1 / 16

Chapter 10

Chapter 10. Collisions. The Take-Home Message on Chapter 10. If two bodies are exerting STRONG forces on each other over a short period of time, then you have a collision problem. Linear Momentum is ALWAYS conserved. Kinetic Energy is NOT always conserved.

nasya
Download Presentation

Chapter 10

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. Chapter 10 Collisions

  2. The Take-Home Message on Chapter 10 • If two bodies are exerting STRONG forces on each other over ashortperiod of time, then you have a collision problem. • Linear Momentum is ALWAYS conserved. • Kinetic Energy is NOT always conserved. • Kinetic Energy is conserved during an elastic collision. • Kinetic Energy is not necessarily conserved during an inelastic collision. • Impulse-Linear Momentum Theorem • Relates strength and duration of collision force to change in momentum • Similar to Work-Kinetic Energy Theorem

  3. Equations • Δp = pf - pi = J (10-4) • J = F Δt (10-8) • (10-10) • (10-18) • (10-19)

  4. More Equations • . (10-28) • . (10-29) • . (10-34) • . (10-36)

  5. Last One • . (10-30)

  6. Problem Types • Impulse and Linear Momentum • Series of Collisions • Elastic Collision • 1-D • 2-D • Moving Target vs. Stationary Target • Inelastic Collision • 1-D • 2-D

  7. How to Solve Them • Impulse and Linear Momentum • Use Conservation of Momentum • Impulse-Linear Momentum Theorem • Series of Collisions • Use Equation (10-10) • Elastic Collisions • Use Conservation of Momentum • Use Conservation of Kinetic Energy • In Two Dimensions Treat Momentums as Vector Quantities (x,y), and Solve as Usual

  8. Problem #1 • A 283g air-track glider moving at 69 m/s on a 24 m long air track collides elastically with a 467g glider at rest in the middle of the horizontal track. The end of the track over which the struck glider moves is not frictionless, and the glider moves with a coefficient of friction =0.02 with respect to the track. Will the glider reach the end of the track? Neglect the length of the glider.

  9. m1 = 0.283 kg m2 = 0.467 kg Kinetic Energy Conserved (Elastic Collision) Momentum Conserved Ki=Kf (0.283 kg)(0.69 m/s)2 = (0.283 kg)(v1f)2 + (0.467 kg)(v2f)2 0.1445(v1f)2 + 0.2335(v2f)2 = 0.0674 Problem #1 Solution

  10. Pi = Pf m1v1i = m1v1f + m2v2f (0.283 kg)(0.69 m/s) = (0.283 kg)(-v1f) + (0.467 kg)(v2f) v1f = -0.69 m/s + (1.65)v2f Two Equations and Two Unknowns 0.1415 (-0.69 m/s + 1.65v2f)2 + 0.2335(v2f)2 = 0.067 Solve to find that v2f= 0.525 m/s Substitute this to Solve for Kinetic Energy of Struck Glider K = 1/2 m2(v2f)2 = 1/2 (0.467)(0.525)2 KE of struck glider = 0.064 J Problem #1 Solution (cont.)

  11. Problem #1 Solution (cont.) • Ffriction = μN = μm2g = (0.02)(0.467 kg)(9.8 m/s2) = 0.0915 N • Work Done by Friction = (0.0915 N)(1.2 m) = 0.11 J • K < Wfriction • Glider Does Not Reach the End of the Track

  12. Problem #2 • Wile E. Coyote, having just passed Physics 207 at VCU, splurges on a 500 kg ACME rocket. He reads the rocket specs to discover a relative exhaust speed of 600 m/s and a fuel consumption rate of 2 kg/s. • He launches horizontally after the roadrunner and flies for twenty seconds before the road turns and the rocket does not work. He collides with the 4000 kg ACME delivery truck (initially stationary). Neglecting gravity and air resistance, what is the relative velocity of the truck just after the totally inelastic collision? Assume the super genius to have a mass of 25 kg.

  13. Problem #2 (cont.) • Rocket Science is needed for this problem: • Use the second rocket equation to find the rocket speed at impact • vf - vi = u ln • vf = (600 m/s) ln • vf = 47.55 m/s

  14. Problem #2 (cont.) • Now solve for vtruck using conservation of momentum: • Pi = Pf • mrocket ivrocket i + mtruck i vtruck i = mrocket f vrocket f + mtruck f vtruck f • vf = = 5.14 m/s

  15. Recommendations for Further Study • If you have any more questions • Read your expensive textbook • Go see our Web Study Guide • Go harass Carlisle (see attached schedule)

More Related