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Classical Mechanics Lecture 12

Classical Mechanics Lecture 12. Today’s Concepts: a) Elastic Collisions b) Center-of-Mass Reference Frame. Main Points. Main Points. Elastic Collisions: Analogy using spring. Mechanical Energy of system is if the form of kinetic energy moving block.

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Classical Mechanics Lecture 12

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  1. Classical MechanicsLecture 12 Today’s Concepts: a) Elastic Collisions b) Center-of-Mass Reference Frame

  2. Main Points

  3. Main Points

  4. Elastic Collisions: Analogy using spring • Mechanical Energy of system is if the form of kinetic energy moving block. • Moving block experiences a negative acceleration due to spring force. • Kinetic Energy of block is transferred to potential energy of spring. • Stationary block experiences an acceleration due to Spring force acting on formerly stationary block. • Potential Energy of Spring is transferred back to the Kinetic Energy of blocks after the collision. Mechanical Energy is conserved

  5. Elastic Collisions

  6. Elastic Collisions…final state? Quadratic Equation… Solvable !...but “tedious”…

  7. A better way!! Eliminates Quadratic Equation…

  8. Center of Mass Frame When viewed from reference frame moving with Center of Mass

  9. Center-of-Mass Frame In the CM reference frame, VCM= 0 In the CM reference frame, PTOT= 0

  10. Center of Mass Frame

  11. Elastic Collisions in CM frame Initial state Final state

  12. Elastic Collisions in CM frame Equal speeds but opposite velocity

  13. Elastic Collisions in Lab frame

  14. Elastic Collisions in CM frame

  15. Elastic Collisions in CM frame

  16. Elastic Collisions in CM frame What are the speeds of the boxes in the Center of Mass frame after the collision? In the center of mass frame the solution is TRIVIAL!!!! The final speeds of each object are the same as their initial speeds. But in opposite direction…again in the CM frame…

  17. Elastic Collisions in CM frame Need to transform the velocities back into the Lab frame…

  18. Elastic Collisions in CM frame Generalized result…

  19. Checkpoint

  20. 2-d elastic collisions and one object at rest

  21. Clicker/Checkpoint

  22. Summary

  23. Summary In center of mass frame Speeds remain the same Even in 2-d!!!

  24. Summary

  25. Center of Mass Frame & Elastic Collisions m2 v*1,i v*2,i m1 m2 m1 m2 m1 v*2, f v*1, f The speed on an object is the same before and after an elastic collision is viewed in the CM frame:

  26. v*2,i =0 Example: Using CM Reference Frame + vCM x +=CM + + v*1,i m2 m1 m2 m1 m2 m1 v*2, f v*1, f A glider of mass m1= 0.2 kgslides on a frictionless track with initial velocity v1,i= 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?

  27. Example Four step procedure: First figure out the velocity of the CM, VCM. VCM= (m1v1,i+ m2v2,i), butv2,i=0so VCM= v1,i So VCM= 1/5 (1.5 m/s) = 0.3 m/s Step 1: (for v2,i=0only)

  28. Example Now consider the collision viewed from a frame moving with the CM velocity VCM. m2 v*1,i v*2,i m1 m2 m1 m2 m1 v*2, f v*1, f

  29. Example v*1,i= v1,i- VCM= 1.5m/s - 0.3m/s=1.2m/s v*2,i= v2,i-VCM = 0m/s - 0.3m/s= -0.3m/s v*1,i= 1.2 m/s v*2,i= -0.3 m/s Step 2: • Calculate the initial velocities in the CM reference frame(all velocities are in the x direction): v • v=v* +VCM • v*= v - VCM VCM • v*

  30. Example Step 3: Use the fact that the speed of each block is the samebefore and after the collision in the CM frame. v*1, f = -v*1,i v*2, f = -v*2,i m2 v*1,i v*2,i m1 x m2 m1 v*1, f=- v*1,i = -1.2m/s v*2, f= - v*2,i=.3 m/s m2 m1 v*2, f v*1, f

  31. Example Calculate the final velocities back in the lab reference frame: Step 4: v • v=v* +VCM VCM • v* v1, f= v*1, f+ VCM = -1.2m/s+ 0.3 m/s= -0.9 m/s v2, f= v*2, f+ VCM = 0.3 m/s+ 0.3 m/s= 0.6 m/s v1, f= -0.9 m/s v2, f=0.6m/s Four easy steps! No need to solve a quadratic equation!

  32. Collision with Friction

  33. Do same steps for car 2 vcmis the same = final speed Compare 1/2 mv2 before and after for both cases

  34. Bumper Cars 2

  35. Elastic collision 

  36. Spring Loaded Collision

  37. Checkpoint • 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

  38. CheckPoint Response A) Kinitial>Kfinal B) Kinitial=Kfinal C) Kinitial<Kfinal initial A) Since the two boxes stick together, this collision is inelastic, and therefore energy is not conserved. final

  39. Relationship between Momentum & Kinetic Energy since This is often a handy way to figure out the kinetic energybefore and after a collision since p is conserved. initial same p final

  40. Checkpoint • A green block of mass m slides to the right on a frictionless floor and collides elasticallywith a red block of mass M which 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>MB) M=m C) M>mD) Need more information M m Before Collision M m After Collision

  41. CheckPoint B) In order to stop m in the collision, M must be the same as m. If m were smaller, it would bounce back and if m were bigger, it would continue to move forward. M m Before Collision M m After Collision A) m>MB) M=m C) M>mD) Need more information Mechanics Lecture 12, Slide 45

  42. Newton’s Cradle

  43. CheckPoint 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

  44. CheckPoint Response 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 Before Collision 2m m C) The total momentum of the system is zero. This means that the velocity of the center of mass must be zero.

  45. CheckPoint 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) v B) -v C) 2vD) -2v E) zero v 2v 2m m + Before Collision This is the CM frame

  46. CheckPoint Response 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) vB) -vC) 2vD) -2vE) zero + Before Collision V 2V This is the CM frame 2m m A) Since the collision is elastic, and the velocity of the center of mass is zero then the blocks simply travel backwards at their same speeds, (or opposite velocities). B) the magnitude of the velocity stays the same but the direction changes E) in order to conserve momentum, the blocks will both stop.

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