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Chapter 6

Chapter 6. Momentum and Collisions. Momentum. Definition:. Important because it is CONSERVED proof:. Since F 12 =-F 21 ,. for isolated particles never changes!. Vector quantity. Both S p x and S p y are conserved. Example 6.1.

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Chapter 6

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  1. Chapter 6 Momentum and Collisions

  2. Momentum Definition: Important because it is CONSERVEDproof: Since F12=-F21, for isolated particles never changes!

  3. Vector quantity • Both Spx and Spy are conserved

  4. Example 6.1 An astronaut of mass 80 kg pushes away from a space station by throwing a 0.75-kg wrench which moves with a velocity of 24 m/s relative to the original frame of the astronaut. What is the astronaut’s recoil speed? 0.225 m/s

  5. Center of mass does not accelerate

  6. Do the back-to-back demo

  7. Example 6.2 Ted and his ice-boat (combined mass = 240 kg) rest on the frictionless surface of a frozen lake. A heavy rope (mass of 80 kg and length of 100 m) is laid out in a line along the top of the lake. Initially, Ted and the rope are at rest. At time t=0, Ted turns on a wench which winds 0.5 m of rope onto the boat every second. a) What is Ted’s velocity just after the wench turns on? b) What is the velocity of the rope at the same time? c) What is the Ted’s speed just as the rope finishes? d) How far did the center-of-mass of Ted+boat+rope move e) How far did Ted move? f) How far did the center-of-mass of the rope move? 0.125 m/s -0.375 m/s 0 0 12.5 m -37.5 m

  8. Example 6.3 A 1967 Corvette of mass 1450 kg moving with a velocity of 100 mph (= 44.7 m/s) slides on a slick street and collides with Mr.D’s Hummer of mass 3250 kg which is parked on the side of the street. The two vehicles interlock and slide off together. What is the speed of the two vehicles immediately after they join? 13.8 m/s =30.9 mph

  9. Impulse Useful for sudden changes when not interested in force but only effects of force

  10. Bunjee Jumper Demo

  11. Total Impulse Graphical Representation of Impulse For complicated force, p is area under F vs. t curve F F t Dt

  12. Example 6.4 A pitcher throws a 0.145-kg baseball so that it crosses home plate horizontally with a speed of 40 m/s. It is hit straight back at the pitcher with a final speed of 50 m/s. a) What is the impulse delivered to the ball? b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0 x 10–3 s. c) What is the acceleration experienced by the ball? a) 13.05 kgm/s b) 6,525 N c) 45,000 m/s2

  13. Elastic & Inelastic Collisions • ELASTIC: • Both energy & momentum are conserved • INELASTIC • Momentum conserved, not energy • Perfectly inelastic -> objects stick • Lost energy goes to heat

  14. Catching a baseball • Football tackle • Cars colliding and sticking • Bat eating an insect Examples of Perfectly Elastic Collisions • Superball bouncing • Electron scattering

  15. Ball Bounce Demo

  16. Example 6.5a A superball bounces off the floor, A) The net momentum of the earth+superball is conserved B) The net energy of the earth+superball is conserved C) Both the net energy and the net momentum are conserved D) Neither are conserved

  17. Example 6.5b A astronaut floating in space catches a baseball A) Momentum of the astronaut+baseball is conserved B) Mechanical energy of the astronaut+baseball is conserved C) Both mechanical energy and momentum are conserved D) Neither are conserved

  18. Example 6.5c A proton scatters off another proton. No new particles are created. A) Net momentum of two protons is conserved B) Net kinetic energy of two protons is conserved C) Both kinetic energy and momentum are conserved D) Neither are conserved

  19. Example 6.6 A 5879-lb (2665 kg) Cadillac Escalade going 35 mph =smashes into a 2342-lb (1061 kg) Honda Civic also moving at 35 mph=15.64 m/s in the opposite direction.The cars collide and stick. a) What is the final velocity of the two vehicles? b) What are the equivalent “brick-wall” speeds for each vehicle? a) 6.73 m/s = 15.1 mph b) 19.9 mph for Cadillac, 50.1 mph for Civic

  20. Example 6.7 An proton (mp=1.67x10-27 kg) elastically collides with a target proton which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target proton bounces forward, what are a) the final velocity of the projectile proton? b) the final velocity of the target proton? 0.0 3.0x106 m/s

  21. 2. Problem is easy to solve in this frame: Work in center-of-mass frame (elastic) • Find c.o.m. velocity and subtract it from both v1 and v2 Note:

  22. Working out answer in center-of-mass 3. Relative velocity switches sign For elastic head-on collisions: Equivalent to energyconservation

  23. Example 6.8 An proton (mp=1.67x10-27 kg) elastically collides with a target deuteron (mD=2mp) which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target deuteron bounces forward, what are a) the final velocity of the projectile proton? b) the final velocity of the target deuteron? vp =-1.0x106 m/s vd = 2.0x106 m/s Head-on collisions with heavier objects always lead to reflections

  24. Example 6.9a The mass M1 enters from the left with velocity v0 and strikes the mass M2=M1 which is initially at rest. The collision is perfectly elastic. a) Just after the collision v2 ______ v0. A) > B) < C) =

  25. Example 6.9b The mass M1 enters from the left with velocity v0 and strikes the mass M2=M1 which is initially at rest. The collision is perfectly elastic. Just after the collision v1 ______ 0. A) > B) < C) =

  26. Example 6.9c The mass M1 enters from the left with velocity v0 and strikes the mass M2=M1 which is initially at rest. The collision is perfectly elastic. Just after the collision P2 ______ M1v0. A) > B) < C) =

  27. Example 6.9d The mass M1 enters from the left with velocity v0 and strikes the mass M2=M1 which is initially at rest. The collision is perfectly elastic. At maximum compression, the energy stored in the spring is ________ (1/2)M1v02 A) > B) < C) =

  28. Example 6.9e The mass M1 enters from the left with velocity v0 and strikes the mass M2<M1 which is initially at rest. The collision is perfectly elastic. Just after the collision v2 ______ v0. A) > B) < C) =

  29. Example 6.9f The mass M1 enters from the left with velocity v0 and strikes the mass M2<M1 which is initially at rest. The collision is perfectly elastic. Just after the collision v1 ______ 0. A) > B) < C) =

  30. Example 6.9g The mass M1 enters from the left with velocity v0 and strikes the mass M2<M1 which is initially at rest. The collision is perfectly elastic. Just after the collision P2 ______ M1v0. A) > B) < C) =

  31. Example 6.9h The mass M1 enters from the left with velocity v0 and strikes the mass M2<M1 which is initially at rest. The collision is perfectly elastic. At maximum compression, the energy stored in the spring is ________ (1/2)M1v02 A) > B) < C) =

  32. Example 6.9i The mass M1 enters from the left with velocity v0 and strikes the mass M2>M1 which is initially at rest. The collision is perfectly elastic. Just after the collision v2 ______ v0. A) > B) < C) =

  33. Example 6.9j The mass M1 enters from the left with velocity v0 and strikes the mass M2>M1 which is initially at rest. The collision is perfectly elastic. Just after the collision v1 ______ 0. A) > B) < C) =

  34. Example 6.9k The mass M1 enters from the left with velocity v0 and strikes the mass M2>M1 which is initially at rest. The collision is perfectly elastic. Just after the collision P2 ______ M1v0. A) > B) < C) =

  35. Example 6.9l The mass M1 enters from the left with velocity v0 and strikes the mass M2>M1 which is initially at rest. The collision is perfectly elastic. At maximum compression, the energy stored in the spring is ________ (1/2)M1v02 A) > B) < C) =

  36. Example 6.10 Ballistic Pendulum: used to measure speed of bullet. 0.5-kg block of wood swings up by height h = 65 cm after stopping 8.0-g bullet.What was bullet’s velocity? 227 m/s

  37. Example 6.11 A 5-g bullet traveling at 500 m/s embeds in a 1.495 kg block of wood resting on the edge of a 0.9-m high table. How far does the block land from the edge of the table? 71.4 cm

  38. Example 6.12 Tarzan (M=80 kg) swings on a 12-m vine by letting go from an angle of 60 degrees from the vertical. At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing? 35.8 degrees (indepedent of L or g)

  39. Example 6.12a Tarzan (M=80 kg) swings on a 12-m vine by letting go from an angle of 60 degrees from the vertical. At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing? To calculate Tarzan’s speed just before he picks up Jane, you should apply: A) Conservation of Energy B) Conservation of Momentum

  40. Example 6.12b Tarzan (M=80 kg) swings on a 12-m vine by letting go from an angle of 60 degrees from the vertical. At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing? To calculate Tarzan&Jane’s speed just after their collision (given the previous answer) you should apply: A) Conservation of Energy B) Conservation of Momentum

  41. Example 6.12c Tarzan (M=80 kg) swings on a 12-m vine by letting go from an angle of 60 degrees from the vertical. At the bottom of his swing, he picks up Jane (m=50 kg). To what angle do Tarzan and Jane swing? To calculate Tarzan&Jane’s final height (given the previous answer) you should apply: A) Conservation of Energy B) Conservation of Momentum

  42. (Perfectly) Inelastic Collisions in Two Dimensions SKIP LAST 4 SLIDES Two Equations : Two unknowns (vfx, vfy)

  43. Elastic Collisions in Two Dimensions Three Equations : Four Unknowns (v1f,v2f,q1f,q2f) = 0

  44. Example 6.14 A 1200-kg vehicle moving at 25.0 m/s east collides with a vehicle of mass 1500 kg moving northward at 20.0 m/s. After they join, what is their final speed and direction? vf = 15.7 m/s qf = 45

  45. Example 6.15 A projectile proton moving with v0=120,000 m/s collides elastically with a second proton at rest. If one proton leaves at an angle q=30, a) what is its speed? b) what are the speed and direction of the second proton? a) v1 = 103,923 m/s b) v2 = 60,000 m/s q2 = 60

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