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1. Midterm Exam #2 Second midterm is this Friday, February 28(75 minutes!!!) Units 6-11: Friction – Conservation of momentum Things to bring:Pencil (for bubble sheets), calculator, 1 sheet of notes (8.5 X 11 in., front & back) Bring your U of U ID card: you will not receive a grade unless you show it upon turning in your exam No makeups!!! (Unless you are unavailable due to University/military business or documented medical emergency) Cheating: You will receive a zero and referred to the University administration – may result in an “E” for the course and expulsion Second attempt online: You will have 11:59 PM on Friday until 11:59 on Sunday to complete the online exam. 2 hours once begun! Locations:

2. Lecture 12: Elastic Collisions

   Today’s Concepts: a) Elastic Collisions b) Center-of-Mass Reference Frame
3. Center of Mass & Collisions so far: The CMbehaves just like a point particle then Ifthen momentum is conserved (usually the case during a collision) If you are in a reference frame moving along with the CMthen the total momentum you measure is 0.
4. CheckPoint: Identical Box Collision A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest. Compare the initial and final kinetic energies of the system. A) Kinitial>Kfinal B) Kinitial=Kfinal C) Kinitial<Kfinal initial final About 50% of you got this correct the first time.
5. CheckPoint: Discuss & Re-vote A) Kinitial>Kfinal B) Kinitial=Kfinal C) Kinitial<Kfinal A) Due to this being an inellastic collision the kinetic energy will be lost as work will be done on deformation of the two objects. initial B) The kinetic energy would be equal before and after the collision. The initial K = 1/2mv2 and the final K = 1/2*2m(v/2)2. These equations are equal. final C) Set up equation mvi = 2mvf and find that vf= vi/2
6. Relationship between Momentum & Kinetic Energy since This is often a handy way to figure out the kinetic energybefore and after a collision since Ptotis conserved if Fnet,ext= 0. initial These are the same! final
7. Review: Inelastic Collisions Inelastic collisions: Momentum is conserved, but mechanical energy is not.
8. CheckPoint: Unknown Box Collision A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass Mwhich is initially at rest. After the collision the green block is at rest and the red block is moving to the right. How does M compare to m? A) m > M B) m= MC) m<M M m Before Collision M m After Collision
9. Elastic collisions: momentum & energy conserved A) m>M B) m=M C) m<M A) If the first block does not have less mass than the second block than the first block would keep moving with second block when it hit it instead of coming to rest. M m Before Collision B) If m were greater than M it would continue moving, if m were less than M it would bounce back. C) if mass m had equal or greater mass it would move with mass M instead of stopping M m After Collision
10. Review: Inelastic vs. Elastic Collisions Inelastic collisions: Momentum is conserved, but mechanical energy is not. Elastic collisions: Momentum and mechanical energy are both conserved.
11. Newton’s Cradle A great example of elastic collisions: The only way to understand this is to conserve both P and E.
12. Solving 2-body Collision Problems There are 6 variables: m1, m2, v1i, v2i, v1f, v2f If we know 4 of the 6, then we need 2 equations to solve for remaining 2 unknowns One equation comes from conservation of momentum: If collision is inelastic: Then Ktot before and after are not the same, but… Often v1 = v2 after collision (if they stick together) If collision is elastic: Then Ktot before and after are the same: Eqn. 2 Eqn. 2 Eqn. 1
13. The problem with elastic collisions… Conservation of momentum: Conservation of energy: YIKES!!! Very tedious!!! Use Center of Mass frame!
14. Center-of-Mass Frame In the CMreference frame, VCM= 0 In the CMreference frame, Ptot= 0
15. Center of Mass Frame & Elastic Collisions The speed of each object is the same before and after an elastic collision as viewed in the CMframe: Before: m2 v*1,i v*2,i m1 v*1 = v*2= 0 During: m2 m1 After: m2 v*2, f = -v*2,i m1 v*1, f = -v*1,i
16. v2i= 0 Example: Using CM Reference Frame A glider of mass m1= 0.2 kgslides on a frictionless track with initial velocity v1i= 1.5 m/s. It hits a stationary glider of mass m2=0.8kg. A spring attached to the first glider compresses and relaxes during the collision, but there is no friction (i.e., energy is conserved). What are the final velocities? x v1i + m2 m1 vCM m2 +:CM m1 + m2 m1 v1f v2f +
17. Example Four step procedure: First figure out the velocity of the CM, VCM. VCM= (m1v1,i+ m2v2,i), but v2,i =0so VCM= v1,i So VCM= 1/5 (1.5 m/s) = 0.3 m/s Step 1: (for v2,i =0only)
18. Example Now consider the collision viewed from a frame moving with the CMvelocity VCM. m2 v*1i v*2i m1 m2 m1 v*1,f m2 v*2,f m1
19. Example v*1i= v1i– VCM = 1.5m/s - 0.3m/s=1.2m/s v*2i= v2i- VCM = 0m/s - 0.3m/s= -0.3m/s v*1i= 1.2 m/s v*2i = -0.3 m/s Step 2: Calculate the initial velocities in the CMreference frame(all velocities are in the xdirection): v v = v* +VCM v*= v - VCM VCM v*
20. Example Step 3: Use the fact that the speed of each block is the samebefore and after the collision in the CMframe. v*1f =-v*1i v*2f = -v*2i v*1i v*2i m2 m1 x m2 m1 v*1f= -v*1i = -1.2 m/s v*2f = -v*2i= 0.3 m/s m2 v*2f m1 v*1f
21. Example Calculate the final velocities back in the lab reference frame: Step 4: v v = v* +VCM VCM v* v1f= v*1f+ VCM = -1.2m/s+ 0.3 m/s= -0.9 m/s v2f= v*2f+ VCM = 0.3 m/s+ 0.3 m/s= 0.6 m/s v1f= -0.9 m/s v2f= 0.6m/s Four easy steps! No need to solve a quadratic equation!
22. Review: Elastic Collisions in the CM Frame Step 1: Figure out the velocity of the CM: Step 2: Find the initial velocities of each object in the CMframe: Step 3:The speed of each block is the same before and after the collision in the CMframe: Step 4:Transform back to the Labframe: for only! 1D: 3D:
23. CheckPoint: Double Mass Half Speed 1 Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other. The velocity of the center of mass of this system before the collision is. A) Toward the left B) Toward the right C) zero v 2v 2m m Before Collision
24. CheckPoint Results: Double Mass Half Speed 1 The velocity of the center of mass of this system before the collision is A) Toward the left B) Toward the right C) zero This is the CM frame! v 2v 2m m
25. CheckPoint: Double Mass Half Speed 2 Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other. Suppose the blocks collide elastically. Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place? A) +2v B) +v C) 0 D) -v E) -2v v 2v 2m m This is the CM frame + Before Collision
26. CheckPoint Results: Double Mass Half Speed 2 Suppose the blocks collide elastically. Picking the positive direction to the right, what is the velocity of the bigger blockafterthe collision takes place? A) +2vB) +vC) 0 D) -vE) -2v + Before Collision v 2v This is the CM frame 2m m