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Marrying Kinematics and Dynamics

Marrying Kinematics and Dynamics. 3.1.1 Impulse & Momentum. Definitions. momentum: property of MOVING objects Depends on MASS and VELOCITY vector : same direction as VELOCITY. Equation. Units. Example #1.

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Marrying Kinematics and Dynamics

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  1. Marrying Kinematics and Dynamics 3.1.1 Impulse & Momentum

  2. Definitions • momentum: property of MOVING objects • Depends on MASSandVELOCITY • vector : same direction as VELOCITY Equation Units

  3. Example #1 • 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) Ff = +10000 kg·m/s

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

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

  6. Definitions • impulse: A CHANGE IN MOMENTUM • Caused by a change in VELOCITY • vector : same direction as NET FORCE Equation Units

  7. Example #2 - Starting • 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? pi = 0 pf = (1000 kg)(148 m/s) pf = 148000 kg·m/s J = Δp = 148000 kg·m/s

  8. Example #2 - Starting • 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 = Fnet t 148000 kg·m/s = Fnet (5.0 s) Fnet = 29600 N

  9. Example #2 - Starting • 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/s2 J = Fnet t J = (29600 N)(5.0 s) J = 148000 N·s a = Fnet / m 29.6 m/s2 = Fnet / 1000 kg Fnet = 29600 N

  10. Example #3 – Slowing Down • An F-4 aircraft with a mass of 2.7 x 104 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? a = Δv / t a = 60 m/s / 8.0 s a = 7.5 m/s2 a = Fnet / m 7.5 m/s2 = Fnet / 2.7E4 kg Fnet = 2.0E5 N

  11. Example #3 – Slowing Down • An F-4 aircraft with a mass of 2.7 x 104 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? pi = 2.7E6 kg·m/s pf = 1.08E6 kg·m/s J = Δp = 1.62E6 kg·m/s J = 1.62E6 kg·m/s J = Fnet t 1.62E6 kg·m/s = Fnet (8.0 s) Fnet = 2.0E5 N

  12. End of 3.1.1 - PRACTICE

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