Elastic Collisions & Center of Mass

In summary, the two bumper cars have an elastic collision with the velocity of the center of mass remaining constant throughout. The final velocity of car 1 in the ground reference frame can be calculated using the equation V1f=V1i*(m1-m2/m1+m2), and the same can be applied for car 2 in the ground reference frame. In an inelastic collision, the final velocity of the two cars can be calculated using the same equation, but with the added velocity of the second car. The reason for the difference in kinetic energy between elastic and inelastic collisions is due to the conservation of energy in elastic collisions, while energy is not conserved in inelastic collisions.
  • #1
kiki_havoc
2
0
1. A bumper car with mass m1 = 105 kg is moving to the right with a velocity of v1 = 4 m/s. A second bumper car with mass m2 = 98 kg is moving to the left with a velocity of v2 = -3.8 m/s. The two cars have an elastic collision. Assume the surface is frictionless.
2.Vcm= (m1v1+m2v2)/m1+m2
V1 in CM frame=V1i-Vcm


1. What is the velocity of the center of mass of the system? 0.234 m/s, solved using velocity of center of mass eqn Vcm.

2. What is the initial velocity of car 1 in the center-of-mass reference frame? 3.77 m/s, solved using V1 in CM frame=V1i-Vcm.

3. What is the final velocity of car 1 in the center-of-mass reference frame? I guessed this one, since it is 'perfectly' elastic, the velocity in the reference frame would be the same magnitude, opposite direction (I think...) so = -3.77m/s

4. What is the final velocity of car 1 in the ground (original) reference frame?
this one I'm having issues with: I've been following an example for this type of question in my textbook, and it gives the eqn V1f=V1i*(m1-m2/m1+m2). I have tried this several times and am not getting the answer, and so am very confused.

5. What is the final velocity of car 2 in the ground (original) reference frame? I assume this is a similar question to #4.

6. In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide.
What is the final speed of the two bumper cars after the collision? haven't attempted yet.

7. why would the |ΔKE elastic| > |ΔKE inelastic| ? I guessed this one and got it right, but don't understand why.Any help would be greatly appreciated! Thanks for reading!
 
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  • #2
kiki_havoc said:
1. A bumper car with mass m1 = 105 kg is moving to the right with a velocity of v1 = 4 m/s. A second bumper car with mass m2 = 98 kg is moving to the left with a velocity of v2 = -3.8 m/s. The two cars have an elastic collision. Assume the surface is frictionless.



2.Vcm= (m1v1+m2v2)/m1+m2
V1 in CM frame=V1i-Vcm





1. What is the velocity of the center of mass of the system? 0.234 m/s, solved using velocity of center of mass eqn Vcm.

2. What is the initial velocity of car 1 in the center-of-mass reference frame? 3.77 m/s, solved using V1 in CM frame=V1i-Vcm.

3. What is the final velocity of car 1 in the center-of-mass reference frame? I guessed this one, since it is 'perfectly' elastic, the velocity in the reference frame would be the same magnitude, opposite direction (I think...) so = -3.77m/s

4. What is the final velocity of car 1 in the ground (original) reference frame?
this one I'm having issues with: I've been following an example for this type of question in my textbook, and it gives the eqn V1f=V1i*(m1-m2/m1+m2). I have tried this several times and am not getting the answer, and so am very confused.

5. What is the final velocity of car 2 in the ground (original) reference frame? I assume this is a similar question to #4.

6. In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide.
What is the final speed of the two bumper cars after the collision? haven't attempted yet.

7. why would the |ΔKE elastic| > |ΔKE inelastic| ? I guessed this one and got it right, but don't understand why.


Any help would be greatly appreciated! Thanks for reading!


The key here is that the c of m has the same velocity throughout - before, during and after the collision.
 
  • #3
kiki_havoc said:
4...I've been following an example... V1f=V1i*(m1-m2/m1+m2). I have tried this several times and am not getting the answer
that formula applies when m2 is at rest: v'1→ = 4 x 0.034482758, you should add v'2
or use the http://wikipedia.org/wiki/Elastic_collision" [Broken]
 
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  • #4
logics said:
that formula applies when m2 is at rest: v'1→ = 4 x 0.034482758, you should add v'2
or use the http://wikipedia.org/wiki/Elastic_collision" [Broken]

I liked the look of parts 1,2&3. Your reasoning was correct for part 3.

4, involves the reverse of the process you used for part 2.

5 is indeed the same working as 4. having lead you through the three steps [2,3,4] they have just asked for the final answer - which you do by using the equivalent three steps.

6. Think about my first comment, and your answer to part 1.

7. In an elastic collision the energy is conserved. In an inelastic collision it is not - it is different. You can never get more energy after, than you had before - there is only one alternative left.
 
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  • #5
Thank you! Your replies really helped clarify the concept for me.
 

1. What is an elastic collision?

An elastic collision is a type of collision in which the total kinetic energy of the system is conserved. This means that the total energy before the collision is equal to the total energy after the collision. In other words, there is no loss of energy in an elastic collision.

2. What is the difference between elastic and inelastic collisions?

In an elastic collision, the total kinetic energy of the system is conserved, while in an inelastic collision, some of the kinetic energy is lost in the form of heat, sound, or deformation. In other words, the total energy after an inelastic collision is less than the total energy before the collision.

3. How is the center of mass related to elastic collisions?

The center of mass is the point at which the total mass of a system is considered to be concentrated. In an elastic collision, the center of mass of the system remains the same before and after the collision, assuming there are no external forces acting on the system.

4. Can the center of mass move in an elastic collision?

No, the center of mass cannot move in an elastic collision because the total momentum of the system is conserved. This means that the total mass and velocity of the system remains the same before and after the collision, and therefore the center of mass remains in the same position.

5. How is the coefficient of restitution related to elastic collisions?

The coefficient of restitution is a measure of how much kinetic energy is lost in a collision. In an elastic collision, the coefficient of restitution is equal to 1, meaning that there is no loss of kinetic energy. In other words, the objects bounce off each other with the same speed they had before the collision.

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