1 / 30

Lecture 11

Lecture 11. Goals:. Chapter 9: Momentum & Impulse Understand what momentum is and how it relates to forces Employ momentum conservation principles In problems with 1D and 2D Collisions In problems having an impulse (Force vs. time) Chapter 8: Use models with free fall.

hannaj
Download Presentation

Lecture 11

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 11 Goals: • Chapter 9: Momentum & Impulse • Understand what momentum is and how it relates to forces • Employ momentum conservation principles • In problems with 1D and 2D Collisions • In problems having an impulse (Force vs. time) • Chapter 8: Use models with free fall Assignment: • Read through Chapter 10 • MP HW5, due Wednesday 3/3

  2. Problem 7.34 Hint Suggested Steps • Two independent free body diagrams are necessary • Draw in the forces on the top and bottom blocks • Top Block • Forces:  1. normal to bottom block 2. weight 3. rope tension and 4. friction with bottom block (model with sliding) • Bottom Block • Forces:  1. normal to bottom surface 2. normal to top block interface 3. rope tension (to the left) 4. weight (2 kg) 5. friction with top block 6. friction with surface 7. 20 N Use Newton's 3rd Law to deal with the force pairs (horizontal & vertical) between the top and bottom block.

  3. Locomotion: how fast can a biped walk?

  4. How fast can a biped walk? What about weight? A heavier person of equal height and proportions can walk faster than a lighter person A lighter person of equal height and proportions can walk faster than a heavier person To first order, size doesn’t matter

  5. How fast can a biped walk? What about height? A taller person of equal weight and proportions can walk faster than a shorter person A shorter person of equal weight and proportions can walk faster than a taller person To first order, height doesn’t matter

  6. How fast can a biped walk? What can we say about the walker’s acceleration if there is UCM (a smooth walker) ? Acceleration is radial ! So where does it, ar, come from? (i.e., what external forces are on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, sideways

  7. Orbiting satellites vT = (gr)½

  8. Geostationary orbit

  9. Geostationary orbit • The radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth. • Fgravity is proportional to r-2 and so little g is now ~10 m/s2 / 50 • vT = (0.20 * 42000000)½ m/s = 3000 m/s • At 3000 m/s, period T = 2p r / vT = 2p 42000000 / 3000 sec = = 90000 sec = 90000 s/ 3600 s/hr = 24 hrs • Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary) • Great for communication satellites (1st pointed out by Arthur C. Clarke)

  10. Impulse & Linear Momentum • Transition from forces to conservation laws Newton’s Laws  Conservation Laws Conservation Laws  Newton’s Laws They are different faces of the same physics NOTE: We have studied “impulse” and “momentum” but we have not explicitly named them as such Conservation of momentum is far more general than conservation of mechanical energy

  11. Collisions are a fact of life

  12. Forces vs time (and space, Ch. 10) • Underlying any “new” concept in Chapter 9 is • A net force changes velocity (either magnitude or direction) • For any action there is an equal and opposite reaction • If we emphasize Newton’s 3rd Law and look at the changes with time then this leads to the Conservation of Momentum Principle

  13. 10 F (N) 0 2 - + time (s) Example 1 A 2 kg block, initially at rest on frictionless horizontal surface, is acted on by a 10 N horizontal force for 2 seconds (in 1D). What is the final velocity? • F is to the positive & F = ma thus a = F/m = 5 m/s2 • v = v0 + a Dt = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction) Notice: v - v0 = a Dt  m (v - v0) = ma Dt  m Dv = F Dt If the mass had been 4 kg … now what final velocity?

  14. 10 F (N) Twice the mass Before • Same force • Same time • Half the acceleration (a = F / m’) • Half the velocity ! ( 5 m/s ) 0 2 Time (sec)

  15. 10 F (N) 0 2 Time (sec) Example 1 • Notice that the final velocity in this case is inversely proportional to the mass (i.e., if thrice the mass….one-third the velocity). • Here, mass times the velocity always gives the same value. (Always 20 kg m/s.) Area under curve is still the same ! Force x change in time = mass x change in velocity

  16. 10 F (N) 0 2 Time (sec) Example 1 • There many situations in which the sum of the products “mass times velocity” is constant over time • To each product we assign the name, “momentum” and associate it with a conservation law. (Units: kg m/s or N s) • A force applied for a certain period of time can be graphed and the area under the curve is the “impulse” Area under curve : “impulse” With: m Dv = FavgDt

  17. Force curves are usually a bit different in the real world

  18. 10 A on B F (N) 0 B on A -10 0 2 Time (sec) Example 1 with Action-Reaction • Now the 10 N force from before is applied by person A on person B while standing on a frictionless surface • For the force of A on B there is an equal and opposite force of B on A MA x DVA = Area of top curve MB x DVB = Area of bottom curve Area (top) + Area (bottom) = 0

  19. Example 1 with Action-Reaction MADVA + MBDVB = 0 MA [VA(final) - VA(initial)] + MB [VB(final) - VB(initial)] = 0 Rearranging terms MAVA(final) +MB VB(final) = MAVA(initial) +MB VB(initial) which is constant regardless of M or DV (Remember: frictionless surface)

  20. Example 1 with Action-Reaction MAVA(final) +MB VB(final) = MAVA(initial) +MB VB(initial) which is constant regardless of M or DV Define MV to be the “momentum” and this is conserved in a system if and only if the system is not acted on by a netexternal force (choosing the system is key) Conservation of momentum is a special case of applying Newton’s Laws

  21. Alpha Decay 238U 234Th 4He v2 v1 Applications of Momentum Conservation Radioactive decay: Explosions Collisions

  22. Impulse & Linear Momentum • Definition: For a single particle, the momentum p is defined as: (p is a vector since v is a vector) p≡ mv So px = mvxand so on (y and z directions) • Newton’s 2nd Law: F = ma • This is the most general statement of Newton’s 2nd Law

  23. Momentum Conservation • Momentum conservation (recasts Newton’s 2nd Law when F = 0) is an important principle • It is a vector expression (Px, Py and Pz) . • And applies to any situation in which there is NO net external force applied (in terms of the x, y & z axes).

  24. Momentum Conservation • Many problems can be addressed through momentum conservation even if other physical quantities (e.g. mechanical energy) are not conserved • Momentum is a vector quantity and we can independently assess its conservation in the x, y and z directions (e.g., net forces in the z direction do not affect the momentum of the x & y directions)

  25. Exercise 1 Momentum is a Vector (!) quantity • A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction • In regards to the block landing in the cart is momentum conserved? • Yes • No • Yes & No • Too little information given

  26. Exercise 1 Momentum is a Vector (!) quantity • x-direction: No net force so Px is conserved. • y-direction: Net force, interaction with the ground so depending on the system (i.e., do you include the Earth?) Py is not conserved (system is block and cart only) Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart What is the final velocity of the cart? 2 kg 5.0 m 30° 10 kg 7.5 m

  27. Inelastic collision in 1-D: Example 2 • A block of massMis initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M,andV : What is the momentum of the bullet with speed v ? x v V before after

  28. Inelastic collision in 1-D: Example 2 What is the momentum of the bullet with speed v ? • Key question: Is x-momentum conserved ? Before After v V x before after

  29. Example 2Inelastic Collision in 1-D with numbers Do not try this at home! ice (no friction) Before: A 4000 kg bus, twice the mass of the car, moving at 30 m/s impacts the car at rest. What is the final speed after impact if they move together?

  30. 2 1 Exercise 2Momentum Conservation • Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. • The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks. • Which box ends up moving fastest ? • Box 1 • Box 2 • same

More Related