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 Hk = 0.4. Object 2, m2 = 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 01 2 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? c) [2 points] What are the velocities of object 1 and object 2 after the collision if they stick together? d) [2 points] What is the total kinetic energy of the system just before the collision? e) [2 points] What is the total kinetic energy of the system just after the collision? f) [2 points] How much energy got dissipated during the collision? g) [4 points) How far will the objects travel on the rough surface before coming to a stop?

Answer 1

O. 2. р a) According to conservation of energy. KE; + PEi KE & + P Es 0 + mgh = 1 mv + 0 Vo = √2gh √2x9.864 V = 8.85 m/s W = 2 mv ² - 1 mv. ² um.ge Id = 1 mvg - 1 mvi² 2 ugd = Vg² - Vi² (a て て е 3x2x0.4x9.8*= V2-8.85% IV = 10.09m/s Acc to conservation of energy & m,v +0 - (m. + m.) vs 2x10.09² = (2+5) Vg² IV = 5.39 m/s d KE befove collision Im, v. + 1 m. Vz - 11 x 2x 10.09 KE 101.8I J 2. て

e) KE after collision 2 4) ₂ (m, + ha) ug ² (2+5)*5.392 [101.685 W = Change in kinetic energy ummg d Imus - Imv: 2 Vj = 0 a gulm, + m2) = £ (m, + ma ) vi? dg = V2 28u d 5.39 2x9.8×0.4 d = 3.70m KE das Energy dissipated DE = 1 m, vr - 1 (m, + m2) ve ² * x2x10.09?- < (2+5) 5.34 2 2 0.125