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)

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 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 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 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

2𝑚𝑣1

2 =1

2𝑚𝑣′12 +

1

2𝑚𝑣′12

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= ?

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