1. CHAPTER-9 Momentum & Its Conservation

4. 9.1 Impulses And Momentum How is Velocity Affected by Force? The force changes over time. What is the physics, instant by instant, behind serving a tennis ball?

6. The force acting on the tennis ball quickly increases, maxes out, and then quickly decreases. Just after contact the ball deforms or is “squeezed”, the strings are stretched and the force acting on the ball is increased. The force reaches its maximum, the ball begins to snap away from the strings and the ball will begin to regain its shape.

7. The force “F” acting on the tennis ball begins at 0N, then rapidly increases to a force >1000 times the weight of the ball, then the F will quickly drop back to 0N. The entire event last only thousandths of a second. What would the force vs time graph look like? GRAPH……………..

9. Relating Impulse and Momentum Newton’s Second Law equation =???? F = ma Where a = Δv/Δt (t is time of contact)  F = m Δv/Δt Multiplying each side by Δt yields FΔt = mΔv Impulse = FΔt (units Ns)

10. Impulse = ? Impulse is the product of the average force (F net ) exerted on an object and the time interval over which it acts on the object. Impulse = FΔt Back to: mΔv m(v 2 – v 1 ) mv 2 – mv 1  FΔt = mv 2 – mv 1

11. Linear momentum = ρ ρ = mv ρ = mv Impulse = Δρ Impulse-Momentum Theorem = FΔt = Δρ = ρ1 – ρ 2 = mΔv = mv 2 – mv 1 Demo: I need one volunteer to jump from a lab table, then discuss jump in detail. I need one volunteer to jump from a lab table, then discuss jump in detail.

12. Q: Why did _________ bend his/her knees when landing on the floor? A: To lesson the force of the impact on the knees, back, feet, … so he/she would not be hurt. Bending of the knees “absorbs” the shock of the jump. Q: How/why does bending the knees “absorb” the shock? A: By extending the time it takes to stop. Q: What does time have to do with it?

13. Draw sketch on board and explain why lengthening time decreases the amount of force exerted on the knees, back, feet, … FΔt = Δρ = ρ 2 – ρ 1 = mΔv = mv 2 – mv 1 FΔt = mv 2 – mv 1 mv 1 = just before hits ground, max speed, maximum momentum mv 2 = stopped, no motion, no momentum

14. FΔt = Δρ  Δρ (mv 2 -mv 1 ) is a fixed value and cannot change and if Δt increases, then the force (F) exerted on the knees, back, feet … must decrease.

15. Using the Impulse-Momentum Theorem What is the Δρ of the tennis ball if the area under the graph is 1.8? What units would apply to area? The units = Ns The area = Impulse = Δρ = 1.8Ns(kgm/s)

16. The ball began at rest,ρ = 0, and then gained impulse/speed while in contact with the racket to a final ρ = 1.8Ns (kgm/s) Q: If the mass of the tennis ball is 0.055kg, what is the final velocity of the ball?

17. Δρ = ρ 2 – ρ 1 Δρ = mv 2 – mv 1 1.8Ns = 0.055kg(v 2 ) – 0 1.8Ns = 0.055kg(v 2 ) V 2 = 1.8(kgm/s)/0.055kg V 2 = 32.7m/s Because velocity is a vector quantity, so is Impulse/momentum  the signs (+/-) are important

18. Using the Impulse-Momentum Theorem to Save Lives A large Δρ occurs when there is a large Impulse. A large impulse occurs when: 1. large force acting over a short period of time 2. small force acting over a long period of time

19. EGG EXAMPLE If you remember nothing else about this class, remember this example, it could save your life some day.

20. Car Accident – seat belt/air bag vs windshield/tree/pole An 80kg person is traveling in a car at 60m/s, the car hits a tree and stops in 0.04 seconds. The driver “A” is wearing a seat belt  increasing the stopping time to 0.95 seconds. What force will the driver experience during the stop?

21. FΔt = mΔv Δv is the same with/without the air bag and/or seat belt FΔt = mv 2 – mv 1 FΔt = 0 – mv1 FΔt = mv 1 F = mv 1 /Δt F = (80kg)(60m/s)/0.95s F = 5052.6kgm/s 2 F = 5052.6N

22. An 80kg person is traveling in a car at 60m/s, the car hits a tree and stops in 0.04 seconds. The driver “B” is not wearing a seat belt. What force will the driver experience during the stop? F = (80kg)(60m/s)/0.04s F = 120,000kgm/s 2 F = 120,000N vs 5052.6N (24X greater F) REMEMBER THIS EXAMPLE FOREVER!

23. Example-Stopping an SUV A 2200kg SUV traveling at 94km/hr (___m/s) can be stopped in a)21s by gently applying the brakes, b)5.5s by a panic stop and c)0.22s if it hits a wall. What is the average force exerted on the SUV for each of the stops?

24. SOLUTION Draw a sketch, table known/unknown, equations, steps…

25. FΔt = Δρ FΔt = ρ 2 – ρ 1 Ρ2 = 0  FΔt = - ρ 1 F a = - ρ 1 /Δt F a = -mv 1 /Δt F a = -(2200kg)(26m/s)/21s F a = -2724kgm/s 2 F a = -2724N

26. F b = -mv 1 /Δt F b = -(2200kg)(26m/s)/5.5s F b = -10,400kgm/s 2 F b = -10,400N F c = -mv 1 /Δt F c = -(2200kg)(26m/s)/0.22s F c = -260,000kgm/s 2 F c = -260,000N

27. 9.2 THE CONSERVATION OF MOMENTUM Law of Conservation of Momentum = The total momentum of a system before a collision/explosion is equal to the total momentum of the system after the collision/explosion as long as there is no net external force acting on the system. p 1 = p 2 p A1 + p B1 = p A2 + p B2

28. Momentum in a Closed System Closed System = A system (collection of objects) that does not loose or gain mass. Internal Forces = All the forces within a closed system. External Forces = All the forces outside the system.

29. Isolated System = When the net external force on a closed system is zero. No system on Earth is truly closed and isolated. There is always some interaction between a system and its environment. Usually these interactions are small and can be neglected by physics problems.

31. EXAMPLE A 6050kg truck traveling at 36m/s rear ends a car with a mass of 852kg traveling 21m/s on ice in the same direction. The two vehicles become entangled after the collision. How fast does the wreckage move immediately after the collision?

32. Draw Sketch TableEquations Plug in numbers Show all work/steps

33. Law of Conservation of Momentum p 1 = p 2 p T1 + p c1 = p T2 + p C2 m T v T1 + m C v C1 = m T v T2 + m C v C2 Where v T2 = v C2 ∴ m T v T1 + m C v C1 = (m T + m C )v 2 v 2 = (m T v T1 +m C v C1 )/(m T + m C ) v 2 = (6050kg)(36m/s) + (852kg)(21m/s) (6050kg + 852kg) (6050kg + 852kg)

34. 217,800kgm/s + 17,892kgm/s 217,800kgm/s + 17,892kgm/s V 2 = 6902kg 235,692kgm/s 235,692kgm/s V 2 = 6902kg V 2 = 34.15m/s

35. Explosions & Explosions in Space Two skaters are at rest facing each other. They push off of each other and move in opposite directions. What are their velocities immediately after the push? v A2 = v B2 =

36. Law of Conservation of Momentum p 1 = p 2 p A1 + p B1 = p A2 + p B2 0 = p A2 + p B2 p A2 = -p B2 m A v A2 = -m B v B2 Solving for v A2 v A2 = -m B v B2 /m A Solving for v B2 v B2 = m A v A2 /(-m B )

37. Example Problem #2 An astronaut in space fires a pistol, the thrust expells 35g of gas at -875m/s. the combined mass of the astronaut/pistol is 84kg. What is the velocity and direction of the astronaut after firing the pistol?

38. Two-Dimensional Collisions A 2kg ball (A) moving at a speed of 5m/s collides with a stationary ball (B) that also has a mass of 2kg. After the collision, ball-A moves off 30 o to the left, ball-B moves off 90 o to the right of ball-A. How fast are they moving after the collision? Possible Bonus/white board/…?