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Marrying Kinematics and Dynamics

This resource covers key concepts related to impulse and momentum in physics. Momentum is defined as the property of moving objects that depends on mass and velocity, represented as a vector in the same direction as velocity. The impulse is the change in momentum caused by a change in velocity, also a vector in the direction of net force. Examples include calculations of momentum and impulse for a drag racing car and an aircraft with a parachute. Learn how to apply the formulas for momentum (p = mv) and impulse (J = Δp) through practical scenarios and comparisons.

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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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