1. 10/30 Momentum in 2D  Text: Chapter7  HW: 10/28 HW Handout “2-D Momentum” Due Thursday, 10/31  HW: 10/30 HW Handout “2-D Momentum and Energy” Due Friday, 11/1  Exam 3: Thursday, 11/14  Exam 4: Thursday, 12/5  Final Exam: Monday, 12/16, 8:00 am

2. 10/28 HW part a) p sys,i = p 1,i = 40kg m/s p 1,f = 28kg m/s @ 45  p 1,f + p 2,f = p sys,i since no net force on the system p 2,f Components of p 1,f are 20kg m/s for both up and left. So the components of p 2,f must be 20kg m/s for both down and left. p 2,f is then 28kg m/s at 45° below horizontal. v 2,f is then 14m/s at 45° below horizontal.

3. Methods for Momentum  p sys,i = p sys,f  Best when things stick together   p A =  p B  Good when you know velocity before and after for one object  Components for 2-D  Can find components first

4. p t,i,y = ?24,000 kg m/s A car and a truck, velocities and masses shown, collide and lock bumpers. Find the final velocity (magnitude and direction) of the pair.  vfvf 37 0 v t = 5 m/s v c = 15 m/s M t = 8000 kg M c = 2000 kg x y p c,i,y = ? p sys,f,x = ? p sys,f,y = ? p t,i,x = ? p c,i,x = ? 0 24,000 kg m/s 32,000 kg m/s 30,000 kg m/s 62,000 kg m/s

5. A car and a truck, velocities and masses shown, collide and lock bumpers. Find the final velocity (magnitude and direction) of the pair.  vfvf 37 0 v t = 5 m/s v c = 15 m/s M t = 8000 kg M c = 2000 kg x y p sys,f,y = 24,000 kg m/s p sys,f,x = 62,000 kg m/s p sys,f = ? v sys,f = ?  = ? 66,500 kg m/s 6.65 m/s tan -1 24,000/62,000 = 21°