Download presentation

Presentation is loading. Please wait.

Published byNya Denny Modified over 2 years ago

1
Ch6.1 – Work and Energy Work - force applied over a distance W = Fnet.s If the force is applied at some angle: W = Fnet.s.cosθ - If the force and direction of motion are in the same direction (+) work - If force opposes motion (–) work Ex1) An engine exerts a force of 400kN on a train, pulling it 500m, at constant speed. a) How much work is done by the engine on the train? b) The brakes apply a force of 100kN to bring the train to a stop in 1000m. How much work do they do on the train?

2
**Ex2) The Egyptians placed 500 kg stones on pyramids by making **

ramps out of dirt, and using round logs as wheels. If a ramp 100m long is used to place a stone 25m above the ground: a) How much work is done overcoming gravity? b) How much work is done overcoming friction? 100m 25m

3
HW#11) A nurse pushes someone in a wheelchair , with her arms at 30°, 100m down a hallway, doing 400 J of work in the process. What average force did the nurse exert in the direction of motion? 16. A pickup truck is hauling a barge along a canal at a constant speed. The truck, driving parallel to the waterway at a 30º angle with the forward direction. If the truck exerts a force of 1000N on the cable, how much work is done in overcoming friction as the barge is moved 10 km? Ch6 HW#1 p205+ 1,5,11,12,14,16,19,20,22,23

4
Ch6 HW#1 p205 1,5,11,12,14,16,19,20,22,23 While floating out in space, a constant force of 500N is applied to a 542.3kg robot by a small rocket motor. The robot moves along a straight line in the direction of the thrust of the motor. How much work is done on it by the rocket for every 10.0m traveled? 5. A book is slowly slid in a straight line, at a constant speed, 1.5m across a level table by a 15N horizontal force. How much work is done on the book-table system and what is it done against?

5
Ch6 HW#1 p205 1,5,11,12,14,16,19,20,22,23 While floating out in space, a constant force of 500N is applied to a 542.3kg robot by a small rocket motor. The robot moves along a straight line in the direction of the thrust of the motor. How much work is done on it by the rocket for every 10.0m traveled? 5. A book is slowly slid in a straight line, at a constant speed, 1.5m across a level table by a 15N horizontal force. How much work is done on the book-table system and what is it done against?

6
Ch6 HW#1 p205 1,5,11,12,14,16,19,20,22,23 While floating out in space, a constant force of 500N is applied to a 542.3kg robot by a small rocket motor. The robot moves along a straight line in the direction of the thrust of the motor. How much work is done on it by the rocket for every 10.0m traveled? 5. A book is slowly slid in a straight line, at a constant speed, 1.5m across a level table by a 15N horizontal force. How much work is done on the book-table system and what is it done against?

7
(11,16 in class) 19. A 100N box filled with books is slid along a 13m long ramp up to a platform 5.0m above the ground. How much work is done if friction is negligible? How much work would have been done if the box were lifted straight up to the platform?

8
**If friction was negligible, how much work was done? **

19. A 100N box filled with books is slid along a 13m long ramp up to a platform 5.0m above the ground. How much work is done if friction is negligible? How much work would have been done if the box were lifted straight up to the platform? Fg=100N 20. Several crates, having a total weight of 400N, are loaded into a 10.0kg wagon that is then pulled up a wooden ramp 10.0m long making an angle of 30.0°. If friction was negligible, how much work was done? Wg = Fg||.s.cos0 or Wg = Fg.sy

9
**= Fgsinθ.s = Fg.(5m) = 500Nsin30°.10m = 500N.5m = 2500 J = 2500 Nm**

19. A 100 N box filled with books is slid along a 13 m long ramp up to a platform 5.0 m above the ground. How much work is done if friction is negligible? How much work would have been done if the box were lifted straight up to the platform? Fg=100N 20. Several crates, having a total weight of 400N, are loaded into a 10.0kg wagon that is then pulled up a wooden ramp 10.0m long making an angle of 30.0°. If friction was negligible, how much work was done? Wg = Fg||.s.cos0 or Wg = Fg.sy = Fgsinθ.s = Fg.(5m) = 500Nsin30°.10m = 500N.5m = 2500 J = 2500 Nm sy = s.sin30° = 10m.(.5) = 5m

10
**22. A 50kg keg of beer slides upright down a 3**

22. A 50kg keg of beer slides upright down a 3.0m long plank leading from the back of a truck 1.5m high to the ground. Determine the amount of work done on the keg by gravity. 3m 1.5m Fg=500N

11
**= 750 J (usually considered negative work)**

22. A 50kg keg of beer slides upright down a 3.0m long plank leading from the back of a truck 1.5m high to the ground. Determine the amount of work done on the keg by gravity. Wg = Fg.sy 3m 1.5m = 500N.1.5m = 750 J (usually considered negative work) Fg=500N

12
**FN Fg|| F c) Wg = Fg.sy Ff Fg Fg┴**

23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. It travels 10 km on level streets. a) How much work does he do in overcoming friction? b) If he pulls his cart 25m along a road inclined at 10°, how much work does he do to overcome road friction? c) How much work overcoming gravity? a) Wf = Ff.s b) Wf = Ff.s FN Fg|| F c) Wg = Fg.sy Ff Fg Fg┴

13
**FN Fg|| F c) Wg = Fg.sy Ff Fg Fg┴**

23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. It travels 10 km on level streets. a) How much work does he do in overcoming friction? b) If he pulls his cart 25m along a road inclined at 10°, how much work does he do to overcome road friction? c) How much work overcoming gravity? a) b) Wf = Ff.s FN Fg|| F c) Wg = Fg.sy Ff Fg Fg┴

14
**= µ.FN.25m = (.02)(Fg┴)25m = (.02)(Fgcos10°)25m = 123 Nm FN**

23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. It travels 10 km on level streets. a) How much work does he do in overcoming friction? b) If he pulls his cart 25m along a road inclined at 10°, how much work does he do to overcome road friction? c) How much work overcoming gravity? a) b) Wf = Ff.s = µ.FN.25m = (.02)(Fg┴)25m = (.02)(Fgcos10°)25m = 123 Nm FN Fg|| F c) Wg = Fg.sy Ff = = Fg Fg┴

15
**= µ.FN.25m = (.02)(Fg┴)25m = (.02)(Fgcos10°)25m = 123 Nm FN**

23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. It travels 10 km on level streets. a) How much work does he do in overcoming friction? b) If he pulls his cart 25m along a road inclined at 10°, how much work does he do to overcome road friction? c) How much work overcoming gravity? a) b) Wf = Ff.s = µ.FN.25m = (.02)(Fg┴)25m = (.02)(Fgcos10°)25m = 123 Nm FN Fg|| F c) Wg = Fg.sy Ff = 250N.(25m.sin10˚) = Fg Fg┴

16
Ch6.2 - Work Graphs Ex1) A force of 5N is applied to an object, and is uniformly increased to 10N while the object moves a total distance of 8m, as shown. How much work is done on the object? 10 F (N) 5 5 10 s (m)

17
**If the force is not uniform finding the area is tricky.**

10 F (N) 5 5 10 s (m)

18
**Ex2) A mass is attached to a spring, on a frictionless surface .**

The mass is initially at 0.3m, the spring’s natural length. The mass is pulled back to the 0.0m mark and released. a) How much work is done from ? b) How much work is done from ? 4 3 2 F (N) 1 -1 (m) -2 - 0.6 - 0.3 - 0.0 - -3 -4 Ch6 HW#2 Practice Graphs

19
Ch6 HW#2 Practice Graphs A force is gradually placed on an object, uniformly increased to 20 N over a distance of 5 m. It then stays constant for the next 5 m. How much work was done on this object? 20 F (N) 10 5 10 s (m)

20
Ch6 Practice Graphs WS A force is gradually placed on an object, uniformly increased to 20 N over a distance of 5 m. It then stays constant for the next 5 m. How much work was done on this object? 20 A = ½.b.h = 50 Nm A = b.h = 100 Nm Atotal = 150 Nm F (N) 10 5 10 s (m)

21
2. A force of 10 N is applied to an object over a distance of 3 m, and then is uniformly increased to 15 N while the object moves a distance of 4 m. How much work is done on the object? 20 F (N) 10 5 10 s (m)

22
2. A force of 10 N is applied to an object over a distance of 3 m, and then is uniformly increased to 15 N while the object moves a distance of 4 m. How much work is done on the object? 20 A = ½.b.h = ½(4m)(5N) = 10 Nm A = b.h = (7m)(10N) = 70 Nm Atotal = 80 Nm F (N) 10 5 10 s (m)

23
**Determine each of the following. **

3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 F (N) s (m)

24
**Determine each of the following. **

3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 F (N) s (m)

25
**Determine each of the following. **

3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at = 0 + ½(20)t2 t = 1.1s F (N) s (m)

26
**Determine each of the following. **

3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at = 0 + ½(20)t2 t = 1.1s c. W = F.s W = (4N)(12m) = 48 J F (N) s (m)

27
**Determine each of the following. **

3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at = 0 + ½(20)t2 t = 1.1s c. W = F.s W = (4N)(12m) = 48 J d. vf2 = vi2 + 2as or vf = vi +at = 0 + (20)(1.1) = 21.9 m/s F (N) s (m)

28
**Determine each of the following. **

3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m f. The change in momentum of the object as it is displaced from x = 12m to x = 20m. a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at = 0 + ½(20)t2 t = 1.1s c. W = F.s W = (4N)(12m) = 48 J d. vf2 = vi2 + 2as or vf = vi +at = 0 + (20)(1.1) = 21.9 m/s e. Work = ∆KE (Area) = KEf – KEi + = ½mvf2 – 0 48J+ 16J = ½(.2kg)vf2 vf2 = 25.3 m/s 5 4 3 2 1 F (N) s (m)

29
**Determine each of the following. **

3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m f. The change in momentum of the object as it is displaced from x = 12m to x = 20m. a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at = 0 + ½(20)t2 t = 1.1s c. W = F.s W = (4N)(12m) = 48 J d. vf2 = vi2 + 2as or vf = vi +at = 0 + (20)(1.1) = 21.9 m/s e. Work = ∆KE (Area) = KEf – KEi + = ½mvf2 – 0 48J+ 16J = ½(.2kg)vf2 vf2 = 25.3 m/s 5 4 3 2 1 F (N) s (m) f. ∆p = mvf – mvi = = (.2kg)(25.3m/s) – (.2kg)(21.9m/s) = 0.68 kg.m/s

30
Ch6.3 – Energy Kinetic Energy: Potential Energy: Work Energy Theorem: Ex1) A 2,200,000N jet starts at rest and takes off at 268 m/s. a) What is it’s initial KE? b) What is it’s final KE? c) How much work was done to bring it up to speed?

31
Ex2) Spiderman (60 kg) is on top of a 30 m building when he decides to scale a 10m tall flagpole, at constant speed. a) How much work did he do? b) What was his initial PE? c) What was his final PE?

32
**HW #34) Suppose that a 0.149 kg baseball is traveling at 40m/s/ **

a) How much work is done to stop it? b) If it is brought to rest in .02m, what average force is acted on it? Ch6 HW#3 p209 28,34,38,41,44,47,55

33
Ch6 HW#3 p209 28,34,38,41,44,47,55 28. The record average speed for the men’s 10km walk is 4.4m/s. How much KE would a 70kg athlete have at that speed? 38. A video of a 70kg male sprinter shows him going from 0m/s to 3.0 m/s in the 1st step, reaching 4.2m/s on 2nd step, and 5.1m/s on 3rd. Compare speed to KE in each step. 41. A 0.046kg golf ball is driven from rest to 70m/s in about 1 ms. Determ KE and approx what dist the club in contact with the ball.

34
**The energy content of beer is about 1.8x106 J/kg. If that energy**

could be turned completely into PEG, how much beer is needed to raise a 1kg mass 1km into the air? 47. A car with a mass of 1000kg …

35
**Ch6.2 - Conservation of Energy **

- The total amount of mechanical energy stays constant Ex1) A 5kg mass is dropped from a 2 meter height. What speed right before it hits the ground? Ex2) A 5kg mass slides down an incline plane, as shown. How fast is it going at bottom? 10 m

36
**Ex3) A cannon is fired at 50 m/s at 40° degrees to horizon. **

What height does it reach? How fast is it going when it comes back down? Ch6 HW#4 p209 59,60,61,62,63

37
Ch6 HW#4 p209 59,60,61,62,63 59. What is the kinetic energy of a 10.0kg piece of concrete after it has fallen for 2.00sec from the side of an old building? 60. A 60kg stuntperson runs off a cliff at 5.0m/s and lands safely in the river 10.0m below. What is the splashdown speed?

38
Ch6 HW#4 p209 59,60,61,62,63 59. What is the kinetic energy of a 10.0kg piece of concrete after it has fallen for 2.00sec from the side of an old building? vf = vi + at = 19.6 m/s vf KEf = ½mvf2 = ½(10)(19.6) = 98 J 60. A 60kg stuntperson runs off a cliff at 5.0m/s and lands safely in the river 10.0m below. What is the splashdown speed?

39
**= 19.6 m/s ½mvi2 + mgh = ½mvf2 + 0 ½(5)2 + (9.8)(10) = ½(vf)2**

Ch6 HW#4 p209 59,60,61,62,63 59. What is the kinetic energy of a 10.0kg piece of concrete after it has fallen for 2.00sec from the side of an old building? vf = vi + at = 19.6 m/s vf KEf = ½mvf2 = ½(10)(19.6) = 98 J 60. A 60kg stuntperson runs off a cliff at 5.0m/s and lands safely in the river 10.0m below. What is the splashdown speed? KEi + PEi = KEf + PEf ½mvi2 + mgh = ½mvf2 + 0 ½(5)2 + (9.8)(10) = ½(vf)2 vf = 14.9 m/s

40
**61. Two automobiles of weight 7. 12 kN and 14**

61. Two automobiles of weight 7.12 kN and kN are traveling along horizontally at 96 km/h (26.7 m/s) when they both run out of gas. Luckily, there is a town in a valley not far off, but it’s just beyond a 33.5 mhigh hill. Neglect friction, which of the cars will make it to town? Is KEi greater than PEf ? 63. A kid in a wagon is traveling at 10 m/s just as she reaches the bottom of a hill and begins to climb a second hill. How high up it will she get before the wagon stops, assuming negligible friction losses?

41
**Ch6.5 Conservation of Energy with Friction **

Ex1) If the 5kg is actually going 5 m/s when it reaches the bottom, what is the coefficient of friction? 10 m

42
**Ch6 HW#5 p209 74 + 2 Bonus questions **

Bonus #1. A 5kg object starts at rest, slides down an incline plane with a kinetic coefficient of friction of 0.10, as shown. What is the speed at the bottom of the incline? 2.9m 1m

43
**PEi = KEf + Wfric mgh = ½mv2 + Ff.s : mgh = ½mv2 + µmgcosθ .s**

Ch6 HW#5 p Bonus questions Bonus #1. A 5kg object starts at rest, slides down an incline plane with a kinetic coefficient of friction of 0.10, as shown. What is the speed at the bottom of the incline? 2.9m 1m PEi = KEf + Wfric mgh = ½mv2 + Ff.s : mgh = ½mv2 + µmgcosθ .s gh = ½v2 + µgcosθ .s (9.8)(1) = ½ v2 + (.1)(9.8)cos20°(2.9) v = 3.8 m/s

44
**Bonus #2. A 2kg object starts at rest, slides down an incline plane, **

as shown. If its speed at the bottom is 2.78 m/s, what is the kinetic coefficient of friction? 0.87m 0.5m

45
**PEi = KEf + Wfric mgh = ½mv2 + Ff.s : mgh = ½mv2 + µmgcosθ .s**

Bonus #2. A 2kg object starts at rest, slides down an incline plane, as shown. If its speed at the bottom is 2.78 m/s, what is the kinetic coefficient of friction? 0.87m 0.5m PEi = KEf + Wfric mgh = ½mv2 + Ff.s : mgh = ½mv2 + µmgcosθ .s gh = ½v2 + µgcosθ .s (9.8)(0.5) = ½(2.78)2 + µ(9.8)cos35°(0.87) µ = 0.15

46
**74. An inclined plane at 30. 0º is 6. 40 m long**

74. An inclined plane at 30.0º is 6.40 m long. A book, which has a kinetic coefficient of friction with the incline of 0.20, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline? 6.4 m

47
**MEi = MEf PEi = KEf + Wfric mgh = ½mv2 + Ff.s mgh = ½mv2 + µFN.s**

74. An inclined plane at 30.0º is 6.40 m long. A book, which has a kinetic coefficient of friction with the incline of 0.20, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline? MEi = MEf PEi = KEf + Wfric mgh = ½mv2 + Ff.s mgh = ½mv2 + µFN.s mgh = ½mv2 + µFg┴.s mgh = ½mv2 + µmgcosθ .s gh = ½v2 + µgcosθ .s v = 6.4 m/s s = ½(vi + vf)t t = 2 sec 6.4 m

48
**Ch6.4 – Power - rate a which work is done**

Units: Common substitutions:

49
**Common substitutions:**

1.

50
**Common substitutions:**

1. 2. Overcoming gravity:

51
**Common substitutions:**

1. 2. Overcoming gravity: 3. Overcoming friction: Ex1) The express elevator in Sears Tower average 9.14 m/s to climb to 103 floor at 408.4m high. Assuming an ave load of 1000kg, what average power supplied by motor?

52
Alternate Formula: Ex2) A rocket produces 889.6N of thrust while traveling at its max speed of m/s. How much power?

53
Ex3) A rocket produces 1000 N of thrust as it accelerates uniformly from rest to 200 m/s, then as it runs out of fuel, its thrust drops off to 0 N as it reaches 300 m/s. How much power is required to reach 300 m/s? Ex4) My neighbors leave six 100W outside lights on at night. If they’re on 11 hours, how much energy is consumed? Ch6 HW#6 p210 78,79,81,87,93

54
Ch6 HW#6 p210 78,79,81,87,93 78. How much power does it take to raise an object weighing 100N a distance of 20.0m in 50s? 79. A runner traverses a 50m long stretch on a horizontal track in 10s at a fairly constant speed. All the while she experiences a retarding force of 1.0 N due to air friction. What power was developed overcoming friction?

55
**81. The per capita power consumption in the United States is around**

10 kJ each and every second. At what speed would you have to push a car exerting a force of 1.0 kN on it all year, day in and day out, to be equivalent to your share? P = F.v = 87. A 2.5 hp motor drives a hoist that can raise a load of 50 kg to a height of 20 m. How long will it take to do it?

56
**A 2x103 kg spacecraft has a ion engine that delivers 150 kW**

of propulsive power via its six thrusters. Assuming it to be far from any massive object, how much time would it take to accelerate from “rest” to 268 m/s (i.e. 600 mi/h)? m = 2000 kg vi = 0 vf = 265 m/s

57
**a and s found with motion eqns P = 150,000W vf = vi + at s = ½at2**

A 2x103 kg spacecraft has a ion engine that delivers 150 kW of propulsive power via its six thrusters. Assuming it to be far from any massive object, how much time would it take to accelerate from “rest” to 268 m/s (i.e. 600 mi/h)? 1. P = W/t m = 2000 kg P = Fs/t vi = 0 vf = 265 m/s P = mas/t a and s found with motion eqns P = 150,000W vf = vi + at s = ½at2

58
**Ch 7.1 - Momentum and Collisions**

Linear momentum Ex p212) Rich Gossage set a fastball record by hurling a .14 kg baseball at a speed of 46.3 m/s, w.r.t. the earth. What was the ball’s momentum? Ex p212) A kg baseball traveling at 28 m/s due south approaches a batter, it’s hit, momentarily crushed, springs back, and sails away at 46 m/s due north. Determine its initial and final momentum, and the change in momentum.

59
**Impulse and Momentum Change.**

in momentum Ex p213) A rocket fires its engine, which exerts an average force of 1000N for 40s in a fixed direction. What is the magnitude of the rocket’s momentum change?

60
Ex pg 216) On 9/12/66. a Gemini spacecraft docked with an orbiting Agena launch vehicle. NASA decided to determine the mass of the Agena, by firing the Gemini’s motor, exerting a constant thrust of 890N in a fixed direction for 7 sec. They sped up by .93 m/s. Assuming Gemini to be a constant mass of 34x102 kg, compute the mass of the Agena. Ch7 HW#1 p235 2,3,4,6,7,11,13,14

61
Ch7 HW#1 p235 2,3,4,6,7,11,13,14 2. What is the momentum of the Earth (m=5.975 x 1027 g) as it moves through space if its orbital speed is about 66,600mi/hr? (66,600x104 x m/s) 3. A 1.0 kg wad of clay is slammed straight into a wall at a speed of 10 m/s. If the clay sticks in place, what was the impulse that acted on it via the wall?

62
**4. What is the impulse provided by the racket on a tennis ball (0**

4. What is the impulse provided by the racket on a tennis ball (0.058 kg) served at 50 m/s? 6. A dried pea fired from a plastic drinking straw has a mass of 0.50g. If the force exerted on the pea is an average lb (0.070 x N) over the 0.10 s flight through the straw, at what speed will it emerge?

63
7. A 1.0kg body initially traveling in the positive x-direction at 10 m/s is acted upon for 2.0s by a force in the same direction of 20N. It then experiences a force acting in the negative direction for 20s equal to 2.0N. Draw a force-time curve and determine the final momentum. Total Area = ∆p 11. A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 10.0 ms, what is her momentum just as she hits the floor? What average force acts on her during the subsequent deceleration? vi = = s = 1m a = 9.8 m/s2 vf = ? vf2 = vi2 + 2as vf = 4.47 m/s 20 F (N) 10 t (s) 20 -20

64
7. A 1.0kg body initially traveling in the positive x-direction at 10 m/s is acted upon for 2.0s by a force in the same direction of 20N. It then experiences a force acting in the negative direction for 20s equal to 2.0N. Draw a force-time curve and determine the final momentum. Total Area = ∆p (20.2) + (–2)(20) = ∆p 0 = no change in p! pi = m.v = (1)(10) = 10 kgm/s 11. A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 10.0 ms, what is her momentum just as she hits the floor? What average force acts on her during the subsequent deceleration? vi = = s = 1m a = 9.8 m/s2 vf = ? vf2 = vi2 + 2as vf = 4.47 m/s 20 F (N) 10 t (s) 20 -20

65
**13. A 47 g golf ball is hit in the air at 60 m/s**

13. A 47 g golf ball is hit in the air at 60 m/s. It lands in sand at the same elevation and comes to rest in 10 ms. Ignoring air friction, what was the average force the sand exerted on the ball? vi = 60m/s vf = 0m/s 14. A golf ball with a mass of 47.0 g can be blasted from rest to a speed of 70.0 m/s during the impact with a club head. Taking that impact to last only about 1.00 ms, calculate the change in momentum of the ball.

66
Ch7.2 – Varying Force Exs: Graphs of force v time Constant force 20 F (N) 5 t (s) 10

67
Ch7.2 – Varying Force Exs: Graphs of force v time Constant force 20 F (N) 5 t (s) 10 Area under the curve = or =

68
Ch7.2 – Varying Force Exs: Graphs of force v time Constant force Shot put 20 F (N) F (N) 5 t (s) 10 t (s) Bow & Arrow (HW#22) F (N) Area under the curve = or = t (s)

69
Ch7.2 – Varying Force Exs: Graphs of force v time Constant force Shot put 20 F (N) F (N) 5 t (s) 10 t (s) Bow & Arrow (HW#22) F (N) Area under the curve = or = t (s)

70
**If the curve is too crazy, find the average force:**

F (N) t (s) Ex1) A 47g golf ball is hit by a golf club, in a collision that lasts 1ms. If the max forces reaches 5640N, what is the impulse and final velocity? 6000 F (N) t (s)

71
**Ex2) A 0.2kg kick ball is rolled toward a 1st grader at 5m/s.**

The 1st grader kicks the ball, in a collision that lasts 10ms. If the max force reaches 500N, what is the impulse and final velocity of the ball? 200 F (N) t (s) Ch7 HW#2 p236 19, Bonus Questions set up #22 B4 class ends…

72
**Ch7 HW#2 p236 19,22 + 3 Bonus Questions**

19. A modern nuclear aircraft carrier weighs 90,000 tons (82 million Newtons) and can travel at 30 knots (15m/s). If it plows into a pier and comes to rest in 0.50min, what average force does it exert on the pier?

73
**22. A typical tubular aluminum arrow 70-cm long has a mass of 25.0 g.**

The force exerted by a bow varies in a fairly complicated fashion from t he initial draw force of, say, 175 N (39 lb) to zero, as the arrow leaves the string. Assume a full draw (70 cm) and a launch time of 16.0 ms. If the arrow when fired straight up reaches a maximum height of m above the point of launch, then (a) What is the average force exerted on it bythe bow? (b) What’s the area under the force-time curve? (c) What impulse is applied to the shooter? vf = 0 h = 143.3m a = -9.8 vi = vf F = 0N s = .70m t = 0.016s F = 175N

74
**Bonus #1. A 0.149kg baseball is traveling south at 35 m/s,**

when it is hit by a bat, exerting a varying force lasting 3ms, as shown in the diagram. What is the ball’s final speed. Impulse = ∆p 10,000 F (N) t (s)

75
**Bonus #1. A 0.149kg baseball is traveling south at 35 m/s,**

when it is hit by a bat, exerting a varying force lasting 3ms, as shown in the diagram. what is the ball’s final speed. Impulse = ∆p (Area) = mvf – mvi 15 = (.149)∆v ∆v = 100m/s 10,000 F (N) t (s)

76
**Bonus #2. A 10kg curling puck starts at rest and is pushed by a ‘curler’,**

for 3 seconds, by a force that follows a fairly complicated fashion, as shown. What is the release speed of the puck? Impulse = ∆p (Area) = mvf – mvi 30 F (N) 1 t (s)

77
**B#3. For each of the graphs shown, describe any physical meaning for:**

1) slope of each line and/or 2) area under each curve F (N) F (N) t (s) d (m) ∆p F (N) vel (m/s) t (s)

78
**Ch7.3 Conservation of Momentum in Explosions**

- work by Newton’s 3rd Law:

79
**FRocket on Exhaust Gases FExhaust Gases on Rocket**

Ch7.3 Conservation of Momentum in Explosions - work by Newton’s 3rd Law: FRocket on Exhaust Gases FExhaust Gases on Rocket

80
**FRocket on Exhaust Gases FExhaust Gases on Rocket**

Ch7.3 Conservation of Momentum in Explosions - work by Newton’s 3rd Law: FRocket on Exhaust Gases FExhaust Gases on Rocket - can be explained by conservation of momentum: Conservation of Linear momentum

81
**FRocket on Exhaust Gases FExhaust Gases on Rocket**

Ch7.3 Conservation of Momentum in Explosions - work by Newton’s 3rd Law: FRocket on Exhaust Gases FExhaust Gases on Rocket - can be explained by conservation of momentum: Conservation of Linear momentum Ex1) A rocket (m=1000kg) expels 5kg of exhaust gases at 1.2 km/s. By how much does it increase it’s speed?

82
**Ex p223) A 8g bullet is fired at a speed of 352 m/s.**

If the mass of the pistol is .90kg, what is the recoil velocity? Ch7 HW#3 p236 25,26,27,28,30,37

83
Ch7 HW#2 p236 25,26,27,28,30,37 25. A rocket engine blasts out 1000 kg of exhaust gas at an average speed of 2 km/s every second. Calculate the resulting average thrust developed by the engine. 26. Water squirting from a hose emerges with a speed of 10 m/s at a rate of 100 kg/s. Compute the average reaction force exerted on the hose. Have you ever seen firefighters working to control a large hose? 27. A 50 kg person at the southernmost end of a 150 kg rowboat at rest in the water begins to walk to the northern end. If, at a given instant, her speed with respect to the water is 10 m/s, what is the velocity of the boat with respect to the water at that moment? pi = pf

84
28. While constructing a space platform a 100 kg robot finds himself standing on a 30 m long 200 kg steel beam that is motionless with respect to the platform. Using his magnetic feet he walks along the beam traveling south at 2.00 m/s. What is the velocity of the beam with respect to the platform as the robot walks? pi = pf m1v1i + m2v2i = m1v1f + m2v2f 0 = 30. While floating in space a 100 kg robot throws a kg wrench at 12.0 m/s toward his partner working on the spaceship. How fast will the robot move away from the ship? Mv = mV

85
**37. An ice skater (with a mass of 55**

37. An ice skater (with a mass of 55.0 kg) throws a snowball while standing at rest. The ball has a mass of 200g and moves straight out with a horizontal speed of 20.0 km/h. Neglecting friction, and assuming the skate blades are parallel to the direction of the throw, describe the skater’s resulting motion in detail. Mv = mV

86
**Ch7.4 – Conservation of Momentum in Collisions**

Elastic collisions- (2 objects bounce off each other) (No deformations)

87
**Ch7.4 – Conservation of Momentum in Collisions**

Elastic collisions- (2 objects bounce off each other) (No deformations) Inelastic collisions- (2 objects stick together) Ex1) An 854N train car at rest is hit and sticks to a 1281N train car that was traveling at 6.1 m/s. a) What speed do they travel off at? b) Calculate the total energy before and after the collision.

88
Ex2) A 10g steel marble traveling at 5m/s West, collides with an 8g steel marble traveling at 6 m/s East. What is the speed of the 10g marble? Ex3) A 2500 kg train car traveling at 10 m/s collides with, but does not stick to a 4500 kg train car traveling at 5 m/s in the same direction. After the collision, the 4500kg car moves off at 11 m/s (same direction). What is the speed of the 2500kg car?

89
Ex4) 2500kg Car 1 traveling at 20m/s South collides with 1450kg Car2 traveling at 20m/s East. At impact, they become entangled and move off at an angle south easterly. Find their speed and direction after the collision. Ch7 HW#4 p ,44,45,46,47,51,54

90
Ch7 HW#4 p ,44,45,46,47,51,54 39. A railroad flatcar having a mass of 10,000 kg is coasting along at 20m/s. As it passes under a bridge, 10 men (having an average mass of 90 kg) drop straight down onto the car. What is the speed as it emerges with its new passengers from beneath the bridge? pi = pf

91
44. Two identical cars of mass 2000 kg each drive toward one another, both traveling at 20.0 m/s. They collide head-on and smash together; what is the final speed of the wreckage? pi = pf 45. A glider on an air track has a mass of 1.00 kg and floats on compressed air, so it can move frictionlessly. Someone shoots a 20.0 g lump of clay at it. The clay strikes the glider, sticks to it, and both move away with a speed of 20.0 cm/s. What was the speed of the clay as it hit the glider?

92
46. Two wads of putty are propelled horizontally directly toward each other. Wad-1 has a mass of 5.00 kg and is traveling at 21.0 m/s south. Wad-2, which is 6.00 kg, is moving at 12.0m/s north. They collide and stick together. What is the velocity of the joint mass of putty? pi = pf 47. A 1263 kg Triumph TR-8 sports car traveling south at 40 km/h (11.6 m/s) crashes head-on into a 1742 kg Checker cab moving north at 90 km/h (25 m/s). If the two cars remain tangled together but free to coast, describe their motion immediately after collision. What is their final speed? How much kinetic energy is lost?

93
51. A 90 kg signal relay floating in space is struck by a 1000 g meteoroid. The latter imbeds itself in the craft and the two sail away at 5.0 m/s. What was the initial speed of the meteoroid? pi = pf

94
**a. What is speed of m1 at bottom of incline? **

54. Two identical blocks, each of mass 10.0 kg, are to be used in an experiment on a frictionless surface. The first is held at rest on a 20.0° inclined plane 10.0m from the second, which is at rest at the foot of the plane. a. What is speed of m1 at bottom of incline? b. m1 slams into and sticks to m2 what is their speed immediately after impact. c. In exp, they don’t stick, m2 = 6.1 m/s. m1 speed? a) PEi = KEf b) pi = pf h = ? m1v1i + m2v2i = (m1 + m2)vf c) pi = pf m1v1i + m2v2i = m1v1f + m2v2f h = 10m.sinθ = 3.42m 10 m m1 m2

95
**a. What is speed of m1 at bottom of incline? **

54. Two identical blocks, each of mass 10.0 kg, are to be used in an experiment on a frictionless surface. The first is held at rest on a 20.0° inclined plane 10.0m from the second, which is at rest at the foot of the plane. a. What is speed of m1 at bottom of incline? b. m1 slams into and sticks to m2 what is their speed immediately after impact. c. In exp, they don’t stick, m2 = 6.1 m/s. m1 speed? a) PEi = KEf mgh = ½mv2 v = 8.3 m/s b) pi = pf h = ? m1v1i + m2v2i = (m1 + m2)vf c) pi = pf m1v1i + m2v2i = m1v1f + m2v2f h = 10m.sinθ = 3.42m 10 m m1 m2

96
**a. What is speed of m1 at bottom of incline? **

54. Two identical blocks, each of mass 10.0 kg, are to be used in an experiment on a frictionless surface. The first is held at rest on a 20.0° inclined plane 10.0m from the second, which is at rest at the foot of the plane. a. What is speed of m1 at bottom of incline? b. m1 slams into and sticks to m2 what is their speed immediately after impact. c. In exp, they don’t stick, m2 = 6.1 m/s. m1 speed? a) PEi = KEf mgh = ½mv2 v = 8.3 m/s b) pi = pf m1v1i + m2v2i = (m1 + m2)vf (10)(8.3) + 0 = (20)vf c) pi = pf vf = 4.15m/s m1v1i + m2v2i = m1v1f + m2v2f h = 10m.sinθ = 3.42m 10 m m1 m2

97
**a. What is speed of m1 at bottom of incline? **

54. Two identical blocks, each of mass 10.0 kg, are to be used in an experiment on a frictionless surface. The first is held at rest on a 20.0° inclined plane 10.0m from the second, which is at rest at the foot of the plane. a. What is speed of m1 at bottom of incline? b. m1 slams into and sticks to m2 what is their speed immediately after impact. c. In exp, they don’t stick, m2 = 6.1 m/s. m1 speed? a) PEi = KEf mgh = ½mv2 v = 8.3 m/s b) pi = pf h = ? m1v1i + m2v2i = (m1 + m2)vf (10)(8.3) + 0 = (20)vf c) pi = pf vf = 4.15m/s m1v1i + m2v2i = m1v1f + m2v2f (10)(8.3) + 0 = (10)v + (10)(6.1) v1f = 2.2m/s h = 10m.sinθ = 3.42m 10 m m1 m2

98
**Ch7.5 Conservation of Momentum and Energy**

Ex1) A 10g marble rolls at 5 m/s towards a 5g marble rolling at 8 m/s the opp direction. After the collision, the 5g marble rebounds backward at 1.5 m/s. What is the speed and direction of the 10g marble after the collision? How much energy is lost in the collision?

99
**Ex2) Mass 2M is travelling at speed +3v, when it collides elastically**

with mass M traveling at speed +v, in the same direction. After the collision, mass 2M has a speed +v. What is the speed of mass M in terms of v? Write an expression for the loss of kinetic energy, in terms of M and v.

100
**HW#53. An 8.0 kg puck floating on an air table is traveling east at 15 cm/s.**

Coming the other way at 25 cm/s is a 2.0 kg puck on which is affixed a wad of bubblegum. The two slam head-on into each other and stick together, find their velocity after the impact. How much kinetic energy is lost? Ch7 HW#5 p235+ 5,9,29,33,53,55

101
Ch7 HW#5 p235+ 5,9,29,33,53,55 5. During a game, a soccer ball having a mass of 0.425kg is kicked from rest to 26m/s in a collision with the player’s foot lasting 8 ms. Impulse? 9. If a 0.061kg handball goes from rest to 20m/s during a serve and the player feels a force of 100N, how long did the impact last? 29. Glider on air track is 30cm long and has a mass of 0.500kg. Floats frictionlessly on compressed air. A 100g wind-up toy car is placed on it and immediately rolls due east at 0.50m/s. Glider? pi = pf

102
**33. 2 kids in small boat at rest wrt water, play catch with a 1kg lead ball.**

Total mass is 250kg. Speed of boat when ball leaves one kid with a speed of 5m/s? pi = pf m1v1i + m2v2i = m1v1f + m2v2f = 53. 8kg puck floating on air table traveling east at 15cm/s collides with a 2kg puck traveling at 25cm/s the other way. Slam together and stick. Speed? ∆KE? KEi KEf pi = pf ½m1v1i2 + ½m2v2i2 ½mTvf2 m1v1i + m2v2i = (m1 + m2)vf 55. 8kg puck floating on air table traveling east at 15cm/s collides with a 2kg puck traveling at 25cm/s the other way. Slam together and 2kg bounces back at 20 cm/s. Speed and dir of 8kg? pi = pf m1v1i + m2v2i = m1v1f + m2v2f

103
Ch6,7 Lab Quizzes - Each lab group is assigned one of two ‘Discovery Labs.’ You are required to design a lab that solves for your lab’s BIG Question - Ch7 HW#5 due at beginning of period

104
**Lab6.1 – Conservation of Energy**

- due in 2 days - Ch6 Rev p205+ 4,21,35,42,50,68,80 due tomorrow

105
**Lab7.1 Conservation of Momentum in Inelastic Collisions**

- due tomorrow - Ch7 Rev p235 5,9,29,53,55 due tomorrow

106
**Ch7.6 Conservation of Momentum in 2 Dimensions**

Ex1) 2 cars enter an icy intersection and skid into each other, the 2500kg sedan heading south at 20m/s and the 1450kg coupe heading east at 30m/s. The 2 vehicles entangle. What speed and direction do they slide off at?

107
**Ex2) A 0.1kg ball slides at 1.4 m/s across a table and collides with **

another 0.1kg stationary ball. The second ball moves off at 30°, with a speed of 0.3 m/s, as shown. What is the speed and direction of the 1st ball after the collision? 0.3 m/s 1.4 m/s ° Ch7 HW#6 p238 57,58,59 1 2

108
Ch7 HW#6 p238 58,59,60 58. A skater with a mass of 75kg is travelling east at 5 m/s when he collides with another skater of mass 45kg heading 60° south of west at 15 m/s. If they stay tangled, what is their final speed?

109
**59. A soft clay block is suspended so as to form a ballistic pendulum.**

A bullet is fired into the clay, raising the pendulum to a height h. Write an expression for the muzzle speed in terms of g, h, and their masses. 60. Derive an expression for the percentage of kinetic energy converted into internal energy during the clay-bullet impact.

110
Ch6 Rev p205+ 4,21,35,42,50,68,80 What is the work done by gravity when a 2.0kg ball falls from a height of 1.50m? PEi = mgh W = ∆PE PEf = 0 Starting from rest, a 25.0kg kid runs up a 5.0m long slide. At the end he is 3.0m higher than at the start. How much work did he do? 3m kg shot put launched at 14 m/s might go 23m. 149g baseball thrown at record speed of 45m/s. Compare KE’s. Shot put Baseball

111
Ch6 Rev p205+ 4,21,35,42,50,68,80 What is the work done by gravity when a 2.0kg ball falls from a height of 1.50m? PEi = mgh W = ∆PE = PEf – PEi = 0 – mgh = -30J PEf = 0 Starting from rest, a 25.0kg kid runs up a 5.0m long slide. At the end he is 3.0m higher than at the start. How much work did he do? W = ∆PE 3m kg shot put launched at 14 m/s might go 23m. 149g baseball thrown at record speed of 45m/s. Compare KE’s. Shot put Baseball

112
Ch6 Rev p205+ 4,21,35,42,50,68,80 What is the work done by gravity when a 2.0kg ball falls from a height of 1.50m? PEi = mgh W = ∆PE = PEf – PEi = 0 – mgh = -30J PEf = 0 Starting from rest, a 25.0kg kid runs up a 5.0m long slide. At the end he is 3.0m higher than at the start. How much work did he do? W = ∆PE = PEf – PEi = mgh – 0 = 750J 3m kg shot put launched at 14 m/s might go 23m. 149g baseball thrown at record speed of 45m/s. Compare KE’s. Shot put Baseball

113
Ch6 Rev p205+ 4,21,35,42,50,68,80 What is the work done by gravity when a 2.0kg ball falls from a height of 1.50m? PEi = mgh W = ∆PE = PEf – PEi = 0 – mgh = -30J PEf = 0 Starting from rest, a 25.0kg kid runs up a 5.0m long slide. At the end he is 3.0m higher than at the start. How much work did he do? W = ∆PE = PEf – PEi = mgh – 0 = 750J 3m kg shot put launched at 14 m/s might go 23m. 149g baseball thrown at record speed of 45m/s. Compare KE’s. Shot put Baseball KE = ½(7.26)(14)2 KE = ½(0.146)(45)2 = 711J = 150J

114
**42. What is the KE and PE of Boeing 747 jet weighing 2.22x106N**

flying at 268m/s at an altitude of 6.1km? Total energy = PE + KE = 50. A 10.0kg package is raised from rest by an elevator at a constant accl of 2.00 m/s2 for 20.0 sec. a. What is its KE at t = 10s? vf = vi + at b. What is its increase in PEG after 10sec? = 20 m/s c. How much work done by elevator? s = ½at2 = 100m

115
**42. What is the KE and PE of Boeing 747 jet weighing 2.22x106N**

flying at 268m/s at an altitude of 6.1km? Total energy = PE + KE = mgh + ½mv2 = 1.35x1010J x109J 50. A 10.0kg package is raised from rest by an elevator at a constant accl of 2.00 m/s2 for 20.0 sec. a. What is its KE at t = 10s? vf = vi + at b. What is its increase in PEG after 10sec? = 20 m/s c. How much work done by elevator? s = ½at2 = 100m

116
**42. What is the KE and PE of Boeing 747 jet weighing 2.22x106N**

flying at 268m/s at an altitude of 6.1km? Total energy = PE + KE = mgh + ½mv2 = 1.35x1010J x109J 50. A 10.0kg package is raised from rest by an elevator at a constant accl of 2.00 m/s2 for 20.0 sec. a. What is its KE at t = 10s? vf = vi + at b. What is its increase in PEG after 10sec? = 20 m/s c. How much work done by elevator? s = ½at2 = 100m a. KE = ½mv2 = ½(10)(20)2 = 2000J b. PE = mgh = (10)(10)(100) = 10,000J c. Total work = 12,000J

117
**68. A 100 kg car racing towards a mountain at 50m/s runs out of gas**

68. A 100 kg car racing towards a mountain at 50m/s runs out of gas. Cleverly, the driver shifts into neutral and coasts onward. Neglecting all friction losses, will he clear the 65 m peak? Speed at top? KEi > PEf ? ½mv2 > mghf KEi = PEtop + KEtop ½mv2 = mghtop + ½mv2

118
**68. A 100 kg car racing towards a mountain at 50m/s runs out of gas**

68. A 100 kg car racing towards a mountain at 50m/s runs out of gas. Cleverly, the driver shifts into neutral and coasts onward. Neglecting all friction losses, will he clear the 65 m peak? Speed at top? KEi > PEf ? ½mv2 > mghf ½(100)(50)2 > (100)(9.8)(65) 125,000 > 65,000 YES KEi = PEtop + KEtop ½mv2 = mghtop + ½mv2 125,000 = 65,000 + ½(100)v2 v = 34.6 m/s

119
**80. A small hoist can raise 100kg of bricks to the top of a construction project**

30m above the street in ½ minute. Determ power.

120
**80. A small hoist can raise 100kg of bricks to the top of a construction project**

30m above the street in ½ minute. Determ power. PEi = KEf + Wfric

121
**Ei = Ef PEi = KEf + Wfric mgh = ½mv2 + Ff.s mgh = ½mv2 + µFN.s**

74. An inclined plane at 30.0º is 6.40 m long. A book, which has a kinetic coefficient of friction with the incline of 0.20, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline? Ei = Ef PEi = KEf + Wfric mgh = ½mv2 + Ff.s mgh = ½mv2 + µFN.s mgh = ½mv2 + µFg┴.s mgh = ½mv2 + µmgcosθ .s gh = ½v2 + µgcosθ .s v = 6.4 m/s s = ½(vi + vf)t t = 2 sec

122
Ch7 Rev p235 12,37,56 12. A 20.0 kg ball of clay drops off a scaffold and falls for 4.0 s before hitting the ground. If it comes to rest in 5.0 ms, what average force did the floor exert on the clay? Use momentum to solve the problem. vi PEi vf = vi + at F.t = ∆p F.t = mvf – mvi a = 9.8 t = 4sec vf KEf 37. An ice skater with a mass of 55.0 kg throws a snowball while standing at rest. The ball has a mss of 200 g and moves straight out with a horizontal speed of 20.0 km/h, Neglecting friction, and assuming the skate blades are parallel to the direction of the throw, describe the skater’s resulting motion in detail. pi = pf m1v1i + m2v2i = m1v1f + m2v2f

123
Ch7 Rev p235 12,37,56 12. A 20.0 kg ball of clay drops off a scaffold and falls for 4.0 s before hitting the ground. If it comes to rest in 5.0 ms, what average force did the floor exert on the clay? Use momentum to solve the problem. vi PEi vf = vi + at F.t = ∆p = 0 + (9.8)(4) F.t = mvf – mvi a = = 39.2 m/s t = 4sec vf KEf 37. An ice skater with a mass of 55.0 kg throws a snowball while standing at rest. The ball has a mss of 200 g and moves straight out with a horizontal speed of 20.0 km/h, Neglecting friction, and assuming the skate blades are parallel to the direction of the throw, describe the skater’s resulting motion in detail. pi = pf m1v1i + m2v2i = m1v1f + m2v2f

124
Ch7 Rev p235 12,37,56 12. A 20.0 kg ball of clay drops off a scaffold and falls for 4.0 s before hitting the ground. If it comes to rest in 5.0 ms, what average force did the floor exert on the clay? Use momentum to solve the problem. vi PEi vf = vi + at F.t = ∆p = 0 + (9.8)(4) F.t = mvf – mvi a = = 39.2 m/s t = 4sec vf KEf 37. An ice skater with a mass of 55.0 kg throws a snowball while standing at rest. The ball has a mss of 200 g and moves straight out with a horizontal speed of 20.0 km/h, Neglecting friction, and assuming the skate blades are parallel to the direction of the throw, describe the skater’s resulting motion in detail. pi = pf m1v1i + m2v2i = m1v1f + m2v2f = (55) v1f + (.2)(5.6) v1f = 0.02 m/s (opp dir)

125
56. As seen from the window of a space station, a 100kg satellite sailing along at 10.0 m/s collides head on with a small 300kg asteroid, which was initially at rest. Taking the collision to be inelastic, and neglecting their mutual gravitational interaction, what is the final velocity of the two bodies? pi = pf m1v1i + m2v2i = (m1+ m2)vf (100)(10) + 0 = (400) vf vf =

Similar presentations

Presentation is loading. Please wait....

OK

The Physics Formula review game.

The Physics Formula review game.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on reuse of wastewater Ppt on solid figures Ppt on magic tee Structural analysis in reading ppt on ipad Ppt on benefits of drinking water Ppt on steve jobs biography pdf Ppt on our environment for class 7th Ppt on automobile related topics based Ppt on job rotation meaning Ppt on duty roster army