Question

Problem 1: An object with mass m1 = 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 μk = 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.

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 haum 0=250 -a let ul let u be the velocity of object 0 Mp=0.4 marang 25kg u My hea + Ao - d=m* be the welocity of object at bottom of plane, from Conservation of Energy, mgham, ut as u,afgh (2x9.884 = 8.85 mis just sebhe collision, Worredone boy frichios change in Ke. -Wym, gd=122 m, ²-12 m, 4,2 may U2² = migh-Mem, gd U = 2gn = 2 Mugd = 7.41 m/s from conservation of momentum mua=(m, +ma)v 2x 7.41 = 2012 m/s (2+5) d Total Kit before collision Ki= 12 m, 4,2 = 1/3+2/70 nig? K;=54001 J rodal K.E. after collision (2+5X2012 1 =15.737 6 loss of KE during collision Korkg = 39.18 J Me (m, tmag x = Ź(m + my v2 v2 u=2x Kos 6 ale= - at things 0057m