Question

Sphere A, of mass 0.600 kg, is initially moving to the right at 4.00 m/s. Sphere B, of mass 1.80 kg, is initially to the right of sphere A and moving to the right at 2.00 m/s. After the two spheres collide, sphere B is moving at 3.00 m/s in the same direction as before. (a) What is the velocity (magnitude and direction) of sphere A after this collision? (b) Is this collision elastic or inelastic? (c) Sphere B then has an off-center collision with sphere C, which has mass 1.60 kg and is initially at rest. After this collision, sphere B is moving at 19.0∘ to its initial direction at 1.40 m/s. What is the velocity (magnitude and direction) of sphere C after this collision? (d) What is the impulse (magnitude and direction) imparted to sphere B by sphere C when they collide? (e) Is this second collision elastic or inelastic? (f) What is the velocity (magnitude and direction) of the center of mass of the system of three spheres (A, B, and C) after the second collision? No external forces act on any of the spheres in this problem.

Use the definition of impulse to find the magnitude of the impulse J Con imparted to sphere B by sphere C. Remember that impulse is a vector. ...... + x PB P.com) = 0 PBGW Con B B ---19 .......... +x (116 PC Express your answer with the appropriate units. View Available Hints) LH THẢ + + 0 = ? Joon B = 4.29 kg. Submit Previous Answers

Answer 1

Page No - Solution Given Mass of A MA= 0.6 kg Mass mo= 1.80 kg I velocity of a before collision UA= 4.0 m/s. - Velocity of B before collision NB = 2 mise velocity of B after collision VB = 3 mise from principle of conservation of angular momentum . Initial momentum = final momentum - MAVA & MBUR = MANA +mova A VA = MAVA & MB VB - MBUB = (0.600)(4.0) + (1180) x (2:00 ) - 1-80 x 3 0.600 I VA = 1:00 mis 9f the Kinetic energy of the system remains conserved before and after collision then the collision is elastic othey wise thelastic We he check Initial K.E. = I MANA + I wave - x0.600x[4.00] + bal8x(3) 8.4 J "Education is the manifestation of perfection already in a man." --Swami Vivekananda

Page No.: - K. E final = 1 MANA2 + 1 mg UR2 - 1x0.600x (1.) 2 + 1 x 1.8 x (312 W = 8.4 I toor guitial kiE. E final K.E. Thue collision is inelastic -C) Griven, me = 1.60 kg Uc=0 and v₂ = 1-40 mic (now after collision with c.) Again applying the law of conservation of momentum in two dimensions. Taking initial direction of B as positive al- direction. Then initially Pa = mB Ug = 1.80x3.0 = 5.40 kgmis and py=0

We can now white bx = b + bex D - 0 and 0 = PRy + pcu 1 Where Pro = PB caso - 1.20x cos Lg e t - 1.13 kgmis PRy = PB sino = 1.20x0.32 = 0.39 kg bex = PR-pro = 5.40-1.13 = 4.27 Kgmis Dey = pc-picy = 0-0.39 = -0.39 kgwiam The magnitude of momentum of C - Pc = + Pray = = J (4.27)2+(-0.39/2 =4.29 km 7. Vc - be - 4.29 - 1.66 = 9.6% mis T and direction is a = tant by DAL - tan - /-0.397 5.2 - 14.27 "Education is the manifestation of perfection already in a man. --Swami Vivekananda

Page : The impulse equal to of c imparted by B to C is the change in momentum The initical momentum of C is zero. pero final momentum of c pe = 4.29 kg mis .. gimpulse I = 4.290 = 4.29 kg*m/s and the the direction is the direction between final and initial momentum 0 = -5:20