Elastic Collision in One Dimension Part A This collision is elastic. What quanti

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Elastic Collision in One Dimension
Part A This collision is elastic. What quantities, if any, are conserved in this collision? This collision is elastic. What quantities, if any, are conserved in this collision? kinetic energy only
momentum only
kinetic energy and momentum SubmitHintsMyAnswersGiveUpReviewPart Part A This collision is elastic. What quantities, if any, are conserved in this collision? This collision is elastic. What quantities, if any, are conserved in this collision? kinetic energy only
momentum only
kinetic energy and momentum SubmitHintsMyAnswersGiveUpReviewPart This collision is elastic. What quantities, if any, are conserved in this collision? kinetic energy only
momentum only
kinetic energy and momentum SubmitHintsMyAnswersGiveUpReviewPart Part B SubmitHintsMyAnswersGiveUpReviewPart Part C SubmitHintsMyAnswersGiveUpReviewPart Elastic Collision in One Dimension Block 1, of mass, moves across a frictionless surface with speed. It collides elastically with block 2, of mass, which is at rest ().(Figure 1)After the collision, block 1 moves with speed, while block 2 moves with speed. Assume that, so that after the collision, the two objects move off in the direction of the first object before the collision.
Part A This collision is elastic. What quantities, if any, are conserved in this collision? This collision is elastic. What quantities, if any, are conserved in this collision? kinetic energy only
momentum only
kinetic energy and momentum SubmitHintsMyAnswersGiveUpReviewPart Part B What is the final speedof block 1? Expressin terms of,, and. = SubmitHintsMyAnswersGiveUpReviewPart Part C What is the final speedof block 2? Expressin terms of,, and. = SubmitHintsMyAnswersGiveUpReviewPart

Explanation / Answer

Linear momentum is always conserved.

since, collision is elastic threfore, kinetic energy is also conserve.

hence,

Part A: kinetic energy and momentum

Part B.

from the linear momentum conservation,

m1ui = m1uf + m2vf ....(1)

from the energy conservation,

(1/2)*m1ui2 = (1/2)m1uf2 +(1/2)m2vf2

m1ui2 = m1uf2 +m2vf2 ..(2)

solving equation (1) and (2),

m1ui2 = m1uf2 +m2((m1ui-m1uf)/m2)2 = [m2m1uf2 +m12(ui-uf)2]/m2

m2m1ui2 = m2m1uf2 +m12(ui-uf)2

m2(ui2-uf2)=m1(ui-uf)2

solving this, equation, we can get,

uf = [(m1-m2)/(m1+m2)]*vi

and

Part C

after puting the value of uf in equation (1),we can get;

vf = [2m1/(m1+m2)]vi

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