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Lecture 12

Lecture 12. Goals:. Chapter 9: Momentum & Impulse Solve problems with 1D and 2D Collisions Solve problems having an impulse (Force vs. time) Chapter 10 Understand the relationship between motion and energy Define Potential & Kinetic Energy

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Lecture 12

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  1. Lecture 12 Goals: • Chapter 9: Momentum & Impulse • Solve problems with 1D and 2D Collisions • Solve problems having an impulse (Force vs. time) • Chapter 10 • Understand the relationship between motion and energy • Define Potential & Kinetic Energy • Develop and exploit conservation of energy principle Assignment: • HW6 due Wednesday 3/3 • For Tuesday: Read all of chapter 10

  2. Momentum Conservation • Momentum conservation (recasts Newton’s 2nd Law when net external F = 0) is an important principle (usually when forces act over a short time) • It is a vector expression so must consider Px, Py and Pz • if Fx (external) = 0 then Px is constant • if Fy (external) = 0 then Py is constant • if Fz (external) = 0 then Pz is constant

  3. Inelastic collision in 1-D: Example • 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 final speed V. • In terms of m, M,andV : What is the momentum of the bullet with speed v ? x v V before after

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

  5. Exercise Momentum is a Vector (!) quantity • Yes • No • Yes & No • Too little information given • 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?

  6. Exercise 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

  7. j i Exercise Momentum is a Vector (!) quantity 1) ai = g sin 30° = 5 m/s2 2) d = 5 m / sin 30° = ½ aiDt2 10 m = 2.5 m/s2Dt2 2s = Dt v = aiDt= 10 m/s vx= v cos 30° = 8.7 m/s • x-direction: No net force so Px is conserved • y-direction: vy of the cart + block will be zero and we can ignore vy of the block when it lands in the cart. Initial Final Px: MVx + mvx = (M+m) V’x M 0 + mvx = (M+m) V’x V’x = m vx / (M + m) = 2 (8.7)/ 12 m/s V’x = 1.4 m/s N 5.0 m mg 30° 30° 7.5 m y x

  8. A perfectly inelastic collision in 2-D • Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). V v1 q m1 + m2 m1 m2 v2 before after • If no external force momentum is conserved. • Momentum is a vector so px, py and pz

  9. A perfectly inelastic collision in 2-D • If no external force momentum is conserved. • Momentum is a vector so px, py and pz are conseved V v1 m1 + m2 q m1 m2 v2 before after • x-dir px : m1 v1 = (m1 + m2 ) Vcos q • y-dir py : m2 v2 = (m1 + m2 ) Vsin q

  10. Before After Elastic Collisions • Elastic means that the objects do not stick. • There are many more possible outcomes but, if no external force, then momentum will always be conserved • Start with a 1-D problem.

  11. Billiards • Consider the case where one ball is initially at rest. after before paq pb vcm Paf F The final direction of the red ball will depend on where the balls hit.

  12. Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11) • Conservation of Momentum • x-dir Px : m vbefore = m vafter cos q + m Vafter cos f • y-dir Py :0 = m vafter sin q + m Vafter sin f after before pafterq pb Pafterf F

  13. Force and Impulse (A variable force applied for a given time) • Gravity: At small displacements a “constant” force t • Springs often provide a linear force (-kx) towards its equilibrium position (Chapter 10) • Collisions often involve a varying force F(t): 0  maximum  0 • We can plot force vs time for a typical collision. The impulse, J, of the force is a vector defined as the integral of the force during the time of the collision.

  14. t t ti tf Force and Impulse (A variable force applied for a given time) • J a vector that reflects momentum transfer F ImpulseJ= area under this curve ! (Transfer of momentum !) Impulse has units ofNewton-seconds

  15. F t Force and Impulse • Two different collisions can have the same impulse since J depends only on the momentum transfer, NOT the nature of the collision. same area F t t t t big, F small t small, F big

  16. F t Average Force and Impulse Fav F Fav t t t t big, Fav small t small, Fav big

  17. F F heavy light Exercise 2Force & Impulse • heavier • lighter • same • can’t tell • Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has the most momentum after the force acts ?

  18. Boxing: Use Momentum and Impulse to estimate g “force”

  19. Back of the envelope calculation (1) marm~ 7 kg (2) varm~7 m/s (3) Impact time t ~ 0.01 s  Question: Are these reasonable? Impulse J= p ~ marm varm ~ 49 kg m/s  Favg ~ J/t ~ 4900 N (1) mhead ~ 6 kg ahead = F / mhead ~ 800 m/s2 ~ 80 g ! • Enough to cause unconsciousness ~ 40% of a fatal blow • Only a rough estimate!

  20. Chapter 10: Energy (Forces over distance) • We need to define an “isolated system” ? • We need to define “conservative force” ? • Recall, chapter 9, force acting for a period of time gives an impulse or a change (transfer) of momentum • What if a force acts over a distance: Can we identify another useful quantity?

  21. Lecture 12 • Assignment: • HW6 due Wednesday 3/3 • For Tuesday: Read all of chapter 10

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