html5-img
1 / 21

Impulse and Momentum

Impulse and Momentum. Chapter problems Serway 5,6,10,13,16,17,18,27,29,33,43,44,52,54,59,60 cw.prenhall.com/~bookbind/pubbooks/giancoli. Linear momentum & impulse. Linear momentum is defined as the product of mass and velocity p=mv, p x =mv x , p y = mv y units of momentum are kgm/s

salome
Download Presentation

Impulse and Momentum

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. Impulse and Momentum • Chapter problems Serway • 5,6,10,13,16,17,18,27,29,33,43,44,52,54,59,60 • cw.prenhall.com/~bookbind/pubbooks/giancoli

  2. Linear momentum & impulse • Linear momentum is defined as the product of mass and velocity • p=mv, px=mvx , py= mvy • units of momentum are kgm/s • From Newtons 2nd law • F= ma F=mdv/dt F= dp/dt • The rate of momentum change with respect to time is equal to the resultant force on an object • The product of Force and time is known as IMPULSE • J= Fdt • units of impulse are Ns

  3. Linear momentum & impulse Examples of impulses being applied on everyday objects

  4. Impulse Momentum Theorem Fdt=mdv You apply an impulse on an object and you get an equal change in momentum Area under a Force vs time graph

  5. Impulse Graph

  6. Linear Momentum and Impulse Example problems 1,2,3 Chapter questions 5,6,10,13,16

  7. Conservation of momentum2 particle system For gravitational or electrostatic force F12 is force of 1 on 2 F21 is force of 2 on 1 m2 m1 F12 F21 F12 =dp1/dt F21 = dp2/dt

  8. Conservation of momentum2 particle system From Newton’s 3rd Law F12 = - F21 or F12 + F21 = 0 F12 is force of 1 on 2 F21 is force of 2 on 1 m2 m1 F12 F21 F12 + F21 =dp1/dt + dp2/dt = 0 d(p1 + p2)/dt= 0 Since this derivative is equal to 0

  9. Conservation of momentum2 particle system Since this derivative is equal to 0 d(p1 + p2)/dt= 0 then integration yields p1 + p2 = a CONSTANT F12 is force of 1 on 2 F21 is force of 2 on 1 m2 m1 F12 F21 Thus the total momentum of the system of 2 particles is a constant.

  10. Conservation of linear momentum Provided the particles are isolated from external forces, the total momentum of the particles will remain constant regards of the interaction between them F12 m1 F21 m2 Simply stated: when two particles collide,their total momentum remains constant. pi = pf p1i + p2i = p1f + p2f (m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

  11. Conservation of linear momentum Serway problems 9.2 17 & 18

  12. Collisions

  13. Collisions Event when two particles come together for a short time producing impulsive forces on each other., No external forces acting. Or for the enthusiast: External forces are very small compared to the impulsive forces Types of collisions 1) Elastic- Momentum and Kinetic energy conserved 2) Inelastic- Momentum conserved, some KE lost 3) Perfectly(completely) Inelastic- Objects stick together

  14. Collisions in 1 d Perfectly Elastic 1) Cons. of mom. 2) KE lost in collision 3) KE changes to PE

  15. Elastic Collision Calculation2 objects

  16. Collisions - Examples Computer Simulations example 2, problems 5,24,29 Serway Problems 27,29,33,37

  17. Collisions in 2 dimensions After Collision x momentum before collision equals x momentum after the collision mavaf 1 mavafx mavax Before collision mb vel=0 p=0 mbvbxf 2 mbvbf

  18. Collisions in 2 dimensions mavax= mavafx + mbvbxf or mavax= mavaf cos1 + mbvbf cos2

  19. Collisions in 2 dimensions After Collision y momentum before collision equals y momentum after the collision mavaf 1 mavax mavayf Before collision mb vel=0 p=0 Velocity y axis =0 py=o 2 Mbvbyf mbvbf

  20. Collisions in 2 dimensions 0=mavafy - mbvbfy or 0= mavaf sin1-mbvbf sin2

  21. Collisions in 2 dimensions 0= mavaf sin1-mbvbf sin2 mavax= mavaf cos1 + mbvbf cos2 Problems ex 9.9 43,44

More Related