In a head on collision between two particles, is it necessary that the two particles will acquire the same/common velocity for at least one instant?

Bonus question : Same question but now one of the particles is rear ending the second one.

Extra Info : It is not mentioned if the collision is elastic, semi elastic or inelastic. So I think that we need to cover all three cases.

What I'm thinking : I think that their velocities will indeed be zero (for both) for an instant.. because only then can the direction of velocity reverse. But what I'm not clear on is if they will have zero velocity at the SAME instant.

BOOK : Concepts of Physics volume 1 by H.C. Verma (Indian book). Page 156 question 8.

  • $\begingroup$ Momentum is always conserved. $\endgroup$ – Rob Jeffries Dec 24 '16 at 10:00

I assume you're considering a one dimensional collision i.e. the particle motion is always along the same straight line. If so the simple way to consider your question is to work in the centre of momentum frame where the total momentum is zero and the collision looks like:

COM frame

At the moment of collision (the middle picture) the velocities of the colliding objects are the same because they are both zero. Regardless of what initial velocities you give the particles you can always find the COM frame and the collision will always resemble the diagram above. Therefore in every collision of this type there is an instant at which the velocities of the objects are equal.

| cite | improve this answer | |
  • 3
    $\begingroup$ And the original poster should remember that when the speeds are zero, the energy is stored as potential energy (elastic deformation of balls, or electrostatic potential energy, etc). Because maybe that is what is behind this question? $\endgroup$ – Pieter Dec 24 '16 at 10:54

The velocities will be the same at one instant, but that velocity will not necessarily be zero.

Suppose particle A is moving to the right (+ve velocity) while particle B is moving to the left (-ve velocity). See velocity-time graph below. At time $t_1$ they come into contact. Their velocities change in some way. At time $t_2$ they break contact. A and B separate with A now moving to the left (-ve velocity) and B to the right (+ve velocity).

Because the velocities change continuously, there will always be some instant between $t_1$ and $t_2$ at which the velocity-time graphs intersect. At this instant the velocities of A and B are equal.

It makes no difference if the velocities change in a non-linear manner. They will still intersect. If they did not intersect, the particles would pass each other.

The second case, in which the particles are both travelling to the right with particle A "rear-ending" particle B, is just the same graph translated vertically upwards so that all velocities are above the time axis.

This is a consequence of the Intermediate Value Theorem in mathematics.

enter image description here

| cite | improve this answer | |

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.