Your SlideShare is downloading. ×
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
1 2012 ppt semester 1 word problems review
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

1 2012 ppt semester 1 word problems review

252

Published on

0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
252
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
1
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide
  • Welcome to the physics review for semester one. Hello everybody, this is Dr. Fiala. The tutorial you are about to begin is comprehensive. Material reviewed includes Kinematics, force, projectiles, momentum and impulse. To receive the most benefit from this video there are two important points that should be noted.
  • Point one. I have color coded the slides so you can more easily practice skills with which you have difficulty. This is a good way for you to roll up your sweatpants. Word problems are presented on red slides. Data tables can be found on olive slides. Green slides work you graph reading skills and tan slides match concepts. Solutions are presented for each problem on the following slide. I will give you an audio cue that the slide will be changing. That way if you want to stop the video before you see the solution you can do so at that time. You may choose to view the entire video or simply move to the slide color representing your weakness.
  • Point two. You must have all of the formulas handy while going though the semester on physics review. If you have them written down in my order it would be helpful because I refer to the formulas as formula one, formula two, formula three and so on according to the order in which they appear on my card. Because all students must use the formula given out on final exam day, it is important that you are skilled at finding any variable in any formula. You can stop this video at any time and I encourage you do do so. I suggest that you complete this video review and tutorial when you can think about what we are talking about without distraction. Stay positive . You can do this!
  • Vi&f = 13.67 m/s ti = 0 s tf = 7.32 s Xi = 0 m a = 0 m/s2
    Xf=
  • So now that we have the problem KUEd we can look at the index card and choose an equation. The first equation assumes a constant velocity which we have. We know the change in time is 7.32 seconds, so we can use that formula. The second, third, and fifth formulas do not have final position in so they are of no use to us. The seventh equation assumes a velocity change and since there is no acceleration, it cannot be used here.
  • We can use the fourth, and sixth formulas to find final horizontal position with the information we are given. You should try them all to practice your math skills. The most simple equation to use is the first one. Manipulating the first equation gives us final position equals constant velocity times time.
  • Although Harley Davidson motorcycles are made in Milwaukee, Wisconsin, they still come equipped with a speedometer that registers speed in both the English and the metric systems of measurement. When I look down I see I am traveling at 49.21 km/h. How far will I travel in that same 7.32 second time period?
  • The answer lies in the ability to convert within the metric system using dimensional analysis. An hour has 60 minutes in it so in essence I am multiplying my first fraction by one therefore not changing the relative value of the 49.21 only its units. I repeated the process because I know that there are 60 seconds in each minute. The units I have now is km/s so I need to change the kilometers to meters. There are 1000 meters in a kilometer. When all the multiplication is done I find out that 49.21 km/h is equal to 13.67 m/s. That means I travel exactly the same distance, 100.07 meters. If you did not remember there are 1000 meters in a kilometer, review the pneumonic I gave you at the beginning of the year. Many kids have dropped over dead converting metrics. I have placed a one under the K for kilometers. To convert to meters on the chart I need to move three places to my right. Meters is located at 0 for over. All metric system base units, meters, liters, and grams, are located at 0. So simply place a zero in for each place moved to the right and you convert one to one thousand.
  • vi = 0 m/s ti = 0 s tf = 8 s a = 4 m/s2 Xi = 0 m
    Xf = vf
  • Looks like the beast is going to be the winner. The first three formulas and the fifth formula do not contain final position in them so they are of no use to us. We cannot use the fourth formula because we do not know final velocity right now. It does not mean we cannot currently find final velocity, but there is no need to do that work if we can find the final position without it.
  • Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s
    Vf= tf =
  • You cannot use formula one, two, four, five, and six because you do not know the final time, You cannot use formula three because you do not know average velocity. That leaves the last and most ignored of the kinematic formulas.
  • Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s
    tf= Vf =
  • Here you again have a bunch of ways to solve the problem. Once again I encourage you to practice each of these formulas. The more comfortable you become with using the formulas the easier physics becomes as well. The first formula is for constant velocity which could be provided by using the third formula first but there is no reason to make this a two step problem!
  • Remember that the only math we ever perform with vectors is addition. Sometimes the addition is as simple as adding two positive numbers. Sometimes it can become increasingly complex such as adding negative numbers, of numbers at right angles. Here identify the vector magnitude by its value and the direction of the vector by the arrow. The first one is adding two positive numbers. The second problem is solved by adding a positive number with a negative number. The final problem is solved by using the pythagoreum theorem.
  • Positive integer plus a positive integer. Positive integer plus a negative integer. A2 + B2 = C2
  • W = 153.08 N g = -9.8 m/s2
    m =
  • Remember that weight is a force so we can use the first or third formulas. But since we dealing with weight here I would go with the simplest formula possible.
  • m = 51.62 kg a = 0 m/s2 g = -9.8 m/s2 Fa = 6 N

    Ff =
    Here you must recognize that you are dealing with more than one force in the horizontal. Seek out a formula that contains the sum of all forces in the horizontal.
  • Combine the second and fourth formulas to calculate the force of friction. The sum of the horizontal forces is what is happening to the object. This object is not accelerating so the max side of the equation is equal to zero. That means that the applied force and the force of friction have to be equal in magnitude but opposite in direction.
  • Here is the motion map of the object moving at constant velocity to the right. Predict what the force diagram would look like.
  • m = 1315 kg a = 3.23 m/s2
    F =
  • Again this is a problem with only one force in it. Go with the simplest formula that you can.
  • m = 1315 kg F = 4247.45 N a = 3.00 m/s
    Ff =
  • Notice now that we have two forces in the horizontal for which to account. Lets use the sum of all forces in the x formula. Remember that the sum of all forces in the x is what is actually happening to the car so use the 3 m/s2 and the mass of the car and occupants for that side of the equation.
  • m = .448 kg g = -9.8 m/s2
    Fs =
  • The force diagram of a projectile is easy to understand. The only force acting on the projectile is gravity. The sum of all forces is equal to what is happening to the apple. The .448 kg apple is falling at -9.8 m/s2. Notice that the force of gravity in this case will also be the product of the .448 kg apple is falling at -9.8 m/s2. Since the force of support acting on an object is equal to its weight and the force of support is zero, the apple is “weightless” during freefall.
  • m = 44.10 kg a = 1.1 m/s2
    F =
  • Since the problem stated that the surface was frictionless, we really do not need to use a formula with more than one force in it. However, you can combine formula two and four together and plug in zero for your force of friction.
  • m1 = 42 kg m2 = 43.25 kg m3 = 44 kg m4 = 44.23 kg m5 = 44.77 kg m6 = 45.01 kg m2 = 45.45 kg g = -9.8 m/s2
    Fs Fg
  • We needed to add the masses of everyone on the elevator to solve this problem. Since elevators travel in the vertical, we will use the sum of all forces in the y. Since the elevator is not accelerating, the force of support and the force of gravity will be equal in magnitude but opposite in direction.
  • m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .75 m/s2
    Fs=
    Experience tells us that if we are speeding up on the way up in an elevator we feel pressure in the bottom of our feet.
  • This indicates that the force of support is larger than the force of gravity at that time. We will again use the formula for the sum of all forces in the y. This time the elevator is accelerating so the force of support and the force of gravity cannot be equal in magnitude because there has to be an unbalanced force to cause this acceleration. The sum of all forces in the y is what is actually happening to the elevator. The elevator is accelerating at .75 m/s2 . Remember that the force of gravity cannot change since we stay on planet Earth and no one falls out of the elevator. So the only force that changes in an elevator is the force of support.
  • m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .-50 m/s2
    Fs=
    Recall once again your experience about slowing down on the way up in an elevator. Should the force of support be larger or smaller than the force of gravity? I think I feel less pressure on the bottom of my feet at this time.
  • We will again use the formula for the sum of all forces in the y. Again the elevator is accelerating so the force of support and the force of gravity cannot be equal in magnitude because there has to be an unbalanced force to cause this acceleration. The force diagram shows us that the force of support is smaller than the force of gravity.
  • mt = 2000 kg mb = .0002 kg g = -9.8 m/s2 F = -50 N
    at =
    ab =
  • Both the truck and the bug experience the same amount of force at the same time. Remember we have spoken in the past about the effect of g-forces on the body. You may recall that around 4-5 g’s most people will black out. To calculate the g-force on the body simply divide the acceleration experienced by the object by 9.8. So the bug has over 25,000 g-forces acting on it! The outside of the bug stopped moving when it hit the truck, however, the insides of the bug kept accelerating until they hit the truck.
  • T1 = 375.4 N g = -9.8 m/s2 Θ = 7.5°
    m=
  • The weight of an object in equilibrium must equal the total force of support. Notice that the two cables are hung at the same angle. How do we determine the angle? You can see that there is 165° between the two cables. In a straight line there is 180°. If you subtract 165° from 180° it means there is 15° left over to be divided up for the cable angles. Each side therefore is hung at 7.5°. Force of support is in the vertical so if you measure from positive X to the vector you use the sine of 7.5°. Remember there are two cables so you need to double your answer to account for the opposite cable’s support.
  • Vi = 30 m/s Vi = 5 m/s a2 + b2 = c2
    Vf=
  • Here we are given two velocity vectors that are directly on the y and x-axis. When we want to create a resultant vector out of the components we use the pythagoreum theorem. In this case you can see that I am being pushed southeast by the wind.
  • Remember that the formulas for finding angles will give you the angles on either side of the vector inside the quadrant in which the vector is located. Once you have made your calculation you need to assign that angle to the correct location within the quadrant.
  • We have two choices to find the angle. One will find theta and one will find phi. You can see that the 9.46° is located between the 270° y-axis and the resultant vector. So you can add 9.46° to 270° to find the angle as measured from positive x. You can also see that the 80.54° angle is located between the 360° x-axis and the resultant vector. You can subtract 80.54° from 360° to find the angle as measured from positive x.
  • Yf = -15 m Yi = 0 m m = 15 kg g = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/s
    ti = 0 s ti = 4 s
    a =
    F =
  • First we need to find the acceleration of the box to find the force acting on the box. We can use the change in velocity over the change in time formula to find this acceleration. We can now use Newton’s 2nd law to find the applied force on the box. Since this is a horizontal problem and there is more than one force acting on the box we will use formulas two and four to get the sum of all forces in the horizontal.
  • Yf = -15 m Yi = 0 m m = 15 kg g = -9.8 m/s2 Vi = 0 m/s ti = 0 s
    tf =
    Vf =
  • This package is now in free fall. It had no initial velocity in the vertical. What does the motion map look like?
  • V = 27 m/s Θ= 27° g = -9.8 m/s2
    Viy =
    Vix =
  • The same concept that we have used in force is also true for velocity because they are both vector quantities. We resolve a vector into its horizontal and vertical components using trigonometry. If the angle is measured from positive x then we use cosine for the horizontal component and sine for the vertical component.
  • g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m
    tf =
  • Using the beast in the vertical will give us the overall time that the baseball will spend in the air. Remember to set your launch and land positions at 0 meters since the baseball will be launch and land at the same elevation. You will need to factor out time in the equation. If you worried about doing that, you can find the time to reach maximum height and then multiply by two. Remember that you can only do that if the projectile is launched and lands at the same elevation. At maximum height the baseball will have a vertical velocity of 0 m/s and you would use formula two on this slide.
  • g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 mYf = 0 m
    Xi = 0 m tf = 2.5 s
    Xf =
  • Range is the product of horizontal velocity and time in the air.
  • g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Vix = 24.06 m/s ti = 0 s Yi = 0 mYf = 0 m
    Xi = 0 m tf = 2.5 s Xf = 60.15 m
    Δy =
  • Since we are looking for maximum height and we have an initial vertical velocity we have three formula choices. We can average the velocity and multiply by half the time in the air, use the beast in the vertical using half the time in the air or formula four on the card where you do not need to worry about time.
  • m = .448 kg Vi = 30 m/s g = -9.8 m/s2
    p =
  • Momentum is the product of mass times velocity.
  • m = .448 kg Vi = 30 m/s g = -9.8 m/s2 ti = 0 s tf = 3.06 s Δp = -13.44 kgm/s

    F =
  • That should make sense because gravity is what changes the apples momentum from positive to zero and them to negative. At zero velocity the apple has no momentum abut is still under the influence of gravity. The acceleration due to gravity is -9.8 m/s2 and the mass of the apple is .448 kg. So when the apple is a projectile it has a force of gravity of -4.9 N acting on it at all times!
  • m1 = .448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s
    p =
    Vf =
  • To find the final velocity of the two apples stuck together we need to know the momentum of the system before the collision. Since the second apple was at rest it contributed no momentum to the system. The momentum of the first apple ends up being 4.66 kgm/s2. Since momentum is a conserved quantity and the momentum of the system initially was 4.66 kgm/s2 it has to remain 4.66 kgm/s2. Now you need to combine the masses to find the new velocity because the apples are stuck together and have to be moving at the same rate.
  • m1 = .448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0 m/s p = 4.66 kgm/s
    F =
  • Here it is important to remember that impulse equals the change in momentum. If we set these two formulas equal to each other we can find force because we know the time of contact with the wall.
  • This completes this comprehensive review for the semester one final exam. Thank you for taking the time to review with me. If you have studied hard you deserve to do well. Good luck and have a nice day.
  • Transcript

    • 1. The slides used in this video are color coded. If you are experiencing difficulty with one aspect of your understanding than another you might find this coding… useful! Slides with Slides with tan red backgrounds backgrounds involve involve word matching Slides with green problems. concepts. backgrounds Slides with olive involve backgrounds graphing. involve reading data tables.
    • 2. Dr. Fiala is traveling on his Harley at a constant 13.67 m/s. What is the distance traveled by Doc in 7.32 seconds?
    • 3. SOLUTION: Find the distance traveled. Vi&f = 13.67 m/s Xf ti = 0 s tf = 7.32 s Xi = 0 m Xf = m a = 0 m/s2
    • 4. SOLUTION: Find the distance traveled. Vi&f = 13.67 m/s Xf ti = 0 s tf = 7.32 s Xi = 0 m Xf= 100.07 m a = 0 m/s2
    • 5. Dr. Fiala notices he is now traveling at a constant 49.21 km/h. What is the distance in meters traveled by Doc in 7.32 seconds?
    • 6. SOLUTION: Dimensional analysis. M K H D 1 x x X 0 d c x X X 49.21 km/h = 13.67 m/s So the Xf remains 100.07 m m
    • 7. Dr. Fiala jumps in his unstarted car. He accelerates at a rate of 4 m/s2 for 8 seconds. How far did Doc travel?
    • 8. Doc’s final position. vi = 0 m/s ti = 0 s tf = 8 s a = 4 m/s2 Xi = 0 m vf = Xf = Xf = 128 m
    • 9. If Dr. Fiala starts from a full stop and accelerates at 2.25 m/s2, how incredibly fast will he be traveling when he has traveled 1530 meters?
    • 10. SOLUTION: Calculate final velocity without knowing time. Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf tf Vf= 82.98 m/s
    • 11. If Dr. Fiala starts from a full stop and accelerates at 2.25 m/s2, how long will it take him to drive 1530 meters?
    • 12. SOLUTION: Solve for time. Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf = 82.98 m/s tf Vf t = 36.88 s
    • 13. 2.25 + 3.25 = 2.25 + 3.25 = 2.25 + 3.25 =
    • 14. SOLUTION: Adding vectors. + 2.25 + 3.25 = 5.5 + 2.25 + 3.25 = 1.00 + 2.25 + 3.25 = 3.95 + - + + - +
    • 15. ? kg 153 N Determine the mass of a 153.08 N object.
    • 16. SOLUTION: Calculate mass. W = 153.08 N g = -9.8 m/s2 m m = 15.62 kg
    • 17. Determine the force of friction on a 15.62 kg object traveling at a constant horizontal velocity of 3.62 m/s while experiencing an applied force of 6 N.
    • 18. SOLUTION: Calculate force of friction. m = 51.62 kg a = 0 m/s2 g = -9.8 m/s2 Fa = 6 N Ff Ff = -6 N
    • 19. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 20. Determine the force needed to accelerate Dr. Fiala’s car and its occupants at a rate of 3.23 m/s2 if the total mass of car and occupants is 1315 kg and there is no friction force.
    • 21. SOLUTION: Find applied force. m = 1315 kg a = 3.23 m/s2 F F = 4247.45 N (kg)(m/s ) 2
    • 22. This time, when we apply that 4247.45 N force to Dr. Fiala’s car and its occupants, the resulting acceleration is actually lower. It registers at a rate of only 3.00 m/s2. What is the magnitude for the force of friction causing the acceleration to be decreased?
    • 23. SOLUTION: Find applied force. m = 1315 kg F = 4247.45 N a = 3.00 m/s2 Ff Ff = - 302.45 N
    • 24. When an object is freefalling it is weightless. Prove mathematically that a .448 kg apple is weightless during its freefall from a tree. Draw a force diagram of the apple during its fall from the tree.
    • 25. SOLUTION: Find force of support. m = .448 kg g = -9.8 m/s2 Fs Fs = 0 N Fg = -4.39 N
    • 26. Assuming a perfectly frictionless surface, ideal for launching students in a game of faculty bowling, Dr. Fiala uses a brand new gizmo that automatically applies a force that results in an acceleration of 1.1 m/s2. Experimentation resulted in a student with a mass of 44.10 kg, accelerating at 1.1 m/s2. Find the force generated by the gizmo for that student.
    • 27. SOLUTION: Find force in the horizontal. m = 44.10 kg a = 1.1 m/s2 F F = 48.51 N
    • 28. All of the students from the previous problem (combined mass) step into an elevator at the same time. Draw a force diagram of this situation including the magnitude of Fg and Fs.
    • 29. SOLUTION: Find force of gravity and force of support. m1 = 42 kg Fg m2 = 43.25 kg Fs m3 = 44 kg m4 = 44.23 kg m5 = 44.77 kg m6 = 45.01 kg m2 = 45.45 kg Fg= -3025.36 N g = -9.8 m/s2 Fs= 3025.36 N
    • 30. This same elevator accelerates at a rate of .75 m/s2 towards the second floor. Draw a force diagram of this situation including the magnitude of Fg and Fs.
    • 31. SOLUTION: Find force of support. m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .75 m/s2 Fs Fs = 3256.89 N Fs= N Fg = 3025.36 N 3256.89
    • 32. This same elevator accelerates 2 at a rate of .50 m/s as it begins its stop for the second floor. Draw a Force diagram of this situation including the magnitude of Fg and Fs.
    • 33. SOLUTION: Find force of support. m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .-50 m/s2 N Fs = 2871.01 N Fg = 3025.36 N Fs Fs= 2871.01
    • 34. According to Newton’s 3rd law, an action force causes an equal on opposite reaction force. It is no wonder a truck windshield squashes a bug and not vice versa. A 2000 kg truck and a .0002 kg bug hit with a 50 N force. Take a closer look at why the truck wins the collision by calculating the acceleration exerienced by the bug and by the truck.
    • 35. SOLUTION: Why the bug doesn’t survive. mt = 2000 kg at mb = .0002 kg ab g = -9.8 m/s2 F = -50 N at = -.025 m/s2 ab = -250,000 m/s2
    • 36. .40 N 375 These cables will snap if the mass of the trafffic light exceeds 10.1 kg. Does the traffic light exceed 10.1 kg?
    • 37. SOLUTION: The cable does not break. T1 = 375.4 N g = -9.8 m/s2 Θ = 7.5° m T1y m= 10 kg
    • 38. Dr. Fiala attempts to walk due east at 5 m/s at the same time as a 30 m/s cold, winter wind is blowing due south. What is the magnitude of Dr. Fiala’s velocity.
    • 39. SOLUTION: Resultant velocity magnitude. Vi = 30 m/s Vi = 5 m/s Vx = 5 m/s Vy = 30 m/s Vf Vf= 30.41 m/s 2 2 a +b =c 2
    • 40. Vx = 5 m/s V =3 0.4 1m /s Vy = 30 m/s If Dr. Fiala continues his velocity and the wind continues to blow steadily, at what angle, as measured from positive “X”, is Dr. Fiala’s velocity.
    • 41. SOLUTION: Resultant velocity angle measured from positive x. tan Θ = x y Θ = 9.46° Vx = 5 m/s Vy = 30 m/s tan Φ = y x Φ = 80.54° Θ (from +x) = 279.46°
    • 42. Because of this wind, a 15 kg package is blown from Dr. Fiala’s arms and onto the ground. The 15 kg package reaches a velocity of 30.41 m/s in a time of 4 seconds. Find the force acting on the box horizontally if there is no friction.
    • 43. SOLUTION: Find applied force. Yf = -15 m a Yi = 0 m F m = 15 kg g = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/s ti = 0 s a = 7.85 m/s2 ti = 4 s F = 117.79 N
    • 44. If the package is blow horizontally at 30.41 m/s off a ledge onto a parking lot that is 15 meters below how much time will it spend in the air before striking the ground? What does the motion map look like?
    • 45. SOLUTION: Find time package spends in the air. Yf = -15 m Yi = 0 m m = 15 kg g = -9.8 m/s2 Vi = 0 m/s ti = 0 m/s Vf tf tf = 1.75 s
    • 46. Dr. Fiala throws a baseball in the air with an initial velocity of 27 m/s at an angle of 27° to the horizon. Create a velocity vector diagram and show, by parallelogram method, the “X” and “Y” components of the baseball’s velocity.
    • 47. SOLUTION: Resolve velocity vector into “x” and “y” components just like force or any other vector. V = 27 m/s Θ= 27° g = -9.8 m/s2 m/s 27 27° Viy Vix Viy = 12.26 m/s Vix = 24.06 m/s Vx = V Vy = V
    • 48. How much time will it take for the baseball to reach the same height from which it was thrown?
    • 49. SOLUTION: Find time in the air. g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m tf tf = 2.5 s
    • 50. How far will the baseball travel in 2.5 seconds?
    • 51. SOLUTION: Find range. g = -9.8 m/s2 Xf Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m tf = 2.5 s Xf = 60.15 m
    • 52. What is the maximum height the baseball attained during its flight?
    • 53. SOLUTION: Find Δy. g = -9.8 m/s2 Δy Θ= 27° Viy = 12.26 m/s Vix = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m Δy = 7.67 m tf = 2.5 s
    • 54. If it was a .448 kg apple that was thrown into the air at 30 m/s what was the apple’s intial momentum?
    • 55. SOLUTION: Find momentum of apple. m = .448 kg Vi = 30 m/s g = -9.8 m/s2 p p = 13.44 kgm/s
    • 56. What constant force is needed to get a change in the apple’s momentum from 13.44 kgm/s to 0 In 3.06 seconds?
    • 57. SOLUTION: Find force necessary to change momentum. m = .448 kg F Vi = 30 m/s g = -9.8 m/s2 ti = 0 s tf = 3.06 s Δp = -13.44 kgm/s F = 4.39 N
    • 58. After falling to the ground the .448 kg apple rolled at a constant 10.4 m/s where collided with a stationary .577 kg apple. If the two apples stuck together, at what velocity would they roll?
    • 59. SOLUTION: Find the velocity of two apples stuck together. m1 = .448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s Vf p p = 4.66 kgm/s2 Vf = 4.55 m/s
    • 60. Determine the force applied if the rolling apples strike a wall and a come to a stop in .311 seconds.
    • 61. SOLUTION: Find force needed to stop apples. m1 = .448 kg m2 = .577 kg ti = 0 s tf = .311 s g = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0 m/s p = 4.66 kgm/s F F = 14.98 N

    ×