1. How many significant figures are there in 0.0143 10 -4 cm ? 9 5 4 3

2. What does the prefix “M” in front of a unit mean? Multiply by 10 -6 Multiply by 10 -3 Multiply by 10 3 Multiply by 10 6

3. How many professors are there at Otterbein? 10 100 1000 10000

4. Somebody claims the correct equation to calculate the position of an object from its (constant) velocity v and the elapsed time t is x(t) = x(t=0) + v t 2. Can this be correct? No Yes Not enough information

5. A red and an orange car make a race. When do they have the same velocity? At point A Between point A and B At point B Never t x A B

6. A red and an orange car make a race. Who wins? Red car Yellow car Insufficient information t x

7. Motion at constant acceleration If (and only if) acceleration is constant, we can calculate x,v,a in any instant by using a(t) = a(t=0) v(t) = v(t=0) + a(t=0) t x(t) = x(t=0) + v(t=0) t + ½ a(t=0) t 2 Green represent independent variable, darker hues initial values. Additionally, we may use v 2 (t) = v 2 (t=0) + 2 a(t=0) [x(t) - x(t=0)]

8. In the equation below, the following is NOT true… v 2 (t) = v 2 (t=0) + 2 a(t=0) [x(t) -x(t=0)] We need to know the position of the object in the moment at which we want to calculate its velocity We need to know the initial velocity of the object to calculate its velocity at time t This equation is only true if the acceleration of the object is constant. The equation is only true for positive velocities, since we have to take the square root eventually.

9. Example for derivation Constant acceleration means linear rising (or falling) velocity  velocity at later time is velocity at earlier time plus slope times time elapsed. Slope of velocity curve is acceleration  v(t 2 ) = v(t 1 ) + a (t 2 -t 1 ) t 1 t 2 t v(t) v(t 2 ) v(t 1 ) ∆v ∆t

10. A car accelerates at constant (non- zero) rate. Which of the following motion diagrams is NOT correct? v v t a t x tt

11. An dropped object accelerates downward at 9.8 m/s 2. If instead you throw it downward, its downward acceleration after release is … less than 9.8 m/s 2 exactly 9.8 m/s 2 more than 9.8 m/s 2 Insufficient information

12. Solving Kinematic Problems 1.Read problem carefully 2.Draw diagram 3.List knowns and unknowns 4.What physics principles do apply? 5.Find equations that do apply 6.Solve algebraically (with variables, not values!) 7.Calculate numerically 8.Check results: Numbers reasonable? Units correct?

13. Vectors “Directions with magnitudes” that can be shifted around

14. Addition: Tail-to-Tip or Parallelogram

15. Subtraction by adding negative vector a-b = a + (-b)

16. 3D Vectors 3D vector as a sum of multiples of the three unit vectors i,j,k Example: a=1.5 i + 2.5j +3 k

17. To which quadrant does the following vector belong? A = - 4 i + 6.5 j 1 st Quadrant 2 nd Quadrant 3 rd Quadrant 4 th Quadrant

18. To which quadrant does the following vector belong? |B| = 4.5, φ B = -45º 1 st Quadrant 2 nd Quadrant 3 rd Quadrant 4 th Quadrant

19. To which quadrant does the following vector belong? |c| = 4.5m/s, φ c = 3.10 1 st Quadrant 2 nd Quadrant 3 rd Quadrant None of the above

20. To which quadrant does the following vector belong? 1 st Quadrant 2 nd Quadrant 3 rd Quadrant None of the above

21. To which quadrant does the following vector belong? 1 st Quadrant 2 nd Quadrant 3 rd Quadrant Not enough information

22. What is the length of the following vector? -1 m/s 1 m/s 5 m/s 25 m 2 /s 2

23. What is the length of this vector? Upper formula Middle formula Bottom formula None of the above

24. Which is a correct statement concerning this vector? |B| = 4.5, φ B = -45º Its magnitude is negative Its x component is negative Its y component is positive Its x and y component have the same absolute value

25. Projectile Motion

26. By aiming a gun higher (increasing the angle of the barrel w.r.t. the x- axis), I achieve the following: Higher initial position Larger initial y-component of velocity Higher initial velocity None of the above

27. A gun is accurately aimed at a dangerous terrorist hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed v 0, the terrorist lets go and drops to the ground. What happens? The bullet hits the terrorist regardless of the value of v 0 The bullet hits the terrorist only if v 0 is large enough The bullet misses the terrorist Not enough information v0v0 gun gutter

28. A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic paths shown, which ship gets hit first? A B Both at the same time Need more information A B

29. Circular Motion

30. An object at the end of an r=1m string in circular motion completes 10 revolutions in one second. How long does each revolution take? (2 π r)(1s) 1s/(2π r) 0.1 s Need more information

31. An object at the end of an r=1m string in circular motion completes 10 revolutions in one second. What is its speed? 6.3 m/s 3.1 m/s 63 m/s 0.31 m/s

32. An object at the end of an r=1m string in circular motion completes 10 revolutions in one second. What is its frequency? 10 Hz 1 Hz 1/10 Hz 3.14 Hz

33. Newton’s first Law In the absence of a net external force, a body either is at rest or moves with constant velocity.force body –Motion at constant velocity (may be zero) is thus the natural state of objects, not being at rest. Change of velocity needs to be explained; why a body is moving steadily does not.

34. Mass & Weight Mass is the property of an object Weight is a force, e.g. the force an object of certain mass may exert on a scale

35. Mass: On the surface of the Moon a standard block of lead … … has a different mass … has a different weight Its weight is unchanged, but g has a different value Need more information

36. Newton’s first law states that objects remain at rest only when they no net force acts on them. A book on a table is subject to the force of gravity pulling it down. Why doesn’t it move? Newton’s first law does not apply (obstacle!) There must be another force opposing gravity Table shelters book from force of gravity Not enough information

37. Newton’s second Law The net external force on a body is equal to the mass of that body times its acceleration F = ma. Or: the mass of that body times its acceleration is equal to the net force exerted on it ma = F Or: a=F/m Or: m=F/a

38. Newton II: calculate Force from motion The typical situation is the one where a pattern of Nature, say the motion of a projectile or a planet is observed: –x(t), or v(t), or a(t) of object are known, likely only x(t) From this we deduce the force that has to act on the object to reproduce the motion observed

39. Calculate Force from motion: example We observe a ball of mass m=1/4kg falls to the ground, and the position changes proportional to time squared. Careful measurement yields: x ball (t)=[4.9m/s 2 ] t 2 We conclude v=dx/dt=2[4.9m/s 2 ]t a=dv/dt=2[4.9m/s 2 ]=9.8m/s 2 Hence the force exerted on the ball must be F = 9.8/4 kg m/s 2 = 2.45 N –Note that the force does not change, since the acceleration does not change: a constant force acts on the ball and accelerates it steadily.

40. Newton II: calculate motion from force If we know which force is acting on an object of known mass we can calculate (predict) its motion Qualitatively: –objects subject to a constant force will speed up (slow down) in that direction –Objects subject to a force perpendicular to their motion (velocity!) will not speed up, but change the direction of their motion [circular motion] Quantitatively: do the (vector) algebra!

41. Newton II: calculate motion from force Say a downward force of 4.9N acts on a block of 1 kg (we call the coordinate y and start at y=0) We conclude that the block will be accelerated: a y = F/m= -4.9N / 1kg = -4.9 m/s 2 We use this constant acceleration to calculate y = -1.225m/s 2 t 2 + b t + c, where b &c are constants, c=0 Q: What physical situation is this?

42. Newton II: calculate mass from force and acceleration Pretty simple, if you know force and acceleration we have m=F/a