Elastic Collisions & Center of Mass


by kiki_havoc
Tags: bumper cars, center of mass, elastic collision
kiki_havoc
kiki_havoc is offline
#1
Nov4-11, 11:40 PM
P: 2
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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PeterO
PeterO is offline
#2
Nov5-11, 02:25 AM
HW Helper
P: 2,316
Quote Quote by kiki_havoc View Post
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.
logics
logics is offline
#3
Nov5-11, 05:40 AM
P: 137
Quote Quote by kiki_havoc View Post
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 general formula

PeterO
PeterO is offline
#4
Nov5-11, 06:58 AM
HW Helper
P: 2,316

Elastic Collisions & Center of Mass


Quote Quote by logics View Post
that formula applies when m2 is at rest: v'1→ = 4 x 0.034482758, you should add v'2
or use the general formula
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.
kiki_havoc
kiki_havoc is offline
#5
Nov5-11, 09:57 PM
P: 2
Thank you! Your replies really helped clarify the concept for me.


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