1. Fall Semester Physics Review Key

2. Intro Chapter 1 #1 Problem/Question- What curiosity do you have?
Background Info- what do you already know?
Hypothesis- What do you think the outcome will be?
Experiment/Data collection- Methodically collect data.
Process Data- Examine, calculate using data to discover if hypothesis was correct.
Conclusion-discuss results

3. Chapter 1 #2
Independent variable- the variable that the experimenter chooses to manipulate.
Dependent variable- the variable that is the outcome of manipulating the independent variable.

4. Chapter 1 #3 Variable Unit Dist./Displacement Meters (m) Mass
Kilogram (kg)
Time
Second (s)
Speed
Meters/second (m/s)
Velocity
Acceleration
Meters/second2 (m/s2 )
Force
Newtons (N)
Work
Joules (J)
Energy
Power
Watts (W)
Momentum
Kilogram·meter/second (kgm/s)
Impulse
Newton·second (Ns) or (kgm/s)

5. Chapter 1 #4
Accuracy-

6. Chapter 1 #5 7 Things Every Physics Graph Can't Live Without Title
Y-axis label
X-axis label
Y-axis units in parenthesis
X-axis units in parenthesis
*Correctly numbered axes. (No squiggles!)
Data! (Sometimes connect with line of best fit)

7. Chapter 1 #6 y = mx + b y= y-coordinate m= slope x = x-coordinate
b= y-intercept

8. Chapter 1 #7
When the real test is impossible (hurricane modeling), inhumane (virus transmission models), too small (atom models), or too large (architecture models), etc.
Allows time to be manipulated through computer modeling if takes too long.

9. Chapter 1 #8

10. Chapter 1 #9 Prefix Symbol Power of ten Tera T 1012 Giga G 109 Mega M
106
Kilo K
103
Hecto/Hecta H
102
Deka/Deca D
101
Base unit (m, g, s)
100
Deci d
10-1
Centi c
10-2
Milli m
10-3
Micro μ
10-6
Nano n
10-9
Pico p
10-12
Femto f
10-15

11. Chapter 2 #1 Distance- total path taken.
Displacement- straight line between start and end points, includes direction.
Dist= 50km + 20km = 70km

12. Chapter 2 #2 Distance- total path taken.
Displacement- straight line between start and end points, includes direction.
Disp= 50km + -20km = 30km, East

13. Chapter 2 #3
Scalar- a variable that has magnitude only (no direction is associated with it).
Ex: distance, speed, mass, time.

14. Chapter 2 #4
Vector- A variable that has both magnitude and direction. (need to put negatives for negative directions: south, west, backwards, down, left, etc.)
Ex: displacement, velocity, acceleration, force.

15. Chapter 2 #5
Displacement= 50km North
This is a silly question.

16. Chapter 2 #6 Distance= 20km + 70km= 90km
Displacement= 20km + -70km = -50km or
50km west.

17. Chapter 2 #7 Instantaneous velocity example:
Your speed the cops write on your ticket.
Average velocity example:
Your total displacement divided by total time.

18. Chapter 2 #8
50km= 50000m
2h= 7200s
V= x/t= 50000m/7200s= 6.94m/s

19. Chapter 2 #9 X= 80m a= -7m/s2 V= 0m/s V0= ? V2= Vo2 + 2ax
0= Vo2 + (2)(-7m/s2)(80m)
33m/s = V0

20. Chapter 2 #10 V0 = 0m/s a = 0.25m/s2 X = 3.26m V = ? V2= Vo2 + 2ax
V2 = 0 + 2(0.25m/s2)(3.26m)
V = 1.3 m/s

21. Chapter 2 #11 V0 = 0 m/s V = 11.5m/s X= 15.0m a = ? V2= Vo2 + 2ax
132 = 0 + 2(a)(15.0m) a = 4.4m/s2
t = ?
v = v0 + at
11.5 = 0 + (4.4m/s2)(t) t = 2.6s

22. Chapter 2 #12 V0 = 90km/hr = 25m/s V = 0m/s X = 0.80m a = ?
V2= Vo2 + 2ax
0 = (a)(0.80m)
-391m/s2

23. Chapter 2 #13

24. Chapter 2 #14 Slope = rise/run = x/t = velocity.
This graph would have a -1m/s where the blue triangle is.

25. Chapter 2 #15 Average velocity = total distance/total time
Vavg = 50m/2s = 25m/s

26. Chapter 2 #16 Average velocity = total distance/total time
Vavg = 200m/5s = 40m/s

27. Chapter 2 #17

28. Chapter 2 #18 Find the area under the graphed line and the x-axis.
x =Velocity x time (rectangle)
x = (5m/s)(10s) = 50m

29. Chapter 2 #19
Slope = rise/run = v/t = acceleration

30. Chapter 2 #20 Acceleration- the rate of change of velocity.
Another way to say it: How fast your speed is changing.
If you have 3m/s2 acceleration, you are speeding up 3m/s every second.

31. Chapter 2 #21

32. Chapter 2 #22 Vo= 0m/s V = 5m/s t= 5s a = ? V = V0 + at
5m/s = 0 + a(5s)
1m/s2 = a

33. Chapter 2 #23 V0 = 0m/s V = 15 m/s t = 2s a = ? V = V0 + at
15m/s = 0 m/s + a(2s)
7.5m/s2 = a

34. Chapter 2 #24 V0 = 0m/s V = 10 m/s a = 2m/s2 t = ? V = V0 + at
10 m/s = 0 + (2m/s2)t
5s = t

35. Chapter 2 #25 V0 = 5m/s V = 25 m/s t = 10s x = ? V = V0 + at a= 2m/s2
x = V0t + (.5)(a)(t2)
X = (5m/s)(10s) + (.5)(2m/s2)(102) = 150m

36. Chapter 2 #26 V0 = 5m/s a = 2m/s2 t = 10s x = ? x = V0t + (.5)(a)(t2)
X = (5m/s)(10s) + (.5)(2m/s2)(102) = 150m
*Pretty much the same as 25……oops.

37. Chapter 2 #27 a= 10m/s2 V0= 0m/s X = 180m V = ? V2= Vo2 + 2ax
V = 60m/s *Don’t forget to square-root at the end!

38. Chapter 2 #28 x = -30m a = -10m/s2 V0 = 0m/s t= ?
x = V0t + (.5)(a)(t2)
-30 = 0 + (.5)(-10)(t2)
2.4s = t

39. Chapter 3 #1 Add vectors head to tail.
Add them in any order you want to.
Their magnitude is proportionate to their length.

40. Chapter 3 #2 Scalar and scalar make a scalar.
Scalar and vector make a vector.
Vector and vector make a scalar.
Think of a vector like a negative and a scalar like a positive and the outcome will make sense.

41. Chapter 3 #3
x direction y-direction +140m m 0 -23m x total=+140m y total= +100m R= = = 172m θ= tan-1(ytot/xtot)= tan-1(100/140)= 36° N of E.

42. Chapter 3 #4
Same process as #3- get totals, do Pythagorean, do inverse tangent.
x direction y-direction
-0.57km
-3.82km(cos25°) km(sin25°)
5.8km(cos15 °) km(sin15 °)
x tot=1.57km y tot= -0.11km
R= = = 1.57km
θ= tan-1(ytot/xtot)= tan-1(.11/1.57)= 4° S of E.

43. Chapter 3 #5 10m is the vertical height.
3m is the horizontal distance.
You are looking for the hypotenuse.
R= = = 10.4m
θ= tan-1(y/x)= tan-1(10/3)= 73° to the horizontal

44. Chapter 3 #6
A. Something that moves two dimensionally in space due to a force launching it out.
B. A parabola
C. Free fall is just vertical motion, projectile has a horizontal component too.

45. Projectile Motion Formula Notes Sheet Original Form of Equations
X-Direction Version
Y-Direction Version
a= 0
a= -10
Vx = const
Vy = -V0y for angled lauches
Δx= Vo·Δt + 1/2·a·Δt²
x= Vxt
y = Voyt + 1/2(a)t2
V= Vo + aΔt
Vy = Voy + at
V² = Vo² + 2aΔx
Vy² = Voy² + 2ay

46. Chapter 3 #7
Horizontal launch:
Vertical launch:

47. Chapter 3 #8 A. Since launched horizontally, the V0y= 0m/s.
We will use
to solve.
Vx= 8m/s
y= -10m
a= -10m/s2
t = 1.4s (from second formula)
B. x = (8m/s)(1.4s)= 11.2m (from first formula)
x= Vxt
y = Voyt + 1/2(a)t2

48. Chapter 3 #9
Since it’s an angle launch, you must use the triangle to get Vx and V0y: also Vy=-V0y
Vx = Vocosθ= 20m/s(cos35°) = 16.4m/s
V0y = Vosinθ= 20m/s(sin35°) = 11.5m/s
Find t from:
t= 2.3s
Find x from x=Vxt= (16.4m/s)(2.3s)= 38m
Vy = Voy + at

49. Chapter 3 #10 Farthest (biggest range) is 45°.
highest (biggest vertical distance) is 90°.
Stay in the air the longest is vertical 90°.

50. Chapter 3 #11 A. Since launched horizontally, the V0y= 0m/s.
We will use
to solve.
Vx= 3.1m/s
y= -1.5m
a= -10m/s2
t = .55s (from second formula)
x = (3.1m/s)(.55s)= 1.7m (from first formula)
x= Vxt
y = Voyt + 1/2(a)t2

51. 2-D Motion #11 and ¾
Remember, the horizontal velocity is always constant since there is no horizontal gravity and vertical motion is pulled down 10m/s every second. A B C D E
V0y = +50 m/s        V0x = 25 m/s

52. …..11 and ¾ continued t (sec) Vx (m/s) Vy (m/s) 25 50 6 -10 1 25 40 725 50 6
-10 1
 25 40 7
 -20 2 30 8
 -30 3 20 9
 -40 4 10
 -50 5

53. Chapter 4 Forces #1
When all forces are balanced, the net force is zero, so the acceleration must be zero. (The object will continue its motion, whether at rest or at constant speed).
ΣF = ma
0 = m0

54. Chapter 4 Forces #2
When the forces are not balanced, the net force is a nonzero number, meaning that you must have some number for acceleration once you divide out the mass, so you will have to be changing your velocity.
EX:
ΣF = ma
10N = (5kg)a a=2m/s2

55. Chapter 4 Forces #3 Contact forces result from 2 objects touching.
EX: friction, normal, tension, applied.

56. Chapter 4 Forces #3 continued.
Field forces act over a distance- objects can be pushed/pulled from a distance. Gravity, electrical, magnetism.

57. Chapter 4 Forces #4 Tension force, FT Gravitational force, Fg
Applied force, Fapp
Normal force, FN or N
Friction force, Ff or Fk or Fs

58. Chapter 4 Forces #5
Any object on an incline will have the same free body diagram if it
isn’t being acted on by an applied force.

59. Chapter 4 Forces #6

60. Chapter 4 Forces #7
Constant speed tells us that the net force is zero, since there is no acceleration.
If the box is moving at constant speed then the friction and applied force are balanced.
so the friction force, Ff= 30N
Ff = μ Fn
Fn= Fg, so
30N = μ 100N
μ = 0.3 (no units!)

61. Chapter 4 Forces #8
Our problem is the same as the diagram below except we are using 35° instead.
Σfx= Fax -Ff= So, Ff= Fax = Fa(cos θ) = 30N (cos35 °) = 27N

62. Chapter 4 Forces #9
a = 2m/s2
Since the weight, Fg= 20000N, it’s mass is 2000kg.
ΣF = ma
ΣF = (2000kg)(2m/s2) = 4000N

63. Chapter 4 Forces #10 51kg means Fg = 510N
Which means Fgx = Fgsinθ = 510N (sin 42°) = 341N
Fn= Fgy = Fg(cos θ) = 510N(cos 42°) = 379N
Ff = μ Fn = 0.10(379N) = 37.9N
Sum of forces along ramp
is the net force
= Fgx-Ff
= 341N – 37.9N = 303N

64. Chapter 4 Forces #11 Fg = (mass)(gravity)
Your weight is your force of gravity!! You MUST MUST MUST know this fact for all physics!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

65. Chapter 4 Forces #12
Weight is how much force your body exerts downward due to the gravitational pull. Mass is a measure of the amount of matter that makes up your body. Mass doesn’t change if gravity changes.

66. Chapter 4 Forces #13 If the mass is 20kg, the weight is 200N.
That means the normal force is 200N

67. Chapter 4 Forces #14 ΣFy= 0 Fapy – Fg +FN = 0 FN = Fg – Fapp sinθ
FN = 200N – 10N(sin 25°)
FN = 196N

68. Forces #15
mass = 11g = 0.011kg V0 = 0m/s V = 180mm/min= 0.003m/s x = 3cm = 0.03m ΣF = ma = ? You can get acceleration from the equation: V2= Vo2 + 2ax a= m/s2 (1.5x10-4m/s2) ΣF = ma = (0.011kg)(1.5x10-4m/s2) = 1.65x10-6N

69. Forces #16
Same process as vectors in chapter 3- get totals, do Pythagorean, do inverse tangent.
x direction y-direction
20N N N
-40N
x tot=-20N y tot= -9N
R= = = 22N
θ= tan-1(ytot/xtot)= tan-1(9/20)= 24° S of W.

70. Forces #17
The solution of the problem involves substituting known values of G (6.673 x 10-11 N m2/kg2), m1 (5.98 x 1024 kg), m2 (70 kg) and d (6.38 x 106 m) into the universal gravitation equation and solving for Fgrav. The solution is as follows:

71. Forces #18
The solution of the problem involves substituting known values of G (6.673 x 10-11 N m2/kg2), m1 (5.98 x 1024 kg), m2 (70 kg) and d (6.39 x 106 m) into the universal gravitation equation and solving for Fgrav. The solution is as follows:

72. Forces #19 Fgrav = 980 N Fgrav = 392 N Fgrav = 641 N

73. Chapter 5- Energy and Power #1
Energy of height = gravitational potential energy, PEg = mgh
Energy of motion = kinetic energy, KE= ½ mv2
Energy stored in compression or stretched object = elastic potential energy
Friction changes mechanical energy into thermal energy. (environment becomes warmer)
When a force causes displacement, work has been done, W= Fd cosθ
Rate of energy usage is power, P= W/t or Fv

74. Chapter 5- Energy and Power #2
P = W/t = Fd/t = mgd/t
m= 500kg
g= +10m/s2
d= 1.5m
t = 1 min = 60s
P = (500kg)(10m/s2)(1.5m)/(60s)
P= 125W

75. Chapter 5- Energy and Power #3
For work to have been done by a force, there must be displacement in the direction that force is pushing. Ex: No work done in pushing on a wall all day.

76. Chapter 5- Energy and Power #4
The rock starts with gravitational potential energy, PEg and it converts to KE as it falls. The ending KE is the same as the starting PEg!
PEgi = KEf
mgh = KE
(5.75kg)(10m/s2)(7.25m)= KEf
417J = KEf

77. Chapter 5- Energy and Power #5
The rock starts with gravitational potential energy, PEg and it converts to KE as it falls. The ending KE is the same as the starting PEg!
PEgi = KEf
mgh = ½ mv2
gh = ½ v2
= v
14.1m/s = Vf
Mass doesn’t matter!!!!!!!!
Mass cancels when solving for v final. An elephant and a pebble fall at the same rate of acceleration!

78. Chapter 5- Energy and Power #6
First, find the net force (ΣF).
ΣF = +1140N N = 190N
Then, use the work kinetic energy theorem:
W= ΔKE (remember ΔKE is KEf- KEi)
Fnetd = ½ mvf2 – ½ mvi2
(190N)(d) = ½ (2000kg)((2.0m/s)2)- 0
d= 21m

79. Chapter 5- Energy and Power #7
PEg = mgh
PEg = (8kg)(10m/s2)(2m)
PEg = 160J

80. Chapter 5- Energy and Power #8
KE= ½ mv2
KE = ½ (1500kg)((200m/s)2)
KE = 3.0 x 107 J
A little unrealistic….but okay.

81. Chapter 6 #1
Momentum, p=mv
p=mv
p= (2250kg)(25m/s)
p= 56250kgm/s

82. Chapter 6 #2 Δp= pf - pi Δp= mvf – mvi Δp= m(vf- vi)
Δp= (0.55kg)(-20m/s – 10m/s)
Δp= -16.5kgm/s
Don’t forget- rebound velocity
needs to be negative!

83. Chapter 6 #3 Use the conservation of momentum! p1i + p2i = p1f + p2f
mv1i + mv2i = mv1f + mv2f
(0.005kg)(0m/s) + (0.52kg)(0m/s)= (0.005kg)(v1f) + (0.52kg)(2.1m/s)
0= (0.005kg)(v1f) + (1.092kgm/s)
-218.4m/s = v1f

84. Chapter 6 #4 Use the conservation of momentum! p1i + p2i = p1f + p2f
mv1i + mv2i = mv1f + mv2f
(70kg)(0m/s) + (2.5kg)(0m/s)= (70kg)(v1f) + (2.5kg)(25m/s)
0= (70kg)(v1f) + (62.5kgm/s)
-0.89m/s = v1f

85. Chapter 6 #5
p1i + p2i = p1f + p2f
mv1i + mv2i = mv1f + mv2f (*make one of your initial velocities negative!)
(0.3kg)(15m/s) + (0.25kg)(-15m/s)= (0.3kg)(vf) + (0.25kg)(vf)
You can combine the left side:
4.5kgm/s kgm/s = (0.55kg)(vf)
±1.4m/s = vf
ΔKE = KEf – KEi = (½ mv1f2 + ½ mv2f2)- (½ mv1i2 + ½ mv2i2)
ΔKE = 0.539J – 62J = -61J

86. Chapter 6 #6
A. perfectly inelastic indicates they stick together upon colliding.
(25kg)(5m/s) + (97kg)(-3m/s)= (25kg)(vf) + (97kg)(vf)
125kgm/s kgm/s = 122kg(vf)
-1.4m/s = vf
B. ΔKE = KEf – KEi = (½ mv1f2 + ½ mv2f2)- (½ mv1i2 + ½ mv2i2)
ΔKE = 120J – 749J = -629J

87. Chapter 6 #7 Use FΔt= Δp for each. m1: F(1s) = (.5kg)(Vf)
Being that they have the same force, F, you can see that the m1 mass will have a much much bigger velocity at the end and therefore, will have a much much bigger momentum.

88. Chapter 6 #8 A long deceleration time so that the force is less
FΔt = Δp

89. Chapter 6 #9
A. FΔt = Δp
Impulse is the same as change in momentum- calculate impulse if they give you force and time. Calculate change in momentum if they give you mass and velocity.
Δp= pf – pi = mvf- mvi = m(vf – vi)
Δp=2500kg(35m/s – 0) = 87500kgm/s
B. FΔt = Δp
F(0.03s)= 87500kgm/s
F = N

90. Chapter 6 #10 Variable Scalar or Vector? Mass Scalar Time Distance
Displacement
Vector
Velocity
Acceleration
Force
Power
Momentum/Impulse
Work/Energy