Question

Unsure on all parts of this question

Pre-lab Assignment- Experiment 11 Conservation of Momentum Using the Air Track 1. Suppose that before collision, slider A (MA-200 g) is traveling at 3.6 m/s and slider B (Ma 210 g) is at rest. a) (1 pt) What is the total momentum before collision? (Find the momentum of each slider befone the collision and then add them together.) b). (1 pt) After the collision, presume that the sliders move in the same direction with v m/s and vir = 3.05 m/s. What is the total momentum following the collision? (Find the momentum of each slider after the collision and combine them appropriately.) 0.398 e). (1 pt) What is the total kinetic energy before collision? (Find the kinctic energy of each slider before the collision and add them together.) d), (I pt) What is the total kinetic energy after collision? e) (2 pts) Is this an elastic collision? Why or why not? (Look in Chapter 7 of the text to find what is meant by an elastic collision. What quantities are conserved in an elastic collision?)

Answer 1

(a) K.E. 7 slider A = 1/2 mv^2 = 1/2 times 200 100/1000 10 times (3.6)^2 = (3.6)^2/10 = 1.296 J Momentum 7 slides A = mv = 200/1000 times 3.6 = 72 Momentum 7 slider B = 0 (since it is at rest) Total momentum before collision = .72 + 0 = .72 (b) After collision momentum 7 slider A = 200/1000 times .398 = .0796 kg m/sec momentum f slides B 210/1000 times 3.05 = .64 kg m/sec Total momentum = .0796 + .64 = .7196 kg m/sec (c) Total kinetic energy before collision = 1.296 J + 0 = 1.296 J (d) Total kinetic energy after collision = 1/2 times 200/1000 times (398)^2 + 1/2 210/1000 times (3.05)^2 = .01584_.97676 = .9926 J (e) It is not a can f elastic collision. In it we find that momentum is same or it is conserve during collision but there is loss of kinetic energy