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# Linear Momentum

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1. Linear Momentum • p = mass X velocity • p = mv • Unit = kg-m/s • Vector Quantity

2. What is the momentum of a 100.0 kg football player running at 6.0 m/s? p = mv = (100.0 kg)(6.0 m/s) = 600 kg-m/s

3. If a freshman running at 2 m/s has a momentum of 80 kg m/s, what is his mass?

4. Momentum and Force Second Law of Motion “rate of change of momentum of a body is equal to the net force applied on it.” SF = Dp (this is the more general form) Dt

5. Derivation SF = Dp = mv-mvo Dt Dt SF = m(v-vo) Dt SF = mDv Dt SF = ma

6. Impulse • Impulse = Dp Dp = FDt • Usually occurs over a very short timeframe

7. Force area = impulse time

8. Impulse: Example 1 A 50-g golf ball is struck with a force of 2400 N. The ball flew off with a velocity of 44 m/s. How long was the club in contact with the ball? Dp = pf – pi Dp = mvf – 0 (ball was still) Dp = (0.050 kg)(44 m/s) = 2.2 kg-m/s SF = Dp/Dt Dt = Dp/SF Dt = 2.2 kg-m/s/(2400 kg-m/s2) = 9.1 X 10-4 s

9. Impulse: Example 2 In a crash test, a 1500 kg car moving at –15.0 m/s collides with a wall for 0.150 s. The bumpers on the car cause it to rebound at +2.60 m/s. Calculate the impulse, and the average force exerted on the car. Impulse = Dp pi = mvi = (1500 kg)(-15.0 m/s) = -22,500 kg-m/s pf = mvf = (1500 kg)(+2.60 m/s) = +3900 kg-m/s

10. Dp = pf – pi Dp = 3900 kg-m/s – (-22,500 kg-m/s) Dp = 26,400 kg-m/s To calculate the force: SF = Dp Dt SF = 26,400 kg-m/s = 176,000 N 0.150 s

11. Impulse and Area • Impulse = Area under Force vs. Time graph

12. Can use the average Force to approximate an answer Average Force Area = FaveDt

13. Connection to Calculus • You integrate to get area Dp = FDt dp = Fdt p = F dt = J tf to

14. Calculate the impulse (Dp) given the following graph.

15. Calculate the impulse (Dp) given the following graph.

16. Calculate the impulse (Dp) given the following graph.

17. A 150.0 g baseball is thrown with a speed of 20.0 m/s. The interaction force vs. time is shown in the following graph. • Calculate the impulse. (4.50 kg m/s) • Calculate the return speed of the baseball. (10.0 m/s)

18. A 100.0 g golf ball is dropped from a height of 2.00 meters from the floor. • Calculate the speed of the ball just as it hits the ground. (6.26 m/s) • Calculate the impulse. (1.2 kg m/s) • Calculate the return speed of the ball just as it bounces back.(5.74 m/s) • Calculate the return height of the ball. (1.68 m)

19. The Law of Conservation of Momentum “The total momentum of an isolated system of bodies remains constant m1v1 + m2v2 = m1v1’ + m2v2’ momentumbefore = momentumafter

20. A 0.015 kg bullet is fired with a velocity of 200 m/s from a 6 kg rifle. What is the recoil velocity of the rifle (consider it’s direction)?

21. A 10g and a 30 g ball are placed inside a tube with a compressed spring between them. When the spring is released, the 10 g ball exits the tube at 6.0 m/s. • Calculate the speed of the 30 g ball. • Does it have to be in the opposite direction?

22. A 10,000 kg railroad car moving at 24.0 m/s strikes an identical car that is stationary. They lock together. What will be there common speed afterwards?

23. Mr. West (75 kg) sees a stationary cart (25.0 kg) 8.00 m ahead of him in the hallway. He accelerates at 1.00 m/s2 and jumps on the cart. • Calculate his speed just as he jumps on the cart • Calculate his speed as he and the cart zoom down the hall.

24. Rocket Ship • Can use the Third Law (“equal and opposite forces”) • Can also use Law of Conservation of momentum • Dp = 0 (since neither the gas or rocket were moving initially) • Which moves more, the rocket or the gas?

25. A 0.145 kg ball is thrown at +40.2 m/s. The bat has a mass of 0.840 kg. • How fast must you swing the bat to return the ball at 20.1 m/s? (-10.4 m/s) • How fast must you swing the bat to return the ball at 40.2 m/s? (-13.9 m/s)

26. Types of Collisions • Elastic • KE and momentum conserved • Inelastic • Momentum conserved • KE not conserved • Also, if energy is added (like in a chemical reaction) • Perfectly inelastic • Two objects stick together after the collision

27. Perfectly Inelastic Collision: Ex 1 An 1800-kg Cadillac is stopped at a traffic light. It is struck in the rear by a 900-kg Ion moving at 20.0 m/s. The cars become entangled. Calculate their velocity after the collision. KE is not conserved m1v1 + m2v2 = m1v1’ + m2v2’ m1v1 + m2v2 = (m1 + m2)vf 0 + (900 kg)(20.0 m/s) = (1800kg + 900kg)vf vf= +6.67 m/s

28. How much KE was lost as a result of the collision? KEi = ½ (900 kg)(20 m/s)2 = 180,000 J KEf = ½ (2700 kg)(6.67 m/s)2 = 60,000 J DKE = 60,000 J – 180,000 J = -120,000J

29. Perfectly Inelastic Collision: Ex 2 Two balls of mud collide head on and stick together. The first ball of mud had a mass of 0.500 kg and was moving at +4.00 m/s. The second had a mass of 0.250 kg and was moving at –3.00 m/s. Find the velocity of the ball after the collision. m1v1 + m2v2 = m1v1’ + m2v2’ m1v1 + m2v2 = (m1 + m2)vf (0.500 kg)(4.00 m/s)+(0.250 kg)(-3.00 m/s) = (0.750 kg)(vf) vf = +1.67 m/s

30. A lab experiment, a 200 g air track glider and a 400 g glider collide and stick with velcro. The 200 g glider was moving to the right at 3.00 m/s. Afterwards, the system is moving to the left at 0.40 m/s. Calculate the initial speed of the 400 g glider. (-2.1 m/s)

31. A uranium atom (238 amu) decays into a small fragment and a “daughter” nucleus. The small fragment is ejected at 1.50 X 107 m/s and the daughter nucleus at 2.56 X 105 m/s. Calculate the mass of both products. (234 amu, 4 amu)

32. Collisions in 2 or 3 Dimensions (Glancing collisions) • A moving marble collides with a stationary marble. • Which of the following situations is/are not possible after:

33. 2-D Collisions: Example 1 At an intersection, a 1500-kg car travels east at 25 m/s. It collides with a 2500-kg van traveling north at 20 m/s. If the vehicles stick together afterwards, calculate the magnitude and velocity of the cars after the collision.

34. Let’s first work with the initial components Spix = (1500 kg)(25 m/s) Spix = 37,500 kg-m/s Spiy = (2500 kg)(20 m/s) Spiy = 50,000 kg-m/s After the collision: v vy = v sinq q vx = v cosq

35. After the collision: Spix = Spfx 37,500 kg-m/s = mvx 37,500 kg-m/s = (1500 kg + 2500 kg)vx 37,500 kg-m/s = (4000 kg)vcosq vcosq = 9.375 m/s Spiy = Spfy 50,000 kg-m/s = mvy 50,000 kg-m/s = (1500 kg + 2500 kg)vy 50,000 kg-m/s = (4000 kg)vsinq vsinq = 12.5 m/s

36. Divide the two equations vsinq = 12.5 m/s vcosq = 9.375 m/s sinq = 1.33 tanq = 1.33 q = 53o cosq Now solve for v vsinq = 12.5 m/s v = (12.5 m/s)/sin (53o) v = 16 m/s

37. A 5000 kg van travelling south at 10.0 m/s collides with a 2500 kg SUV travelling west. After the collision, the vehicles stuck together and made a 45o angle with the horizontal in a south-west direction. a. Calculate the initial speed of the 2500 kg car. b. Calculate the speed just after the collision. (Remember, two equations, two unknowns).

38. 2-D Collisions: Example 3 A pool ball moving at +3.0 m/s strikes another pool ball (same mass) that is initially at rest. After the collision, the two balls move off at + 45o. Calculate the speeds of the two balls. (2.1 m/s) vred q = +450 q = -450 vgreen

39. Let’s first work with the initial components Spix = (m)(3 m/s) Spiy = 0 After the collision (x-direction): vredx = vredcos(45o) vgreenx = vgreencos(-45o) Spix = Spfx Spix = predfx + pgreenfx (m)(3 m/s) = mvredcos(45o) + mvgreencos(-45o) 3 m/s = vredcos(45o) + vgreencos(-45o)

40. After the collision (y-direction): vredy = vredsin(45o) vgreeny = vgreensin(-45o) Spiy = Spfy Spiy = predfy + pgreenfy 0 = mvredsin(45o) + mvgreensin(-45o) mvredsin(45o) = -mvgreensin(-45o) vredsin(45o) = -vgreensin(-45o)

41. Here are the two equations we will solve: 3 m/s = vredcos(45o) + vgreencos(-45o) vredsin(45o) = -vgreensin(-45o) vred = -vgreensin(-45o) sin(45o) vred = vgreen 3 m/s = vredcos(45o) + vredcos(-45o) 3 m/s = vred(cos(45o) + cos(-45o)) 3 m/s = 1.414vred vred = 2.1 m/s = vgreen

42. Two zombie heads of unequal mass sit on an icerink. The first head (m1 = 4.00 kg) is propelled toward the stationary second head (m2 = 2.00 kg) at a velocity of v1= 4.00 m/s. After the collision, both heads move off at 35o to the x-axis. Calculate their final speeds. 1.63 m/s, 3.25 m/s

43. Two different pucks are placed on an air hockey table. A 15.0 g red puck is pushed to the right at 1.00 m/s. A 20.0 g green puck is pushed to the left at 1.20 m/s. After the collision, the green puck travels at 1.10 m/s at an angle of 40.0o south of the horizontal. • Calculate the x and y components of the green puck’s velocity after the collision. (-0.843 m/s, -0.707 m/s) • Calculate the x and y components of the green puck’s momentum after the collision. (-0.0169 kg m/s, -0.0141 kg m/s) • Find the speed and direction of the red puck. (1.08 m/s, 60.7o north of east) • Can you express your answer to (c) in “i and j” notation? ((0.527 i + 0.940j) m/s)