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Marrying Kinematics and Dynamics 3.1.1 Impulse & Momentum.

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Presentation on theme: "Marrying Kinematics and Dynamics 3.1.1 Impulse & Momentum."— Presentation transcript:

1 Marrying Kinematics and Dynamics 3.1.1 Impulse & Momentum

2 momentum: property of MOVING objects –Depends on MASS and VELOCITY –vector : same direction as VELOCITY Definitions EquationUnits

3 What is the momentum of a 2000 kilogram car moving east with a speed of 5.0 meters per second? p = mv p = (2000 kg)(+5.0 m/s) F f = +10000 kg·m/s Example #1

4 Which has more momentum? 2,000 kg car with v = 5m/s 10,000 kg bus with v = 1m/s Both have p = 10,000 kg·m/s Neither Comparing Momentum

5 Which has more momentum? 30 kg player with v = 2 m/s 60 kg referee with v = 0 m/s The player has p = 60 kg·m/s The referee has p = 0. Player Comparing Momentum

6 impulse: A CHANGE IN MOMENTUM –Caused by a change in VELOCITY –vector : same direction as NET FORCE Definitions EquationUnits

7 A 1000 kilogram drag racing car accelerates from 0 to 330 miles per hour (148 meters per second) in 5.0 seconds. –What is the impulse experienced by the drag racer? p i = 0 p f = (1000 kg)(148 m/s) p f = 148000 kg·m/s J = Δp = 148000 kg·m/s Example #2 - Starting

8 A 1000 kilogram drag racing car accelerates from 0 to 330 miles per hour (148 meters per second) in 5.0 seconds. –What is the net force experienced by the drag racer? J = 148000 kg·m/s J = F net t 148000 kg·m/s = F net (5.0 s) F net = 29600 N Example #2 - Starting

9 A 1000 kilogram drag racing car accelerates from 0 to 330 miles per hour (148 meters per second) in 5.0 seconds. ALTERNATIVE SOLUTION a = Δv / t a = 148 m/s / 5.0 s a = 29.6 m/s 2 Example #2 - Starting a = F net / m 29.6 m/s 2 = F net / 1000 kg F net = 29600 N J = F net t J = (29600 N)(5.0 s) J = 148000 N·s

10 An F-4 aircraft with a mass of 2.7 x 10 4 kilograms lands while moving at 100 meters per second. The aircraft deploys a parachute that slows the aircraft to a speed of 40 meters per second in 8.0 seconds. –What force does the parachute exert on the F-4? Example #3 – Slowing Down a = Δv / t a = 60 m/s / 8.0 s a = 7.5 m/s 2 a = F net / m 7.5 m/s 2 = F net / 2.7E4 kg F net = 2.0E5 N

11 An F-4 aircraft with a mass of 2.7 x 10 4 kilograms lands while moving at 100 meters per second. The aircraft deploys a parachute that slows the aircraft to a speed of 40 meters per second in 8.0 seconds. –What force does the parachute exert on the F-4? Example #3 – Slowing Down p i = 2.7E6 kg·m/s p f = 1.08E6 kg·m/s J = Δp = 1.62E6 kg·m/s J = 1.62E6 kg·m/s J = F net t 1.62E6 kg·m/s = F net (8.0 s) F net = 2.0E5 N

12 End of 3.1.1 - PRACTICE


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