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=?