UNIT 4 MOMENTUM & IMPULSE

Transcription

1 UNIT 4

2 UNIT 4 MOMENTUM & IMPULSE

3 IMPULSE-MOMENTUM THEOREM Remember, means final initial p = p f p i v = v f v i J = F( t) = p = m v = (mv f mv i ) The impulse, J, that acts on an object is equal to the change in the object s momentum, p.

4 EXAMPLE A baseball player bunts (hits softly) a kg baseball thrown at 43.0 m/s. If the bat exerts an average force of 6500 N on the ball for s, what is the final speed of the ball? Given: m = kg v i = m/s F = 6500 N t = s Unknown: v f =? Equation: F( t) = p = (mv f mv i ) Substitute: (6500 N)( s) = 0.144kg(v m/s) Solve: v f = 12.1 m/s

5 YOU TRY A soccer ball has a momentum of 2.5 kg*m/s. You then apply and impulse of 10 N*s. What is the soccer ball s final momentum? Given: p i = 2.5 kg*m/s J = 10 N*s Unknown: p f =? Equation: J = p = (p f p i ) Substitute: 10 N*s= p f 2.5 kg*m/s Solve: p f = 12.5 kg*m/s

6 UNIT 4 CONSERVATION OF MOMENTUM

7 CONSERVATION OF MOMENTUM m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Before collision After collision

8 ELASTIC COLLISIONS The objects bounce off each other! m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Before collision After collision

9 INELASTIC COLLISIONS The objects stick together! m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f = (m 1 +m 2 ) v f Before collision After collision

10 EXAMPLE A 0.014kg red bouncy ball is traveling at a velocity of 1.2 m/s to the right. It collides elastically with a initially stationary blue bouncy ball (mass of kg). If the red ball has a final velocity of 0.5 m/s to the left, what is the final velocity of the blue ball? G: m1 = 0.014kg v1i = 1.2 m/s v1f = -0.5 m/s m2 = kg v2i = 0 m/s U: v2f =? E: m1v1i + m2v2i = m1v1f + m2v2f S: v2f = 2.16 m/s

11 #5 FROM THE MOCK EOC WITH YOUR PARTNER

12 UNIT 4 WORK & POWER

13 WORK, W (Units of joules (J)) Work is ONLY done when a force moves an object a distance.

14 WORK, W (units of joules (J)) W = F (d) cosθ W = work (N*m or J) F = Force (N) d = displacement (m) θ = angle between the force and displacement vectors ( ) θ = 0 θ = 90 θ = θ F F F θ d d d

15 #32 FROM THE MOCK EOC WITH YOUR PARTNER

16 EXAMPLE Your mom tells you to get to work cleaning your room. Being the smart-alec you are and using your knowledge of physics, you lift up one sock (mass = 0.04kg) straight up 1m. This is done at a constant speed. How much work did you do? G: Fg +Fa = m * ay (0.04kg * -9.8) + Fa = 0 Fa = 0.392N d = 1m θ = 0 U: W =? E: W = F*d*cosθ = 0.392*2*cos(0) S: W = J

17 YOU TRY Your mom tells you to get to work cleaning your room. Being the smart-alec you are and using your knowledge of physics, you drag your laundry basket with a force of 5N at an angle of 45 with the horizontal for 5m. How much work do you do? G: F = 5N d = 5m θ = 45 U: W =? F =? E: W = F*d*cosθ = 5*5*cos(45) S: W = J

18 POWER, P (units of watts (W)) P = W/t P = Power (J/s or W) W = work (N*m or J) t = time

19 UNIT 4 ENERGY

20 KINETIC ENERGY, KE (Units of joules (J)) KE = ½mv 2 Mass (in kg) Velocity (in m/s)

21 KE EXAMPLE A constant 15,000 N force accelerates a 1700kg Maserati from rest. What is the kinetic energy of the car after traveling 50 m? G: F = 15000N d = 50m m = 1700kg vi = 0 m/s U: KE =? vf =? a =? E: F = m*a a = 8.82 m/s 2 a = (vf 2 vi 2 ) / (2Δd) vf = m/s KE = ½ mv 2 S: KE = J

22 WORK & KINETIC ENERGY (Units of joules (J)) W net = ΔKE Kinetic Energy (in J)

23 #22 FROM THE MOCK EOC WITH YOUR PARTNER

24 KE EXAMPLE A worker does 1000 J of work on a 100 kg box. If the box loses 505 J of heat to the floor through the friction between the box and the floor, what is the velocity of the box after the work has been done on it? G: W net = J m = 100kg vi = 0 m/s U : W net = ΔKE E: W net = ½ mv f 2 - ½ mv i 2 S: v f = 3.15 m/s

25 GRAVITATIONAL POTENTIAL ENERGY, PE g (Units of joules (J)) PE = mgh Mass (in kg) m/s 2 Height off the ground (in m)

26 PE g EXAMPLE A 10kg goat is in a tree, 15m above the ground. How much potential energy does it have? G: m = 10 kg h = 15m U: PE =? E: PE = mgh = 10*9.8*15 S: PE = 1470 J

27 CONSERVATION OF ENERGY KE i + PE i = KE f + PE f In a closed system, the total mechanical energy before (initial) must be equal to the total mechanical energy after (final).

28 #12, 29, 39 FROM THE MOCK EOC WITH YOUR PARTNER

29 CONSERVATION OF ENERGY Your friend is upstairs (10m above you) and calls for you to toss them your 2kg physics textbook. How fast do you have to throw it so it reaches them?

30 CONSERVATION OF ENERGY Your friend is upstairs (10m above you) and calls for you to toss them your 2kg physics textbook. How fast do you have to throw it so it reaches them? G: m = 2kg hi = 0m hf = 10m vf = 0m/s U: vi =? E: KE i + PE i = KE f + PE f ½ mv i2 + mgh i = ½ mv f2 + mgh f ½ (2)v i2 + (2)(9.8)(0) = ½ (2)(0) 2 + (2)(9.8)(10) S: v i = 14m/s

31 CONSERVATION OF ENERGY YOU TRY 80kg Rapunzel rests on top of a 50m hill. What will be the KE of Rapunzel be when she is 5m from the bottom?

32 CONSERVATION OF ENERGY 80kg Rapunzel rests on top of a 50m hill. What will be the KE of the Rapunzel be when she is 5m from the bottom? G: m = 80kg vi = 0m/s hi = 50m hf = 5m U: KE f =? E: KE i + PE i = KE f + PE f ½ mv i2 + mgh i = KE f + mgh f ½ (80)(0) 2 + (80)(9.8)(50) = KE f + (80)(9.8)(5) S: KE f = 35,280 J

33 CONSERVATION OF ENERGY With what minimum speed must the cat leave the top of the car in order to lift its mass up 0.75 m to the roof with a speed of 3m/s?

34 CONSERVATION OF ENERGY With what minimum speed must the cat leave the top of the car in order to lift its mass up 0.75 m to the roof with a speed of 3m/s? G: hi = 0m hf = 0.75m vf = 0.25m/s U: vi =? E: KE i + PE i = KE f + PE f ½ mv i2 + mgh i = ½ mv f2 + mgh f ½ mv i2 + mgh i = ½ mv f2 + mgh f ½ v i2 + (9.8)(0) = ½ (3) 2 + (9.8)(0.75) S: v i = 4.87m/s

35 EXIT TICKET 5 question quiz

36 EXIT TICKET 5 question quiz