Why Do Objects in Elastic Collisions Sometimes Share or Transfer Velocities?

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Homework Help Overview

The discussion revolves around elastic collisions, specifically addressing why objects can sometimes share velocities or transfer momentum during such interactions. The original poster expresses confusion regarding the conditions under which two objects of equal mass behave similarly in collisions, particularly when one is stationary.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the conditions of elastic collisions, questioning how initial and final velocities relate to the outcomes. Some inquire about the implications of mass and momentum transfer when one object is stationary.

Discussion Status

Participants are actively discussing the principles of conservation of momentum and kinetic energy in elastic collisions. Some have suggested clarifying initial and final conditions to better understand the scenarios presented. There is an ongoing exploration of the equations governing these interactions, but no consensus has been reached.

Contextual Notes

There are mentions of specific conditions such as equal mass and one object being stationary, which are central to the discussion. Participants are also considering the implications of these conditions on the outcomes of the collisions.

Biker
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Homework Statement


So I have just studied this topic and it seems a bit confusing to me.
Lets just say we have elastic collision, Why sometimes both of the objects move at the same speed? and sometimes if one object is stationary and the other object hits it ( They have the same mass) why does it give all its velocity to the 2nd objects.
I have found something that says:
If m1 = m2 and one of them are stationary (m2) then m2 takes m1's momentum.
Like I just want to know why? Is there is something that I can know about this?

2. Homework Equations .
P = mv
 
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Biker said:
Why sometimes both of the objects move at the same speed? and sometimes if one object is stationary and the other object hits it ( They have the same mass) why does it give all its velocity to the 2nd objects.
Could you state the initial, and final conditions clearly?
 
All of these results can be obtained from the respective equations. The equations for the conservation of energy and conservation of kinetic energy give you these results. The best thing would be to obtain the final velocities of the two objects. This will solve most of your doubts in this regard. This is easy for one-dimensional collisions. The results hold in general.
 
Biker said:

Homework Statement


So I have just studied this topic and it seems a bit confusing to me.
Lets just say we have elastic collision, Why sometimes both of the objects move at the same speed?
I'm not sure what you mean by that. You have to be given the initial speeds of the two objects and, of course, they might happen to have the same. If you are referring to after a collision, the speeds are given by "conservation of mass" and "conservation of (kinetic) energy".

and sometimes if one object is stationary and the other object hits it ( They have the same mass) why does it give all its velocity to the 2nd objects.
I have found something that says:
If m1 = m2 and one of them are stationary (m2) then m2 takes m1's momentum.
Like I just want to know why? Is there is something that I can know about this?

2. Homework Equations .
P = mv
That's the equation for momentum. If object 1 has mass m and speed v, then it has momentum mv but also has kinetic energy (1/2)mv^2. If object 2 has the same mass but initial speed 0, it has momentum and kinetic energy 0. The total momentum is mv and the total kinetic energy is (1/2)mv^2. Now suppose that, after the collision, they have speeds v1 and v2. The total momentum is now mv1+ mv2= m(v1+ v2) and the total kinetic energy is (1/2)mv1^2+ (1/2)mv2^2= (1/2)m(v1^2+ v2^2). By "conservation of momentum" we have m(v1+ v2)= mv and by "conservation of energy" we have (1/2)m(v1^2+ v2^2)= (1/2)mv^2.

Those are two equations to solve for the two speeds, v1 and v2. An obvious first step in the momentum equation is to divide both sides by m to get v1+ v2= v and divide both sides of the energy equation by (1/2)m to get v1^2+ v2^2= v^2. From the first equation, v2= v- v1 and replacing v2 by that in the second equation, v1^2+ (v^2- 2vv1+ v1^2)= 2v1^2- (2v)v1+ v^2= v^2 so we have the quadratic equation 2v1^2- (2v)v1= 2v1(v1- v)= 0. That two solutions- either v1= 0 so that v2= v or v1= v so that v2= 0. The first is the situation in which the first object hits the second, stopping and the second object moves off with the same speed. The second is the situation in which the first object doesn't hit the second object at all!
 

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