Finding the final velocity using momentum and KE

Click For Summary

Homework Help Overview

The discussion revolves around an elastic collision involving two masses, where one mass is initially at rest and the other is moving. Participants are exploring the implications of momentum and kinetic energy conservation in determining the final velocities of the masses post-collision.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the correctness of the momentum and kinetic energy equations presented, particularly regarding the direction of motion after the collision. There is exploration of whether mass 2 changes direction and how mass ratios affect outcomes. Some participants suggest that the equations should be solved to determine the final velocities rather than assuming directions beforehand.

Discussion Status

There is an ongoing exploration of the relationships between mass, velocity, and direction. Some participants have provided guidance on how to approach the conservation equations, emphasizing the need to solve for final velocities before determining direction. Multiple interpretations of the collision dynamics are being discussed, with no explicit consensus reached yet.

Contextual Notes

Participants are working under the constraints of an elastic collision scenario and are considering the implications of mass ratios on the outcomes. There is an acknowledgment of the ambiguity in determining directionality of velocities prior to solving the equations.

lonewolf219
Messages
186
Reaction score
2
Before the collision, there are two masses. Mass 1 is at rest and mass 2 is moving. No friction. It is an elastic collision and mass 2 is moving in positive x direction.

After the collision, is it correct that mass 1 moves in positive x direction but mass 2 rebounds and moves in negative x direction? If so, is this the equation for momentum:

m_{2}v_{2i}=m_{1}v_{1f}-m_{2}v_{2f}

And the equation for KE:

_{2}(v_{2i})^2=m_{1}(v_{1f})^2+m_{2}(v_{2f})^2

And if solving for v_{1f} :

v_{1f}=\sqrt{\frac{m_{2}(v_{2i})^2}{m_{1}}}

And the velocity of this mass should be in the positive x direction.

Does everything look correct?
 
Physics news on Phys.org
lonewolf219 said:
is this the equation for momentum:

m_{2}v_{2i}=m_{1}v_{1f}-m_{2}v_{2f}
Does the RHS represent total momentum after collision? Doesn't look like it.
 
Thanks Haruspex... Does mass 2 then not change direction and not travel along negative x? Mass 2 is twice the mass as mass 1.
 
Last edited:
lonewolf219 said:
Thanks Haruspex... Does mass 2 then not change direction and not travel along negative x? Mass 2 is twice the mass as mass 1.
What do the equations tell you? You shouldn't need to be taking any square roots, thereby introducing ambiguity in sign. If you handle the conservation equations correctly you can obtain a simple ratio between relative velocities. Combined with the mass ratio you can then determine all velocities in terms of the initial velocity of mass 2.
 
Thanks for pointing out the proper method. So since mass 1 has half the mass of mass 2, mass 2 does not rebound and change directions? What if the masses were switched? If mass 1 were moving and collided with the more massive mass 2 at rest, would mass 1 change direction in this case?
 
lonewolf219 said:
Thanks for pointing out the proper method. So since mass 1 has half the mass of mass 2, mass 2 does not rebound and change directions? What if the masses were switched? If mass 1 were moving and collided with the more massive mass 2 at rest, would mass 1 change direction in this case?
Is that what the equations tell you? Please post your working.
 
lonewolf219 said:
Thanks Haruspex... Does mass 2 then not change direction and not travel along negative x? Mass 2 is twice the mass as mass 1.
It is not a matter of "changing direction" or "not changing direction". Total momentum is always m_1v_1+ m_2v_2. Whether or not a mass changes direction is signaled by the sign of its v. If mass 2 is twice mass 1 and has initial velocity v_i then "conservation of momentum" is 2mv_i= mv_1+ 2mv_2 where v_1 is the velocity after the collision of mass 1 and v_2 is the velocity after the collision of mass 2.
"Conservation of kinetic energy" is (1/2)2mv_i^2= (1/2)mv_1^2+ (1/2)(2m)v_2^2. Of course,, you can divide both equations by "m" eliminating that. You will have v_1+ 2v_2= 2v_i and v_1^2+ 2v_2^2= 2v_i^2. You could then, for example, solve the first equation for v_1= 2v_i- 2v_2 and replace v_1 in the second equation by that to get a quadratic equation for v_2.
 
  • Like
Likes   Reactions: 1 person
Thanks Haruspex and HallsofIvy, I appreciate your help!

Haruspex has asked me more than once "Is that what the equation tells you?" Now I realize what you are saying (thanks HallsofIvy for your explanation)... you are saying that we do not know if the mass will rebound or not until we solve the equation. So, we can't choose a direction for the final velocity before we have solved for it. But if the final velocities are in opposite directions, then we would conclude that the moving mass rebounded. Is this right, Haruspex?
 
lonewolf219 said:
Thanks Haruspex and HallsofIvy, I appreciate your help!

Haruspex has asked me more than once "Is that what the equation tells you?" Now I realize what you are saying (thanks HallsofIvy for your explanation)... you are saying that we do not know if the mass will rebound or not until we solve the equation. So, we can't choose a direction for the final velocity before we have solved for it. But if the final velocities are in opposite directions, then we would conclude that the moving mass rebounded. Is this right, Haruspex?
Yes. It often happens when trying to find a vector (force, velocity, whatever) that you don't know in advance which direction it will be in. So you just choose a direction as positive, and if the equations give you a negative answer then you know it was the other way.
 
  • #10
Awesome! Thanks for your help, Haruspex... This concept is very important!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
Replies
9
Views
954
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K