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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.

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Momentum

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  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

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