4
$\begingroup$

Okay, suppose 2 identical bodies/particles(i.e., of equal masses) moving with same velocities approach each other and have an elastic collision.

Now, as there is elastic collision in between them, both of them will rebound and move in opposite direction with same velocity.(Source: Directs to Quora(Refer to Mohd. Faiz Answer))

enter image description here

Here's the doubt: I don't think so there could be any elastic collision when 2 identical bodies with same speed moves towards each other.

Why there is such doubt in my mind? :-

I know I stated Elastic Collision and thus, so to conserve Kinetic Energy, they should rebound with same velocity.

Now, please read this clearly.

Consider the same identical bodies/particles, A and B approaching each other with same velocity and have an elastic collision. As there is elastic collision, during time of collision, there will be conservative forces acting on the bodies.

Now equal and opposite conservative forces $F_{AB}$ and $F_{BA}$ acts on bodies A and B respectively. The conservative forces will make the bodies to rebound and thus, the bodies will go away from each other, i.e, the bodies will move in direction of conservative forces.

Hence, both the conservative forces $F_{AB}$ and $F_{BA}$ will do Positive Work on particles A and B respectively.

As per Work-Energy Theorem, Work(Force)=Change in Kinetic Energy of the Body

So, Positive Work By Conservative Forces = Increase in Kinetic Energies of bodies

In other words, $u_A < v_A$ and $u_B < v_B$ such that Initial Total KE < Final Total KE

But this couldn't be possible if the collision is Elastic as, in Elastic Collision Total Kinetic Energy remains conserved.

So, where am i going wrong?

Please correct me.

Note: I know in reality there cannot be any Elastic Collision at macroscopic level. But here i have assumed that the bodies are perfectly rigid and no ext. force is acting on them. Also the mechanical energy do not converts into any forms

$\endgroup$
  • $\begingroup$ You realize that those conservative forces that you are talking about are applied on body for a Rally really short time, that's the reason we call the impact of such large forces with such a short interval of time by the name "Impulse" $\Delta (p)$ $\endgroup$ – The Dead Legend May 13 '17 at 11:28
  • 2
    $\begingroup$ Note: You don't need to assume perfectly rigid object to have ideal elastic collisions. A tennis ball hitting a wall is also an elastic collision, since the deformation that happens in the ball is not permanent but returns to exactly the initial state. $\endgroup$ – Steeven May 13 '17 at 13:12
12
$\begingroup$

Conservative forces don't necessarily do positive work. Gravity is a conservative force, and if you throw a ball upwards, gravity does negative work by pulling the ball down, thus slowing it down until it reaches its maximum height. Only on the way down does gravity do positive work to speed the ball up. A conservative force is one that, if an object under its influence starts at a certain location with a certain kinetic energy, it will have the same kinetic energy every time it returns to that same location. In the gravity example, the ball will have the same kinetic energy with which it was initially thrown when the thrower catches it because the ball has returned to its initial height.

Same thing with the elastic collision. The collision happens over a finite time $\Delta t$. During the first half of the collision--from $t = 0$ to $t = \Delta t/2$--the force between the balls does negative work to bring the balls to a stop. At this point, the kinetic energy has been converted into the elastic deformation of the balls (elastic in the spring sense of the word). For the second half of the collision--from $t = \Delta t/2$ to $t = \Delta t$--the forces between the balls do positive work to accelerate the balls away from each other. Since the collision force is elastic, the balls have the same kinetic energy going out that they did going in.

The diagram in your question only shows the second half of the collision.

If we're talking about particles, the replace the elastic deformation energy with the potential energy of the relevant forces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.