MOMENTUMConservation of Momentum Perfectly Elastic Collision Momentum Conserved...
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LINEAR MOMENTUM Momentum Impulse Conservation of Momentum Inelastic Collisions Elastic Collisions Momentum In 2 Dimensions
Transcript of MOMENTUMConservation of Momentum Perfectly Elastic Collision Momentum Conserved...
LINEAR MOMENTUMMomentum
Impulse
Conservation of Momentum
Inelastic Collisions
Elastic Collisions
Momentum In 2 Dimensions
MOMENTUM
β’ Quantity of Motion
β’ Product of Mass and Velocity
β’ π = ππ£ = kg βm
s
β’ Vector Quantity
IMPULSE
β’ Change in Momentum
β’ To change momentum, apply a force for a period of time.
β’ π½ = βπ = ππ£2 βππ£1 = πβπ£ = πΉ β π‘ = (π β π )
IMPULSEβ’ Follow Through Example (Bunt vs. Swing)
β’ Apply force for longer period of time = larger momentum change
Impulse (Follow Through)
Nordic Ski Racing Slap shot
Impulse (reduce force)
F*t = mΞv = F*tβ’Helmets
β’Padding
Impulse (reduce force) F*t = mΞv = F*tβ’ Air Bag β’ Crumple Zone
Impulse Examplesβ’ A soccer player kicks a 0.43 kg ball with a force of 50N
for a time of 0.15s. What is the final velocity of the ball?
Impulse Examplesβ’ A Car is moving at 15 m/s, when it collides with a tree.
The 75 kg driver comes to rest in a time of 0.3 seconds. What is the force exerted on the driver.
β’ What if he was not wearing a seat belt and came to rest in a time of 0.05s?
β’ What distance is required to stop?
Impulse Examples
β’ A baseball moving at 40 m/s is hit back towards the pitcher with a speed of 35m/s. If the force exerted on the ball is 350N. What is the force exerted on the ball?
Conservation of Momentumβ’ Total momentum of a closed system remain constant
β’ Closed System: no net external forces
β’ p1 = p2 mv1+mv2 = mvβ1 + mvβ2
β’ Kick back or explosions
Conservation of Momentumβ’ p1 = p2 mv1+mv2 = mvβ1 + mvβ2
β’ Mass of Bullet = 50 g
β’ Mass of gun = 4kg
β’ Both start from rest
β’ Bullet velocity =500m/s
β’ Velocity of Gun =?
Conservation of Momentum
p1 = p2 mv1+mv2 = mvβ1 + mvβ2
Before Collisionβ’ Mass of receiver = 75kg
β’ Velocity of Receiver = -5m/s
β’ Mass of defender = 85kg
β’ Velocity of Defender = +8m/s
After Collision β’ Velocity of Receiver = ? m/s
β’ Velocity of Defender = +2m/s
Conservation of MomentumPerfectly Inelastic Collision
β’ Objects stick together and travel at same velocity after collision
β’ Momentum Conserved
β’ π1π£1 +π2π£2 = π1 +π2 π£β²2β’ Mass of QB= 85kg
β’ Velocity of QB = -0m/s
β’ Mass of defender = 110kg
β’ Velocity of Defender = +6m/s
β’ Velocity of Both After = ? m/s
Conservation of MomentumPerfectly Elastic CollisionMomentum Conserved
β’ π1π£1+π2π£2 = π1π£β²1+π2π£2β²
Kinetic Energy Conserved
β’1
2ππ£1
2+1
2ππ£1
2 =1
2ππ£β²12 +
1
2ππ£β²12
Relative Velocity same before and after collision, but in opposite direction
β’ π£2β π£1 = π£β²1β π£β²2
Perfectly Elastic CollisionBefore Collision
β’ m1 = 1.0kg, v1 = 3m/s
β’ m2 = 2.0kg, v2 = -2 m/s
Velocity of Each ball After Collision?
π1π£1 +π2π£2 = π1π£β²1+π2π£2β² π£2β π£1 = π£β²1β π£β²2
m1 = 1.0kg
v1 = 3m/s
m2 = 2.0kg
v2 = -2m/s
v1 = ? v2= ?
Before Collision
β’ m1 = 2.0kg, v1 = 3m/s
β’ m2 = 2.0kg, v2 = -6 m/s
Velocity of Each ball After Collision?
π1π£1 +π2π£2 = π1π£β²1+π2π£2β² π£2β π£1 = π£β²1β π£β²2
m1 =2.0kg m2 = 2.0kg
v1 = 3m/s v2 = -6m/s
v1 = ? v2= ?
Collisions in 2Dβ’ Vector Sum of momentum before collision is equal to vector sum after collision.
m=750kgv=15 m/s
m=650kgv=20m/s
v= ? m/s
Collisions in 2Dβ’ Vector Sum of momentum before collision is equal to vector sum after collision.
m = 3kgv=5m/s
m = 2kgv=0m/s m = 2kg
ΞΈ = 30o
v= 1.5m/s
m = 2kgΞΈ = ?o
v= ? m/s
