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Momentum

Momentum. Overview. Kinematics Motion of an object under constant acceleration 1-D and 2-D motion Laws of motion Newton’s concept of forces, F =m a (chapter 4) Apply kinematics to determine object’s motion Energy Concept of work from a force, W=F x *dx+F y *dy

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Momentum

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  1. Momentum PHY231

  2. Overview • Kinematics • Motion of an object under constant acceleration • 1-D and 2-D motion • Laws of motion • Newton’s concept of forces, F=ma (chapter 4) • Apply kinematics to determine object’s motion • Energy • Concept of work from a force, W=Fx*dx+Fy*dy • Solve problems applying energy conservation • Momentum • Concept of momentum p=mv • Solve problems applying momentum conservation PHY231

  3. Momentum • Momentum for an object defined as its mass times its velocity • SI unit kg.m/s • P and v point in the same direction • Large velocity and large mass = large momentum PHY231

  4. Momentum - example • A 750 kg car moves initially at 20.0 m/s • What is its momentum? • The car slows down and finally comes to rest, what is its momentum then? • What was the change in momentum between the two instants considered? PHY231

  5. Momentum – other relations • Momentum and KE • Momentum in 2-D • A man of mass = 70 kg has a momentum p=700 kg.m/s PHY231

  6. Newton revisited • Newton’s second law stated • But acceleration is the change in velocity divided by the time elapsed • We found that A force is needed to change p !! F is the average force during Dt PHY231

  7. example • A bullet with an initial momentum of 5.0 kg.m/s is shot into water where it is decelerated to rest in 0.010 s. What is the magnitude of the average force that acted on the bullet ? A) 0 B) 50 N C) 5.0 * 102 N D) 5.0 * 103 N PHY231

  8. example • A bullet with an initial momentum of 5.0 kg.m/s is shot into water where it is decelerated to rest in 0.010 s. What is the magnitude of the average force that acted on the bullet ? A) 0 B) 50 N C) 5.0 * 102 N D) 5.0 * 103 N PHY231

  9. Impulse • We have • We can rewrite it as an impulse • Impulse = Force*time = change in momentum PHY231

  10. Impulse and Fav • A block of 1.3 kg is sliding on a table with some kinetic friction. The speed of the block decreases from 3.5 m/s to 1.2 m/s in 2.0 s. What is the average force that acted on the block? PHY231

  11. Child safety • A friend claims that it is safe to go on a car trip with your child on his laps since he can hold your 24 kg child even if the car makes a frontal collision (lasting 0.05s and causing the vehicle to stop completely) with vi=14 m/s (~30 mph). Is he to be trusted? Equivalent mass to lift (mg=6700 N) m~1500 lbs !!! He won’t be able to hold your child. USE A CARSEAT!!! PHY231

  12. Dt matters • Imagine a car is hitting a soft haystack or a wall and suppose the car loses all its momentum in both cases Larger forces occurred in the second case because the time interval is smaller PHY231

  13. Conservation of momentum • Total momentum of an isolated system is conserved • Isolated system means no net external force acts on the system (no gravity for example) • Total momentum = sum of momentum for each constituents of the system • Total momentum is conserved, but momentum for each constituent can change!! PHY231

  14. Collision of two objects, isolated system • Impulse = change of momentum • And using Newton • Thus • Total momentum of an isolated system is conserved PHY231

  15. Golf club-ball isolated system • The head of a golf club hits a ball at rest on a tee. The club has a mass mc= 0.20 kg, and travels at vc,i=50 m/s initially. The ball has mb=40g. The club still has vc,f=40 m/s after hitting the ball. What is the speed of the ball just after impact? (neglect gravity) A) 10 m/s B) 20 m/s C) 40 m/s D) 50 m/s PHY231

  16. A) 10 m/s B) 20 m/s C) 40 m/s D) 50 m/s PHY231

  17. Elastic/inelastic collisions • Two main types of collisions • Inelastic • Elastic • Most collisions are in between Elastic and inelastic • Total momentum is conserved for both elastic and inelastic collisions • Total kinetic energy is also conserved for elastic collisions but not for inelastic collisions (some kinetic energy can be transformed into thermal energy for example ) • You will be told for each problem what type of collision is happening PHY231

  18. Inelastic collisions (1D) • Inelastic collision between 2 objects • Only conservation of momentum applies • Perfectly inelastic if the objects stick together after collision (objects end-up with same velocity in this case) PHY231

  19. Elastic collisions (1D) • Elastic collision between 2 objects • Conservation of momentum applies • Conservation of kinetic energy applies • After some math… PHY231

  20. Summary • Total momentum conserved for an isolated system • Collisions of two objects (1D) • Perfectly inelastic • Elastic (KEi=KEf) PHY231

  21. Perfectly inelastic Collision demo (1) • Two carts are on a frictionless rail system • m1=m2=m and both moving at same speed but one to the right the other to the left v2i=-v1i • They collide, both carts stick together (velcro) • This is a perfectly inelastic collision !! • What is the final velocity vf of the system ? vi -vi m m ? BEFORE AFTER PHY231

  22. Perfectly inelastic Collision demo (1) vf=0 vi -vi m m 2m BEFORE AFTER PHY231

  23. Elastic Collision demo (1) • Two carts are on a frictionless rail system • m1=m2=m and both moving at same speed but one to the right the other to the left v2i=-v1i • They collide and the collision is elastic • What are the final velocities v1f and v2f ? vi -vi m m ? BEFORE AFTER PHY231

  24. vi -vi -vi vi m m m m BEFORE AFTER PHY231

  25. Elastic Collision demo (2) • Collision is elastic between m and 3m • Both are moving at same speed, head-on collision • What are their velocities after collision? vi -vi m 3m ? BEFORE AFTER PHY231

  26. vi -vi -2vi 0 m 3m m 3m BEFORE PHY231 AFTER

  27. Summary • Total momentum conserved for an isolated system • Collisions of two objects (1D) • Perfectly inelastic • Elastic (KEi=KEf) PHY231

  28. Girl on a plank (homework) • Conservation of momentum can be applied to the girl+plank system • Trick is that the velocity of the girl with respect to the plank is given, not v with respect to the ice • Thus, relative motion is also needed PHY231

  29. Girl: G Plank: P Ice: I w.r.t.=with respect to • Remember relative motion notation and rule • Velocity of the girl w.r.t. the ice : vGI • Velocity of the girl w.r.t. the plank : vGP • Velocity of the plank w.r.t. the ice : vPI • vGI = vGP + vPI PHY231

  30. Example: • mG = 46.3 kg • mp = 127 kg • vGI = 1.59 m/s mG=46.3 kg VGP=0 m/s VGI=0 m/s mG=46.3 kg VGP=1.59 m/s VGI=1.165 m/s mP=127 kg VPI=0 m/s mP=127 kg VPI=-0.4248 m/s plank plank ICE PHY231

  31. Car collision (homework) • Collision is inelastic, you can find the velocity of the cars right after impact. Use this as initial instant. • Work done by friction force is equal to the change in kinetic energy Wfriction = -FkDx = KEf-KEi • You know masses and initial and final velocities so you know KEi and KEf • You know expression for Fk = mkn = mkMg with M=total mass • In Wfriction equation, only thing you don’t know is mk !! PHY231

  32. Initial M2=2970 kg M1=1820 kg V2 = 0 m/s V1 = 12.1 m/s M = M1+M2 = 4790 kg After perfectly Inelastic collision V= 4.60 m/s Final, Friction force took all KE Vf= 0 m/s PHY231

  33. Block down and up a slide • A block of mass m1=5.0 kg is released from a height h=5.0 m on a frictionless slide. It collides elastically with the block m2=10 kg initially at rest. What height H does m1 rises after the collision ? A) 0.0 m B) 0.56 m C) 1.7 m D) 2.5 m E) 5.0 m PHY231

  34. Before sliding • After sliding • Before ellastic collision • After ellastic collision (special case v2i=0) • Before sliding back up • After sliding back up B) 0.56 m PHY231

  35. Slide/collision/fly • A block of mass m1=0.500 kg is released from a height h1=2.50 m on a frictionless slide posed on a table itself h2=2.00 m high. It collides elastically with the block m2=1.00 kg initially at rest. What distance x from the right edge of the table does m1eventually land on the floor ? PHY231

  36. Before sliding • After sliding • Before ellastic collision • After ellastic collision (special case v2i=0) • m1 slides back up, then back-down, at that time • Parabolic motion • time to fall height h2 • Horizontal distance PHY231

  37. Large Hadron Collider (LHC) • Located at CERN (France-Switzerland border) • 27 km (17 miles) circumference • Proton-proton collider • What is the origin of mass? Hunting for a particle, the Higgs boson which is theorized to give mass to particles • To try to create Higgs bosons, proton beams with extreme energies are used • Head-on collision, one beam clockwise, the other counter-clockwise….Why?? PHY231

  38. LHC • Reason: Conservation of momentum !! • Proton beams have same KE and collide head-on • The total momentum is therefore zero! • All the kinetic energy of the proton beams can be used to create a particle (Higgs boson) without “wasting” energy in the conserved momentum • Lot’s of energy is needed to create the massive new particle because of Mr. Enstein’s E=mc2 PHY231

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