Question

A track consists of a frictionless arc XY, which is a quarter-circle of radius R, and a rough horizontal section YZ, as shown above.

In trial 1, Block A of mass M is released from rest at point X, slides down the curved section of the track, and collides instantaneously and inelastically with block B, with mass M, at point Y. The two blocks move together to the right, sliding past point P, which is a distance L from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is µ.

In trial 2, Block A of mass M is released from rest at point X, slides down the curved section of the track, and collides instantaneously and inelastically with block B, with mass 2M, at point Y. The two blocks move together to the right, sliding past point P, which is a distance L from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is 2µ

Indicate whether the speed of the blocks relative to the rough horizontal track section YZ is greater in trial 1 or in trial 2. Justify your answer in a clear, coherent, paragraph length explanation.

M.Block A Block B HK = 0 Hk = Side View n Block A Block B Uk = 0 2M Hk = 24 - Side View

Answer 1

ANSWER

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 before 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 total Mass M , Let y velocity just when they stast nocing. By lave of conservatwo of momentum. Momentum before a Momentum after Mr = Many, My = 2MV. - N = Y = Sage KE at the start = 2x2mW: CK E) when both the blockes nove together for a distance, L on a track with coefficient of friction I, the frictional Къto IX 2 м) х = 2 Mма, work 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 energy therem, change 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 done = 64 ML. Final Kinetic Energy = ₂ x 3 mx 4 = 3 By woule - endgy throum, KEI~K Ez a work done. KE - KE work 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 final velocity is considerably reduced. This is due to the reason that, the fictional force is increased due to the increase is omass and coefficient of friction. Hence the average velocity as the second trial is less when compared to the Frost Frial.