Question

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include free body diagram!!

block of mass m is positioned on the frictionless semi-circular surface with radius R as shown. It slides down the surface and undergoes a completely inelastic collision with an identical block which is at rest at the bottom. After the collision, both blocks slide up ramp that makes an angle θ with the horizontal until they come to rest. The coefficient of kinetic friction between the blocks and the ramp is μk. Find expressions for the following: (a) the block's speed immediately before the collision (in terms of g and R only), (b) the combined speed of the two blocks immediately after the collision (in terms of g and R only), of 11k, m, g, h and a trigonometric function of θ only) should be fully simplified, and don't forget to do a consistency check of some sort (c) the thermal energy generated by the blocks as they slide up the ramp (in terms (d) the height h of the blocks on the ramp when they come to rest. Your expression

Answer 1

coefficient of kinetic friction = mu_k Decrease in potential energy = Increase in kinetic energy mgR = 1/2 mv^2 v = squareroot 2gR = speed of block just before collision Momentum conservation mv = 2mv prime v prime = v/2 Friction (kinetic) on the ramp f_k = mu_k N f_k = (2 mg cos theta) mu_k length travelled on ramp l = h cosec theta work-energy theorem initial kinetic energy + work done by friction = potential energy at higher point = > 1/2 (2 m) (v prime)^2 - mu_k (2 mg cos theta) middot l = (2m) middot g middot h \r\n = > m middot v^2/4 - mu_k middot 2 mg cos theta middot h cosec theta = 2m middot gh = > 2gR/4 - 2 mu_k gh cot theta = 2gh = > h = gR/2/2g + 2 mu_k g cot theta h = R/4 (1 + mu_k cot theta)