1 / 24

Momentum

Momentum. Momentum.  product of mass and velocity. m = mass (kg). p = m v. v = velocity (m/s). p = momentum ( kg m/s ). Unlike energy, momentum is a vector. Direction matters. EX. What is the momentum of a 3.5 kg medicine ball traveling 4.3 m/s north?. p = m v.

avye-burt
Download Presentation

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. Momentum

  2. Momentum  product of mass and velocity m = mass (kg) p = m v v = velocity (m/s) p = momentum (kg m/s) Unlike energy, momentum is a vector. Direction matters. EX. What is the momentum of a 3.5 kg medicine ball traveling 4.3 m/s north? p = m v = 15 kg m/s north = 3.5 kg (4.3 m/s N)

  3. CHANGES OBJECT’S MOMENTUM The Impulse-Momentum Theorem F Dt = m Dv impulse: a net force multiplied by the time it acts Derived from Newton’s 2nd Law:

  4. 1320 kg car traveling 40.0 m/s. 68 kg passenger applies brakes 8.5 s to bring car (and himself) to rest. Find avg. force seatbelt must exert on person over this time. = –320 N Suppose Dt = 0.12 s instead. = –2.3 x 104 N

  5. 255 g ball traveling 12 m/s hits wall and rebounds at 12 m/s . If ball contacts wall for 0.18 s, find force of wall on ball. + =  = ??? = –34 N (v is a vector!)

  6. Three Types of Collisions 1. elastic: objects bounce; no energy lost -- mechanical energy is conserved  SMEf = SMEi -- momentum is conserved  Spf = S pi -- best approximate examples: billiards; a super ball -- best nearly-perfect example: gas particles at high temp. and low pressure

  7. “energy not easily recovered,” “energy disappeared.” “Energy lost” means NOT Three Types of Collisions (cont.) objects stick together 2. perfectly inelastic: -- some energy is “lost” as heat or deformation -- momentum is conserved  Spf = S pi -- e.g., railroad cars coupling; a “good” tackle in football

  8. Three Types of Collisions (cont.) 3. inelastic: objects bounce, but some energy is “lost” -- momentum is conserved  Spf = S pi In any closed system, momentum is ALWAYS conserved. Spf = S pi

  9. 14 kg toddler throws 0.38 kg football 4.7 m/s . Find recoil velocity of child. Spf = S pi mchildvchild + mballvball = 0 14 kg (vchild) + 0.38 kg (4.7 m/s) = 0 vchild = –0.13 m/s (i.e., 0.13 m/s )

  10. 8.0 ton rocket travels at 3.0 mi/s . After firing engines, rocket’s mass drops to 7.0 tons and it travels at 5.0 mi/s . Has momentum been conserved? Explain. m = 8.0 T v = 3.0 mi/s  m = 7.0 T v = 5.0 mi/s  (engines fire) Ans. #1: YES; p is ALWAYS conserved. ??? Ans. #2: YES, and don’t forget the exhaust gas. vgas = –11 mi/s

  11. pf pi Train Car A has mass 25,000 kg and init. vel. 5.0 m/s  . Car B, w/mass 15,000 kg, is initially at rest. Cars collide and couple. Find vel. at which cars move off together. B B A A = = 40,000 kg (vf) 25,000 kg (5 m/s ) vf = 3.1 m/s 

  12. “HA!” Is kinetic energy conserved? Ans. #1: NO; ME isn’t conserved for perfectly inelastic collisions. Ans. #2: NO. (Show me!) How much energy was lost as heat (and/or used to deform coupling mechanism)? “Lost” energy = (3.1 x 105 – 1.9 x 105) J = 1.2 x 105 J

  13. 55 kg shopping cart travels at 3.0 m/s . 19 kg child, moving 5.0 m/s , jumps onto end of cart. Find final vel. of cart/child system. pf pi + =  = 74 vf 19(5) + 55(–3) vf = –0.95 m/s (i.e., )

  14. Handcarts A and B are on train track. “A” w/passenger has mass 185 kg and moves at 3.5 m/s . “B” w/passenger has mass 150 kg and moves at 2.0 m/s . B starts out 15 m ahead of A. 15 m 150 kg 185 kg A B 2.0 m/s  3.5 m/s  a. How long before A catches B? 15 m to make up; 1.5 m/s “closing speed”  10. s b. How far do A and B go in this time? A: 3.5 m/s (10. s) = 35 m B: 2.0 m/s (10. s) = 20. m

  15. pi pf c. What % of initial KE is “lost” if handcarts couple? 335 kg 185 kg 150 kg vf 3.5 m/s  2.0 m/s  (2.83 m/s) = 335 (vf) 185 (3.5) + 150 (2.0) = 1433 J = 1341 J 6.4% lost % lost = 92 J / 1433 J

  16. vf = ? Conservation of Momentum in 2-D Car A, w/mass 1340 kg, travels east at 21.3 m/s. Car B, w/mass 2150 kg, travels north at 18.4 m/s. Vehicles collide and couple. Find resultant vel. of coupled vehicles. A For a system, total momentum is conserved, and so are components of momentum. 1340 kg 21.3 m/s 2150 kg 18.4 m/s B Spf = S pi  Spxf = Spxi ANDSpyf = Spyi

  17. vf = ? vxf = 8.18 m/s vf 11.34 m/s Q vyf = 11.34 m/s 8.18 m/s A 1340 kg 21.3 m/s 2150 kg Spxi = Spxf 18.4 m/s B = 3490 (vxf) 1340 (21.3) Spyi = Spyf = 3490 (vyf) 2150 (18.4) vf = 14.0 m/s @ 54.2o N of E

  18. 63o A B vxf = –2.30 m/s (i.e., 2.30 m/s ) Car A, w/mass 1500 kg, goes 22 m/s at 63o S of E. Car B, w/mass 1900 kg, goes 12 m/s W. Vehicles collide and couple. Find resultant vel. of two-car object. + =  1500 kg 1900 kg 22 m/s 12 m/s vf = ? Spxi = Spxf 3400 (vxf) + 1900(–12) 1500 (22 cos 63) = –7818.3 = 3400 (vxf)

  19. + =  63o A vyf = 8.65 m/s 2.30 m/s B Q 8.65 m/s vf 1500 kg 1900 kg 22 m/s 12 m/s Spyi = Spyf = 3400 (vyf) 1500 (22 sin 63) vf = 9.0 m/s @ 75o S of W

  20. 44 m 18o 30.o Car A, w/mass 1260 kg, goes E. Car B, w/mass 1630 kg, goes 30.o N of E. Vehicles collide and couple. The skid marks are 44 m long at 18o N of E. Coeff. of friction is 0.80. Find initial speeds of A and B. vAi = ? 1260 kg A mk = 0.80 B 1630 kg vBi = ?

  21. 0 Draw FBD of coupled cars… v Fn = Fg = 2890 (9.81) = 28351 N Fk 22681 N Fk = mk Fn = 0.80 (28351) = 22681 N Fn Fg 28351 N Wnet = DKE Fnet d = ½ m (vf2 – vi2) –22681 (44) = ½ (2890) (–vi2) vi = 26.3 m/s (@ 18o) This is vel. of (A+B) immediately after impact.

  22. 44 m 18o vAi = ? 1260 kg 26.3 m/s A mk = 0.80 1630 kg 30.o vBi = ? B Spyf = Spyi 2890 (26.3 sin 18) = 1630 (vBi sin 30) vBi = 28.8 m/s vBi = 29 m/s (~64 mph) Spxf = Spxi = 1630 (28.8 cos 30) + 1260 vAi 2890 (26.3 cos 18) (~56 mph) vAi = 25 m/s vAi = 25.1 m/s

  23. 44 m 18o vAi = ? 1260 kg 26.3 m/s A mk = 0.80 1630 kg 30.o vBi = ? B Spyf = Spyi 2890 (26.3 sin 18) = 1630 (vBi sin 30) vBi = 29 m/s (~64 mph) Spxf = Spxi = 1630 (29 cos 30) + 1260 vAi 2890 (26.3 cos 18) (~56 mph) vAi = 25 m/s

  24. p = m v F Dt = m Dv KE = ½ m v2

More Related