A Realistic Impact: A _________ Impact: F net Think of how a baseball bat _________ (comes into contact with) a ball as a function of _________. bat first _________ ball ball _________ the bat t ___________ force of bat on ball Area = _______ F net t F net  t impacts time maximum touches leaves Simplified

The ________ of and ____________ (time) of impact determine the future __________ of the ball. The quantity _______ is called the _____________ It is a ______________ quantity. magnitude :J = _______ direction : same as the dir. of ______ units of J: [ ] [ ] = _______ (derived) J J F net t direct t F net F net t N · s vector forceduration motion impulse, J F net t F net

Ex 1: A net force of 25 N to the right acts on a 40-kg snowman for 3.0 s. Calculate the impulse exerted on Frosty. J = = mag. of J same as F net 40 kg F net = 25 N F net t (25 N, right)(3.0 s) 75 N·s, right = + 75 N·s

Ex 2: Give the direction of the impulse for a: A/ ball moving up during free fall B/ ball falling down during free fall C/ ball fired up at an angle at the four points shown below: down All J's are ___________ because that is the direction of ___________ downward gravity FgFg FgFg FgFg FgFg ignore air resistance

 _____________ changes ________________ pp Newton's Second Law: F net = Rewrite a as  v/t: F net = Multiply both sides by t: F net = But m  v =  p, so write: F net = Since F net t = ____, the last line can be written: mama m m  v t pp t  v t ==F net tJ Impulsemomentum J (Historical note: Newton actually first wrote his second law using _____, and not ____.) pp a

[J] = [ ][ ] = [ ] [ ] J = F net t = From this last equation, the _______ of impulse J can be written two ways: This is true because: 1 N·s = 1 ( ) s = 1 If you re-write  p = p f - p i and substitute in, you get: J = F net t = or: m f v f – m i v i p f – p i units [J] = ( ) ( ) = ( ) ( ) N s kg m/s = kg·m/sN·s kg·m/s 2 kg·m/s __________   ____________ derived fundamental F net t m v

Ex: An impulse of 24 N·s north is applied to a 0.15-kg baseball initially moving at an initial speed of 40 m/s south. What is the change in momentum of the baseball? Equation: Given: J = m = v = Unknown: Answer:  p = + 24 Ns 0.15 kg -40 m/s pp J = F net t =  p + 24 Ns = + 24 kg·m/sSame as J!

Ex: A 0.5-kg ball is moving at 4.0 m/s to the right when it hits a wall. Afterwards, it moves 2.0 m/s to the left. Determine the impulse exerted on the ball by the wall. v i = 4.0 m/s wall v f = -2 m/s  p = p f – p i = m f v f – m i v i = (0.5)(-2) – = - 3 kg·m/s = - 3 N·s J = F net t =  p To find J, find  p m = 0.5 kg (0.5)(4)

p f – p i p i + J This can be written: J = And can be rearranged to: p f = This says, "J is what you add to ___ to get ___." p i = mv i = (0.5)(4) = 2 kgm/s Before the impulse: J = -3 pfpf Ex. The last example found  p = J = -3 N·s = ___ kg·m/s pp ==F net tJ p i = 2 Adding the impulse of -3 N·s from the wall to p i : = -1 kgm/s -3 pfpf pipi

The impulse J is ______________ (to the left) in the previous example because _____ from the wall is. The wall in the previous example exerts its force for a time of 0.12 seconds. Calculate the net force that acts on the ball during that time. J = F net t =  p F net ( ) = F net = -3 N·s / 0.12 s = -25 N negative F net 0.12 s -3 N·s

The equation: Ft =  p has many applications in sports and collisions…. 1. To ______________ (make the most of)  p, you can:  apply a ____________ F ___t (hit harder)  ____________ the impact time: F___ (follow through)   Both of these help you to take a ball moving in one direction and allow you to send it in another direction with a _____________________ velocity. F greater t pp much different maximize increase

2. Suppose 2 identical cars (m=1000 kg), traveling at the same initial v i (30 m/s) both come to rest: a/ Car A hits a _________ wall and stops in 1 s. b/ Car B hits _________ barrels and stops in 4 s.  For both cars:  p = m f v f – m i v i = = A : F t =  p F ____ = -30,000 F = ________ __________ time to stop _________ force of impact brick water -(1000)(30) -30,000 Apply Ft =  p to each car to find force on car: 1 -30,000 less more B : F t =  p F ____ = -30,000 F = ________ __________ time to stop _________ force of impact 4 -7,500 more less 0