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dynamics question help

Capture.PNG Why is the answer D? How would we use the kinetic energy formula to derive the answeR?
Original post by Revision99
Capture.PNG 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
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
Original post by Revision99
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Original post by shadnic117
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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:

12mvA12=12mvA22+12mvB22[br]  ⟹  vA12=vA22+vB22.\dfrac{1}{2}mv_{A1}^2 = \dfrac{1}{2}mv_{A2}^2 + \dfrac{1}{2}mv_{B2}^2[br]\implies v_{A1}^2 = v_{A2}^2 + v_{B2}^2.

Consider the conservation of momentum:

mvA1=mvA2+mvB2[br]  ⟹  vA1=vA2+vB2[br]  ⟹  vA12=vA22+2vA2vB2+vB22.mv_{A1} = mv_{A2} + mv_{B2}[br]\implies v_{A1} = v_{A2} + v_{B2}[br]\implies v_{A1}^2 = v_{A2}^2 + 2v_{A2}v_{B2} + v_{B2}^2.

Equate the two expressions:

vA22+vB22=vA22+2vA2vB2+vB22[br]  ⟹  vA2vB2=0.v_{A2}^2 + v_{B2}^2 = v_{A2}^2 + 2v_{A2}v_{B2} + v_{B2}^2[br]\implies v_{A2}v_{B2} = 0.

Now, vB2≠0v_{B2} \neq 0, as particle B is experiencing an impulse from the collision of particle A. Hence, vA2=0v_{A2} = 0.

Return to the conservation of momentum:

vA1=vA2+vB2[br]vA1=0+vB2[br]  ⟹  vB2=vA1.v_{A1} = v_{A2} + v_{B2}[br]v_{A1} = 0 + v_{B2}[br]\implies v_{B2} = v_{A1}.

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