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Impulse and Momentum

Section. Impulse and Momentum. 9.1. In this section you will:. Define the momentum of an object. Determine the impulse given to an object. Section. Impulse and Momentum. 9.1. Impulse and Momentum.

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Impulse and Momentum

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  1. Section Impulse and Momentum 9.1 In this section you will: • Define the momentum of an object. • Determine the impulse given to an object.

  2. Section Impulse and Momentum 9.1 Impulse and Momentum • The right side of the equation FΔt = mΔv, mΔv, involves the change in velocity: Δv = vf−vi. Therefore, mΔv = mvf−mvi • 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.

  3. Section Impulse and Momentum 9.1 Impulse and Momentum • 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. The left side describes the impulse, I = FΔt. Thus, the impulse on an object is equal to the change in its momentum, which is called the impulse-momentum theorem.

  4. Section Impulse and Momentum 9.1 Impulse and Momentum • 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.

  5. Section Impulse and Momentum 9.1 Impulse and Momentum • 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.

  6. Section Impulse and Momentum 9.1 Using the Impulse-Momentum Theorem • 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 also is 13.1 N·s. • 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.

  7. Section Impulse and Momentum 9.1 Using the Impulse-Momentum Theorem • What is the momentum of the ball after the collision? • Solve the impulse-momentum theorem for the final momentum. pf=pi+FΔt

  8. Section Impulse and Momentum 9.1 Using the Impulse-Momentum Theorem • 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 =−5.5 kg·m/s+13.1 kg·m/s = +7.6 kg·m/s

  9. Section Impulse and Momentum 9.1 Using the Impulse-Momentum Theorem • What is the baseball’s final velocity? Because pf =mvf,solving for vf yields the following:

  10. Section Impulse and Momentum 9.1 Using the Impulse-Momentum Theorem to Save Lives • 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. • 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.

  11. Section Impulse and Momentum 9.1 Using the Impulse-Momentum Theorem to Save Lives • 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.

  12. Section Impulse and Momentum 9.1 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 average force is exerted on the vehicle in each of these stops?

  13. Section Impulse and Momentum 9.1 Average Force Step 1: Analyze and Sketch the Problem

  14. Section Impulse and Momentum 9.1 Average Force Sketch the system. Include a coordinate axis and select the positive direction to be the direction of the velocity of the car.

  15. Section Impulse and Momentum 9.1 Average Force Draw a vector diagram for momentum and impulse.

  16. Section Impulse and Momentum 9.1 Average Force Identify the known and unknown variables. Known: m = 2200 kg Δtgentle braking= 21 s vi = +26 m/s Δthard braking= 3.8 s vi = +0.0 m/s Δthitting a wall= 0.22 s Unknown: Fgentle braking= ? Fhard braking= ? Fhitting a wall= ?

  17. Section Impulse and Momentum 9.1 Average Force Step 2: Solve for the Unknown

  18. Section Impulse and Momentum 9.1 Average Force Determine the initial momentum, pi, before the crash. pi = mvi

  19. Section Impulse and Momentum 9.1 Average Force Substitute m = 2200 kg, vi = +26 m/s pi = (2200 kg) (+26 m/s) = +5.7×104 kg·m/s

  20. Section Impulse and Momentum 9.1 Average Force Determine the initial momentum, pi, before the crash. pf = mvf

  21. Section Impulse and Momentum 9.1 Average Force Substitute m = 2200 kg, vf = +0.0 m/s pf = (2200 kg) (+0.0 m/s) = +0.0×104 kg·m/s

  22. Section Impulse and Momentum 9.1 Average Force Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle. FΔt = pf−pi

  23. Section Impulse and Momentum 9.1 Average Force Substitute pf = 0.0 kg·m/s, vi = 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

  24. Section Impulse and Momentum 9.1 Average Force Substitute Δtgentle braking = 21 s = −2.7×103 N

  25. Section Impulse and Momentum 9.1 Average Force Substitute Δthard braking = 3.8 s = −1.5×104 N

  26. Section Impulse and Momentum 9.1 Average Force Substitute Δthitting a wall = 0.22 s = −2.6×105 N

  27. Section Impulse and Momentum 9.1 Average Force Step 3: Evaluate the Answer

  28. Section Impulse and Momentum 9.1 Average Force • 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.

  29. Section Impulse and Momentum 9.1 Average Force • 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.

  30. Section Impulse and Momentum 9.1 Average Force • 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:

  31. Section Impulse and Momentum 9.1 Average Force • 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:

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