Chapter 2 - Worked Solutions

to questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

Solutions-Ch2-RW.pdf

Question 60

RACE 1: Chris runs 100 m in time t; Sandy runs 90 m in time t
v(Chris) = 100/t; v (Sandy) = 90/t
RACE 2: s (Chris) = 110 m; s (Sandy) = 100 m
t2(Chris) = s/v = 110t/100 = 1.10t; t2(Sandy) = s/v = 100t/90 = 1.11t
Chris is still faster.


Question 61

Mach 2.5 = 825 m/s; Mach 1.2 = 396 m/s
v = u + at
396 = 825 + a x 30
a = - 14.3 m s-2


Question 62

s1 = 300 m; s2 = 300 m; s3= 300 m; s4 = 300 m
(1) v12 = u 2 + 2as
= 0 + 2 x 0.95 x 300
= 23.87 m s-1
v = u + at
23.87 = 0 + 0.95 t
t1 = 25.1 s

(2) v2 = s2/t2
t2= s2/ v2 = 300/23.87 = 12.57 s;v2 = 23.87 m s-1

(3) t3= 12.57 s; v3 = 23.87 m s-1

(4) v 2 = u 2 + 2as
02 = 23.872 + 2 x a x 300
a = - 0.95 m s-2
t4 = t1 = 25.1 s
Answers: (a) Maximum speed = 23.87 m s-1
(b) Time = t1 + t2 + t3 + t4 = 75.3 s
(c)


Question 63

The four intervals are shown in the following diagram


At top of jump v4 = 0
v2 = u2 + 2as
02 = u2 + 2 x -10 x +0.76
u (take-off speed (v4)) = 3.9 m/s

BOTTOM 15 cm:
v22 = u2 + 2 as2
= 3.92 + 2 x -10 x 0.15
v2 = 3.49 m/s
v2 = u + at2
t2 = (v2 - u)/a = 0.04 s

TOP 15 cm:
v3 = u2 + 2as
= 3.92 + 2 x -10 x 0.61
v3 = 1.73 m/s
t3 = (v3 - u)/a = 0.22 s
t4 = (v4 - u)/a = 0.39 s

Answers:
(a) Bottom 15 cm: t1 to t2 = 0.04 s
(b) Top 15 cm: t3 to t4 = 0.17 s
(c) Therefore, time at bottom is about 1/4 of time at top.

Question 64

First toss: s1, t1, u1, a
Second toss: s2, 2t1, u2, a
At top:
First toss: v = u1 + at1 = 0
u1 = - at1
Second toss: v = u2 + at2 = 0
u2 = - at2which equals -2at1
Hence: u1 = -at1 = u2/2

s1 = u1t1 + ½ at12

s2 = u2t2 + ½ at22
s2 = 2u12t1 + ½ a(2t1)2
s2 = 4u1t1 + 2at12

Ratio s2/s1 = (4u1t1 + 2at12)/ (u1t1 + ½ at12)
= (4u1 + 2 at1)/(u1 + ½ at1)
Replace at1 by -u1
Ratio s2/s1 = (4u1 + -2 u1)/(u1 + ½ u1)
= 2u1/½ u1
Therefore: s2 is 4 times s1 or second toss goes four times as high.

Question 65

(a) s = -50m; t2 = t1 - 1; u1 = 0; u2 = ?

Race 1:

s = ut1 + ½ at12
-50 = ½ x -10 t12
t1 = 3.2 s; hence t2 = 2.2 s
v1 = u + at = 0 + -10 x 3.2 = -32 ms-1

Race 2:

s = u2t2 + ½ at22
-50 = u2 x 2.2 + ½ x -10 x 2.22
u2 = -11.7 ms-1; v2 = u +at = -11.7 + -10 x 2.2 = -33.7 ms-1

(b)

 

Question 66

Acceleration is more thrilling!
a (Kitty) = (v-u)/t = (628 x 1000/3600)/3.72 = 46.9 ms-1
a (Eli) = (v-u)/t = (116 x 1000/3600)/0.04 = 805 ms-1
 

Question 67

Case 1: starting velocity = +u

v12 = u2 + 2as
v1 = Ö (u2 + 2as)

Case 2: starting velocity = -u

v22 = u2 + 2as
v2 = Ö (u2 + 2as)


The formula is the same for both because when u is squared it doesn't matter if the value for u is + or -. Hence, the velocity is the same in both cases.
 

Question 68

s1 = ut1 + ½ at12
s1 = -5t12
s2 = u(t1 - 1) + ½ a(t1 - 1)2
s2 = -5 (t1 - 1)2
s2 = -5t12 + 10t1 - 5

(a) s1 - s2 = 5 - 10t1   (or, alternatively:  s2 - s1 = 10t1 - 5). This is actually the difference in displacement. The difference in "distance" (as the question asks) is the absolute value of the answer. Thank you to Vic Braun for pointing out the difference.

(b) v1 = u1 + at
v1 = -10t1
v2 = u2 + a(t1 - 1)
v2 = -10t1 + 10
Ratio v1/v2 = -10t1/(-10t1 + 10) = t1/(t1 - 1)

  Question 69

Case 1

s = ut + ½ at2
400 = 0 + ½ a x 7.22
a = 15.4 ms-2

Case 2

v = u + at
305 x 1000/3600 = 0 + a x 7.2
a = 11.8 ms-2
The rider probably reached top speed before the 400 m so in Case 2 where he reached 305 kmh-1 at the finish line, he probably reached this after 5 seconds and kept constant speed from then on. In case 1, we are assuming acceleration is constant and that the top speed is irrelevant. The v/t graphs probably look like this:


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    Question 70

    At current mass: t1 = 6.92s and s = 400 m

    s = ut + ½ at12
    400 = 0 + ½ a x 6.922
    a = 16.7 ms-2

    At heavier mass: 60 pounds - 60/7.2 = 27.3 kg
    Time lost = 20/27.3 x 0.3 s = 0.22 s
    t2 = 6.92 + 0.22 s = 7.14 s
    s = ut +½ at2
    400 = 0 + ½ a x 7.142
    a = 15.7 ms-2
     

    Question 71

    (a) 30 knots = 30 x 100 feet in 1 minute = 30 x 100 x 60 ft per hr
    = 180000 ft/hr = 180000/6080 nautical miles/hr = 30 naut mi/hr

    (b) 30 knots - 180000 ft/hr = 180000 x 0.305 metres/hr = 180 x 0.305 km/h
    = 54.9 km h-1
     

    Question 72

    t (s)

    s (m) Earth

    s (m) Moon

    s (m) Mars

    s (m) Sun

    0.0

    0.00

    0.00

    0.00

    0.00

    0.2

    0.20

    0.03

    0.08

    5.40

    0.4

    0.80

    0.13

    0.30

    21.60

    0.6

    1.80

    0.29

    0.68

    48.60

    0.8

    3.20

    0.51

    1.22

    86.40

    1.0

    5.00

    0.80

    1.90

    135.00

    1.2

    7.20

    1.15

    2.74

    194.40

    1.4

    9.80

    1.57

    3.72

    264.60

    1.6

    12.80

    2.05

    4.86

    345.60

    1.8

    16.20

    2.59

    6.16

    437.40

    2.0

    20.00

    3.20

    7.60

    540.00

     

    Question 73 - The 30-Year Problem

    There are at least five solutions to this problem. Perhaps the easiest (?) one for Senior Physics students is one using internal triangles.

    1. Drop perpendiculars from points B and D to line AC (see figure below):


    2. Let the constant speed = v (miles/minute). Then: distance (miles) = time (minutes) x speed (v)
    The distance AC (directly) = 30v; the distance ADC = 40v; the distance ABC = 35v.
    Hence: AD = 40v -10; AB = 35v - 10
    Let x be the distance from C to the perpendicular on AC from B.
    Let y be the distance from C to the perpendicular on AC from D.
    The two lower triangles are similar, hence the perpendiculars have the lengths x and y as shown in the figure.

    3. Using Pythagorean theorem on the top two triangles and one of the two identical (similar) lower triangles:
    Equation 1: (40v-10)2 = x2 + (30v-y) 2
    Equation 2: (35v-10)2 = y2 + (30v-x) 2
    Equation 3: 102 = x2 + y2

    4. Expand Equation 1 and substitute 102 - y2 for x 2
    1600v2 - 800 v + 100 = 100 - y2 + 900v2 - 60 vy + y2
    700v2 - 800 v = -60vy
    y = 13.33 - 11.67v

    5. Repeat for Equation 2 substituting 102 - x2 for y 2:
    1225v2 - 700v +100 = 100 - x2 + 900v2 - 60vx 2
    325v - 700v = -60vx
    x = 11.67 - 5.42v

    6. Substitute these values for x and y into equation 3:
    100 = (11.67 - 5.42v) 2 + (13.33 - 11.67v) 2
    100 = 136.1 - 126.4v + 29.38v2 + 177.8 - 311.1v + 136.1v
    165.5v 2 - 437.5v + 213.9 = 0

    7. Use quadratic formula to solve for v
    Two solutions: v = 0.648 miles/minute and v = 1.996 miles/minute.

    8. Convert to miles/hour (x 60): 38.8 mph and 119.8 mph.

    9. Discard the second speed as too fast (break the speed limit?).
    Answer: 38.8 mph

    ALTERNATIVE SOLUTION:
    Use the cos rule: the cosine of any angle of a triangle equals the sum of the squares of the adjacent sides less the square of the other side, all divided by twice the product of the adjacent sides.

    Applying this to the figure (and simplifying) shows that:
    Cosine of angle ABC = (7 - 3.25v)/6
    Cosine of angle ACD = (8-7v)/6
    When these two values are squared and added together they equal 1.
    This follows from the rule that: when two angles add up to 90°, the sum of squares of their cosines = 1.
    Simplifying and collecting terms gives a quadratic which produces two roots as in the solution above.

    ANOTHER SOLUTION
    Construct a rectangle as shown in the figure below:

    1. Write square-of-the-hypotenuse equations for the three right triangles of which AC, AD and AB are the hypotenuses.
    Equation 1 (for AC): (30v)2 = (10 + x) 2 + (y + 10)2
    Equation 2: (for AD) (40v-10)2 = x2 + (y + 10)2
    Equation 3: (for AB): (35v - 10)2 = y2 + (x + 10) 2

    2. Subtract Equation 2 from Equation1 and find x in terms of v.
    30v2 - (40v - 10) 2 = (x + 10) 2 - x2
    x = -35v2 + 40v - 10

    3. Subtract Equation 3 from Equation 1 and find y in terms of v.
    30v2 - (35v - 10) 2 = (y + 10) 2 - y2
    y = -16.25v2 + 35v - 10

    4. Substitute these values of x and y into Equation 1.
    (30v) 2 = (-16.25v2 + 35v) 2 + (-35v2 + 40v) 2
    900v2 = 16.252v4 - 16.25 x 35 x 2 v3 + 352v2 + 352v4 - 35 x 40 x 2 v3 + 402v2

    5. Simplify and divide by v2, (allowable since v can't be equal to zero).
    1489.1v2 - 3937.5v + 1925 = 0

    6. You should have a quadratic. Solve using the quadratic formula and you should get roots as in the methods above.
    Pretty simple huh?
     

  • Question 74

    First statement: s µ t2 is correct as our formula is s = ½ at2 which is the same as s µ t2
    Second statement: v µ s is incorrect as v2 = 2as or v2 µ s
     

    Question 75


    tan 4.3° = 1000/s therefore s = 1000/tan 4.3° = 13300 m
    298 km/h = 298 x 1000/3600 ms-1 = 82.8 ms-1
    t = s/v = 13300/82.8 = 160 s
    NOTE: the answer in the back of the book is wrong. It is not 44.6 s but 160 s.
     

    Question 76

    Impact speeds:
    Of a child in a car: v = 50 x 1000/3600 = 13.9 ms-1
    Falling from a building: (assume each story to be about 3 m high):
    v2 = u2 + 2as
    v = Ö (2 x 10 x 6) = 10.9 ms-1
    Hence there is a lower impact speed in falling from a building but the concrete is harder than a car dashboard. Probably about equal impact damage.
     

    Question 77

    s = ut + ½ at2
    64 = 0 + ½ x 2 x t2
    t = 8.0 s
    s (t=0) = 0
    s(t=2) = ½ at2 = ½ a 22 = 4 m
    s(t=4) = ½ a x 42 = 16 m
    s(t=6) = ½ a x 62 = 36 m
    s(t=8) = ½ a x 82 = 64 m
     

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