Chapter 9: Momentum and Its Conservation

1. PHYSICS Principles and Problems Chapter 9: Momentum and Its Conservation

2. Momentum and Its Conservation CHAPTER9 BIG IDEA If the net force on a closed system is zero, the total momentum of that system is conserved.

3. Table Of Contents CHAPTER9 Section 9.1 Impulse and Momentum Section 9.2 Conservation of Momentum Click a hyperlink to view the corresponding slides. Exit

4. Impulse and Momentum SECTION9.1 MAIN IDEA An object’s momentum is equal to its mass multiplied by its velocity. • What is impulse? • What is momentum? • What is angular momentum? Essential Questions

5. Impulse Momentum Impulse-momentum theorem Angular momentum Angular impulse-angular momentum theorem Impulse and Momentum SECTION9.1 Review Vocabulary • Angular velocity the angular displacement of an object divided by the time needed to make the displacement New Vocabulary

6. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem Click image to view the movie.

7. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • The right side of the equation FΔt = mΔv, mΔv, involves the change in velocity: Δv = vf−vi. • Therefore, mΔv = mvf−mvi

8. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • The product of the object’s mass, m, and the object’s velocity, v, is defined as the momentum of the object. Momentum is measured in kg·m/s. An object’s momentum, also known as linear momentum, is represented by the following equation: Momentump = mv • The momentum of an object is equal to the mass of the object times the object’s velocity.

9. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • Recall the equation FΔt = mΔv = mvf−mvi. Because mvf= pf and mvi = pi, you get: FΔt = mΔv = pf−pi • The right side of this equation, pf−pi, describes the change in momentum of an object. Thus, the impulse on an object is equal to the change in its momentum, which is called the impulse-momentum theorem.

10. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • The impulse-momentum theorem is represented by the following equation. Impulse-Momentum Theorem FΔt = pf−pi • The impulse on an object is equal to the object’s final momentum minus the object’s initial momentum.

11. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • If the force on an object is constant, the impulse is the product of the force multiplied by the time interval over which it acts. • Because velocity is a vector, momentum also is a vector. • Similarly, impulse is a vector because force is a vector. • This means that signs will be important for motion in one dimension.

12. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • Let’s discuss the change in momentum of a baseball. The impulse that is the area under the curve is approximately 13.1 N·s. The direction of the impulse is in the direction of the force. Therefore, the change in momentum of the ball is also 13.1 N·s.

13. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • Because 1 N·s is equal to 1 kg·m/s, the momentum gained by the ball is 13.1 kg·m/s in the direction of the force acting on it.

14. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • What is the momentum of the ball after the collision? • Solve the impulse-momentum theorem for the final momentum. pf=pi+FΔt

15. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • The ball’s final momentum is the sum of the initial momentum and the impulse. Thus, the ball’s final momentum is calculated as follows. pf=pi+13.1 kg·m/s • Mass of the ball is 0.145kg and velocity, before collision, is -38m/s. Therefore, the baseball’s pi = (0.145kg)(-38m/s) = -5.5 kg·m/s =−5.5 kg·m/s+13.1 kg·m/s = +7.6 kg·m/s

16. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • What is the baseball’s final velocity? Because pf =mvf,solving for vf yields the following:

17. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • What happens to the driver when a crash suddenly stops a car? • An impulse is needed to bring the driver’s momentum to zero. • A large change in momentum occurs only when there is a large impulse.

18. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • A large impulse can result either from a large force acting over a short period of time or from a smaller force acting over a long period of time.

19. Impulse and Momentum SECTION9.1 Impulse-Momentum Theorem (Cont.) • According to the impulse-momentum equation, FΔt = pf − pi, the final momentum, pf, is zero. The initial momentum, pi, is the same with or without an air bag. • Thus, the impulse, FΔt, also is the same.

20. Impulse and Momentum SECTION9.1 Average Force • A 2200-kg vehicle traveling at 94 km/h (26 m/s) can be stopped in 21 s by gently applying the brakes. It can be stopped in 3.8 s if the driver slams on the brakes, or in 0.22 s if it hits a concrete wall. What is the impulse exerted on the vehicle? What average force is exerted on the vehicle in each of these stops?

21. Impulse and Momentum SECTION9.1 Average Force (Cont.) Step 1: Analyze and Sketch the Problem • Sketch the system. • Include a coordinate axis and select the positive direction to be the direction of the velocity of the car. • Draw a vector diagram for momentum and impulse.

22. Impulse and Momentum SECTION9.1 Average Force (Cont.) Identify the known and unknown variables. Known: m = 2200 kg Δtgentle braking= 21 s vi = +26 m/s Δthard braking= 3.8 s vf = +0.0 m/s Δthitting a wall= 0.22 s Unknown: Fgentle braking= ? Fhard braking= ? Fhitting a wall= ?

23. Impulse and Momentum SECTION9.1 Average Force (Cont.) Step 2: Solve for the Unknown

24. Impulse and Momentum SECTION9.1 Average Force (Cont.) Determine the initial momentum, pi, before the crash. pi = mvi

25. Impulse and Momentum SECTION9.1 Average Force (Cont.) Substitute m = 2200 kg, vi = +26 m/s pi = (2200 kg) (+26 m/s) = +5.7×104 kg·m/s

26. Impulse and Momentum SECTION9.1 Average Force (Cont.) Determine the final momentum, pf, before the crash. pf = mvf

27. Impulse and Momentum SECTION9.1 Average Force (Cont.) Substitute m = 2200 kg, vf = +0.0 m/s pf = (2200 kg) (+0.0 m/s) = +0.0 kg·m/s

28. Impulse and Momentum SECTION9.1 Average Force (Cont.) Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle. FΔt = pf−pi

29. Impulse and Momentum SECTION9.1 Average Force (Cont.) Substitute pf = 0.0 kg·m/s, pi = 5.7×104 kg·m/s FΔt = (+0.0×104 kg·m/s) − (− 5.7×104 kg·m/s) = −5.7×104 kg·m/s

30. Impulse and Momentum SECTION9.1 Average Force (Cont.) Substitute Δtgentle braking = 21 s = −2.7×103 N

31. Impulse and Momentum SECTION9.1 Average Force (Cont.) Substitute Δthard braking = 3.8 s = −1.5×104 N

32. Impulse and Momentum SECTION9.1 Average Force (Cont.) Substitute Δthitting a wall = 0.22 s = −2.6×105 N

33. Impulse and Momentum SECTION9.1 Average Force (Cont.) Step 3: Evaluate the Answer

34. Impulse and Momentum SECTION9.1 Average Force (Cont.) Are the units correct? Force is measured in newtons. Does the direction make sense? Force is exerted in the direction opposite to the velocity of the car and thus, is negative.

35. Impulse and Momentum SECTION9.1 Average Force (Cont.) Is the magnitude realistic? People weigh hundreds of newtons, so it is reasonable that the force needed to stop a car would be in thousands of newtons. The impulse is the same for all three stops. Thus, as the stopping time is shortened by more than a factor of 10, the force is increased by more than a factor of 10.

36. Impulse and Momentum SECTION9.1 Average Force (Cont.) Step 1: Analyze the Problem Sketch the system. Include a coordinate axis and select the positive direction to be the direction of the velocity of the car. Draw a vector diagram for momentum and impulse. The steps covered were:

37. Impulse and Momentum SECTION9.1 Average Force (Cont.) Step 2: Solve for the Unknown Determine the initial momentum, pi, before the crash. Determine the final momentum, pf, after the crash. Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle. Step 3: Evaluate the Answer The steps covered were:

38. Impulse and Momentum SECTION9.1 Angular Momentum • The angular velocity of a rotating object changes only if torque is applied to it. • This is a statement of Newton’s law for rotating motion, τ = IΔω/Δt.

39. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • This equation can be rearranged in the same way as Newton’s second law of motion was, to produce τΔt = IΔω. • The left side of this equation is the angular impulse of the rotating object and the right side can be rewritten as Δω = ωf− ωi.

40. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • The angular momentum of an object is equal to the product of a rotating object’s moment of inertia and angular velocity. Angular MomentumL = Iω • Angular momentum is measured in kg·m2/s.

41. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • Just as the linear momentum of an object changes when an impulse acts on it, the angular momentum of an object changes when an angular impulse acts on it. • Thus, the angular impulse on the object is equal to the change in the object’s angular momentum, which is called the angular impulse-angular momentum theorem.

42. Impulse and Momentum Angular Impulse-Angular Momentum Theorem τΔt = Lf − Li SECTION9.1 Angular Momentum (Cont.) • The angular impulse-angular momentum theorem is represented by the following equation.

43. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • If there are no forces acting on an object, its linear momentum is constant. • If there are no torques acting on an object, its angular momentum is also constant. • Because an object’s mass cannot be changed, if its momentum is constant, then its velocity is also constant.

44. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • In the case of angular momentum, however, the object’s angular velocity does not remain constant. This is because the moment of inertia depends on the object’s mass and the way it is distributed about the axis of rotation or revolution.

45. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • Thus, the angular velocity of an object can change even if no torques are acting on it. • Observe the animation.

46. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • How does she start rotating her body? • She uses the diving board to apply an external torque to her body. • Then, she moves her center of mass in front of her feet and uses the board to give a final upward push to her feet. • This torque acts over time, Δt, and thus increases the angular momentum of the diver.

47. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • Before the diver reaches the water, she can change her angular velocity by changing her moment of inertia. She may go into a tuck position, grabbing her knees with her hands.

48. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • By moving her mass closer to the axis of rotation, the diver decreases her moment of inertia and increases her angular velocity.

49. Impulse and Momentum SECTION9.1 Angular Momentum (Cont.) • When she nears the water, she stretches her body straight, thereby increasing the moment of inertia and reducing the angular velocity. • As a result, she goes straight into the water.

50. Section Check SECTION9.1 Define the momentum of an object. A. Momentum is the ratio of change in velocity of an object to the time over which the change happens. B. Momentum is the product of the average force on an object and the time interval over which it acts. C. Momentum of an object is equal to the mass of the object times the object’s velocity. D. Momentum of an object is equal to the mass of the object times the change in the object’s velocity.