LINEAR MOMENTUMMomentum
Impulse
Conservation of Momentum
Inelastic Collisions
Elastic Collisions
Momentum In 2 Dimensions
Center of Mass
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)
Impulse
• The derivative of momentum over time is Force.
• ∆𝑝 = 𝐹 ∗ 𝑡 → 𝐹 =∆𝑝
𝑡→ 𝐹 =
𝑑𝑝
𝑑𝑡
• The integral of a Force Time graph is impulse
• 𝐽 = ∆𝑝 = 𝐹 ∗ 𝑡 = 𝑭 𝑑𝑡
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 150N
for a time of 0.50s. 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 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?
• The force exerted on a ball by a baseball bat is given by the equation:
• 𝐹 = 1.6 ∗ 107𝑡 − 6.0 ∗ 109𝑡2
• The Force act for 2.5s. Determine the final velocity of the 0.145kg baseball.
Determine the change in momentum given by the graph for the following intervals:
0-2s2-4s4-8sThe total impulse
If the mass of the object is 2.0kg, what is the velocity after2.0s.
What is the final velocity of the object?
Conservation of Momentum• Total momentum of a closed system remain constant
• Closed System: no net external forces
• p1 = p2 mv1+mv2 = mv’1 + mv’2
• Kick back or explosions
Conservation of Momentum• p1 = 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
p1 = p2 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 MomentumΣp1 = Σ p2 = m1v1+m2v2 = m1v’1 + m2v’2
• mblue = 50kg
• vblue = 3 m/s
• msilver = 40kg
• vsilver = ?
Conservation of Momentum• Perfectly 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
2𝑚1𝑣1
2+1
2𝑚2𝑣2
2 =1
2𝑚1𝑣′12 +
1
2𝑚2𝑣
′22
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 = 60.0kg, v1 = 0m/sm2 = 50kg, 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 = 3kgv=5m/s
m = 2kgv=0m/s m = 2kg
θ = 30o
v= 1.5m/s
m = 2kgθ = ?o
v= ? m/s
CENTER OF MASS
𝑥𝑐𝑚 = 𝑚1𝑥1+𝑚
2𝑥2+𝑚
3𝑥3+⋯
𝑚1+𝑚
2+𝑚
3…
M=40kgX=3m
M=50kgX=8m
Mbeam=20kgL=10m
CENTER OF MASS 𝑥𝑐𝑚 = 𝑚1𝑥1+𝑚
2𝑥2+𝑚
3𝑥3+⋯
𝑚1+𝑚
2+𝑚
3=⋯
m = 195g, L=100cm
m=200gx=10cm
m=500gx=70cm
Xcm =?
𝑥𝑐𝑚 = 𝑚1𝑥1+𝑚
2𝑥2+𝑚
3𝑥3+⋯
𝑚1+𝑚
2+𝑚
3=⋯
CENTER OF MASS
𝑦𝑐𝑚 = 𝑚1𝑦1+𝑚
2𝑦2+𝑚
3𝑦3+⋯
𝑚1+𝑚
2+𝑚
3=⋯
MOMENTUM OF CENTER OF MASS
𝑣𝑐𝑚 = 𝑚1𝑣1+𝑚
2𝑣2+𝑚
3𝑣3+⋯
𝑚1+𝑚
2+𝑚
3…
𝑝𝑐𝑚 = 𝑚1𝑣1 +𝑚2𝑣2 +𝑚3𝑣3 +⋯ .
Problem Solving w/ MomentumA force applied to a 20kg object is given by (F = 6t2 +4t)
• If the force is applied for 2.0 seconds, what is the final velocity of the object.
• How much work was done on the object?
• If the coefficient of friction is μ=.25, what is the stopping distance?
F
Problem Solving w/ Momentum• A 50 kg snowboarder is unable to stop at the bottom of a 10m hill. He
collides and holds onto a 60kg skier waiting for the lift. If the coefficient of friction between the people an the snow is μ=.25, How much time is required for them to come to rest?
Problem Solving w/ Momentumm=2.0 kgv=2.5 m/s
m=2.0 kgv=5 m/s m=10.0 kg
v=0 m/s
m=10.0 kgv= ?ϴ=?
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