LINEAR MOMENTUM Momentum 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 = mv 2 mv 1 = m v = F t = (N s)
Impulse The derivative of momentum over time is Force. p = F t F = p t dp F = 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 10 7 t 6.0 10 9 t 2 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: 0-2s 2-4s 4-8s The total impulse If the mass of the object is 2.0kg, what is the velocity after 2.0s. 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 p 1 = p 2 mv 1 +mv 2 = mv 1 + mv 2 Kick back or explosions
Conservation of Momentum p 1 = p 2 mv 1 +mv 2 = 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 p 1 = p 2 mv 1 +mv 2 = 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 Momentum Σp 1 = Σ p 2 = m 1 v 1 +m 2 v 2 = m 1 v 1 + m 2 v 2 m blue = 50kg v blue = 3 m/s m silver = 40kg v silver =?
Conservation of Momentum Perfectly Inelastic Collision Objects stick together and travel at same velocity after collision Momentum Conserved m 1 v 1 + m 2 v 2 = m 1 + m 2 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 Momentum Perfectly Elastic Collision Momentum Conserved m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 Kinetic Energy Conserved 1 2 m 1v 12 + 1 2 m 2v 22 = 1 2 m 1v 12 + 1 2 m 2v 2 2 Relative Velocity same before and after collision, but in opposite direction v 2 v 1 = v 1 v 2
Perfectly Elastic Collision Before Collision m 1 = 1.0kg, v 1 = 3m/s m 2 = 2.0kg, v 2 = -2 m/s m 1 = 1.0kg v 1 = 3m/s v 1 =? v 2 =? m 2 = 2.0kg v 2 = -2m/s Velocity of Each ball After Collision? m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 v 2 v 1 = v 1 v 2
Perfectly Elastic Collision Before Collision m 1 = 2.0kg, v 1 = 3m/s m 2 = 2.0kg, v 2 = -6 m/s m 1 =2.0kg m 2 = 2.0kg v 1 = 3m/s v 2 = -6m/s v 1 =? v 2 =? Velocity of Each ball After Collision? m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 v 2 v 1 = v 1 v 2
Perfectly Elastic Collision Before Collision m 1 = 60.0kg, v 1 = 0m/s m 2 = 50kg, v 2 = 6 m/s Velocity of Each ball After Collision? m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 v 2 v 1 = v 1 v 2
Collisions in 2D Vector Sum of momentum before collision is equal to vector sum after collision. v=? m/s m=750kg v=15 m/s m=650kg v=20m/s
Collisions in 2D Vector Sum of momentum before collision is equal to vector sum after collision. m = 3kg v=5m/s m = 2kg θ =? o v=? m/s m = 2kg v=0m/s m = 2kg θ = 30 o v= 1.5m/s
CENTER OF MASS x cm = m 1 x 1 +m 2 x 2 +m 3 x 3 + m 1 +m 2 +m 3 M beam =20kg L=10m M=40kg X=3m M=50kg X=8m
CENTER OF MASS x cm = m 1 x 1 +m 2 x 2 +m 3 x 3 + m 1 +m 2 +m 3 = m = 195g, L=100cm m=200g x=10cm X cm =? m=500g x=70cm
CENTER OF MASS x cm = m 1 x 1 +m 2 x 2 +m 3 x 3 + m 1 +m 2 +m 3 = y cm = m 1 y 1 +m 2 y 2 +m 3 y 3 + m 1 +m 2 +m 3 =
MOMENTUM OF CENTER OF MASS p cm = m 1 v 1 + m 2 v 2 + m 3 v 3 +. v cm = m 1 v 1 +m 2 v 2 +m 3 v 3 + m 1 +m 2 +m 3
Problem Solving w/ Momentum A force applied to a 20kg object is given by (F = 6t 2 +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/ Momentum m=2.0 kg v=2.5 m/s m=2.0 kg v=5 m/s m=10.0 kg v=0 m/s m=10.0 kg v=? ϴ=?