You are watching: A track consists of a frictionless arc xy
In trial 1, Block A of mass M is released from remainder at pointX, slides down the curved area of the track, and also collidesinstantaneously and also inelastically via block B, through mass M, atpoint Y. The two blocks relocate together to the right, sliding pastsuggest P, which is a distance L from allude Y. The coeffective ofkinetic friction between the blocks and the horizontal part of thetrack is µ.
In trial 2, Block A of mass M is released from rest at pointX, slides dvery own the curved area of the track, and collidesinstantaneously and inelastically with block B, with mass 2M, atsuggest Y. The 2 blocks move together to the appropriate, sliding pastsuggest P, which is a distance L from suggest Y. The coreliable ofkinetic friction between the blocks and the horizontal component of thetrack is 2µ
Indicate whether the rate of the blocks family member to the roughhorizontal track area YZ is better in trial 1 or in trial 2.Justify your answer in a clear, meaningful, paragraph lengthexplanation.
Given Dalas Mass of Block A = M. Radius of circular track = R, ilinitial potential energy of the block = MgR. Trial 1 When block A deaches Y and just prior to collision usthy B, let its velocity be ve By lane of conservatwo of enegy PE at X = KE at Y a. Mg R = I MN² After collision, the blocks sticks together, .. The full Mass M , Let y velocity simply once they stast nocing. By lave of conserva2 of momentum. Momentum prior to a Momentum after Mr = Many type of, My = 2MV. - N = Y = Sage KE at the begin = 2x2mW: CK E) once both the blockes nove together for a distance, L on a track through coefficient of friction I, the frictional Къto IX 2 м) х = 2 Mма, job-related done by frictional force a FXL = 2 MMgXL, Let ₂ the velocity when they reaches P. . Kinetic Energy a 2 x 20 x 2 - M (KE) By work-related energy therem, adjust in KE = Workolone. KEI - KE2 - Work done. .: KEz=KE, - Wook done. = Mgr. - 24mg xL in MV - Mgr - 24 Mg x L
12 = &R - 4 ?. 2 a SR - GUOL Trial 2 v= 2gR After collision total mansa Mtan = 3M 3 MV23MV, >> Ni - Ž - zgle KE at start = I x 3m xv, ² = 1 *3 mx 2g R a Mgr Here coefficrent of friction - 24. Frictional Force = 24x3 Mg = 6 lung. work-related done = 64 ML. Final Kinetic Energy = ₂ x 3 mx 4 = 3 By woule - endgy throum, KEI~K Ez a work-related done. KE - KE job-related done = - LAML ve a Mol - 6MMSL 32 = 3 2 - 6 Mg L = gr - 18 Mg L. 3 . : V2² = 2R - 36 MgL In the sea N2 = 1 / 2g R -36 MBL wad trial, the last velocity is significantly diminished. This is due to the reason that, the fictional force is raised due to the increase is omass and coeffective of friction. Hence the average velocity as the second trial is less as soon as compared to the Frost Frial.
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