Written Home Work Solutions PHYS111.02 SFSU Eradat [WH#1 CHAPTERS 13] 1 Chapter 1 39. Picture the Problem: The lottery winnings are represented either by quarters or paper dollars. Strategy: There are about 5 quarters and about 30 dollar bills per ounce. Solution: 1. (a) Multiply by conversion factors: 4 × 12 × 10 6 quarters ( ) 1 oz 5 quarters ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 lb 16 oz ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 600,000 lb ≅ 10 6 lb 2. (b) Repeat for the dollar bills: 12 × 10 6 dollars ( ) 1 oz 30 dollars ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 lb 16 oz ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 25,000 lb ≅ 10 4 lb Insight: Better go with large denominations or perhaps a single check when you collect your lottery winnings! Even the dollar bills weigh over ten tons! 40. Picture the Problem: This is a dimensional analysis question. Strategy: Manipulate the dimensions in the same manner as algebraic expressions. Solution: 1. (a) Substitute dimensions for the variables: v = at m s = m s 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ s () = m s ∴ The equation is dimensionally consistent. 2. (b) Substitute dimensions for the variables: v = 1 2 at 2 m s ≠ 1 2 m s 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ s () 2 = m ∴ NOT dimensionally consistent 3. (c) Substitute dimensions for the variables: t = a v ⇒ s ≠ ms 2 ms = 1 s ∴ NOT dimensionally consistent 4. (d) Substitute dimensions for the variables: v 2 = 2ax m 2 s 2 = 2 m s 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ m ( ) = m 2 s 2 ∴ dimensionally consistent Insight: The number 2 does not contribute any dimensions to the problem. 46. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: 1. (a) The acceleration must be greater than 14 ft/s 2 because there are about 3 ft per meter. 2. (b) Convert m/s 2 to ft/s 2 : 14 m s 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3.28 ft m ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 46 ft s 2 3. (c) Convert m/s 2 to km/h 2 : 14 m s 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 km 1000 m ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3600 s h ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1.8 × 10 5 km h 2 Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one. They are often helpful in displaying a number in a convenient, useful, or easy tocomprehend fashion.
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Written Home Work Solutions PHYS111.02 SFSU Eradat [WH#1 CHAPTERS 1-‐3] 1 Chapter 1 39. Picture the Problem: The lottery winnings are represented either by quarters or paper dollars.
Strategy: There are about 5 quarters and about 30 dollar bills per ounce.
Solution: 1. (a) Multiply by conversion factors:
4 ×12 ×106 quarters( ) 1 oz5 quarters
⎛⎝⎜
⎞⎠⎟
1 lb16 oz
⎛⎝⎜
⎞⎠⎟= 600,000 lb ≅ 106 lb
2. (b) Repeat for the dollar bills:
12 ×106 dollars( ) 1 oz
30 dollars⎛⎝⎜
⎞⎠⎟
1 lb16 oz
⎛⎝⎜
⎞⎠⎟= 25,000 lb ≅ 104 lb
Insight: Better go with large denominations or perhaps a single check when you collect your lottery winnings! Even the dollar bills weigh over ten tons!
40. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
Solution: 1. (a) Substitute dimensions for the variables:
v = a tms=
ms2
⎛⎝⎜
⎞⎠⎟
s( ) = ms
∴ The equation is dimensionally consistent.
2. (b) Substitute dimensions
for the variables:
v = 12 at2
ms≠ 1
2
ms2
⎛⎝⎜
⎞⎠⎟
s( )2= m ∴ NOT dimensionally consistent
3. (c) Substitute dimensions for the variables:
t = a
v ⇒ s ≠
m s2
m s=
1s
∴ NOT dimensionally consistent
4. (d) Substitute dimensions for the variables:
v2 = 2a xm2
s2= 2
ms2
⎛⎝⎜
⎞⎠⎟
m( ) = m2
s2 ∴ dimensionally consistent
Insight: The number 2 does not contribute any dimensions to the problem. 46. Picture the Problem: This is a units conversion problem. Strategy: Multiply the known quantity by appropriate conversion factors to change the units. Solution: 1. (a) The acceleration must be greater than 14 ft/s2 because there are about 3 ft per meter.
2. (b) Convert m/s2 to ft/s2:
14 ms2
⎛⎝⎜
⎞⎠⎟
3.28 ftm
⎛⎝⎜
⎞⎠⎟= 46 ft
s2
3. (c) Convert m/s2 to km/h2:
14 m
s2
⎛⎝⎜
⎞⎠⎟
1 km1000 m
⎛⎝⎜
⎞⎠⎟
3600 sh
⎛⎝⎜
⎞⎠⎟
2
= 1.8 ×105 kmh2
Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one. They are often helpful in displaying a number in a convenient, useful, or easy-‐to-‐comprehend fashion.
2
Chapter 2 23. Picture the Problem: Following the motion specified in the
position-‐versus-‐time graph, the father walks forward, stops, walks forward again, and then walks backward.
Strategy: Determine the direction of the velocity from the slope of the graph. Then determine the magnitude of the velocity by calculating the slope of the graph at each specified point.
Solution: 1. (a) The slope at A is positive so the velocity is positive. (b) The velocity at B is zero. (c) The velocity at C is positive. (d) The velocity at D is negative.
2. (e) Find the slope of the graph at A:
vav = Δx
Δt= 2.0 m
1.0 s= 2.0 m/s
3. (f) Find the slope of the graph at B:
vav = Δx
Δt= 0.0 m
1.0 s= 0.0 m/s
4. (g) Find the slope of the graph at C:
vav = Δx
Δt= 1.0 m
1.0 s= 1.0 m/s
5. (h) Find the slope of the graph at D:
vav = Δx
Δt= −3.0 m
2.0 s= −1.5 m/s
Insight: The signs of each answer in (e) through (h) match those predicted in parts (a) through (d). With practice you can form both a qualitative and quantitative “movie” of the motion in your head simply by examining the position-‐versus-‐time graph.
29. Picture the Problem: The given position function indicates the particle begins traveling in the positive
direction but is accelerating in the negative direction.
Strategy: Create the x-‐versus-‐t plot using a spreadsheet, or calculate individual values by hand and sketch the curve using graph paper. Use the known x and t information to determine the average speed and velocity.
Solution: 1. (a) Use a spreadsheet to create the plot:
2. (b) Find the average velocity from t = 0.35 to t = 0.45 s:
Written Home Work Solutions PHYS111.02 SFSU Eradat [WH#1 CHAPTERS 1-‐3] 3 4. (d) The instantaneous speed at t = 0.40 s will be closer to 0.56 m/s. As the time interval becomes
smaller the average velocity is approaching 0.56 m/s, so we conclude the average speed over an infinitesimally small time interval will be very close to that value.
Insight: Note that the instantaneous velocity at 0.40 s is equal to the slope of a straight line drawn tangent to the curve at that point. Because it is difficult to accurately draw a tangent line, we usually resort to mathematical methods like those illustrated above to determine the instantaneous velocity.
54. Picture the Problem: The two cars are
traveling in opposite directions.
Strategy: Write the equations of motion based upon equation 2-‐11, and set them equal to each other to find the time at which the two cars pass each other.
Solution: 1. (a) Write equation 2-‐11 for car 1:
x1 = x0,1 + v0,1t +12 a1t
2 = 0 + 20.0 m/s( ) t + 1.25 m/s2( ) t2
2. Write equation 2-‐11 for car 2: x2 = x0,2 + v0,2t +
12 a2t
2 = 1000 m − 30.0 m/s( ) t + 1.6 m/s2( ) t2
3. (b) Set x1 = x2 and solve for t:
20.0 m/s( ) t + 1.25 m/s2( ) t2 = 1000 m − 30.0 m/s( ) t + 1.6 m/s2( ) t2
0 = 1000 − 50t + 0.35t2
t =50 ± 502 − 4 0.35( ) 1000( )
0.70= 24, 119 s ⇒ 24 s
Insight: We take the smaller of the two roots, which corresponds to the first time the cars pass each other. Later on the larger acceleration of car 2 means that it’ll come to rest, speed up in the positive direction, and overtake car 1 at 119 s.
103. Picture the Problem: A youngster bounces straight up and down on a trampoline. The child rises straight
upward, slows down, and momentarily comes to rest before falling straight downward again.
Strategy: Find the time of flight by exploiting the symmetry of the situation. If it takes time t for gravity to slow the child down from her initial speed v0 to zero, it will take the same amount of time to accelerate her back to the same speed. She therefore lands at the same speed v0 with which she took off. Use this fact together with equation 2-‐7 to find the time of flight. The maximum height she achieves is related to the square of v0, as indicated by equation 2-‐12.
Solution: 1. (a) Because the time of flight depends linearly upon the initial velocity, doubling v0 will increase her time of flight by a factor of 2.
2. (b) Because the time of flight depends upon the square of the initial velocity, doubling v0 will increase her maximum altitude by a factor of 4.
3. (c) The time of flight for v0 = 2.0 m/s , using Eq. 2-‐7:
t =
v − v0
−g=
−v0( ) − v0
−g=
2v0
g=
2 2.0 m/s( )9.81 m/s2 = 0.41 s
4. The time of flight for v0 = 4.0 m/s :
t =
2v0
g=
2 4.0 m/s( )9.81 m/s2 = 0.82 s
5. The maximum height for v0 = 2.0 m/s , using Eq. 2-‐
Δx =
v2 − v02
−2g=
02 − v02
−2g=
v02
2g=
2.0 m/s( )2
2 9.81 m/s2( ) = 0.20 m
4
12:
6. The maximum height for v0 = 4.0 m/s :
Δx =
v02
2g=
4.0 m/s( )2
2 9.81 m/s2( ) = 0.82 m
Insight: The reason the answer in step 6 is not exactly four times larger than the answer in step 5 is due to the rounding required by the fact that there are only two significant digits. If you recalculate using 2.00 m/s and 4.00 m/s, the answers are 0.204 and 0.816 m, respectively.
Written Home Work Solutions PHYS111.02 SFSU Eradat [WH#1 CHAPTERS 1-‐3] 5 Chapter 3 26. Picture the Problem: The vectors involved in the problem
are depicted at right.
Strategy: Add the vectors using the component method in order to find the components of the vector sum. Use the components to find the magnitude and the direction of the vector sum.
Solution: 1. (a) Make estimates
from the drawing: A +B +C ≈ 20 m θ ≈ 1.5°
2. (b) Add the vector components:
A +B +C = 0 + 20.0 m( )cos 45° + 7.0 m( )cos −30°( )⎡⎣ ⎤⎦ x +
−10.0 m( ) + 20.0 m( )sin 45° + 7.0 m( )sin −30°( )⎡⎣ ⎤⎦ yA +B +C = 20.2 m( ) x + 0.64 m( ) y
3. Use the components to find the magnitude:
A +B +C = 20.2 m( )2 + 0.64 m( )2 = 20.2 m
4. Use the components to find the angle:
θ = tan−1 0.64 m
20.2 m⎛⎝⎜
⎞⎠⎟= 1.8°
Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answers using this approach than by adding the vectors graphically. Notice, however, that when your calculator returns the angle of 1.8° in step 4, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.
53. Picture the Problem: The vectors involved in this problem are
depicted at right.
Strategy: Let
vpg = velocity of the plane relative to the ground,
vpa = velocity of the plane relative to the air, and
vag = velocity
of the air relative to the ground. The drawing at right depicts the vectors added according to equation 3-‐8,
vpg = vpa + vag .
Determine the angle of the triangle from the inverse sine function.
Solution: 1. (a) Use the inverse sine function to find θ:
θ = sin−1
vag
vpa
⎛
⎝⎜
⎞
⎠⎟ = sin−1 65 km/h
340 km/h⎛⎝⎜
⎞⎠⎟= 11° west of north
2. (b) The drawing above depicts the vectors.
3. (c) If the plane reduces its speed but the wind velocity remains the same, the angle found in part (a) should be increased in order for the plane to continue flying due north.
Insight: If the plane’s speed were to be reduced to 240 km/h, the required angle would become 16°.
C
x
y B
A
+ +A B C
θ
45°
30°
6
75. Picture the Problem: The velocities of the surfer vss and the
waves vws relative to the shore are shown in the diagram at
right. Strategy: Set the y component of the surfer’s velocity equal to the
velocity of the waves, and solve for the angle θ. Then apply equation 3-8 to find the surfer’s velocity relative to the wave.
Solution: 1. (a) Set
vss, y = vws :
vss sinθ = vws
θ = sin−1 vws
vss
⎛
⎝⎜⎞
⎠⎟= sin−1 1.3 m/s
7.2 m/s⎛⎝⎜
⎞⎠⎟= 10°
2. (b) Apply equation 3-‐8: vss =
vsw + vws ⇒ vsw = vss −vws = vss cosθ x + vss sinθ y⎡⎣ ⎤⎦ − vws y
3. Since vss sinθ = vws , the y components cancel out:
vsw = vss cosθ x = 7.2 m/s( )cos10° x = 7.1 m/s( ) x
4. (c) If the y component stays the same, but the vector increases in length, the angle it makes with the x-‐axis must decrease.
Insight: In this problem we assumed that the water is at rest relative to the shore, so that the surfer’s speed relative to the water is the same as the surfer’s speed relative to the shore.