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Download Unit 6 MOMENTUM AND ITS Conservation 1
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Principles of Technology/Physics in Context (PT/PIC) Chapter 6 Momentum and Its Conservation 1 Text p. 111-113 Principles of Technology/Physics in Context (PT/PIC) • During the lecture assessment questions will be asked. • The assessment questions will be collected at the end of the lecture for grading. Key Objectives At the conclusion of this chapter you’ll be able to: • Define the term momentum, and state its SI unit. • Solve problems involving mass, velocity, and momentum. • Define the term impulse, and state its SI unit. Key Objectives At the conclusion of this chapter you’ll be able to: • Relate impulse to change in momentum. • Solve impulse-momentum problems. 6.1 MOMENTUM • Suppose we wanted to measure the “tendency” of an object to remain in motion. • We would find that two factors are necessary to describe this tendency: • the mass and the velocity of the object. 6.1 MOMENTUM • The more mass an object has, the more force is required to bring it to rest. • Similarly, the greater the velocity of the object, the more force is necessary to bring it to rest. • These two factors, mass and velocity, are combined into a single quantity that we call momentum. 6.1 MOMENTUM • We define momentum, symbolized as p. as the product of mass and velocity: • p=mv 6.1 MOMENTUM • Momentum is a vector quantity and its direction is the direction of the velocity of the object. Its unit is the kilogram meter per second (kg∙m/s). • p=mv Assessment Question 1 • All of the following are true EXCEPT: A.Momentum is the “tendency” of an object to remain in motion is. B.The more mass an object has, the more force is required to bring it to rest. C.The greater the velocity of the object, the more force is necessary to bring it to rest. D.Mass and velocity, are combined into a single quantity that we call momentum. E.Momentum is a scalar quantity and its magnitude is the magnitude of the velocity of the object. 6.1 MOMENTUM • PROBLEM • An object whose mass is 3.5 kilograms is traveling at 20. meters per second [east] Calculate the momentum of the object. 6.1 MOMENTUM • SOLUTION Assessment Question 2 • An object whose mass is 35 kilograms is traveling at 42 meters per second [west] Calculate the momentum of the object. • p = mv = 35 kg ∙ 42 m/s [west] = A.0.833 kg∙m/s [west] B.1.2 kg∙m/s [west] C.7 kg∙m/s [west] D.77 kg∙m/s [west] E.1500 kg∙m/s [west] 6.2 NEWTON’S SECOND LAW AND MOMENTUM • When Newton developed his second law of motion, he recognized that the unbalanced force on an object caused a change in the object momentum: 6.2 NEWTON’S SECOND LAW AND MOMENTUM • Since mass is usually constant: 6.2 NEWTON’S SECOND LAW AND MOMENTUM • We can rewrite Newton’s second law in the following form: 6.2 NEWTON’S SECOND LAW AND MOMENTUM • We call the quantity F∙t, the impulse delivered to the object. • Impulse is symbolized by the letter J, and its unit is the newton second (N∙s). • Impulse is a vector quantity, and its direction is the direction of the net force. 6.2 NEWTON’S SECOND LAW AND MOMENTUM • From Newton’s second law it follows that the impulse delivered to the object changes its momentum, and the unit newton second is equivalent to the unit kilogram meter per second. Assessment Question 3 • All of the following are true EXCEPT A. The unbalanced force on an object caused a change in the object momentum B. The quantity F∙t, the impulse delivered to the object. C. Impulse is symbolized by the letter J, and its unit is the newton second (N∙s). D. Impulse is a vector quantity, and its direction is the direction of the net force. E. The impulse delivered to an object does not affect its momentum. 6.2 NEWTON’S SECOND LAW AND MOMENTUM • PROBLEM • A 5.0-kilogram object traveling at 3.0 meters per second [east] is subjected to a force that increases its velocity to 7.0 meters per second [east] • Calculate: (a) the initial momentum of the object, 6.2 NEWTON’S SECOND LAW AND MOMENTUM • SOLUTION • Calculate: (a) the initial momentum of the object, Assessment Question 4 • • A. B. C. D. E. A 15 kg (m) object traveling at 13 m/s [west] (vi) is subjected to a force that increases its velocity to 27 m/s [west] (vf) Calculate the initial momentum of the object. pi = mvi = 15 kg∙13 m/s [west] = 0.087 kg∙m/s [west] 1.2 kg∙m/s [west] 28 kg∙m/s [west] 200 kg∙m/s [west] 350 kg∙m/s [west] 6.2 NEWTON’S SECOND LAW AND MOMENTUM • PROBLEM • A 5.0-kilogram object traveling at 3.0 meters per second [east] is subjected to a force that increases its velocity to 7.0 meters per second [east] • Calculate: (b) the final momentum of the object, 6.2 NEWTON’S SECOND LAW AND MOMENTUM • SOLUTION • Calculate: (b) the final momentum of the object, Assessment Question 5 • • A. B. C. D. E. A 15 kg (m) object traveling at 13 m/s [west] (vi) is subjected to a force that increases its velocity to 27 m/s [west] (vf) Calculate the final momentum of the object, pf = mvf = 15 kg∙27 m/s [west] = 0.55 kg∙m/s [west] 1.8 kg∙m/s [west] 54 kg∙m/s [west] 240 kg∙m/s [west] 410 kg∙m/s [west] 6.2 NEWTON’S SECOND LAW AND MOMENTUM • PROBLEM • A 5.0-kilogram object traveling at 3.0 meters per second [east] is subjected to a force that increases its velocity to 7.0 meters per second [east] • Calculate: (c) the change in momentum of the object, 6.2 NEWTON’S SECOND LAW AND MOMENTUM • SOLUTION • Calculate: (c) the change in momentum of the object, Assessment Question 6 • A 15 kg (m) object traveling at 13 m/s [west] (vi) is subjected to a force that increases its velocity to 27 m/s [west] (vf) • Calculate the change in momentum of the object Δp = pf - pi = mvf - mvi = Δp = 15 kg∙27 m/s [W] - 15 kg∙13 m/s [W] = A. B. C. D. E. 55 kg∙m/s [west] 130 kg∙m/s [west] 210 kg∙m/s [west] 340 kg∙m/s [west] 420 kg∙m/s [west] 6.2 NEWTON’S SECOND LAW AND MOMENTUM • PROBLEM • A 5.0-kilogram object traveling at 3.0 meters per second [east] is subjected to a force that increases its velocity to 7.0 meters per second [east] • Calculate: (d) the impulse delivered to the object. 6.2 NEWTON’S SECOND LAW AND MOMENTUM • SOLUTION • Calculate: (d) the impulse delivered to the object. Assessment Question 7 • A 51 kg (m) object traveling at 31 m/s [west] (vi) is subjected to a force that increases its velocity to 72 m/s [west] (vf). • Calculate the impulse delivered to the object. • J = F∙t = Δp = pf - pi = mvf - mvi = Δp = 51 kg∙72 m/s [W] - 51 kg∙31 m/s [W] = A. 75 kg∙m/s [west] B. 210 kg∙m/s [west] C. 510 kg∙m/s [west] D. 1300 kg∙m/s [west] E. 2100 kg∙m/s [west] 6.2 NEWTON’S SECOND LAW AND MOMENTUM • PROBLEM • A 5.0-kilogram object traveling at 3.0 meters per second [east] is subjected to a force that increases its velocity to 7.0 meters per second [east] • Calculate: (e) If the force acts for 0.20 second, what are its magnitude and its direction? 6.2 NEWTON’S SECOND LAW AND MOMENTUM • SOLUTION • Calculate: (e) If the force acts for 0.20 second, what are its magnitude and its direction? Assessment Question 8 • • A. B. C. D. E. If an impulse of 55 kg∙m/s [w] acts for 15 second, what are its magnitude and its direction of the force of the impulse? F = J/t = 55 kg∙m/s [W] / 15 s = 0.27 N [w] 3.7 N [w] 40 N [w] 70 N [w] 825 N [w] Assessment Question 9 • • A. B. C. D. E. If an impulse of 85 kg∙m/s [W] (J) acts with a force of 98 N [W](F) how long is the impulse (t)? t = J/F = 85 kg∙m/s [W] / 98 N [W] = 0.87 s 1.2 s 13 s 290 s 830 s Conclusion • Momentum is the product of mass and velocity. • It is a vector quantity that measures the tendency of an object to remain in motion. • An impulse is the product of force and time and is responsible for changing the momentum of an object. • An impulse is also a vector quantity. Assessment Question 10 • All of the following are true EXCEPT: A. The product of mass and velocity is momentum. B. Momentum has magnitude and direction. C. Momentum measures the tendency of an object to remain in motion. D. An impulse is the product of force and time and is responsible for changing the momentum of an object. E. An impulse has magnitude only.