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MOMENTUM & IMPULSE. Momentum : product of a system’s mass it’s linear velocity Δ P = m Δ V = m ( V f – V i ) Units: kg·m /s = N·s More massive…..more mo’ Greater velocity….more mo ’ Is a vector Momentum provides inertia. Ex.1
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Momentum: product of a system’s mass it’s linear velocity • Δ P = m ΔV = m (Vf – Vi) • Units: kg·m/s = N·s • More massive…..more mo’ • Greater velocity….more mo’ • Is a vector • Momentum provides inertia
Ex.1 What is the change in momentum of a 555 kg object that changes velocity by 32 m/s? 18800
Ex. 2 Calculate the change in momentum of a 125 kg person in which the velocity increases from 1.10 m/s to 2.50 m/s. 175
Ex. 3 A 125 kg person moving at 1.00 m/s experiences a momentum change of 195 N·s. Calculate the final velocity. 2.56
Impulse: the product of the force exerted on a system and the time interval in which it occurs • I = F ΔT = W ΔT = mg ΔT • Units: N·s • How fast the force is applied • Is a vector
Ex. 4 Calculate the impulse on an 75000N object that changes momentum over 17.0 s. 1300000
Ex. 5 How much time would be needed to experience a 882 N·s impulse for a 22.5 kg object? 4.00
III. Momentum-Impulse Theorem *Relates the change of momentum of an object to the change in impulse of the object ∆p = ∆I m∆V = F∆t
Ex. 6 A 225 kg object accelerates from rest to 25.0 m/s in 8.25 s. Calculate the force required. 682
Ex. 7 A 78 kg cart goes from rest to 5.55 m/s. Find the force needed.
IV. Law of Conservation of Momentum: the momentum of a closed, isolated system does not change • Mo’ is transferred from one system to another • Mo’ is transferred during a collision • Momentum-Impulse obey Newton’s Third Law (action-reaction)
“before” = “after” MAVA + MBVB = MAVA’ + MBVB’ (object do not lock together) MAVA + MBVB = MA+B VAB’ (Objects do interlock)
Ex. 8 (net change in mo’ is 0) A 85.0 kg RB moving at 7.8 m/s collides with a 122 kg MLB moving in the opposite direction. The tackle is made on contact. Calculate the LB’s speed.
Ex. 9 (interlocking) A 42300 kg truck has a head on collision with a 22200 kg car moving at 14.5 m/s in the opposite direction. They interlock and move in the direction of the truck at a speed of 3.33 m/s. Find the initial speed of the smaller car.
Ex. 10 (rebounding, same direction) Two cars are moving in the same direction. The red car (m=22750 kg) is moving at 9.50 m/s. The blue car (m= 28600 kg) is behind the first car and rear-ends it (they do not interlock). The red car speeds up to 12.50 m/s. What is the final speed of the blue car?
Ex. 11 (rebounding, opposite direction) A 0.888 kg baseball is thrown at a 12.54 kg metal jar. The baseball is thrown at 42.0 m/s and bounces back at 12.0 m/s. What velocity is given to the jar?
***the nylon rope stretches as the climber falls, increasing the TIME that the force is applied
VI. Collisions: when one system (object) runs into another’s path A. Elastic Collisions: the two objects bounce after the collision …return to their original shape …work done to deform the object is equal to the work done to return it to its original shape
B. Inelastic: objects deform during the collision such that the total KE decreases …object moves separately after the collision
Perfectly Inelastic Collision: objects collide and move together as one mass ex. • Football players tackling and interlocking • Meteor striking the Earth • Car crash that interlocks • Mo’ is conserved • KE is not…some is “wasted” as sound, light, heat, etc.
Collision terms for CARS: *If a car bounces off another car in a collision, it is called rebounding. Rebounding increases the time of the crash, and the direction of the force is changed. *Most cars are building with CRUMPLE ZONES or areas of the car that collapse in and absorb the force of the crash. Crumpling decreases the time factor for a collision as well.