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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. - PowerPoint PPT Presentation
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Marrying Kinematics and Dynamics
3.1.1 Impulse & Momentum
• momentum: property of MOVING objects
– Depends on MASS and VELOCITY
– vector : same direction as VELOCITY
Definitions
mvp s
mkg Equation Units
• 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
Example #1
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
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
• impulse: A CHANGE IN MOMENTUM
– Caused by a change in VELOCITY
– vector : same direction as NET FORCE
Definitions
ptFJ net sN
Equation Units
s
mkg
• 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
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 t148000 kg·m/s = Fnet (5.0 s)
Fnet = 29600 N
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
Example #2 - Starting
a = Fnet / m
29.6 m/s2 = Fnet / 1000 kg
Fnet = 29600 N
J = Fnet t
J = (29600 N)(5.0 s)
J = 148000 N·s
• 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?
Example #3 – Slowing Down
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
• 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?
Example #3 – Slowing Down
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 t1.62E6 kg·m/s = Fnet (8.0 s)
Fnet = 2.0E5 N
End of 3.1.1 - PRACTICE
