Question

Problem 1: An object with mass m, = 2 kg slides down a frictionless incline that makes a 25° with the horizontal (as in the figure). At the bottom of the incline, object 1 continues to travel along the rough surface with the = 0.4. Object 2, m, = 5 kg is d = 3 m away from the bottom of the incline. Object 2 is initially at rest. The height of the incline is h = 4m. h 0 2 d a) [3 points] What is the velocity of object 1 at the bottom of the incline? b) [3 points) What is the velocity of object 1 right before it collides with object 2?

Answer 1

ANS: m, = 2kq @1=25° Mk=0.4 m2 = 5kg d = 3m h =um. h 2 LO) using conservation of energy Potential energy at top d kinetic energy at botom Imva 2 W.D in friction mgh v=1/29h = √2x9.8×4 V -8.85 mis (6) Loss of energy in friction = "x migd 0.4X2X98X3 = 23.52 Joule : Change in kie 22 m V I MVA - VF + (8-85) 2 x 23.52 W.D Imv.2 23.52 2 "p² = 7.4 m/s Velocity with which I will collide 2.

conserving (0) momentum. m, vp + m2 x 0 = (Mi+m2) v 2X 7.4 = (2+5) V V=201 mis V = (d) Total kinetic energy = l m n + 1 M2 XO before 1x2 x 7.42 2 2 = 54.76J Le Total ke after = I (M1 + M2) v2 = 1.7 x 2.12 2 2 - 15.43 J (f) energy dissipation = 54.76-15.43 = 39.33 J. W.D change in ke. (9) M (m, +M2) g a Bespazos - 0 15:43 15:43 Bereces 0.487 x 9.8 a OL 0.55m.. a=0.56 m