In this section you will: Review the momentum of an object. Determine the impulse given to an object. Section 9.1-1

Momentum Review 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. Momentum p = mv The momentum of an object is equal to the mass of the object times the object’s velocity. Section 9.1-4

Change in momentum is Impulse- Impulse and Momentum Change in momentum is Impulse- FΔt = mΔv = pf − pi pf − pi, describes the change in momentum of an object. The impulse (FΔt) on an object is equal to the change in its momentum. Section 9.1-5

Impulse-Momentum Theorem FΔt = pf − pi 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. Section 9.1-6

Impulse and Momentum Because velocity is a vector, momentum also is a vector. Impulse is a vector because force is a vector. Section 9.1-7

Using the Impulse-Momentum Theorem The impulse on the ball is 13.1 kgm/s What is the momentum of the ball after the collision? Solve the impulse-momentum theorem for the final momentum. pf = pi + FΔt Section 9.1-10

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 Section 9.1-11

Using the Impulse-Momentum Theorem What is the baseball’s final velocity? Because pf = mvf, solving for vf yields the following: Section 9.1-12

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. Section 9.1-13

Using the Impulse-Momentum Theorem to Save Lives 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. Section 9.1-14

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. Section 9.1-15

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? Section 9.1-16

Average Force Sketch the system. Include a coordinate axis and select the positive direction to be the direction of the velocity of the car. Section 9.1-18

Average Force Draw a vector diagram for momentum and impulse. Section 9.1-19

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 vf = +0.0 m/s Δthitting a wall = 0.22 s Unknown: Fgentle braking = ? Fhard braking = ? Fhitting a wall = ? Section 9.1-20

Step 2: Solve for the Unknown Average Force Step 2: Solve for the Unknown Section 9.1-21

Average Force Determine the initial momentum, pi, before the crash. pi = mvi Section 9.1-22

Average Force Substitute m = 2200 kg, vi = +26 m/s pi = (2200 kg) (+26 m/s) = +5.7×104 kg·m/s Section 9.1-23

Average Force Determine the final momentum, pf, before the crash. pf = mvf Section 9.1-24

Average Force Substitute m = 2200 kg, vf = +0.0 m/s pf = (2200 kg) (+0.0 m/s) = +0.0 kg·m/s Section 9.1-25

Average Force Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle. FΔt = pf − pi Section 9.1-26

Average Force 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 Section 9.1-27

Average Force Substitute Δtgentle braking = 21 s = −2.7×103 N Section 9.1-28

Average Force Substitute Δthard braking = 3.8 s = −1.5×104 N Section 9.1-29

Average Force Substitute Δthitting a wall = 0.22 s = −2.6×105 N Section 9.1-30

Step 3: Evaluate the Answer Average Force Step 3: Evaluate the Answer Section 9.1-31

Question 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. Section 9.1-48

Answer 1 Reason: The momentum of an object is equal to the mass of the object times the object’s velocity p = mv. Momentum is measured in kg·m/s. Section 9.1-49

Question 2 Mark and Steve are playing baseball. Mark hits the ball with an average force of 6000 N and the ball snaps away from the bat in 0.2 ms. Steve hits the same ball with an average force of 3000 N and the ball snaps away in 0.4 ms. Which of the following statements is true about the impulse given to the ball in both the shots is true? Reduced line spacing here, to fit the content. Section 9.1-50

Question 2 A. The impulse given to the ball by Mark is twice the impulse given by Steve. B. The impulse given to the ball by Mark is four times the impulse given by Steve. C. The impulse given to the ball by Mark is the same as the impulse given by Steve. D. The impulse given to the ball by Mark is half the impulse given by Steve. Reduced line spacing here, to fit the content. Section 9.1-51

Answer 2 Reason: Impulse is the product of the average force on an object and the time interval over which it acts. Since the product of the average force on the ball and the time interval of the impact in both the shots is the same, the impulse given to the ball by Mark is the same as the impulse given by Steve. Reduced line spacing here, to fit the content. Section 9.1-52

Answer 2 Reason: Impulse given to the ball by Mark = (6000 N) (0.2×10−3 s) = 1.2 N·s Impulse given to the ball by Steve = (3000 N) (0.4×10−3 s) Reduced line spacing here, to fit the content. Section 9.1-52

Question 3 In a baseball game, a pitcher throws a ball with a mass of 0.145 kg with a velocity of 40.0 m/s. The batter hits the ball with an impulse of 14.0 kg·m/s. Given that the positive direction is toward the pitcher, what is the final momentum of the ball? A. pf = (0.145 kg)(40.0 m/s) + 14.0 kg·m/s B. pf = (0.145 kg)(–40.0 m/s) – 14.0 kg·m/s C. pf = (0.145 kg) (40.0 m/s) – 14.0 kg·m/s D. pf = (0.145 kg)(–40.0 m/s) + 14.0 kg·m/s Section 9.1-53

Answer 3 Reason: By the impulse-momentum theorem, pf = pi + Ft where, pi = mvi Ft = impulse pf = mvi + impulse Reduced line spacing here, to fit the content. Section 9.1-54

Answer 3 Reason: Since the positive direction is toward the pitcher, vi is taken as negative as the ball is moving away from the pitcher before the batter hits the ball. The impulse is positive because the direction of the force is toward the pitcher. Therefore, pf = mvi + impulse = (0.145 kg)(−40 m/s) + 14 kg·m/s. Reduced line spacing here, to fit the content. Section 9.1-54