CHAPTER 2

4.) 0.414 s

8.) (a) 40 km/hr

(b) 40 km/hr

(c)

16.) (a) -12 + 6t

(b) It is moving in the negative x direction.

(c) 6m/s

(d)For 0< t< 2s, │v(t)│ decreases with time. At t= 2s, v(t) is zero. Afterwards, │v(t)│increases with time.

(e) Yes, at t = 2s.

(f) t = 3s

20.) (a) -20 m/s²

(b)

58.) First train = 200 m

Second train = 800 m

There is a collision.

68.) (a) 3.70m/s

(b) 1.74m/s

(c) 0.154 m higher

It goes 0.154 higher

70.) (a) 101.4 m

(b) 13 s

76.) For the first throw H₁ = gt²₁/2. For the second throw H₂ = gt²₂/2 = g(2t₁)² = 4H_1.

Thus one has to throw four times as high to make the ball stay in the air twice as

long.

78.) The average acceleration is 1.65 × 10³ m/s^2.

84.) (a) 17s

(b) h = 240m

Thus the fall begins at a height of H = 50m + h = 290m

90.) 2.34m

CHAPTER 3

4.) 40º north of east

16.) (a) 27.8m

(b) 13.4m

18.) 26.1ft

20.) (a) r = -9.0i + 10j

(b) 132 º

24.) a = 9i + 12j

b = 3i + 4j

28.) 21º east of north

32.) a = b (Where a and b are the two vectors.)

36.) (a) 10m (north)

(b) 7.5m (south)

40.) a × a = 0

42.) (a) 30

(b) 52

48.) 22 º

50.) (a) 2k

(b) 26

(c) 46

52.) (a) 2.97 units

(b) (1.51i + 2.67j – 1.36k) units

(c) 48º

54.) 70.5º

56.) (a) 0

(b) The vector a x (b x a) is in the plane of a and b and is perpendicular to a.

58.) (a) 8.66

(b) q = 4.33

p = -6.67

CHAPTER 4

2.) (a) 6.2m

(b)

4.) -2.0i + 6.0j – 10k

18.) (a) 0.18m

(b) 1.9m

22.) (a) 0.50s

(b) 10ft/s

28.) (a) 1.15s

(b) 12.0m

(c) = 4.80m/s

(d) Since v_y> 0 when the ball hits the wall, it has not reach the highest point yet.

30.) (a) tan Ө_o

(b) 27º

38.) 6.7ft/s (not accidental)

42.) (a) 260 m/s

(b) 45s

(c) The presence of the air slows down the bomb. Thus the actual value of v_o has to be greater than the answer given in (a).

48.) 2900 ft

72.) (a) 5 km/hr

(b) -1km/h (The minus sign indicates that the velocity is in the downstream direction)

78.) (a) v_mp = v_m – v_p = (-60j + 80i) (km/h)

(b) (80 km/h)/800m

(c) No, they remain unchanged

80.) (0.96 j)m/s

CHAPTER 5

8.)(a) (-32 – 21j)N

(b) 213 º

14.) (a) 7.4 • 10² N

(b) 2.9 • 10² N

(c) 0

(d) 75 kg

18.) 108 N

22.) 6.8 • 10³ N

30.) 69 lb

36.)(a) 0.62 m/ s²

(b) 0.13 m/s²

(c) 2.6 m

38.) 2.2 • 10^-3 N

40.) (a) 1.1 N

(b) 2.1 N

42.) (a) 4.3 m/s

(b) (15i + 6.4j)m

48.) (a) .970 m/ s²

(b) 34.9 N

58.) (a) 0.735 m/s²

(b) The result is positive, indicating that the acceleration of m₁ is the up the plane and the acceleration of m₂ is downward.

(c) 20.8 N

60.) (a) -1.18 m

(b) 0.674 s

(c) 3.50 m/s. It is down the plane.

62.) (a) 7.3 kg

(b) 89 N

64.) (a) 4.9 m/s²

(b) 120 N

66.) (a) 1.2 •10² m/s²

(b) 12g

(c) 1.4 • 10⁸ N

(d) 4.2 y, which is about 1.3 • 10⁸ s.

68.) (a) 2.18 m/s²

(b) 116 N

(c) 21.0 m/s²

70.) m = 2Ma / (g + a)

72.) (a) 9.4 • 10³ m

(b) 6.1 • 10⁴ m

74.) (a) 466 N

(b) 527 N

(c) 1050 N

(d) case (a): 932 N ; case (b): 1050 N ; case (c): 1860 N ; case (d): 2110 N

CHAPTER 6

6.) 9.3 m/s²

18.) Drawing

22.) Drawing

34.) (a),(b) 13 N

38.) 4.9 • 10² N

40.) Drawing

42.) (a) 19º

(b) 3.3 • 10³ N

58.) 49.5 m/s = 178 km/h

62.) (a) 175 lb

(b) 50 lb

64.) v = (gR) ^(1/2)

66.) 2.2 × 10³ m

70.) (a) Drawing

(b) Drawing

(c) 37.9 N

(d) 6.45 m/s

CHAPTER 7

4.) (a) -5 • 10¹⁴ J

(b) 0.1 megaton TNT

(c) 8

6.) (a) 2.9 • 10⁷ m/s

(b) 1.3 MeV

10.) (a) 590J

(b) The work done by the force of gravity is 0.

(c) 0

(d) 590 J

16.) (a) 1.5 J

(b) Since W > 0 the kinetic energy of the trunk increases.

22.) (a) 2.7 • 10² N

(b) -4.0 •10² J

(c) 4.0 • 10² J

(d) The normal force does not do any work on the block

(e) The work done by the net force F_net must be zero, since F_net itself is zero.

32.) (a) Approximately 13 J.

(b) 13 J

34.) (a) 797 N

(b) The total work done is zero.

(c) -1.55 • 10³ J

(d) The force T exerted by the rope does not do any work on the crate

(e) 1.55 • 10³ J

(f) Since F is the variable force, you cannot use W = F • distance, which is valid only for constant forces

38.) (a) 23 mm

(b) 45 N

40.) (a) 2.9 • 10^-1 J

(b) -1.8 J

(c) 3.5 m/s

(d) 23 cm

44.) (a) 28 W

(b) (6.0 m/s) j

48.) (a) 4.2 J

(b) 5.0 W

50.) 90 hp

CHAPTER 8

4.) (a) 4.31 • 10^-3 J

(b) -4.31 • 10^-3 J

(c) U = +4.31 • 10^-3 J at the top

(d) U = -4.31 • 10^-3 J at the bottom

10.) (a) 4mgR

(b) 3mgR

(c) 5mgR

(d) Uq = mgR

(e) U = 2mgR

22.) (a) 7.84 N/cm

(b) 62.7 J

(c) 62.7 J

(d) 0.80 m

24.) (a) 2.8m/s

(b) 2.7m/s

26.) (a) .35 m

(b) 1.7 m/s

30.) 3.7 J

34.) 0.10 m

36.) 1.25 cm

40.) 930 N (The vine will not break.)

44.) (a) 0.396 m

(b) 3.64 cm

46.) Drawing

48.) (a) +4.8N (positive x direction)

(b) -5.3J

(c) 3.4m/s

50.) (a) 5.6 N

(b) -3.5J

CHAPTER 9
8.) 6.8 • 10^-12 m

10.) (a) 36.8 m

(b) 1.7 • 10¹²J

16.) (a) s_max = L

(b) The condition for s = L is that the cannonballs have to travel the length of the car and that m = M.

(c) After the firing, there is no relative motion in the system, thus the speed of the car is equal to that of the center of mass of the system, which is zero.

22.) (a) Place the origin of a coordinate system at the center of the pulley, with the x axis horizontal and to the right and with the y axis downward. The center of mass is halfway between the containers, at x = 0, and y = l, where l is the vertical distance from the pulley center to either of the containers.

(b) 26 Matthew Murrey

(c) When they are released the heavier container moves downward and the lighter container moves upward, so the center of the mass, which must remain closer to the heavier container, moves downward.

(d) 3.84 m/s²

36.) 4.4 • 10³ km/h

38.)

46.) 0 m/s (It stops.)

48.) (a) 540m/s

(b) 40.4º

66.) (a) 3.4 • 10³ W

(b) 1.8 • 10³ W

72.) (a) 1.07 • 10³ N

(b) 3.6 • 10⁴ W

(c) -.064, the tangent of the angle has nearly the same value so the incline has a 6.4% grade.

74.) 69hp

CHAPTER 10

12.) 10 m/s

34.) (a),(b) 1.23m/s

50.) (a),(b) V _o,i = 937 m/s , V _o,f = 721 m/s

52.) 5.06 kg

56.) 0.33m

62.) (a) (-1 • 10^-19 i + 6.68 • 10^-20 j)(kg•m/s)

(b) 1.19 • 10^-12 J

72.) -2.0m/s; The target ball bounces back in the negative x direction.

CHAPTER 11

6.) (a) 2rad

(b) 0

(c) 130 rad/s

(d) 32rad/s²

(e) The angular acceleration, given by a = 8 + 12t, depends on the time and so is not constant.

8.) 14 rev

34.) W _o = (((mu _s) • g)/R) ^1/2

36.) (a) 40.2 cm/s²

(b) 2.36 • 10³ m/s²

(c) 83.2m

38.) (a) 3.8 • 10³ rad/s

(b) 190 m/s

40.) (a) 7.27 •10^-5 rad/s

(b) 3.5 • 10² m/s

(c) 463 m/s

42.) (a) 1.5 • 10² cm/s

(b) 15 rad/s

(c) 15 rad/s

(d) 75cm/s

(e) 3.0 rad/s

50.) (a) 5ml² + (8/3)Ml²

(b) ((5/2)m + (4/3)M)l²w²

56.) (1/3)M(a² + b²)

58.) (a) R_H = R/(2^(1/2))

(b) k = (I/M)^(1/2)

64.) 12N·m

76.) (a) 2Ө/t²

(b) 2ӨR/t²

(c) Mg – (2Ө/t²) • (MR + (I/R))

CHAPTER 22

16.) q _o = -(4/9)q , x = L/3

22.) (a) x = (L/2) • (1 + (1/(4πε_o)) • (3qQ/W))

(b) h = ((1/(4πε_o) • (3qQ/W)) ^1/2

CHAPTER 23

20.) Proof

28.) 0.51

CHAPTER 24

10.) -πa²E

14.) 3.54 • 10^-6C = 3.54 μ C

44.) (a) see Eq. 24-16

(b) 2.9 • 10⁴ N/C

(c)200 N/C

48.) E=(2e/4πε_o )

CHAPTER 25

14.) (a) V =

(b) V =

(c) V =

(d) Solutions agree at r = r₁ and r = r₂

20.) Drawings

30.) 0.94q

4πε_od

44.) E_x (ab) = -6 V/m ; E_x (bc) = 0 ; E_x (cd) = E_x (de) = 3 V/m ; E_x (ef) = 15 V/m ;

E_x (fg) = 0 ; E_x (gh) = -3 V/m

CHAPTER 26

14.) 9090 = N

56.) 8.1 • 10^-11 F/m = 81 p F/ m

CHAPTER 27

8.) 1.9 •10^-4 m

12.) 5 • 10³

18.) 2.0 • 10 ⁶ (Ω·m)^-1

22.) 57°C

30.) (a) 2.39

(b) It is impossible.

36.) (a) 3.83 • 10^-2 A

(b) 109 A/m²

(c) 1.28 • 10^-2 m/s

(d) 227 V/m

44.) 560W

50.) (a) 4.9 • 10 ⁶ A/m2

(b) 8.3 • 10^-2 V/m

(c) 25V

(d) 6.3 • 10 ² W

CHAPTER 28

30.) (a) 6.67 Ω (three 20.0-Ω resistors in parallel)

(b) 6.67 Ω (three 20.0-Ω resistors in parallel)

(c) 0 (as B and C are connected by a conducting wire)

CHAPTER 29

4.) 4.00 X 10⁵ m/s

32.) v(t) = v_0,xi + v_0,ycos(wt)j + v_0,y(sin wt)k

36.) 3.6 •10^-10

CHAPTER 30

4.) 1.2 x 10^-8 T

40.) (a) -2.5 X 10 ^-6 T • m

(b) 0