Physics Final Review

1. Physics Final Review • We will go through HW questions and Old exams Questions • Ask questions as they come up • Dan C. and Ashwin M. – Exam 1 • Allen W. + Vincent L. – Exam 2 • Collin W. – Final [new material]

2. Final FormatYour exam will cover everything – 2/3 of the exam will be new material and 1/3 of the final will be old material [exam 1+2 stuff].But it is important to know Exam 1+2 stuff really well because the professors will build that material into the new material Questions.

3. Physics 121Midterm 1 Review Ashwin Malhotra and Dan Carrero

4. Vectors Key points • Know all your prefixes: • Micro 10^-6 • Nano 10^-9 • Pico 10^-12 • Etc • Know basic trig • Sin(x)=O/H • Cos(x)=A/H • Tan(x)=O/A • Understand that X can change and that can change the relationships of the sides

5. Vectors

6. Solution 2.8 m/s Explanation: You are give two sides of a right triangle, you can easily find the third side with a2+b2=c2. To find the angle you can use “inverse cosine” and use what they gave you or do “inverse sine” or “inverse tan” involving the y-component: Cos-1(2.2m/s/2.8m/s) = angle The units m/s cancel and you are left with the answer for the angle in degrees 2.2 m/s The Angle in Question Cos-1(X/Hyp) = angle Sin-1(Y/Hyp) = angle Tan-1(X/Y) = angle X2 + Y2 = C2 Hyp aka C = sqt(X2 + Y2)

7. Vectors Key Theory • What is a vector? • A quantity which has magnitude and direction • How to add/substract Vectors? • Two ways  graphically versus components • Graphical lets you quickly see what going to happen • Components helps to get exact values i.e. previous problem • How to multiple/divide Vectors by scalars? • Scalars only alter the magnitude • The direction of the vector remains the same i.e. Arrow * 4 = 4Arrow • Do Not need to know Cross Product or Dot Product

8. Kinematics • Difference between displacement and distance • Difference between velocity and speed • Difference between instantaneous velocity/acceleration and average velocity/acceleration Motion key equations: • V=Vo+at • deltaX=Vot+.5at2 • V2=Vo2+2a[deltaX]

9. Kinematics You can find out that the CS is 5m/s easily. Then you can also find out that in 1.1seconds the car travels 5.5m. But that is not the answer to part A. You have to add 16m to that and get 21.5m ~22m to be correct. Why? Because they ask for the position at 1.1 not the change in position. Similar method for part B.

10. Kinematics

11. The amount of time it takes to get to max speed must be the same as the amount of time it takes to get to rest from max speed and the same amount of distance must be covered in both. They gives us the Max Speed. We can calculate how much distance is covered and how long it takes to do so. Using the max speed/max acceleration we can figure out how long it takes to accelerate and how much distance is covered Decelerating to rest Constant speed  its max velocity Accelerating to max speed This must also be 13.8m This leaves us with this middle distance – we have it is being travelled at 6m/s. To find out how long, we take the distance and divide by speed: [100 – 2(13.8)] / 6 = 21seconds V=Vo+at 6=0+1.3t T=6/1.3 deltaX=Vot+.5at2 deltaX=0+.5(1.3)(6/1.3)2 deltax=13.8m

12. Kinematics

13. Earth Moon .5 = v^2 / (2*9.8) get v then use h = v^2 / (2*(9.8/6)) Earth Moon 1m = v^2 / (2*9.8) get V=4.42 then use that to get h = 4.42^2/ (2*(9.8/6)) Vf=V0+at The Vf at the max point is zero. 0= V0 + (9.8*1.5); get V0=14.7 Same V0 on moon 0 = 14.7 + ((9.8/6)*(t)); get t  9seconds. This T is only half; have to double!

16. Relative Velocity • There two important cases to think about. The dumb boater and the smart boater. • One gets where he needs to be faster than the other. Why? • And the other gets exactly where he needs to be but slower. Why? • Make a good free body diagram for these and read carefully about directions of things travelling. • All kinematics equations are still valid here

17. Relative Velocity

18. The boat travels in one direct faster than the other  why? the current must be helping. So we can figure out the direction of the water flowing. We can also figure out the speed: 29/4=7.25 29/6=4.83 7.25 – x = 4.83 + x x = 1.2m/s the x is helping one and slowing down the other; so we say if there was no x there would be an equal speed for traveling both ways – we can set up the above expression and find what x is and that equal speed

19. The way he will travel The way he travels

20. Relative Velocity

21. This is how they should travel to get where they want to be: 17m/s This how they will travel West South West wind: 13m/s With the picture above it should be easy to figure out what the angle [red star] of travel should be  Sin(angle)=13/17 Angle = 49.88 ~ 50 to be exact the angle is 50 degrees Southwest

22. How he will travel How he travels X velocity comes from the current

23. Motion • There is the simple linear motion. • Then there is projectile motion. • There are horizontal projectiles and there are vertical projectiles • Treat X and Y directions separately • Key thing to keep in mind is that all projectile motion problems have time in common between the X and Y direction. • Basic question would be to give enough things to get time from one direction and to find something using that time in the other direction

24. Motion

25. Sin(A) = Opp/Hyp 29{Sin(31)} = opposite = Velocity in the Y Vf – Vi = a*t  0 – Vy = (9.8/6) * (t) t = 9.14455 [this the time required to reach the max height, half total time. Have to double it to get answer~18s] Cos(A) = Adj/Hyp 29{Cos(31)} = Adjacent = Velocity in X Vx * t = Distance Vx * t = distance [earth] 29{Cos(31)} t = x plug in for t, x= 75.7m Vx * t = distance [moon] 29{Cos(31)} * 18s = 450m Divide 450/75 ~ 6 Time earth comes from Vf – Vi = a * t  0 – 29{Cos(31)} = 9.8 * t t = 3.045…

26. Motion

28. Force • F=ma • The ma is the cause and the F is effect; not the other way around. Think about this: • The mass does not change but acceleration does, together they create a force which can do work • You will have to put together force with work for the final • Drawing a free body diagram is the only way to solve force problems • Best thing to do is to draw sample free body diagrams on your cheat sheet so you do not miss any forces during the exam

29. Force

30. Force

32. Force

33. The Elevator Problem

39. Work, Force, and Energy WARM UPS!! 1st: Convert the velocity to meters/second 100 km/hr = 27.8 m/s 2nd: Once everything is in SI Units, we can start thinking about the problem!! 3rd: There are two ways to think about this problem KINEMATICS or WORK/ENERGY KINEMATICS: Step 1: Vf2 = Vi2 + 2ad Solve for a. Step 2: Since F = ma and friction stops the car F = ma = Ff = ukFn = ukmg Therefore uk = a/g Work – Energy: Step 1: Since the car has speed it has a kinetic energy (KE). Since the car stopped work was done (by friction) to change the kinetic energy KEf – KEi = work = Ff d Step2: 1/2mvf2 – 1/2mvi2 = ukmg d Therefore uk = v2/(gd)

40. Newton’s 2nd law is: The SUM of all the on an object forces is equal to that objects mass multiplied by its acceleration. F1 + F2 + F3 + ….. = ma With forces you need to visualize what’s doing what! This can only be learned through practice. 1. Draw the direction of forces acting on all of the objects in the problem. 2. The direction of your arrows only gets meaning when YOU decide which way is + No force is ever negative Unless YOU decide it is. Your decision is based on The direction you set your Coordinates. If a force arrow you draw Is opposite your Coordinates, then and only then is it negative. T Ty Tx Fwall +y mb g +x

41. Normal Force: FN=ukFf M mg

43. Given: mC = 1200 kg v = 15 m/s h = 5 m Want: v’ = ? Initial Final

44. Allen Wu and Vincent Lin Midterm 2 review

45. Work/energy is scalar meaning that it does not matter on the path it takes, as long as the surface is frictionless. Only focus on the beginning and the end. Initial: Car is moving (kinetic energy) and is 5 m above the “final” ground (potential energy) Final: Car is moving (kinetic energy) on “final” ground (no potential energy) W = W’ KE + PE = KE’ + PE’ ½mv2 + mgh = ½mv’2 ½v2 + gh = ½v’2 ½(15 m/s)2 + 9.81 m/s2(5 m) = ½(v’)2 17.97 m/s = v’

47. vy vx’ vy' vx v'

48. Energy is NOT conserved (inelastic) Momentum is conserved Given mx = 1000 kg my = 800 kg vx = 15 m/s vy= 20 m/s Want v‘ = ? m1v1i + m2v2i = m1v1f + m2v2f x: (1000 kg)(15 m/s) + (800 kg)(0 m/s) = (1000 kg + 800 kg)(v’x) : 15000 kg m/s = (1800 kg)(v’x) 15000 kg * m/s = 8.33 m/s = v’x 1800 kg y: (1000 kg)(0 m/s) + (800 kg)(20 m/s) = (1000 kg + 800 kg)(v’y) : 16000 kg * m/s = (1800 kg)(v’y) 16000 kg * m/s = 8.89 m/s = v’y 1800 kg a2+ b2 = c2 8.332 + 8.892 = c2 12.18 = c

49. Want Angular velocity (ω) = ? Given Radius of disc = 2 m Initial tangential velocity = 3 m/s Angular acceleration (α) = 1 rad/s2 Angular position (θ) = 2 revolution Convert to SI units! 2 revolutions * 2π radians = 4π radians 1 revolution

50. Look at your angular mechanics formulas, their format is identical to linear mechanics Formulas θ = θ0 + ω0t + ½αt2 ω = ω0 + αt ω2 = ω02 + 2α(θ – θ0) Conversions from linear to angular mechanics ω = v/r α = a/r Since we are working with angular stuff, convert tangential velocity to angular velocity ω0 = 3 m/s = 1.5 rad/s 2 m We are looking for angular velocity after the disc accelerated for after a measured distance. Also, notice that there is no time mentioned in the question, so if would be harder to use the first 2 equations. ω2 = ω02 + 2α(θ – θ0) ω2 = (1.5 rad/s)2 + 2(1 rad/s2)(4πrad) ω2 = 27.38 rad2/s2 ω = 5.23 rad/s