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In the figure below, block 1 of mass m 1 slides from rest along a frictionless r

ID: 1777373 • Letter: I

Question

In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.95 m and then collides with stationary block 2, which has mass m2 = 2.00m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction k is 0.545 and comes to a stop in distance d within that region.

(a) What is the value of distance d if the collision is elastic?

(b) What is the value of distance d if the collision is completely inelastic?

Diana Haddad-La Fiesta (, 4) × Y e Horme l Chegg.com x7 Assignment #6-linear mome/X youse C secure https://www.webassign.net/web/Student/Assignment-Responses/last?dep= 1 731 5827#Q 17 ::: Apps Studying tools Shoppingmovies Bideas other temprorary Other Bookmarks 17. 2/4 points | Previous Answers HRW9 9.P.068.GO. My Notes Ask Your T In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.95 m and then collides with stationary block 2, which has mass m2 = 2.00m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction , is 0.545 and comes to a stop in distance d within that region. Frictionless (a) What is the value of distance d if the collision is elastic? (b) What is the value of distance d if the collision is completely inelastic? T GO Tutorial Read the eBook Section 9-10 Elastic Collisions in One Dimension Submit Answer Save Progress Practice Another Version Submit Assignment Save Assignment Progress Home My Assignments Extension Request WebAssign840 1997-2017 Advanced Instructional Syatems, Inc. All rights reserved

Explanation / Answer

a]speed of block 1 just before collision, v1 = sqrt(2gh) = sqrt(2*2.95*9.8) = 7.604 m/s

velocity of second block after collision v2' = 2m1v1/(m1+m2) = 2*1*7.604/(1+2) = 5.07 m/s

d = v^2/(2ug) = 5.07^2/[2*0.545*9.8] = 2.406m answer

b] in inelastic collision, v = m1v1*[m1+m2] = 7.604*1/3 = 2.535 m/s answer

d = v^2/(2ug) = 2.535^2/[2*0.545*9.8] = 0.602 m answer