# dynamics question help

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#1
Why is the answer D? How would we use the kinetic energy formula to derive the answeR?
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4 years ago
#2
(Original post by Revision99)
Why is the answer D? How would we use the kinetic energy formula to derive the answeR?
You can use the following approach which you have used it in another post:
relative speed of approach before collision is equal to the relative speed of separation after collision
0
4 years ago
#3
An elastic collision means that momentum and kinetic energy (KE) are both conserved. Assume particles are mass m then the total KE before the collision is 0.5mv². A and C have a total final KE of 0.25mv² so KE is not conserved so it's between B and D but logically the second particle will move so it must be D
1
4 years ago
#4
(Original post by Revision99)
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The elimination approach by shadnic is technically sound but does not offer a rigorous explanation of the velocities post-impact.

Let the left particle be particle A and the right particle be particle B. Let the initial state be denoted with 1 and the final state be denoted with 2.

Consider the conservation of (kinetic) energy:

Consider the conservation of momentum:

Equate the two expressions:

Now, , as particle B is experiencing an impulse from the collision of particle A. Hence, .

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