1. v = vo + at x = xo + vot + ½ at2 A) 47 = 5 + a(3.8) a = 11.05 m/s/s
B) x = 0 + 5(3.8) + ½ (11.05)(3.8)2 = m
C) v = vo + at
v = 5 + (11.05)(2.7) = m/s

2. A) -24. 9 m/s B) v = vo + at -24. 9 = 24. 9 – 9. 8t t = 5
A) m/s B) v = vo + at = 24.9 – 9.8t t = 5.08 sec C) v = vo + at v = 24.9 – 9.8(1.3) v = m/s

3. D) v = vo + at -12.16 = 24.9 –9.8 t t = 3.78 sec E) y = yo + vot + ½ at2 Part C+ D: y = 24.1 m

4. F) y = yo + vot + ½ at2 Part C) = 24
F) y = yo + vot + ½ at2 Part C) = 24.1 m Part D) Goes from ground to ypeak = m at t =2.54 sec then down to 24.1 m at 3.78 sec Distance = m+( ) = m

5. G) y = yo + vot + ½ at2 ypeak = 31. 63 m at t =2
G) y = yo + vot + ½ at2 ypeak = m at t =2.54 sec H) 0 m/s (That’s why it’s at the peak!!) I & J) – 9.8 m/s/s (g for all locations!)

6. 3. y = yo + voyt + ½ at2 0 = 1.4 –4.9t2 t = 0.53 seconds to floor x = vxt x = 2.25(0.53) x = 1.20 m

7. STEP 0! 4A) Vx= 43cos(30) = 37.24 m/s Voy= 43sin(30) = 21.50 m/sq vx
STEP 0! 4A) Vx= 43cos(30) = m/s Voy= 43sin(30) = m/s
Determines the time in the air

8. A) vy = voy + ayt -21.5 = 21.5 –9.8t t = 4.39 seconds total time in the air

9. 4B) y = yo + vot + ½ at2 y = 0 + 21. 5(2. 20) – 4. 9(2
4B) y = yo + vot + ½ at2 y = (2.20) – 4.9(2.20) 2 ypeak = m at t =2.20 sec
ypeak

10. 4C) x =vxt x = 37.24(4.39) (Remember to use total time in the air to get Da’ Range) x = 163.48 m

11. 5A) Fnet = ma (Fapp – Ff) = ma Fapp – 0 = 4(6.8) Fapp = 27.2 N

12. FN Ff
Fapp Fg
5B) Fnet = ma (Fapp – Ff) = ma Fapp – mk(FN) = 4(6.8) Fapp = mk(FN) Fapp = 0.45(39.2) = N

13. Fapp Fg
5C) Fnet = ma (Fapp – Fg) = ma Fapp – mg = 4(6.8) Fapp = mg Fapp = = N

14. 5D) Fnet = ma (Fapp – Ff) = ma Fapp – 0 = 4(0) Fapp = 0 N Inertia keeps it going!!

15. 5E) Fnet = ma (Fapp – Ff) = ma Fapp – mk(FN) = 4(0) Fapp = mk(FN) Fapp = 0.45(39.2) + 0 = N

16. 5F) Fnet = ma (Fapp – Fg) = ma Fapp – mg = 4(0) Fapp = mg + 0 Fapp = 39.2 + 0 = 39.20 N

17. Phase 2 1A) Fg = mg = (12)(9.8) =117.6N 1B) Fgperpendicular = mgcosq = 109.04 N
q = 22 degrees
Fgparallel
1C) Fgparallel = mgsinq = N

18. Phase 2 1D) Fnet = ma Fnet = Fgpar – Ff Fnet = 44.05 – 0 = 44.05 N
Fgparallel
Fnet = Fgpar – Ff
Fnet = – 0
= N
a = Fnet/m = 44.05/12
a = 3.67 m/s/s
Fgperp

19. 1E) Fnet = ma Fnet = Fgpar – Ff Fnet = 44.05 – mkFN
1A) 1B) and 1C) do not change
1E) Fnet = ma FN Ff
Fnet = Fgpar – Ff
Fnet = – mkFN
= –(.07)(109.04)
= N
a = Fnet/m = 36.42/12
a = 3.04 m/s/s
Fgparallel
Fgperp

20. 1F) Fnet = ma Fnet = Fgpar – Ff 0 = mgsinq – mk mgcosq
Fgparallel
1F) Fnet = ma FN Ff
Fnet = Fgpar – Ff
0 = mgsinq – mk mgcosq
mgsinq = mk mgcosq
Tanq = mk
q = Tan-1(mk) = 4.00 deg
Fgpar
Fgperp

21. Phase 2 #2 2. KEtrans =1/2 mv2 (Energy of Motion)
8,750 Joules
Holey Kinetic Energy Batman!

22. Phase #3
3. PE= mgh
(Energy of Position)
PE =mgh =
4,116 Joules

23. Phase 2 #4 4. PE1 + KE1= PE2 + KE2 (Conservation of Energy)
mgh1 + 1/2mv12= mgh2 + 1/2mv22
gh1 + 1/2v12= gh2 + 1/2v22
9.8(25) + ½(5)2= 9.8(17) + 1/2v22
v2 = m/s

24. 5B) W = DPEelastic = 32.5 Joules stored (Work-Energy Theorem)
Phase 2 #5
5A) Work = (Average Force) times Displacement
This word is key when the force is not constant
Here force goes from 0 N to 250 N which is an average force of 125 N
W = Fd = (125)(.26) =32.5 Joules
5B) W = DPEelastic = 32.5 Joules stored
(Work-Energy Theorem)

25. 6A) Momentum, p = mv p = (95)(15) = + 1,425 kgm/s
Phase 2 #6A
6A) Momentum, p = mv
p = (95)(15) = + 1,425 kgm/s
+ momentum is momentum to the right

26. 6B) Impulse (FDt) equals Change in Momentum (Dp)
Phase 2 #6B
6B) Impulse (FDt)
equals
Change in Momentum (Dp)

27. 6B) FDt = Dp m Dv Dp = (95)(4-15) = - 1,045 kgm/s = Impulse
Phase 2 #6B
6B) FDt = Dp
m Dv
Dp = (95)(4-15) = - 1,045 kgm/s = Impulse
Momentum was lost so Impulse is negative

28. Phase 2 #6C Magnitude of the Force, F = 143.15 N
6C) Impulse (FDt) = kg m/s
Magnitude: FDt = kg m/s
F(7.3) = 1045
Magnitude of the Force, F = N

29. wo = 33rev 1 min 2p rad = 3.46 rad/sec 1 min 60 sec 1 rev
Phase 2 #7A First part
7A) q = qo +wot + ½ at2
wo = 33rev 1 min p rad = rad/sec
1 min 60 sec 1 rev
Then: a = Dw/Dt = (0-3.46)/13 = rad/s/s

30. 7A) q = 0 + 3.46(13) + ½ (-0.27)(13)2 q = 22.17 rad = 3.53 revolutions
Phase 2 #7A Second PArt
7A) q = (13) + ½ (-0.27)(13)2
q = rad = 3.53 revolutions

31. Phase 2 #7B & 7C
7B) a = Dw/Dt = (0-3.46)/13 = rad/s/s
7C) w = wo + at = (-0.27)(9.7)
w = 0.84 rad/sec
v = rw

32. Phase 2 #7D
7D) q = qo +wot + ½ at2
q = (9.7) + ½ (-0.27)(9.7)2
q = rad = 3.32 revolutions

33. Phase #8 a A
12 kg a
3 kg B

34. Phase #8 T A
12 kg a
117.6 N
Fnet = ma
117.6 – T = 12a Equation #1

35. Phase 2 #8 a T Fnet = ma T – 29.4 = 3a or T = (29.4 + 3a) B 3 kg
Equation #2

36. Phase 2 #8A) & 8B) Fnet = ma 117.6 – (29.4 + 3a) = 12a 88.2 = 15a
8A) a = 5.88 m/s/s [Down]
8B) a = 5.88 m/s/s [Up]

37. Phase 2 #8 a T T = (29.4 + 3a) = 29.4 + 3(5.88) = 47.04 N B 3 kg
Back to Equation #2
T = ( a) = (5.88) = N

38. Phase 3 #1 and #2A-B 2A) Force on charge A due to charge B
2B) FAB = kq1q2 (Coulomb’s Law) r2
Force is Repulsive for 2 negative charges:
FAB = (9 x 109)(17 x 10-6 )(3 x 10-6 ) = N
(.07)2
#1 and #2A-B
Electrons are mobile (protons are NOT!)

39. Phase 3 #2C-E FAB =- FBA (Repulsive Forces are equal in magnitude
but Opposite in Direction
by Newton’s 3rd Law of Motion
(Action/Reaction)
#2C-E

40. E = 300,000 N/C [radially inward]
Phase 3 #3
E = F/q = kQ/r2
E = (9 x 109)(3 x 10-6)
(0.30)2
E = 300,000 N/C [radially inward]

41. Phase 3 #4 Number of field lines a charge

42. Phase 3 #5 B
+5mC
-7mC
-9mC C A

43. Phase 3 #5 B
+5mC
FAB
FAC
-7mC
-9mC C A

44. Phase 3 #5 FAC = 157.5 N FAB = 28.125 N B +5mC Using Coulomb’s Law

45. Phase 3 #5 Resultant Vector: FAnet = 159.99 N [169.88 deg]
FAB = N
FAC = N
Tan-1 (28.125/157.5) = degrees

46. Phase 3 #6A A) Current is the rate of flow of charge in Amperes
Current, I = Coulombs/second
I = 357/76.2 = 4.69 Amps

47. Phase 3 #6B B) Key: Each electron carries 1.6 x 10-19 Coulombs
(See Equation Sheet)
4.69 Coulombs electron = 2.93 x 1019
1 sec x Coulombs electrons

48. Phase 3 #7: Very Important Rule
V = IR
V = (2.3)(9) = 20.7 volts
P = VI = Watts
Or P = I2R since V=IR

49. Phase 4 1A Series Resistors 1A) Series Resistors- Up and Add Them!
Req = 50 W
Series Resistors: Same Current
Different Voltage

50. Phase 4 1B 1B) Parallel Resistors- Add Reciprocals and Flip IT!
Req = ( )-1 = W
Parallel Resistors:
Different Current
Same Voltage

51. Phase C)
Req = 25 + ( )-1 = 31 W
10 W
25 W
15 W

52. Phase 4 #2A B-field I Use RHR: Fingers in direction of current
Palm facing B-field
Thumb: Force
Here: Force is into page I
B-field x

53. Phase 4 #2B B-field I • Use RHR: Fingers in direction of current
Palm facing B-field
Thumb: Force
Here: Force is out of page I
B-field •

54. Phase 4 #2C B-field I Use RHR: Fingers in direction of current
Palm facing B-field
Thumb: Force
Here: Force is Left I x x
B-field x x

55. Phase 4 #2D B-field I • • • • Use RHR: Fingers in direction of current
Palm facing B-field
Thumb: Force
Here: Force is Right I • •
B-field • •

56. B-field Phase 4 #2E I Use RHR: Fingers in direction of current
Palm facing B-field
Thumb: Force
Here: Force is zero
whenever I and B are
parallel I

57. B-field Phase 4 #2F I Use RHR: Fingers in direction of current
Palm facing B-field
Thumb: Force
Here: Force is zero
whenever I and B are
parallel I

58. Phase 4 #3 RHR Thumb: In direction of decreasing B-field
Fingers curl in the
direction of induced current

59. Phase 4 #4 • • • • RHR Thumb: In direction of current
Fingers curl in the
direction of B-field
Phase 4 #4 x x x x • • • •

60. Phase #5
T = 2p L g
L = 2.3 m
4.7 sec =
g = 4p2L = 4.11 m/s/s T2

61. vs = 331 + 0.6Temp = 345.4 m/s v = fl f = v/l = 345.4/1.42 = 243.24 Hz
Phase 4 #6 f = 14,000 Hz or 14 kHz
Phase 4 #7
vs = Temp = m/s
v = fl
f = v/l = 345.4/1.42 = Hz

62. Phase 5 #1 A-D 1A) d = 2pr = 3.90 m 1B) v = d/T = 3.90/2.5 = 1.56 m/s
1C) w = 2p = rad/s T
1D) Centripetal Force means “center seeking”

63. 1E) Will fly off Tangentially 1F) Tension Force from string
Phase 5 #1 cont’d
1E) Will fly off Tangentially
1F) Tension Force from string
1G) Fc = mv2 r
= (3.5)(1.56)2 =13.74 N
0.62

64. Torque = Forces times lever arm = [(135)(9.8)](2.10) = [(82)(9.8)](r2)
Phase #2
Condition for Rotational Equilibrium:
Clockwise Torque = Counterclockwise Torque
Torque = Forces times lever arm =
[(135)(9.8)](2.10) = [(82)(9.8)](r2) r2 F1 r1 F2
r2 = 3.46 m

65. Speed of Sound: vs = 331 + 0.6Temp = 347. 2 m/s (at 27 deg C)
Phase #3
Speed of Sound:
vs = Temp = m/s (at 27 deg C)
429 meters
Echolocation

66. d t v Phase 5 #3 t = d/v = (858/347.2) = 2.47 sec
Echoes are a 2-way street!! d
429 meters t v

67. 4A) vs = 331 + 0.6Temp = 344. 2 m/s (at 22 deg C)
Phase 5 #4 A & B
4A) vs = Temp = m/s (at 22 deg C)
4B) v = fl
l = v/f = 344.2/480 = 0.72 meters

68. Phase 5 #4 C & D 4C) l/4 = 0.18 meters for closed-end resonance
4C) l/2 = 0.36 meters for open-tube resonance

69. Phase 5 #5 E = mcDT E = 410,228 Joules of thermal energy
= (3.5)(4186)(28)
Change in
temperature
Specific heat capacity, c for water
m is mass in kg
E = 410,228 Joules of thermal energy

70. Phase 5 #6
Q = CV
V = Q/C
V = 1.98/4.73
V = 0.42 volts

71. 7A) constructive interference creates Beats
Phase 5 #7
7A) constructive interference creates Beats
7B) f = cycles per second
f = 6/2 = 3 Hz

72. Phase 5 #8 1.003sin17◦ = 1.21sinqr sinqr = 0.2424
A) nglass = c = 3 x =
v x 108
B) nisinqi = nrsinqr
1.003sin17◦ = 1.21sinqr
sinqr =
qr = sin-1(0.2424) =14.03◦

73. Phase 5 #9 Lens me an Image Given: ho = 2.5 cm do = 15 cm f = 12 cm
Find: A) hi B) di
B) 1/do + 1/di = 1/f  di = 60 cm
hi = - di  hi = -10 cm
ho do