What Path Do Particles Take in an Inelastic Collision?

[AntMantis]

Homework Statement

The figure shows an overhead view of two particles sliding at constant velocity over a frictionless surface. The particles have the same mass and the same initial speed v = 4.00 m/s, and they collide where their paths intersect. An x-axis is arranged to bisect the angle between their incoming paths, so that q = 40.0°. The region to the right of the collision is divided into four lettered sections by the x-axis and four numbered dashed lines. In what region or along what line do the particles travel if the collision is inelastic?

http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%209.%20Center%20of%20Mass%20and%20Linear%20Momentum/Problems/c09x9_14.xform_files/nw0484-n.gif


Answer:
Here the final speeds are less than they were initially. The total x-component cannot
be less, however, by momentum conservation, so the loss of speed shows up as a
decrease in their y-velocity-components. This leads to smaller angles of scattering.
Consequently, one particle travels through region B, the other through region C; the paths
are symmetric about the x-axis.



How did they get this answer? A step-by-step proof would be greatly appreciated.

 

[tiny-tim]

Welcome to PF!

Hi AntMantis! Welcome to PF!

Answer:
Here the final speeds are less than they were initially. The total x-component cannot be less, however, by momentum conservation, so the loss of speed shows up as a decrease in their y-velocity-components. This leads to smaller angles of scattering. Consequently, one particle travels through region B, the other through region C; the paths are symmetric about the x-axis.

That answer is really confusing.

(Is it the official answer?)

If the final paths are symmetric about the x-axis, then their x-components are fixed, the same as the initial x-components.

Since the total energy is less, the y-components must be less … that's all there is to it.

Except I don't understand why the final paths have to be symmetric.

 

[AntMantis]

Thanks! Yes, that is the answer in the book, which I do not understand. So, does the total x-component have to be the same because the masses are originally traveling with the same x-velocity components? Also, how do I come to the conclusion that the final speeds are less than
they were initially? Is there a way to show this with the law of conservation of linear momentum?

 

[tiny-tim]

So, does the total x-component have to be the same because the masses are originally traveling with the same x-velocity components?

Yes, both the total x-component and the total y-component have to be the same because momentum is always conserved in collisions.

(The total y-component of course is zero.)

Also, how do I come to the conclusion that the final speeds are less than
they were initially? Is there a way to show this with the law of conservation of linear momentum?

Since the x-component is the same, if the y-component was the same or greater, then the total speed would be the same or greater, and so the energy would be the same or greater.

But we know the energy is less.