looking at its velocity in the center of mass frame after the collision and the energy it lost during the collision). The difference is due to work done by the other mass.

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1)Perfectly Elastic Collision: In an elastic collision the bodies just exchange their momentum. If we consider Work = Force x displacement the displacement of bodies during collision time is also zero. So No Work done. The subtraction of kinetic energy which every body gain after collision I think that does not correspond to the work done by collision.

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Perfectly Inelastic Collision: In this case there is a collision time (t) of each body and also a displacement-deformation of each body in the given time t.In each fragment of deformation correspond a Force. The integrate of Force x displacement will give the Work done during collision.

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  • $\begingroup$ Welcome to our site. If you want to change your answer you should "edit", not post a new one... a single question (even if it has multiple parts) should be answered in a single answer. $\endgroup$ – Floris Oct 8 '17 at 11:26

protected by Qmechanic Oct 8 '17 at 12:59

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