MOMENTUM - Warren Consolidated Schoolsvirt.wcskids.net/mmstc/staff_websites/mcmillan files/11... ·...
Transcript of MOMENTUM - Warren Consolidated Schoolsvirt.wcskids.net/mmstc/staff_websites/mcmillan files/11... ·...
LINEAR MOMENTUMMomentum
Impulse
Conservation of Momentum
Inelastic Collisions
Elastic Collisions
Momentum In 2 Dimensions
MOMENTUM
• Quantity of Motion
• Product of Mass and Velocity
• 𝑝 = 𝑚𝑣 = kg ∗m
s
• Vector Quantity
IMPULSE
• Change in Momentum
• To change momentum, apply a force for a period of time.
• 𝐽 = ∆𝑝 = 𝑚𝑣2 −𝑚𝑣1 = 𝑚∆𝑣 = 𝐹 ∗ 𝑡 = (𝑁 ∗ 𝑠) 𝐹 = 𝑚𝑎 = 𝑚∆𝑣
𝑡→ 𝑚∆𝑣 = 𝐹 ∗ 𝑡
IMPULSE• Follow Through Example (Bunt vs. Swing)
• Apply force for longer period of time = larger momentum change
Impulse (Follow Through)
Nordic Ski Racing Slap shot
Impulse (reduce force)
F*t = mΔv = F*t•Helmets
•Padding
Impulse (reduce force) F*t = mΔv = F*t• Air Bag • Crumple Zone
Impulse Examples• 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 Examples• 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.
• What if he was not wearing a seat belt and came to rest in a time of 0.05s?
• What distance is required to stop?
Impulse Examples
• A baseball (m=0.145kg) 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 time force is exerted on the ball?
Impulse in 2D • A 0.160kg hockey puck moving at 25 m/s is deflected 90o toward the net with a speed of 30 m/s.
• Determine the magnitude and direction of the impulse exerted on the puck.
∆𝑝 = 𝑝2− 𝑝1 = 𝑝2+ (−𝑝1)
Conservation of Momentum• Total momentum of a closed system remain constant
• Closed System: no net external forces
• Σp1 = Σp2 mv1+mv2 = mv’1 + mv’2
• Kick back or explosions
Conservation of Momentum• Σp1 = Σ p2 mv1+mv2 = mv’1 + mv’2
• Mass of Bullet = 50 g
• Mass of gun = 4kg
• Both start from rest
• Bullet velocity =500m/s
• Velocity of Gun =?
Conservation of Momentum
Σp1 = Σ p2 = mv1+mv2 = mv’1 + mv’2
• mblue = 50kg
• vblue = 3 m/s
• msilver = 40kg
• vsilver = ?
Conservation of Momentum
Σp1 = Σ p2 = mv1+mv2 = mv’1 + mv’2
• Mass of truck = 1000kg
• Speed of truck = 2m/s
• Mass of person = 120kg
• Speed of person = ?
Conservation of Momentum
Σp1 = Σp2 mv1+mv2 = mv’1 + mv’2
Before Collision• Mass of receiver = 75kg
• Velocity of Receiver = -5m/s
• Mass of defender = 85kg
• Velocity of Defender = +8m/s
After Collision • Velocity of Receiver = ? m/s
• Velocity of Defender = +2m/s
Conservation of MomentumPerfectly Inelastic Collision
• Objects stick together and travel at same velocity after collision
• Momentum Conserved
• 𝑚1𝑣1 +𝑚2𝑣2 = 𝑚1 +𝑚2 𝑣′2• Mass of QB= 85kg
• Velocity of QB = 0m/s
• Mass of defender = 110kg
• Velocity of Defender = +6m/s
• Velocity of Both After = ? m/s
Conservation of MomentumPerfectly Elastic CollisionMomentum Conserved
• 𝑚1𝑣1+𝑚2𝑣2 = 𝑚1𝑣′1+𝑚2𝑣2′
Kinetic Energy Conserved
•1
2𝑚𝑣1
2+1
2𝑚𝑣2
2 =1
2𝑚𝑣′12 +
1
2𝑚𝑣′2
2
Relative Velocity same before and after collision, but in opposite direction
• 𝑣2− 𝑣1 = 𝑣′1− 𝑣′2
Perfectly Elastic CollisionBefore Collision
• m1 = 1.0kg, v1 = 3m/s
• m2 = 2.0kg, v2 = -2 m/s
Velocity of Each ball After Collision?
𝑚1𝑣1 +𝑚2𝑣2 = 𝑚1𝑣′1+𝑚2𝑣2′ 𝑣2− 𝑣1 = 𝑣′1− 𝑣′2
m1 = 1.0kg
v1 = 3m/s
m2 = 2.0kg
v2 = -2m/s
v1 = ? v2= ?
Before Collision
• m1 = 2.0kg, v1 = 3m/s
• m2 = 2.0kg, v2 = -6 m/s
Velocity of Each ball After Collision?
𝑚1𝑣1 +𝑚2𝑣2 = 𝑚1𝑣′1+𝑚2𝑣2′ 𝑣2− 𝑣1 = 𝑣′1− 𝑣′2
m1 =2.0kg m2 = 2.0kg
v1 = 3m/s v2 = -6m/s
v1 = ? v2= ?
Perfectly Elastic Collision
Perfectly Elastic Collision
Before Collisionm1 = 50.0kg, v1 = 0m/sm2 = 60kg, v2 = 6 m/s
Velocity of Each ball After Collision?
𝑚1𝑣1+𝑚2𝑣2 = 𝑚1𝑣′1+𝑚2𝑣2′ 𝑣2− 𝑣1 = 𝑣′1− 𝑣′2
Collisions in 2D• 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 2D• Vector Sum of momentum before collision is equal to vector sum after collision.
m=550kgv=? m/s
m=850kgv=?m/s
v= 10m/s40o
Collisions in 2D• Vector Sum of momentum before collision is equal to vector sum after collision.
m = 3kgv=5m/s
m = 2kgv=0m/s m = 2kg
θ = 30o
v= 1.5m/s
m = 3kgθ = ?o
v= ? m/s
Collisions in 2D• Vector Sum of momentum before collision is equal to vector sum after collision.
m = 3kgv=4m/s
m = 2kgv=0m/s m = 2kg
θ = 30o
v= ? m/s
m = 3kgθ =20o
v= ? m/s