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A Realistic Impact:

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

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A Realistic Impact:

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

  2. 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) force duration motion Fnett impulse, J vector Fnett Fnet Fnet t N · s J J direct direct Fnet t

  3. 40 kg 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. Fnet = 25 N Fnett J = = = (25 N, right) (3.0 s) 75 N·s, right = + 75 N·s mag. of J same as Fnet

  4. 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 down ignore air resistance All J's are ___________ because that is the direction of ___________ downward Fg Fg Fg gravity Fg

  5. Newton's Second Law: Fnet = Rewrite a as Dv/t: Fnet = Multiply both sides by t: Fnet = But mDv = Dp, so write: Fnet = Since Fnet t = ____, the last line can be written: ma Dv t m m Dv t Dp t J J = Fnet t = Dp Impulse momentum • _____________ changes ________________ (Historical note: Newton actually first wrote his second law using _____ , and not ____ .) a Dp

  6. If you re-write Dp = pf - pi and substitute in, you get: J = Fnett = pf – pi or: mfvf – mivi J = Fnett = From this last equation, the _______ of impulse J can be written two ways: units [J] = [ ][ ] = [ ] [ ] v Fnet t m N [J] = ( ) ( ) = ( ) ( ) s m/s kg fundamental derived N·s kg·m/s =  ____________ __________  kg·m/s2 This is true because: 1 N·s = 1 ( ) s = 1 kg·m/s

  7. 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? Given: J = m = v = Unknown: + 24 Ns 0.15 kg -40 m/s Dp J = Fnett = Dp Equation: Answer: Dp = + 24 Ns = + 24 kg·m/s Same as J!

  8. 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. vi = 4.0 m/s wall m = 0.5 kg vf = -2 m/s Dp = pf – pi = mfvf – mivi = (0.5)(-2) – = - 3 kg·m/s = - 3 N·s J = Fnett = Dp To find J, find Dp (0.5)(4)

  9. J = Fnet t = Dp This can be written: J = And can be rearranged to: pf = This says, "J is what you add to ___ to get ___." pf – pi pi + J pi pf Ex. The last example found Dp = J = -3 N·s = ___ kg·m/s -3 Adding the impulse of -3 N·s from the wall to pi: Before the impulse: J = -3 pi = mvi = (0.5)(4) = 2 kgm/s pi = 2 pf = -1 kgm/s

  10. The impulse J is ______________ (to the left) in the previous example because _____ from the wall is. negative Fnet 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 = Fnett = Dp Fnet ( ) = Fnet = -3 N·s / 0.12 s = -25 N -3 N·s 0.12 s

  11. The equation: Ft = Dp has many applications in sports and collisions…. maximize • To ______________ (make the most of) Dp, you can: •  apply a ____________ F ___t • (hit harder) •  ____________ the impact time: F___ • (follow through) F greater Dp t  ________ increase 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. much different

  12. Suppose 2 identical cars (m=1000 kg), traveling • at the same initial vi (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: Dp = mfvf – mivi • = = brick water 0 -(1000)(30) -30,000 Apply Ft = Dp to each car to find force on car: A: F t = Dp F ____ = -30,000 F = ________ B: F t = Dp F ____ = -30,000 F = ________ 1 4 -30,000 -7,500 less more • __________ time to stop • _________ force of impact • __________ time to stop • _________ force of impact more less

  13. Get a Text On a separate sheet of paper: On page 211: #1-3 On page 213: # 3 Show your work

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