Physics Chapter 6. Momentum and Its Conservation

Transcription

1 Physics Chapter 6 Moentu and Its Conservation

2 Linear Moentu The velocity and ass of an object deterine what is needed to change its otion. Linear Moentu (ρ) is the product of ass and velocity ρ =v Unit is kg /s

3 Exaple 1 Exaple 1. Find the agnitude of the oentu of each cart shown below, before the collision.

4 Exaple 1. Find the agnitude of the oentu of each cart shown below, before the collision. Given: Find: ρ A Solution: A = 450 kg B = 550 kg n A = 4.50 /s n B = 3.70 /s and ρ B ρ A = A n A = (450 kg)(4.50 /s) = 2025 = 2.0 x 10 3 kg. /s ρ B = B n B = (550 kg)(3.70 /s) = 2035 = 2.0 x 10 3 kg. /s

5 Exaple 2. The an in this picture has a ass of 85.0 kg. If he strikes the ground at 7.7 /s, what is his change in oentu (Dρ) Assue his ass reains the sae after the collision. Given: Find: Solution: = 85.0 kg n f = 0.0 /s n i = -7.7 /s (negative for down) D ρ D ρ = n f - n i = (n f - n i ) = (85.0 kg)((0.0 /s (-7.7 /s)) = 650 kg. /s

6 If the velocity is changed by an outside force, then the oentu is also changed. F=a= (Dv/Dt) = Dv/Dt = D ρ/dt F=Dρ/Dt FDt = D ρ =D v = v f -v i **Dt is the tie during which the force is applied. ** FDt is called ipulse in N s This is called the Ipulse- Moentu Theore.

7 Exaple 3. A player exerts a force (assue constant) of 12.5 N on a ball over a period of 0.35 s. Find the total ipulse. Given: F = 12.5 N Find: Ipulse Dt = 0.35 s Solution: Ipulse = FDt = (12.5 N)(0.35 s) = 4.4 N. s

8 Exaple 4: Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 /s and is aied at the side of a car, which stops it without splashing it back (kind of a fake proble, but that s OK). What is the force exerted by the water on the car each second. = 1.5 kg v i =20/s v f =0/s Dt=1s

9 F C eans F C Dt =D ρ force on water by F C =D ρ /Dt car (F w is F C = (ρ f ρ i ) / Dt force on car by water F C = (v f v i ) / Dt which is F C = (v f v i ) / Dt equal and opposite) F C = 1.5 kg (0 /s 20 /s) / 1 s F WC,x C = -30 = -30 N N F W F= CW,x 30 = N30 N

10 6A and 6B, page 209 and 211

11 Stopping Distance Stopping distance depends on the Ipulse-Moentu Theore. A truck takes longer to stop than a bicycle Also, A change in oentu over a longer tie requires less force. Ex: nets, air attresses, padding (see fig. 6-5, pg. 214)

12 Calculating tie and distance Stopping tie can be calculated using FDt = D p, and solving for Dt. Stopping distance can be calculated using Dx = ½(v i +v f )Dt

13 Exaple A kg sphere traveling west slows fro /s to /s. How long does it take the sphere to decelerate if the force on the sphere is 840 N to the east? How far does the sphere travel during the deceleration?

14 First find change in tie =720.0kg v i = /s v f =-12.00/s F= +840N FDt=Dp Dt=Dp/F=(v f -v i )/F =[(720kgX-12.00/s)- (720.0kgX-230.0/s)]/840N=186.9s

15 Then find change in x Dx=1/2 (v i +v f )Dt= 1/2(-230.0/s-12.00/s)(186.9s) = = 23,000 to the West

16 Newton s 3 rd Law As two objects collide, the ipulse (FDt) is the sae for each. Newton s 3 rd law tells us that the forces are equal and opposite which leads to conservation of oentu. Forces in real collisions are not constant (see fig. 6-9, pg. 220)

17 Conservation of Moentu in Collisions When two objects collide, the oentu of the individual objects change, but the total oentu reains the sae. p 1,i + p 2,i = p 1,f + p 2,f 1 v 1,i + 2 v 2,i = 1 v 1,f + 2 v 2,f Friction is disregarded

18 Exaple pg. 218 A 76 kg person, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the person oves out of the boat with a velocity of 2.5 /s to the right, what is the final velocity of the boat?

19 List Knowns and Unknowns and forula v i,boat =0 v i,person =0 v f,boat =? v f,person =2.5/s person =76kg boat =45kg p v p,i + b v b,i = p v p,f + b v b,f = (76kg)(2.5/s) + (45kg)(v b,f ) V bf = - (76X2.5)/45 = -4.2/s (negative eans the boat is oving in the opposite direction)

20 Exaple 2. A proton with an initial velocity of 235 /s strikes a stationary alpha particle(he) and rebounds off with a velocity of 188 /s. Find the final velocity of the alpha particle. Mass of proton= 1.67 x kg Mass of alpha particle (He)= 6.64 x kg

21 Given: p = 1.67 x kg He = 6.64 x kg n p,i = 235 /s n He,i = 0.0 /s n p,f = -188 /s Find: n He,f Original Forula: p ν p,i He ν He,i ν p p,f He ν He,f Now, the initial velocity of He (alpha particle) is zero, so cross it out!

22 Given: p = 1.67 x kg He = 6.64 x kg n p,i = 235 /s n He,i = 0.0 /s n p,f = -188 /s Find: n He,f Original Forula: p ν p,i ν p p,f He ν He,f Solve for n He,f

23 Given: p = 1.67 x kg He = 6.64 x kg n p = 235 /s n He = 0.0 /s n p,f = -188 /s Find: n He,f Working Forula: ν He,f p ν p,i - He p ν p, f

24 Given: p = 1.67 x kg He = 6.64 x kg n p = 235 /s n He = 0.0 /s n p,f = -188 /s Find: n He,f Working Forula: The ass of the proton doesn t change, so ν He,f p (ν p,i - He ν p,f )

25 ν He,f 1.67x / s kg(235 /s - (-188 /s)) 6.64 x kg

26 Practice 6C page 213 and 6D page 219 To be copleted in class toorrow no hoework tonight.

27 Perfectly Inelastic Collisions A collision is perfectly inelastic when they collide and ove together as one ass with a coon velocity. 1 v 1,i + 2 v 2,i = ( )v f

28 Exaple 1. A oving railroad car, ass=m, speed=3/s, collides with an identical car at rest. The cars lock together as a result of the collision. What is their coon speed afterward?

29 V i1 V i2 =0 M M before V f? M M after

30 V i1 V i2 =0 x V f? x M M M M before after p 1i = p 2f M(+V i1 ) + 0= (M+M)(V f ) V f = MV i1 / 2M = V i1 / 2 V f = (3/s) / 2 = 1.5/s

31 6E, page 224

32 Kinetic Energy and Perfectly Inelastic Collisions During perfectly inelastic collisions, soe kinetic energy is lost because the objects are defored. Once ass and velocities are known (using conservation of oentu), solve for initial and final kinetic energies (1/2v 2 ) using the forula fro Ch. 5 to deterine energy lost. KE i = KE 1,i + KE 2,i KE f = KE 1,f + KE 2,f DKE = KE f - KE i

33 Exaple 1. A red clay ball with a ass of 1.5 kg and an initial velocity of 4.5 /s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the cobined ass. Given: Find: n f r = 1.5 kg b = 2.8 kg n r,i = 4.5 /s n b,i = 0 /s, KE lost Original Forula: r ν r,i ν b b,i ( r b )νf

34 Exaple 1. A red clay ball with a ass of 1.5 kg and an initial velocity of 4.5 /s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the cobined ass. Then, find the loss of Kinetic Energy. Given: Find: r = 1.5 kg b = 2.8 kg n r,i = 4.5 /s n b,i = 0 /s n f, KE lost Original Forula: The black ball starts at rest, so r ν r,i ν b b,i ( r b )νf

35 Exaple 1. A red clay ball with a ass of 1.5 kg and an initial velocity of 4.5 /s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the cobined ass. Given: Find: r = 1.5 kg b = 2.8 kg n r,i = 4.5 /s n b,i = 0 /s n f, KE lost Original Forula: The black ball starts at rest, so r ν r,i ( r b )νf

36 Exaple 1. A red clay ball with a ass of 1.5 kg and an initial velocity of 4.5 /s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the cobined ass. Given: Find: r = 1.5 kg b = 2.8 kg n r = 4.5 /s n b = 0 /s n f, KE lost Working Forula: ν f ν r r ( r b )

37 Exaple 1. A red clay ball with a ass of 1.5 kg and an initial velocity of 4.5 /s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the cobined ass. Given: Find: Solution: r = 1.5 kg b = 2.8 kg n r,i = 4.5 /s n b,i = 0 /s n f, KE lost ν f ( ν r r r,i b ) (1.5 kg)(4.5 /s) (1.5 kg 2.8 kg) /s

38 Recall the forula for kinetic energy. KE 1 v 2 2 Was the kinetic energy conserved in exaple 1?

39 KE red KE 1 2 v black r 2 r v r (1.5 kg)(4.5 /s) 2 r 1 2 (2.8 kg)(0.0 /s) J 2 KE before = 15 J 0 J

40 KE after 1 2 rb v f (4.3 kg)(1.6 /s) J

41 KE before = 15 J KE after = 5.3 J DKE = 5.3 J-15 J = -9.7 J

42 Where does the issing kinetic energy go? Soe energy is converted to internal energy, soe is converted into sound.

43 6F page 226

44 Elastic Collisions Two objects collide and return to their original shape with no loss of Kinetic Energy. The two objects ove separately after the collision. Total oentu and KE reain constant. See page 230. Most collisions lose energy as sound, internal elastic potential energy and friction we assue perfect collisions.

45 Forula for elastic collisions 1 v 1,i + 2 v 2,i = 1 v 1,f + 2 v 2,f ½ 1 v 1,i2 + ½ 2 v 2,i 2 = ½ 1 v 1,f2 + ½ 2 v 2,f 2

46 Exaple 1. A kg blue arble oves right with a velocity of 17 /s and akes an elastic head-on collision with kg red arble oving left with a velocity of 15 /s. After the collision, the blue arble oves to the left with a velocity of 15.0 /s. Find the velocity of the red arble after the collision.

47 Exaple 1. Given: Find: n r,f b = kg r = kg n b,i = 17 /s n r,i, = -15 /s n b,f = -15 /s Original Forula: b ν b,i ν r r,i ν b b,f r ν r,f ν r,f b ν b,i ν r r r,i - b ν b,f

48 ν r,f b ν b,i ν r r r,i - (0.025 kg)(17 /s) (0.025 kg)(-15 /s) kg b ν b,f - (0.025 kg)(-15 /s) =17/s

49 Now you can plug the nubers into the conservation of kinetic energy forula to check your work which they will ask you to do! (Reeber there is NO LOSS OF KINETIC ENERGY with elastic collisions!!) ½ 1 v 1,i2 + ½ 2 v 2,i 2 = ½ 1 v 1,f2 + ½ 2 v 2,f 2

50 ½ 1 v 1,i2 + ½ 2 v 2,i 2 = ½ 1 v 1,f2 + ½ 2 v 2,f 2 ½ (.025)(17) 2 + ½ (.025)(-15) 2 = ½ (.025)(-15) 2 + ½ (.025)(17) J = 6.425J

51 Hoework Probles 6G, page 229

52 Review Probles pg ,15,16,17,18,24,26,27,29, 31,32,34, 36,38,40,54