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Chapter 12: Momentum

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  1. Chapter 12: Momentum 12.1 Momentum 12.2 Force is the Rate of Change of Momentum 12.3 Angular Momentum

  2. Chapter 12 Objectives Calculate the linear momentum of a moving object given the mass and velocity. Describe the relationship between linear momentum and force. Solve a one-dimensional elastic collision problem using momentum conservation. Describe the properties of angular momentum in a system—for instance, a bicycle. Calculate the angular momentum of a rotating object with a simple shape.

  3. Chapter Vocabulary • angular momentum • collision • law of conservation of • momentum • elastic collision • gyroscope • impulse • inelastic collision • linear momentum • momentum

  4. Inv 12.1 Momentum Investigation Key Question: What are some useful properties of momentum?

  5. 12.1 Momentum • Momentum is a property of moving matter. • Momentum describes the tendency of objects to keep going in the same direction with the same speed. • Changes in momentum result from forces or create forces.

  6. 12.1 Momentum • The momentum of a ball depends on its mass and velocity. • Ball B has more momentum than ball A.

  7. 12.1 Momentum and Inertia • Inertia is another property of mass that resists changes in velocity; however, inertia depends only on mass. • Inertia is a scalar quantity. • Momentum is a property of moving mass that resists changes in a moving object’s velocity. • Momentum is a vector quantity.

  8. 12.1 Momentum • Ball A is 1 kg moving 1m/sec, ball B is 1kg at 3 m/sec. • A 1 N force is applied to deflect the motion of each ball. • What happens? • Does the force deflect both balls equally? • Ball B deflects much less than ball A when the same force is applied because ball B had a greater initial momentum.

  9. 12.1 Kinetic Energy and Momentum • Kinetic energy and momentum are different quantities, even though both depend on mass and speed. • Kinetic energy is a scalar quantity. • Momentum is a vector, so it always depends on direction. Two balls with the same mass and speed have the same kinetic energy but opposite momentum.

  10. p = m v 12.1 Calculating Momentum • The momentum of a moving object is its mass multiplied by its velocity. • That means momentum increases with both mass and velocity. Momentum (kg m/sec) Velocity (m/sec) Mass (kg)

  11. Comparing momentum A car is traveling at a velocity of 13.5 m/sec (30 mph) north on a straight road. The mass of the car is 1,300 kg. A motorcycle passes the car at a speed of 30 m/sec (67 mph). The motorcycle (with rider) has a mass of 350 kg. Calculate and compare the momentum of the car and motorcycle. • You are asked for momentum. • You are given masses and velocities. • Use: p = m v • Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s • Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s • The car has more momentum even though it is going much slower.

  12. 12.1 Conservation of Momentum • The law of conservation of momentum states when a system of interacting objects is not influenced by outside forces (like friction), the total momentum of the system cannot change. If you throw a rock forward from a skateboard, you will move backward in response.

  13. 12.1 Conservation of Momentum

  14. 12.1 Collisions in One Dimension • A collisionoccurs when two or more objects hit each other. • During a collision, momentum is transferred from one object to another. • Collisions can be elastic orinelastic.

  15. 12.1 Collisions

  16. Elastic collisions Two 0.165 kg billiard balls roll toward each other and collide head-on. Initially, the 5-ball has a velocity of 0.5 m/s. The 10-ball has an initial velocity of -0.7 m/s. The collision is elastic and the 10-ball rebounds with a velocity of 0.4 m/s, reversing its direction. What is the velocity of the 5-ball after the collision?

  17. Elastic collisions • You are asked for 10-ball’s velocity after collision. • You are given mass, initial velocities, 5-ball’s final velocity. • Diagram the motion, use m1v1 + m2v2 = m1v3 + m2v4 • Solve for V3 : (0.165 kg)(0.5 m/s) + (0.165 kg) (-0.7 kg)= (0.165 kg) v3 + (0.165 kg) (0.4 m/s) • V3 = -0.6 m/s

  18. Inelastic collisions A train car moving to the right at 10 m/s collides with a parked train car. They stick together and roll along the track. If the moving car has a mass of 8,000 kg and the parked car has a mass of 2,000 kg, what is their combined velocity after the collision? • You are asked for the final velocity. • You are given masses, and initial velocity of moving train car.

  19. Inelastic collisions • Diagram the problem, use m1v1 + m2v2 = (m1v1 +m2v2) v3 • Solve for v3= (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s) (8,000 + 2,000 kg) v3= 8 m/s The train cars moving together to right at 8 m/s.

  20. 12.1 Collisions in 2 and 3 Dimensions • Most real-life collisions do not occur in one dimension. • In a two or three-dimensional collision, objects move at angles to each other before or after they collide. • In order to analyze two-dimensional collisions you need to look at each dimension separately. • Momentum is conserved separatelyin the xand ydirections.

  21. 12.1 Collisions in 2 and 3 Dimensions

  22. Chapter 12: Momentum 12.1 Momentum 12.2 Force is the Rate of Change of Momentum 12.3 Angular Momentum

  23. Investigation Key Question: How are force and momentum related? 12.2 Force is the Rate of Change of Momentum

  24. 12.2 Force is the Rate of Change of Momentum • Momentum changes when a net force is applied. • The inverse is also true: • If momentum changes, forces are created. • If momentum changes quickly, large forces are involved.

  25. F = D p D t 12.2 Force and Momentum Change The relationship between force and motion follows directly from Newton's second law. Force (N) Change in momentum (kg m/sec) Change in time (sec)

  26. Calculating force • You are asked for force exerted on rocket. • You are given rate of fuel ejection and speed of rocket • Use F = Δ ÷Δt • Solve: Δ = (100 kg) (-25,000 kg m/s) ÷ (1s) = - 25,000 N • The fuel exerts and equal and opposite force on rocket of +25,000 N. Starting at rest, an 1,800 kg rocket takes off, ejecting 100 kg of fuel per second out of its nozzle at a speed of 2,500 m/sec. Calculate the force on the rocket from the change in momentum of the fuel.

  27. 12.2 Impulse • The product of a force and the time the force acts is called the impulse. • Impulse is a way to measure a change in momentum because it is not always possible to calculate force and time individually since collisions happen so fast.

  28. F D t = D p 12.2 Force and Momentum Change To find the impulse, you rearrange the momentum form of the second law. Impulse (N•sec) Change in momentum (kg•m/sec) Impulse can be expressed in kg•m/sec (momentum units) or in N•sec.

  29. Chapter 12: Momentum 12.1 Momentum 12.2 Force is the Rate of Change of Momentum 12.3 Angular Momentum

  30. Investigation Key Question: How does the first law apply to rotational motion? Inv 12.3 Angular Momentum

  31. 12.3 Angular Momentum • Momentum resulting from an object moving in linear motion is called linear momentum. • Momentum resulting from the rotation (or spin) of an object is called angular momentum.

  32. 12.3 Conservation of Angular Momentum • Angular momentum is important because it obeys a conservation law, as does linear momentum. • The total angular momentum of a closed system stays the same.

  33. L = Iw 12.3 Calculating angular momentum Angular momentum is calculated in a similar way to linear momentum, except the mass and velocity are replaced by the moment of inertia and angular velocity. Moment of inertia (kg m2) Angular momentum (kg m/sec2) Angular velocity (rad/sec)

  34. 12.3 Calculating angular momentum • The moment of inertia of an object is the average of mass times radius squared for the whole object. • Since the radius is measured from the axis of rotation, the moment of inertia depends on the axis of rotation.

  35. Calculating angular momentum An artist is making a moving metal sculpture. She takes two identical 1 kg metal bars and bends one into a hoop with a radius of 0.16 m. The hoop spins like a wheel. The other bar is left straight with a length of 1 meter. The straight bar spins around its center. Both have an angular velocity of 1 rad/sec. Calculate the angular momentum of each and decide which would be harder to stop. • You are asked for angular momentum. • You are given mass, shape, and angular velocity. • Hint: both rotate about y axis. • Use L= I, Ihoop = mr2, Ibar = 1/12 ml2

  36. Calculating angular momentum • Solve hoop: Ihoop= (1 kg) (0.16 m)2 = 0.026 kg m2 • Lhoop= (1 rad/s) (0.026 kg m2) = 0.026 kg m2/s • Solve bar: Ibar= (1/12)(1 kg) (1 m)2 = 0.083 kg m2 • Lbar = (1 rad/s) (0.083 kg m2) = 0.083 kg m2/s • The bar has more than 3x the angular momentum of the hoop, so it is harder to stop.

  37. 12.3 Gyroscopes angular momentum • A gyroscope is a device that contains a spinning object with a lot of angular momentum. • Gyroscopes can do amazing tricks because they conserve angular momentum. • For example, a spinning gyroscope can easily balance on a pencil point.

  38. 12.3 Gyroscopes angular momentum • A gyroscope on the space shuttle is mounted at the center of mass, allowing a computer to measure rotation of the spacecraft in three dimensions. • An on-board computer is able to accurately measure the rotation of the shuttle and maintain its orientation in space.

  39. Jet Engines • Nearly all modern airplanes use jet propulsion to fly. Jet engines and rockets work because of conservation of linear momentum. • A rocket engine uses the same principles as a jet, except that in space, there is no oxygen. • Most rockets have to carry so much oxygen and fuel that the payload of people or satellites is usually less than 5 percent of the total mass of the rocket at launch.