Formulae, Equations…. in Physics, page 2
Dr Dave Menke, Instructor
12. F = m a (in Newtons)
13. E = F x d (in Joules)
14. PE = m g h (in Joules)
15. KE = ½ m v2 (in Joules)
16. p = m v (in kg m/s)
17. Ff = m m g (friction)
18. ac = v2/r (circular acceleration)
19. a = w2 r (angular acceleration)
20. w = 2 p n = θ/t
21. 1 radian = 57.3°
22. F = - kx (Hooke)
Friday, June 4, 2010
HOMEWORK SET 2
PHYSICS
HOMEWORK SET 2
Problems and Conceptual Exercises
1. An object of mass, m, is initially at rest. After a force of magnitude, F, acts on it for a time, t, the object has a speed of “v.” If the mass of the object is doubled, and the force is quadrupled, How long does it take for the object to accelerate from rest to a speed of “v” now?
2. In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec?
3. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push.
a. Is the force experienced by the child more than, less than, or equal to the force experienced by the parent?
b. Is the acceleration experienced by the child more than, less than, or equal to the force experienced by the parent?
If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration
4. A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction).
5. A baseball player slides into 3rd base with an initial speed of 4.0 m/s. If the coefficient of friction between the player and the found is 0.46, how far does the player slide before coming to rest?
6. A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m.
a. what coefficient of static friction is required between the sprinter's shoes and the track?
b. Explain the strategy used to get this answer.
7. A certain spring has a force constant, k.
a. if this spring is cut in half does the resulting half spring have a force constant that is greater than, less than, or equal to k?
b. If two of the original full length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
8. A 0.15 kg ball is placed in a shallow wedge with an open angle of 120° as shown in figure 6-27 on page 181 in the book. For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume no friction.
9. A car is driven with a constant speed around a circular track. Answer each of these following question with a yes or no.
a. Is the car's velocity constant?
b. Is the car's speed constant?
c. Is the acceleration constant?
d. Is the acceleration direction constant?
10. The International Space Station (ISS) orbits Earth in a circular orbit about 375 km above the surface. Over one complete orbit, is the work done by Earth on the ISS positive, negative, or zero? Explain.
11. To clean a floor, a custodian pushes on a mop handle with a force of 50.0 N.
a. If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop a distance of 0.5 meter?
b. If the angle is increased to 65°, does the work done increase, decrease, or stay the same? Explain.
12. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
13. A pine cone of 0.14 kg mass falls 16 meters to the ground landing at 13 m/s.
a. How much work was done on the pinecone by air resistance?
b. What was the average force of air resistance on the pinecone?
14. A car of 1100 kg coasts on a horizontal road at 19 m/s. After crossing an un-paved sand stretch 32 meters long its speed decreases to 12 m/s.
a. If the sandy portion had been only 16 meters long, would the car speed have decreased by 3.5 m/s, more, or less? Explain.
b. Calculate the change of speed.
15. It takes 180 Joules of work to compress a certain spring 0.15 meter.
a. What is the force constant of the spring?
b. To compress it another 0.15 meter, will it require 180 Joules, more, or less? Explain.
16. Calculate the work done by friction as a 3.7-kg box is slid along a floor from point A to point B as in figure 8-16 on page 244 in the book. Do this for all three paths: 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.
HOMEWORK SET 2
Problems and Conceptual Exercises
1. An object of mass, m, is initially at rest. After a force of magnitude, F, acts on it for a time, t, the object has a speed of “v.” If the mass of the object is doubled, and the force is quadrupled, How long does it take for the object to accelerate from rest to a speed of “v” now?
2. In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec?
3. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push.
a. Is the force experienced by the child more than, less than, or equal to the force experienced by the parent?
b. Is the acceleration experienced by the child more than, less than, or equal to the force experienced by the parent?
If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration
4. A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction).
5. A baseball player slides into 3rd base with an initial speed of 4.0 m/s. If the coefficient of friction between the player and the found is 0.46, how far does the player slide before coming to rest?
6. A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m.
a. what coefficient of static friction is required between the sprinter's shoes and the track?
b. Explain the strategy used to get this answer.
7. A certain spring has a force constant, k.
a. if this spring is cut in half does the resulting half spring have a force constant that is greater than, less than, or equal to k?
b. If two of the original full length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
8. A 0.15 kg ball is placed in a shallow wedge with an open angle of 120° as shown in figure 6-27 on page 181 in the book. For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume no friction.
9. A car is driven with a constant speed around a circular track. Answer each of these following question with a yes or no.
a. Is the car's velocity constant?
b. Is the car's speed constant?
c. Is the acceleration constant?
d. Is the acceleration direction constant?
10. The International Space Station (ISS) orbits Earth in a circular orbit about 375 km above the surface. Over one complete orbit, is the work done by Earth on the ISS positive, negative, or zero? Explain.
11. To clean a floor, a custodian pushes on a mop handle with a force of 50.0 N.
a. If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop a distance of 0.5 meter?
b. If the angle is increased to 65°, does the work done increase, decrease, or stay the same? Explain.
12. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
13. A pine cone of 0.14 kg mass falls 16 meters to the ground landing at 13 m/s.
a. How much work was done on the pinecone by air resistance?
b. What was the average force of air resistance on the pinecone?
14. A car of 1100 kg coasts on a horizontal road at 19 m/s. After crossing an un-paved sand stretch 32 meters long its speed decreases to 12 m/s.
a. If the sandy portion had been only 16 meters long, would the car speed have decreased by 3.5 m/s, more, or less? Explain.
b. Calculate the change of speed.
15. It takes 180 Joules of work to compress a certain spring 0.15 meter.
a. What is the force constant of the spring?
b. To compress it another 0.15 meter, will it require 180 Joules, more, or less? Explain.
16. Calculate the work done by friction as a 3.7-kg box is slid along a floor from point A to point B as in figure 8-16 on page 244 in the book. Do this for all three paths: 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.
LESSON 3
PHYS LESSON 3 FOR
THURSDAY, JUNE 03, 2010
I Introduction
II Logistics: Seating, Syllabus, etc.
III Turn in Assignments Due: Homework 1, Essay 1, etc. and Return of Papers
IV Test 1
V Mind Game 1
VI Review of Lesson 2: Vector Physics/Motion
A. Vectors
1. Virtual arrow: magnitude (size) and direction
a. heads, tails
b. adding vectors ≠ adding algebraically
c. adding vectors
i. align the head of one vector with the tail of another; never put 2 tails together, or 2 heads
together: →→ is
okay; NOT →←
and NOT ←→
ii. →↑ is okay; but ↑→ is not okay; and →↑ is not okay
2. A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:
3. Pythagoras (576 BC – 495 BC)
4. Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
5. Components: see above; the x-component of vector A+B is A; the y-component is B.
6. If a vector is not directly along the x-axis or along the y-axis, it can be broken down into its x- and y- components
7. Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:
8. Projectile Motion
a. Range, height, angle (above)
Lesson 3: Circular motion
a. v = 2pr/P Э 2pr = c (circumference); r = radius; P = period, in seconds, to make one trip around the circle; and P = 1/n, Э n = the frequency in cycles per second (Hz).
b. v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2
c. So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r
d. And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
e. However, 2pn = w in rad/sec, thus a = w2r
f. If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sinθ = s, and if it’s even smaller, then r θ = s because for small θ, sin θ = θr where the angle, θ, is measured in radians, not degrees. 360° = 2p radians, so 1 radian = 57.3°.
g. Since v = dist/time, then v = s/t = r (θ/t) but is another way of writing (θ/t) = w, so
v = wr and v2/r = w2r = a, “acceleration”
V Conclusion
A. HWK Assignment 2: handout. due 6/10
B. Essay 2: Sir Isaac Newton, due 6/10
THURSDAY, JUNE 03, 2010
I Introduction
II Logistics: Seating, Syllabus, etc.
III Turn in Assignments Due: Homework 1, Essay 1, etc. and Return of Papers
IV Test 1
V Mind Game 1
VI Review of Lesson 2: Vector Physics/Motion
A. Vectors
1. Virtual arrow: magnitude (size) and direction
a. heads, tails
b. adding vectors ≠ adding algebraically
c. adding vectors
i. align the head of one vector with the tail of another; never put 2 tails together, or 2 heads
together: →→ is
okay; NOT →←
and NOT ←→
ii. →↑ is okay; but ↑→ is not okay; and →↑ is not okay
2. A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:
3. Pythagoras (576 BC – 495 BC)
4. Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
5. Components: see above; the x-component of vector A+B is A; the y-component is B.
6. If a vector is not directly along the x-axis or along the y-axis, it can be broken down into its x- and y- components
7. Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:
8. Projectile Motion
a. Range, height, angle (above)
Lesson 3: Circular motion
a. v = 2pr/P Э 2pr = c (circumference); r = radius; P = period, in seconds, to make one trip around the circle; and P = 1/n, Э n = the frequency in cycles per second (Hz).
b. v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2
c. So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r
d. And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
e. However, 2pn = w in rad/sec, thus a = w2r
f. If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sinθ = s, and if it’s even smaller, then r θ = s because for small θ, sin θ = θr where the angle, θ, is measured in radians, not degrees. 360° = 2p radians, so 1 radian = 57.3°.
g. Since v = dist/time, then v = s/t = r (θ/t) but is another way of writing (θ/t) = w, so
v = wr and v2/r = w2r = a, “acceleration”
V Conclusion
A. HWK Assignment 2: handout. due 6/10
B. Essay 2: Sir Isaac Newton, due 6/10
Wednesday, June 2, 2010
Solution Set 1
PHYSICS
SOLUTION SET 1
1. The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in scientific notation.
Solution:
I need to pick a number between 1 and 10 from the large number above, so I pick “1.14.” So, What would I have to multiply 1.14 by to get 114,000,000? I think 10^8. So the answer is 1.14 x 10^8.
Solution:
2. The speed of light is 299,792,458 m/s. Express that in scientific notation, and round to 3 significant figures.
Solution:
If I merely lopped off this number with the first three numbers, I'd have 299,000,000. But since the numbers after “299” are greater than 50% of the next number up, I have to round up to 299,800,000. But that's 4 significant figures. So, it must be 300,000,000. And in scientific notation, that is 3.00 x 10^8 m/s.
3. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
Solution: a = 2xtp = (2)(m)(s-2),
Because
(a) x is in units of meters (m)
(b)t is in units of seconds (s)
(c)p is the exponent (a number), and has to be “-2”,
Why? Because (s-2) ≡ 1/s2 (in this case, the symbol “≡” means “defined as” or “is the same as”
4. The irrational number p = 3.14159265358979…..Round this to seven significant figures
Solution:
To lop off the first seven digits would make it 3.141592, but since the next number is past 5, then we have to round off to 3.141593.
5. The largest blue whale observed was 108 feet long. Find that in meters.
Solution:
Since 3 feet = 1.0 yard, and 1 yard almost equals 1.0 meter, then a good answer would be 108 divided by 3, or, 36 (approx)
6. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution:
The acceleration of gravity is 9.8 m/sec^2, or, 32 ft/sec^2. This number is ten times that, so the answer is 320 ft/sec^2.
7. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Solution:
a. A quarter has a mass of about 5.6 grams. $12 million in quarters is 48,000,000 quarters and if we multiply that by 5.6 grams, we get (48 million x 5.6 grams) = 1,537,708,800 grams = 1.538 x 10^6 kg.
b. A dollar bill is about 1.0 gram, so, 12 million grams, or, 12,000 kg.
8. Take a look at the diagram (not) below. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
a. How far did you walk, i.e., total distanced from your house to the Library to the Park?
b. What was your displacement, i.e., net distance from your house to the Park?
Solution: Distance is total amount of length traveled, while displacement is the difference between your starting point and your ending point, so…
a. From the house to the library = 0.75 + 0.60 = 1.35, plus, from the library to the park = 0.6, for a total of 1.35 + 0.6 = 1.95 mi.
b. You started at your house and ended at the park, which is 0.75 mi.
9. The golfer, not seen below, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
Solution: The ball travels past the hole and has to come back, retracing its steps, so to speak
a. The ball traveled 10 + 2.5 + 2.5 = 15 meters.
b. The ball started at the golfer and ended up 10 meters away.
10. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
Solution: Some runner traveled 200 meters in 19.75 seconds. 200 meters is 0.2 kilometers. 1.0 kilometer = 0.6214 mile*.
*http://en.wikipedia.org/wiki/Kilometer
a. (200 m)/(19.75 s) = 10.12658228 m/s, or 10.1 m/s.
b. 10.12658228 m/s = (10.12658228 m/sec)(3600 sec/hour) = 36,455.69621 m/hour = 36.45569621 km/hr, and since 1.0 kilometer = 0.6214 mile, then 36.45569621 km = 22.65356962 mi, so, the answer is 22.65356962mi/hour = 22.7 mi/hr.
11. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
Solution: The distance from Earth to the Moon, and back, is about 480,000 miles. So, the light travels 480,000 miles at 186,000 mi/hr; thus, divide 480,000 by 186,000: (480,000)/(186,000) = (480)/(186) = 2.58 sec.
12. A dog named Fido runs back and forth between Jack and Jill, as not seen in the diagram below. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
Solution: In order to KISMIE, forget about the dog until later. Then, pretend either Jack or Jill is standing still while the other is moving. Relative to each other, it’s the same as if Jack is moving at 2.6 m/s towards a non-moving Jill. How long would it take Jack to travel 10.0 meters at 2.6 m/s? Divide 10 by 2.6 to get approximately 4 seconds: (10.0)/(2.6) = 3.846153846 sec. Now, let’s go back to the dog. How far does Fido travel in if he travels at a constant speed of 3.0 m/s? x = (3.846153846 sec)(3.0 m/s) = 11.53846154 m. or 12 m.
13. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
Solution: Realize that acceleration equals velocity divided by time, or, a = v/t.
a. Since a = v/t, then t = v/a, which means that time and velocity are directly related. Thus, double the velocity and you double the time (factor of two).
b. Okay, I did. Thanks.
c. Ditto.
14. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?
b. How much distance is required for the bacterium to reach a speed of 12 m/s?
Solution: From problem 39, we know that a = v/t, or, t = v/a.
a. t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
b. The standard distance equation, when starting from zero (0.0 m) and with an initial velocity of zero (0.0 m/s) is: x = ½ a t2. Since a = (156 m/s2) and the time we just found to be t = 0.076923076 seconds, then x = ½ a t2 =
½ (156 m/s2)(0.076923076 s)2 = ½ (156 m/s2)(0.005917159621 s2) = 0.46153845 m. Or, just 0.46 m.
15. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. We assume the rocket starts from ground level, i.e., y0 = 0.0 meters, and that it starts from rest, i.e., v0 = 0.0 m/s. So…we can rewrite equation (i) above as:
y = y0 + v0t + ½ a t2. Or, y = 0 + 0 + ½ a t2. Which leaves only y = ½ a t2.
We know “y”, but don’t know “a” even though we know that it is a constant acceleration, so we can’t find “t” just from equation (i). So, let’s use equation (ii) to see if it helps:
v = v0 + a t, but since v0 = 0.0 m/s, that leaves us with v = a t. Again, here we know “v” but, again, don’t know “a” even though we know that it is a constant acceleration.
Let’s use both equations.
First, we found that y = ½ a t2. Which we re-write as t = √ [(2y)/a].
Second, we found that v = a t. Which we re-write as t = v/a.
Since they both equal “t”, we can set them equal to find “a” then go back to get “t.”
Or, √ [(2y)/a] = v/a. Now, square both sides to get rid of the square root:
[(2y)/a] = (v/a)2. = (v2)/(a2). Now, multiply both sides by a2 to get
2ya = v2. Now, divide both sides by “2y” to get: a = (v) 2 / (2y) = (26.0 m/s)2/[(2)(3.2 m)] = (676)/(6.4) = 105.625 m/s2.
Now we use v = a t and rewrite it as t = v/a = (26.0)/(105.625) = 0.246153846 seconds, or 0.25 sec.
b. We found this while doing part (a.), so a =105.625 m/s2 or 106 m/s2. Or 1.06 x 10^2 m/s2.
c. To find the height, we use equation (i): y = ½ a t2 or, y(0.1) = ½ (105.625)(0.1)2 = ½ (105.625)(0.01) = 0.528125 m, or 0.528 m.
To find its speed, we use equation (ii): v = a t = (105.625)(0.10s) = 10.5625 m/s or 10.6 m/s.
16. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill.
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
c.What is Bob’s speed when catches up with Bill?
Solution: The standard distance equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) x = x0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. When Bob starts, his initial distance (x0) and velocity (v0) are zero. Thus, these two equations for Bob now are modified to be:
(i) x = ½ a t2. And we know “a” but don’t know “t”, so we can’t find “x” yet, plus
(ii) v = a t. Here, we also know “a” but not “t” so we can’t find “v” yet.
When Bob starts (t0 = 0.0 s), Bill is at a distance of x = v t = (3.5 m/s)(2.0 s) = 7.0 meters from Bob, and at the end, “t,” Bill is at a distance of zero (0.0 m) from Bob. This means that Bob has to travel some distance, “x” to reach Bill, who is also at the same place, “x.”
For Bill, the two equations are a little different:
(i) (x = x0 + v0t + ½ a t2) becomes (x = 0 + v0t + 0 = v0t) because Bill’s initial distance, x0, is the same as Bob’s, or, zero (0.0 m) and Bill’s initial speed is not zero (0.0 m/s) but 3.5 m/s. Plus, since we are told that Bill has a constant velocity, there is NO acceleration, or, for Bill, acceleration is zero (0.0 m/s2). Thus, for Bill:
x = v0t = (3.5 m/s)t
(ii) v = v0 + a t. Which becomes v = v0 since a = 0.0 m/s2.
So, Bob’s “x” and Bill’s “x” are the same:
½ a t2. = v0t, or ½ a t = v0. Thus, t = (2 v0)/a = (2)(3.5)/(2.4) = 7.0/2.4 = 2.917 s or 2.9 sec.
b. How far has Bob traveled? We use equation (i):
x = x0 + v0t + ½ a t2. = 0.0 m + (0.0 m/s)(2.9 sec) + ½ (2.4)(2.9)2 = 0 + 0 +
½ (2.4)(8.5) = 10.2 meters.
c. What is Bob’s speed at that time? Use equation (ii):
v = a t = (2.4)(2.9) = 7.0 m/s.
17. Review the image of the car falling off a cliff. The driver says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
In this question, y0 = 0.0 meters, or, the edge of the cliff; v0=0.0 m/s as it is falling down not zooming down (it is moving to the left, but that’s the x-axis and not relevant). And we assume that “zero to sixty” means from v0= 0.0 m/s (and 0.0 miles per hours) to v = 60.0 miles/hour = 96.6 km/hour = 96,600 meters/hour = 26.8 m/s. We also know that the acceleration here is ONLY gravity, or, a = - 9.8 m/s2. Also, we don’t seem to need equation (i), so we will use only equation (ii). So,
v = v0 + a t = 0 + (g)(t) = (- 9.8 /s2)(3.0 s) = - 29.4 m/s. (it is negative, as it is falling down, not up). So, the statement is false. It cannot go from 0.0 m/s to 60 miles per hour (- 26.8 m/s) in 3.0 seconds of falling. But it comes close!
PHYSICS
SOLUTION SET 1
1. The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in scientific notation.
Solution:
I need to pick a number between 1 and 10 from the large number above, so I pick “1.14.” So, What would I have to multiply 1.14 by to get 114,000,000? I think 10^8. So the answer is 1.14 x 10^8.
Solution:
2. The speed of light is 299,792,458 m/s. Express that in scientific notation, and round to 3 significant figures.
Solution:
If I merely lopped off this number with the first three numbers, I'd have 299,000,000. But since the numbers after “299” are greater than 50% of the next number up, I have to round up to 299,800,000. But that's 4 significant figures. So, it must be 300,000,000. And in scientific notation, that is 3.00 x 10^8 m/s.
3. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
Solution: a = 2xtp = (2)(m)(s-2),
Because
(a) x is in units of meters (m)
(b)t is in units of seconds (s)
(c)p is the exponent (a number), and has to be “-2”,
Why? Because (s-2) ≡ 1/s2 (in this case, the symbol “≡” means “defined as” or “is the same as”
4. The irrational number p = 3.14159265358979…..Round this to seven significant figures
Solution:
To lop off the first seven digits would make it 3.141592, but since the next number is past 5, then we have to round off to 3.141593.
5. The largest blue whale observed was 108 feet long. Find that in meters.
Solution:
Since 3 feet = 1.0 yard, and 1 yard almost equals 1.0 meter, then a good answer would be 108 divided by 3, or, 36 (approx)
6. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution:
The acceleration of gravity is 9.8 m/sec^2, or, 32 ft/sec^2. This number is ten times that, so the answer is 320 ft/sec^2.
7. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Solution:
a. A quarter has a mass of about 5.6 grams. $12 million in quarters is 48,000,000 quarters and if we multiply that by 5.6 grams, we get (48 million x 5.6 grams) = 1,537,708,800 grams = 1.538 x 10^6 kg.
b. A dollar bill is about 1.0 gram, so, 12 million grams, or, 12,000 kg.
8. Take a look at the diagram (not) below. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
a. How far did you walk, i.e., total distanced from your house to the Library to the Park?
b. What was your displacement, i.e., net distance from your house to the Park?
Solution: Distance is total amount of length traveled, while displacement is the difference between your starting point and your ending point, so…
a. From the house to the library = 0.75 + 0.60 = 1.35, plus, from the library to the park = 0.6, for a total of 1.35 + 0.6 = 1.95 mi.
b. You started at your house and ended at the park, which is 0.75 mi.
9. The golfer, not seen below, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
Solution: The ball travels past the hole and has to come back, retracing its steps, so to speak
a. The ball traveled 10 + 2.5 + 2.5 = 15 meters.
b. The ball started at the golfer and ended up 10 meters away.
10. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
Solution: Some runner traveled 200 meters in 19.75 seconds. 200 meters is 0.2 kilometers. 1.0 kilometer = 0.6214 mile*.
*http://en.wikipedia.org/wiki/Kilometer
a. (200 m)/(19.75 s) = 10.12658228 m/s, or 10.1 m/s.
b. 10.12658228 m/s = (10.12658228 m/sec)(3600 sec/hour) = 36,455.69621 m/hour = 36.45569621 km/hr, and since 1.0 kilometer = 0.6214 mile, then 36.45569621 km = 22.65356962 mi, so, the answer is 22.65356962mi/hour = 22.7 mi/hr.
11. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
Solution: The distance from Earth to the Moon, and back, is about 480,000 miles. So, the light travels 480,000 miles at 186,000 mi/hr; thus, divide 480,000 by 186,000: (480,000)/(186,000) = (480)/(186) = 2.58 sec.
12. A dog named Fido runs back and forth between Jack and Jill, as not seen in the diagram below. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
Solution: In order to KISMIE, forget about the dog until later. Then, pretend either Jack or Jill is standing still while the other is moving. Relative to each other, it’s the same as if Jack is moving at 2.6 m/s towards a non-moving Jill. How long would it take Jack to travel 10.0 meters at 2.6 m/s? Divide 10 by 2.6 to get approximately 4 seconds: (10.0)/(2.6) = 3.846153846 sec. Now, let’s go back to the dog. How far does Fido travel in if he travels at a constant speed of 3.0 m/s? x = (3.846153846 sec)(3.0 m/s) = 11.53846154 m. or 12 m.
13. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
Solution: Realize that acceleration equals velocity divided by time, or, a = v/t.
a. Since a = v/t, then t = v/a, which means that time and velocity are directly related. Thus, double the velocity and you double the time (factor of two).
b. Okay, I did. Thanks.
c. Ditto.
14. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?
b. How much distance is required for the bacterium to reach a speed of 12 m/s?
Solution: From problem 39, we know that a = v/t, or, t = v/a.
a. t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
b. The standard distance equation, when starting from zero (0.0 m) and with an initial velocity of zero (0.0 m/s) is: x = ½ a t2. Since a = (156 m/s2) and the time we just found to be t = 0.076923076 seconds, then x = ½ a t2 =
½ (156 m/s2)(0.076923076 s)2 = ½ (156 m/s2)(0.005917159621 s2) = 0.46153845 m. Or, just 0.46 m.
15. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. We assume the rocket starts from ground level, i.e., y0 = 0.0 meters, and that it starts from rest, i.e., v0 = 0.0 m/s. So…we can rewrite equation (i) above as:
y = y0 + v0t + ½ a t2. Or, y = 0 + 0 + ½ a t2. Which leaves only y = ½ a t2.
We know “y”, but don’t know “a” even though we know that it is a constant acceleration, so we can’t find “t” just from equation (i). So, let’s use equation (ii) to see if it helps:
v = v0 + a t, but since v0 = 0.0 m/s, that leaves us with v = a t. Again, here we know “v” but, again, don’t know “a” even though we know that it is a constant acceleration.
Let’s use both equations.
First, we found that y = ½ a t2. Which we re-write as t = √ [(2y)/a].
Second, we found that v = a t. Which we re-write as t = v/a.
Since they both equal “t”, we can set them equal to find “a” then go back to get “t.”
Or, √ [(2y)/a] = v/a. Now, square both sides to get rid of the square root:
[(2y)/a] = (v/a)2. = (v2)/(a2). Now, multiply both sides by a2 to get
2ya = v2. Now, divide both sides by “2y” to get: a = (v) 2 / (2y) = (26.0 m/s)2/[(2)(3.2 m)] = (676)/(6.4) = 105.625 m/s2.
Now we use v = a t and rewrite it as t = v/a = (26.0)/(105.625) = 0.246153846 seconds, or 0.25 sec.
b. We found this while doing part (a.), so a =105.625 m/s2 or 106 m/s2. Or 1.06 x 10^2 m/s2.
c. To find the height, we use equation (i): y = ½ a t2 or, y(0.1) = ½ (105.625)(0.1)2 = ½ (105.625)(0.01) = 0.528125 m, or 0.528 m.
To find its speed, we use equation (ii): v = a t = (105.625)(0.10s) = 10.5625 m/s or 10.6 m/s.
16. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill.
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
c.What is Bob’s speed when catches up with Bill?
Solution: The standard distance equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) x = x0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. When Bob starts, his initial distance (x0) and velocity (v0) are zero. Thus, these two equations for Bob now are modified to be:
(i) x = ½ a t2. And we know “a” but don’t know “t”, so we can’t find “x” yet, plus
(ii) v = a t. Here, we also know “a” but not “t” so we can’t find “v” yet.
When Bob starts (t0 = 0.0 s), Bill is at a distance of x = v t = (3.5 m/s)(2.0 s) = 7.0 meters from Bob, and at the end, “t,” Bill is at a distance of zero (0.0 m) from Bob. This means that Bob has to travel some distance, “x” to reach Bill, who is also at the same place, “x.”
For Bill, the two equations are a little different:
(i) (x = x0 + v0t + ½ a t2) becomes (x = 0 + v0t + 0 = v0t) because Bill’s initial distance, x0, is the same as Bob’s, or, zero (0.0 m) and Bill’s initial speed is not zero (0.0 m/s) but 3.5 m/s. Plus, since we are told that Bill has a constant velocity, there is NO acceleration, or, for Bill, acceleration is zero (0.0 m/s2). Thus, for Bill:
x = v0t = (3.5 m/s)t
(ii) v = v0 + a t. Which becomes v = v0 since a = 0.0 m/s2.
So, Bob’s “x” and Bill’s “x” are the same:
½ a t2. = v0t, or ½ a t = v0. Thus, t = (2 v0)/a = (2)(3.5)/(2.4) = 7.0/2.4 = 2.917 s or 2.9 sec.
b. How far has Bob traveled? We use equation (i):
x = x0 + v0t + ½ a t2. = 0.0 m + (0.0 m/s)(2.9 sec) + ½ (2.4)(2.9)2 = 0 + 0 +
½ (2.4)(8.5) = 10.2 meters.
c. What is Bob’s speed at that time? Use equation (ii):
v = a t = (2.4)(2.9) = 7.0 m/s.
17. Review the image of the car falling off a cliff. The driver says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
In this question, y0 = 0.0 meters, or, the edge of the cliff; v0=0.0 m/s as it is falling down not zooming down (it is moving to the left, but that’s the x-axis and not relevant). And we assume that “zero to sixty” means from v0= 0.0 m/s (and 0.0 miles per hours) to v = 60.0 miles/hour = 96.6 km/hour = 96,600 meters/hour = 26.8 m/s. We also know that the acceleration here is ONLY gravity, or, a = - 9.8 m/s2. Also, we don’t seem to need equation (i), so we will use only equation (ii). So,
v = v0 + a t = 0 + (g)(t) = (- 9.8 /s2)(3.0 s) = - 29.4 m/s. (it is negative, as it is falling down, not up). So, the statement is false. It cannot go from 0.0 m/s to 60 miles per hour (- 26.8 m/s) in 3.0 seconds of falling. But it comes close!
SOLUTION SET 1
1. The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in scientific notation.
Solution:
I need to pick a number between 1 and 10 from the large number above, so I pick “1.14.” So, What would I have to multiply 1.14 by to get 114,000,000? I think 10^8. So the answer is 1.14 x 10^8.
Solution:
2. The speed of light is 299,792,458 m/s. Express that in scientific notation, and round to 3 significant figures.
Solution:
If I merely lopped off this number with the first three numbers, I'd have 299,000,000. But since the numbers after “299” are greater than 50% of the next number up, I have to round up to 299,800,000. But that's 4 significant figures. So, it must be 300,000,000. And in scientific notation, that is 3.00 x 10^8 m/s.
3. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
Solution: a = 2xtp = (2)(m)(s-2),
Because
(a) x is in units of meters (m)
(b)t is in units of seconds (s)
(c)p is the exponent (a number), and has to be “-2”,
Why? Because (s-2) ≡ 1/s2 (in this case, the symbol “≡” means “defined as” or “is the same as”
4. The irrational number p = 3.14159265358979…..Round this to seven significant figures
Solution:
To lop off the first seven digits would make it 3.141592, but since the next number is past 5, then we have to round off to 3.141593.
5. The largest blue whale observed was 108 feet long. Find that in meters.
Solution:
Since 3 feet = 1.0 yard, and 1 yard almost equals 1.0 meter, then a good answer would be 108 divided by 3, or, 36 (approx)
6. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution:
The acceleration of gravity is 9.8 m/sec^2, or, 32 ft/sec^2. This number is ten times that, so the answer is 320 ft/sec^2.
7. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Solution:
a. A quarter has a mass of about 5.6 grams. $12 million in quarters is 48,000,000 quarters and if we multiply that by 5.6 grams, we get (48 million x 5.6 grams) = 1,537,708,800 grams = 1.538 x 10^6 kg.
b. A dollar bill is about 1.0 gram, so, 12 million grams, or, 12,000 kg.
8. Take a look at the diagram (not) below. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
a. How far did you walk, i.e., total distanced from your house to the Library to the Park?
b. What was your displacement, i.e., net distance from your house to the Park?
Solution: Distance is total amount of length traveled, while displacement is the difference between your starting point and your ending point, so…
a. From the house to the library = 0.75 + 0.60 = 1.35, plus, from the library to the park = 0.6, for a total of 1.35 + 0.6 = 1.95 mi.
b. You started at your house and ended at the park, which is 0.75 mi.
9. The golfer, not seen below, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
Solution: The ball travels past the hole and has to come back, retracing its steps, so to speak
a. The ball traveled 10 + 2.5 + 2.5 = 15 meters.
b. The ball started at the golfer and ended up 10 meters away.
10. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
Solution: Some runner traveled 200 meters in 19.75 seconds. 200 meters is 0.2 kilometers. 1.0 kilometer = 0.6214 mile*.
*http://en.wikipedia.org/wiki/Kilometer
a. (200 m)/(19.75 s) = 10.12658228 m/s, or 10.1 m/s.
b. 10.12658228 m/s = (10.12658228 m/sec)(3600 sec/hour) = 36,455.69621 m/hour = 36.45569621 km/hr, and since 1.0 kilometer = 0.6214 mile, then 36.45569621 km = 22.65356962 mi, so, the answer is 22.65356962mi/hour = 22.7 mi/hr.
11. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
Solution: The distance from Earth to the Moon, and back, is about 480,000 miles. So, the light travels 480,000 miles at 186,000 mi/hr; thus, divide 480,000 by 186,000: (480,000)/(186,000) = (480)/(186) = 2.58 sec.
12. A dog named Fido runs back and forth between Jack and Jill, as not seen in the diagram below. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
Solution: In order to KISMIE, forget about the dog until later. Then, pretend either Jack or Jill is standing still while the other is moving. Relative to each other, it’s the same as if Jack is moving at 2.6 m/s towards a non-moving Jill. How long would it take Jack to travel 10.0 meters at 2.6 m/s? Divide 10 by 2.6 to get approximately 4 seconds: (10.0)/(2.6) = 3.846153846 sec. Now, let’s go back to the dog. How far does Fido travel in if he travels at a constant speed of 3.0 m/s? x = (3.846153846 sec)(3.0 m/s) = 11.53846154 m. or 12 m.
13. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
Solution: Realize that acceleration equals velocity divided by time, or, a = v/t.
a. Since a = v/t, then t = v/a, which means that time and velocity are directly related. Thus, double the velocity and you double the time (factor of two).
b. Okay, I did. Thanks.
c. Ditto.
14. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?
b. How much distance is required for the bacterium to reach a speed of 12 m/s?
Solution: From problem 39, we know that a = v/t, or, t = v/a.
a. t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
b. The standard distance equation, when starting from zero (0.0 m) and with an initial velocity of zero (0.0 m/s) is: x = ½ a t2. Since a = (156 m/s2) and the time we just found to be t = 0.076923076 seconds, then x = ½ a t2 =
½ (156 m/s2)(0.076923076 s)2 = ½ (156 m/s2)(0.005917159621 s2) = 0.46153845 m. Or, just 0.46 m.
15. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. We assume the rocket starts from ground level, i.e., y0 = 0.0 meters, and that it starts from rest, i.e., v0 = 0.0 m/s. So…we can rewrite equation (i) above as:
y = y0 + v0t + ½ a t2. Or, y = 0 + 0 + ½ a t2. Which leaves only y = ½ a t2.
We know “y”, but don’t know “a” even though we know that it is a constant acceleration, so we can’t find “t” just from equation (i). So, let’s use equation (ii) to see if it helps:
v = v0 + a t, but since v0 = 0.0 m/s, that leaves us with v = a t. Again, here we know “v” but, again, don’t know “a” even though we know that it is a constant acceleration.
Let’s use both equations.
First, we found that y = ½ a t2. Which we re-write as t = √ [(2y)/a].
Second, we found that v = a t. Which we re-write as t = v/a.
Since they both equal “t”, we can set them equal to find “a” then go back to get “t.”
Or, √ [(2y)/a] = v/a. Now, square both sides to get rid of the square root:
[(2y)/a] = (v/a)2. = (v2)/(a2). Now, multiply both sides by a2 to get
2ya = v2. Now, divide both sides by “2y” to get: a = (v) 2 / (2y) = (26.0 m/s)2/[(2)(3.2 m)] = (676)/(6.4) = 105.625 m/s2.
Now we use v = a t and rewrite it as t = v/a = (26.0)/(105.625) = 0.246153846 seconds, or 0.25 sec.
b. We found this while doing part (a.), so a =105.625 m/s2 or 106 m/s2. Or 1.06 x 10^2 m/s2.
c. To find the height, we use equation (i): y = ½ a t2 or, y(0.1) = ½ (105.625)(0.1)2 = ½ (105.625)(0.01) = 0.528125 m, or 0.528 m.
To find its speed, we use equation (ii): v = a t = (105.625)(0.10s) = 10.5625 m/s or 10.6 m/s.
16. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill.
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
c.What is Bob’s speed when catches up with Bill?
Solution: The standard distance equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) x = x0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. When Bob starts, his initial distance (x0) and velocity (v0) are zero. Thus, these two equations for Bob now are modified to be:
(i) x = ½ a t2. And we know “a” but don’t know “t”, so we can’t find “x” yet, plus
(ii) v = a t. Here, we also know “a” but not “t” so we can’t find “v” yet.
When Bob starts (t0 = 0.0 s), Bill is at a distance of x = v t = (3.5 m/s)(2.0 s) = 7.0 meters from Bob, and at the end, “t,” Bill is at a distance of zero (0.0 m) from Bob. This means that Bob has to travel some distance, “x” to reach Bill, who is also at the same place, “x.”
For Bill, the two equations are a little different:
(i) (x = x0 + v0t + ½ a t2) becomes (x = 0 + v0t + 0 = v0t) because Bill’s initial distance, x0, is the same as Bob’s, or, zero (0.0 m) and Bill’s initial speed is not zero (0.0 m/s) but 3.5 m/s. Plus, since we are told that Bill has a constant velocity, there is NO acceleration, or, for Bill, acceleration is zero (0.0 m/s2). Thus, for Bill:
x = v0t = (3.5 m/s)t
(ii) v = v0 + a t. Which becomes v = v0 since a = 0.0 m/s2.
So, Bob’s “x” and Bill’s “x” are the same:
½ a t2. = v0t, or ½ a t = v0. Thus, t = (2 v0)/a = (2)(3.5)/(2.4) = 7.0/2.4 = 2.917 s or 2.9 sec.
b. How far has Bob traveled? We use equation (i):
x = x0 + v0t + ½ a t2. = 0.0 m + (0.0 m/s)(2.9 sec) + ½ (2.4)(2.9)2 = 0 + 0 +
½ (2.4)(8.5) = 10.2 meters.
c. What is Bob’s speed at that time? Use equation (ii):
v = a t = (2.4)(2.9) = 7.0 m/s.
17. Review the image of the car falling off a cliff. The driver says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
In this question, y0 = 0.0 meters, or, the edge of the cliff; v0=0.0 m/s as it is falling down not zooming down (it is moving to the left, but that’s the x-axis and not relevant). And we assume that “zero to sixty” means from v0= 0.0 m/s (and 0.0 miles per hours) to v = 60.0 miles/hour = 96.6 km/hour = 96,600 meters/hour = 26.8 m/s. We also know that the acceleration here is ONLY gravity, or, a = - 9.8 m/s2. Also, we don’t seem to need equation (i), so we will use only equation (ii). So,
v = v0 + a t = 0 + (g)(t) = (- 9.8 /s2)(3.0 s) = - 29.4 m/s. (it is negative, as it is falling down, not up). So, the statement is false. It cannot go from 0.0 m/s to 60 miles per hour (- 26.8 m/s) in 3.0 seconds of falling. But it comes close!
PHYSICS
SOLUTION SET 1
1. The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in scientific notation.
Solution:
I need to pick a number between 1 and 10 from the large number above, so I pick “1.14.” So, What would I have to multiply 1.14 by to get 114,000,000? I think 10^8. So the answer is 1.14 x 10^8.
Solution:
2. The speed of light is 299,792,458 m/s. Express that in scientific notation, and round to 3 significant figures.
Solution:
If I merely lopped off this number with the first three numbers, I'd have 299,000,000. But since the numbers after “299” are greater than 50% of the next number up, I have to round up to 299,800,000. But that's 4 significant figures. So, it must be 300,000,000. And in scientific notation, that is 3.00 x 10^8 m/s.
3. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
Solution: a = 2xtp = (2)(m)(s-2),
Because
(a) x is in units of meters (m)
(b)t is in units of seconds (s)
(c)p is the exponent (a number), and has to be “-2”,
Why? Because (s-2) ≡ 1/s2 (in this case, the symbol “≡” means “defined as” or “is the same as”
4. The irrational number p = 3.14159265358979…..Round this to seven significant figures
Solution:
To lop off the first seven digits would make it 3.141592, but since the next number is past 5, then we have to round off to 3.141593.
5. The largest blue whale observed was 108 feet long. Find that in meters.
Solution:
Since 3 feet = 1.0 yard, and 1 yard almost equals 1.0 meter, then a good answer would be 108 divided by 3, or, 36 (approx)
6. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution:
The acceleration of gravity is 9.8 m/sec^2, or, 32 ft/sec^2. This number is ten times that, so the answer is 320 ft/sec^2.
7. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Solution:
a. A quarter has a mass of about 5.6 grams. $12 million in quarters is 48,000,000 quarters and if we multiply that by 5.6 grams, we get (48 million x 5.6 grams) = 1,537,708,800 grams = 1.538 x 10^6 kg.
b. A dollar bill is about 1.0 gram, so, 12 million grams, or, 12,000 kg.
8. Take a look at the diagram (not) below. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
a. How far did you walk, i.e., total distanced from your house to the Library to the Park?
b. What was your displacement, i.e., net distance from your house to the Park?
Solution: Distance is total amount of length traveled, while displacement is the difference between your starting point and your ending point, so…
a. From the house to the library = 0.75 + 0.60 = 1.35, plus, from the library to the park = 0.6, for a total of 1.35 + 0.6 = 1.95 mi.
b. You started at your house and ended at the park, which is 0.75 mi.
9. The golfer, not seen below, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
Solution: The ball travels past the hole and has to come back, retracing its steps, so to speak
a. The ball traveled 10 + 2.5 + 2.5 = 15 meters.
b. The ball started at the golfer and ended up 10 meters away.
10. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
Solution: Some runner traveled 200 meters in 19.75 seconds. 200 meters is 0.2 kilometers. 1.0 kilometer = 0.6214 mile*.
*http://en.wikipedia.org/wiki/Kilometer
a. (200 m)/(19.75 s) = 10.12658228 m/s, or 10.1 m/s.
b. 10.12658228 m/s = (10.12658228 m/sec)(3600 sec/hour) = 36,455.69621 m/hour = 36.45569621 km/hr, and since 1.0 kilometer = 0.6214 mile, then 36.45569621 km = 22.65356962 mi, so, the answer is 22.65356962mi/hour = 22.7 mi/hr.
11. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
Solution: The distance from Earth to the Moon, and back, is about 480,000 miles. So, the light travels 480,000 miles at 186,000 mi/hr; thus, divide 480,000 by 186,000: (480,000)/(186,000) = (480)/(186) = 2.58 sec.
12. A dog named Fido runs back and forth between Jack and Jill, as not seen in the diagram below. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
Solution: In order to KISMIE, forget about the dog until later. Then, pretend either Jack or Jill is standing still while the other is moving. Relative to each other, it’s the same as if Jack is moving at 2.6 m/s towards a non-moving Jill. How long would it take Jack to travel 10.0 meters at 2.6 m/s? Divide 10 by 2.6 to get approximately 4 seconds: (10.0)/(2.6) = 3.846153846 sec. Now, let’s go back to the dog. How far does Fido travel in if he travels at a constant speed of 3.0 m/s? x = (3.846153846 sec)(3.0 m/s) = 11.53846154 m. or 12 m.
13. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
Solution: Realize that acceleration equals velocity divided by time, or, a = v/t.
a. Since a = v/t, then t = v/a, which means that time and velocity are directly related. Thus, double the velocity and you double the time (factor of two).
b. Okay, I did. Thanks.
c. Ditto.
14. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?
b. How much distance is required for the bacterium to reach a speed of 12 m/s?
Solution: From problem 39, we know that a = v/t, or, t = v/a.
a. t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
b. The standard distance equation, when starting from zero (0.0 m) and with an initial velocity of zero (0.0 m/s) is: x = ½ a t2. Since a = (156 m/s2) and the time we just found to be t = 0.076923076 seconds, then x = ½ a t2 =
½ (156 m/s2)(0.076923076 s)2 = ½ (156 m/s2)(0.005917159621 s2) = 0.46153845 m. Or, just 0.46 m.
15. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. We assume the rocket starts from ground level, i.e., y0 = 0.0 meters, and that it starts from rest, i.e., v0 = 0.0 m/s. So…we can rewrite equation (i) above as:
y = y0 + v0t + ½ a t2. Or, y = 0 + 0 + ½ a t2. Which leaves only y = ½ a t2.
We know “y”, but don’t know “a” even though we know that it is a constant acceleration, so we can’t find “t” just from equation (i). So, let’s use equation (ii) to see if it helps:
v = v0 + a t, but since v0 = 0.0 m/s, that leaves us with v = a t. Again, here we know “v” but, again, don’t know “a” even though we know that it is a constant acceleration.
Let’s use both equations.
First, we found that y = ½ a t2. Which we re-write as t = √ [(2y)/a].
Second, we found that v = a t. Which we re-write as t = v/a.
Since they both equal “t”, we can set them equal to find “a” then go back to get “t.”
Or, √ [(2y)/a] = v/a. Now, square both sides to get rid of the square root:
[(2y)/a] = (v/a)2. = (v2)/(a2). Now, multiply both sides by a2 to get
2ya = v2. Now, divide both sides by “2y” to get: a = (v) 2 / (2y) = (26.0 m/s)2/[(2)(3.2 m)] = (676)/(6.4) = 105.625 m/s2.
Now we use v = a t and rewrite it as t = v/a = (26.0)/(105.625) = 0.246153846 seconds, or 0.25 sec.
b. We found this while doing part (a.), so a =105.625 m/s2 or 106 m/s2. Or 1.06 x 10^2 m/s2.
c. To find the height, we use equation (i): y = ½ a t2 or, y(0.1) = ½ (105.625)(0.1)2 = ½ (105.625)(0.01) = 0.528125 m, or 0.528 m.
To find its speed, we use equation (ii): v = a t = (105.625)(0.10s) = 10.5625 m/s or 10.6 m/s.
16. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill.
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
c.What is Bob’s speed when catches up with Bill?
Solution: The standard distance equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) x = x0 + v0t + ½ a t2. And
(ii) v = v0 + a t
a. When Bob starts, his initial distance (x0) and velocity (v0) are zero. Thus, these two equations for Bob now are modified to be:
(i) x = ½ a t2. And we know “a” but don’t know “t”, so we can’t find “x” yet, plus
(ii) v = a t. Here, we also know “a” but not “t” so we can’t find “v” yet.
When Bob starts (t0 = 0.0 s), Bill is at a distance of x = v t = (3.5 m/s)(2.0 s) = 7.0 meters from Bob, and at the end, “t,” Bill is at a distance of zero (0.0 m) from Bob. This means that Bob has to travel some distance, “x” to reach Bill, who is also at the same place, “x.”
For Bill, the two equations are a little different:
(i) (x = x0 + v0t + ½ a t2) becomes (x = 0 + v0t + 0 = v0t) because Bill’s initial distance, x0, is the same as Bob’s, or, zero (0.0 m) and Bill’s initial speed is not zero (0.0 m/s) but 3.5 m/s. Plus, since we are told that Bill has a constant velocity, there is NO acceleration, or, for Bill, acceleration is zero (0.0 m/s2). Thus, for Bill:
x = v0t = (3.5 m/s)t
(ii) v = v0 + a t. Which becomes v = v0 since a = 0.0 m/s2.
So, Bob’s “x” and Bill’s “x” are the same:
½ a t2. = v0t, or ½ a t = v0. Thus, t = (2 v0)/a = (2)(3.5)/(2.4) = 7.0/2.4 = 2.917 s or 2.9 sec.
b. How far has Bob traveled? We use equation (i):
x = x0 + v0t + ½ a t2. = 0.0 m + (0.0 m/s)(2.9 sec) + ½ (2.4)(2.9)2 = 0 + 0 +
½ (2.4)(8.5) = 10.2 meters.
c. What is Bob’s speed at that time? Use equation (ii):
v = a t = (2.4)(2.9) = 7.0 m/s.
17. Review the image of the car falling off a cliff. The driver says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:
(i) y = y0 + v0t + ½ a t2. And
(ii) v = v0 + a t
In this question, y0 = 0.0 meters, or, the edge of the cliff; v0=0.0 m/s as it is falling down not zooming down (it is moving to the left, but that’s the x-axis and not relevant). And we assume that “zero to sixty” means from v0= 0.0 m/s (and 0.0 miles per hours) to v = 60.0 miles/hour = 96.6 km/hour = 96,600 meters/hour = 26.8 m/s. We also know that the acceleration here is ONLY gravity, or, a = - 9.8 m/s2. Also, we don’t seem to need equation (i), so we will use only equation (ii). So,
v = v0 + a t = 0 + (g)(t) = (- 9.8 /s2)(3.0 s) = - 29.4 m/s. (it is negative, as it is falling down, not up). So, the statement is false. It cannot go from 0.0 m/s to 60 miles per hour (- 26.8 m/s) in 3.0 seconds of falling. But it comes close!
Lab Exercise 2
Physics Lab 2 Wednesday, June 2, 2010 Name __________________
Dr Dave Menke, Instructor, Upward Bound at Pima Community College
I Title: Measurements
II Purpose: Practice in Measurements
III Equipment: any typical textbook, meter stick, metric ruler, vernier
IV Procedure:
1. Measure the length, width, and height of the book. Record.
2. Calculate the surface area (approximately) of this 3-D object. Record.
3. Calculate the volume of the book. Record.
4. Use the vernier to estimate the thickness of a page. Record.
5. Think of an alternative way to determine the thickness of a page. Explain your method, and Record.
V Data
1. The length x width x height of object, in cm: _______ x _______ x ________
2. The surface area of the object: ___________ cm2.
3. Volume of the object: ___________cm3.
4. Thickness of one sheet in the book ___________ cm.
5.Alternate method to find thickness (explain).
6.Thickness of one sheet using this method: ______________cm.
VI Results: In this section, explain how successful you were at achieving the Purpose.
VII Error
A. Qualitative – sources of error
1. Personal – what things did you or your partner(s) do to screw things up, if anything at all?
2. Systematic – what was wrong with the equipment that caused the lab to go “bad”, or, what external factors (hurricane, earthquake, etc) contributed to difficulties in this lab?
3. Random – there are always random errors, unless you do this lab exercise 5 times and take an average
B. Quantitative – NA
VIII Questions
1. What is a vernier caliper?
2. What is a micrometer caliper?
Dr Dave Menke, Instructor, Upward Bound at Pima Community College
I Title: Measurements
II Purpose: Practice in Measurements
III Equipment: any typical textbook, meter stick, metric ruler, vernier
IV Procedure:
1. Measure the length, width, and height of the book. Record.
2. Calculate the surface area (approximately) of this 3-D object. Record.
3. Calculate the volume of the book. Record.
4. Use the vernier to estimate the thickness of a page. Record.
5. Think of an alternative way to determine the thickness of a page. Explain your method, and Record.
V Data
1. The length x width x height of object, in cm: _______ x _______ x ________
2. The surface area of the object: ___________ cm2.
3. Volume of the object: ___________cm3.
4. Thickness of one sheet in the book ___________ cm.
5.Alternate method to find thickness (explain).
6.Thickness of one sheet using this method: ______________cm.
VI Results: In this section, explain how successful you were at achieving the Purpose.
VII Error
A. Qualitative – sources of error
1. Personal – what things did you or your partner(s) do to screw things up, if anything at all?
2. Systematic – what was wrong with the equipment that caused the lab to go “bad”, or, what external factors (hurricane, earthquake, etc) contributed to difficulties in this lab?
3. Random – there are always random errors, unless you do this lab exercise 5 times and take an average
B. Quantitative – NA
VIII Questions
1. What is a vernier caliper?
2. What is a micrometer caliper?
LESSON 2
PHYSICS LESSON 2 FOR
WEDNESDAY, JUNE 02, 2010
I Introduction
II Logistics: Seating, Syllabus, website:
http://drdaveupwardboundphysics.blogspot.com
III Review of Lesson 1: Measurements, Ch 1
A. Units – metric (mks; also known as SIU)
B. Meters for length
Length / Distance / Displacement, “x”
C. Kilograms for mass
Mass; 1kg = 2.2 lbs at sea level
D. Seconds for time
Time; second; 3600 sec = 1 hour
E. Coordinates: 3D: x, y, z
F. Speed: v = Dx/Dt; D = “change of” or
(x2 – x1) / (t2 – t1)
Velocity = speed with a vector
G. Acceleration = Dv / Dt or (v2 – v1) / (t2 – t1) ; often a vector
H. Scientific Notation:
5,617 = 5.617 x 103 or 5.617 x 10^3 or 5.617 E3
I. Significant Figures: 5,617 x 27 = 151,659 ? NO! 150,000 or 1.5 x 105
J. Estimating (ball parking; educated guessing): 5,617 x 321 = (5.6 x 3) x 105 = 16.8 x 105 = 1.68 x 106 for a guess. Exact is 1,803,057 = 1.80 x 106
K. Trigonometry: Create a right triangle inside a “unit” circle of radius r = 1. Then the x-component will be r cosine(θ) and the y-component will be r sine(θ) . However, since r = 1, we realize that the x-component will almost always be cosθ while the y-component will almost always be sinθ. Plus, sinθ / cosθ = y/x = tangent of the angle, or, tanθ.
IV Lesson 2: Vector Physics/Motion
A. Vectors
1. Virtual arrow: magnitude (size) and direction
a. heads, tails
b. adding vectors ≠ adding algebraically
c. adding vectors
i. align the head of one vector with the tail of another; never put 2 tails together, or 2 heads
together: →→ is
okay; NOT →←
and NOT ←→
ii. →↑ is okay; but ↑→ is not okay; and →↑ is not okay
2. A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:
3. Pythagoras (576 BC – 495 BC)
4. KISMIE
5. Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
6. Components: see above; the x-component of vector A+B is A; the y-component is B.
7. If a vector is not directly along the x-axis or along the y-axis, it can be broken down into its x- and y- components
8. Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:
V Laboratory Exercise 2: Measurement
VI Conclusion
A. Finish Homework 1 Problems until end of period
B. Prepare for Test 1 (review and problems)
WEDNESDAY, JUNE 02, 2010
I Introduction
II Logistics: Seating, Syllabus, website:
http://drdaveupwardboundphysics.blogspot.com
III Review of Lesson 1: Measurements, Ch 1
A. Units – metric (mks; also known as SIU)
B. Meters for length
Length / Distance / Displacement, “x”
C. Kilograms for mass
Mass; 1kg = 2.2 lbs at sea level
D. Seconds for time
Time; second; 3600 sec = 1 hour
E. Coordinates: 3D: x, y, z
F. Speed: v = Dx/Dt; D = “change of” or
(x2 – x1) / (t2 – t1)
Velocity = speed with a vector
G. Acceleration = Dv / Dt or (v2 – v1) / (t2 – t1) ; often a vector
H. Scientific Notation:
5,617 = 5.617 x 103 or 5.617 x 10^3 or 5.617 E3
I. Significant Figures: 5,617 x 27 = 151,659 ? NO! 150,000 or 1.5 x 105
J. Estimating (ball parking; educated guessing): 5,617 x 321 = (5.6 x 3) x 105 = 16.8 x 105 = 1.68 x 106 for a guess. Exact is 1,803,057 = 1.80 x 106
K. Trigonometry: Create a right triangle inside a “unit” circle of radius r = 1. Then the x-component will be r cosine(θ) and the y-component will be r sine(θ) . However, since r = 1, we realize that the x-component will almost always be cosθ while the y-component will almost always be sinθ. Plus, sinθ / cosθ = y/x = tangent of the angle, or, tanθ.
IV Lesson 2: Vector Physics/Motion
A. Vectors
1. Virtual arrow: magnitude (size) and direction
a. heads, tails
b. adding vectors ≠ adding algebraically
c. adding vectors
i. align the head of one vector with the tail of another; never put 2 tails together, or 2 heads
together: →→ is
okay; NOT →←
and NOT ←→
ii. →↑ is okay; but ↑→ is not okay; and →↑ is not okay
2. A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:
3. Pythagoras (576 BC – 495 BC)
4. KISMIE
5. Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
6. Components: see above; the x-component of vector A+B is A; the y-component is B.
7. If a vector is not directly along the x-axis or along the y-axis, it can be broken down into its x- and y- components
8. Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:
V Laboratory Exercise 2: Measurement
VI Conclusion
A. Finish Homework 1 Problems until end of period
B. Prepare for Test 1 (review and problems)
Tuesday, June 1, 2010
Weekly Workload
PHYSICS Addendum to Course Syllabus, June 2010, Dr Dave Menke, page 3 of 6
Weekly Workload
Week 1, May 31 – June 4, 2010
1. Monday: No School
2. Tuesday
a. Introduction
b. Syllabus
c. Lesson 1: History of Physics, Kinematics
d. Lab Exercise #1: Density of Water
e. Homework Assignment 1: Hand out. Turn in Thursday, 6/3
f. Essay 1: Aristotle, due Thursday, 6/3
3. Wednesday
a. Review of Lesson 1
b. Lesson 2: Vector Physics
c. Lab Exercise #2: Measurement
d. Homework Assignment 1: Hand out. Turn in Thursday, 6/3
e. Essay 1: Aristotle, due Thursday, 6/3
4. Thursday
a. Turn in Homework Assignment 1
b. Turn in Essay 1 on Aristotle
c. Test 1 on Lessons 1 & 2
d. Mind Game 1
e. Lesson 3: Newton’s Laws
f. Homework Assignment 2: Handout. Turn in Thursday, 6/10
g. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
5. Friday: No School
Week 2, June 7-11
1. Monday –
a. Return of papers
b. Running Grades
c. Review of Lesson 3: Newton's Laws
d. Lab Exercise #3: Linear motion
e. Lesson 4: Circular Motion
f. Homework Assignment 2: Handout. Turn in Thursday, 6/10
g. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
2. Tuesday
a. Running Grades
b. Review of Lesson 4: Circular Motion
c. Lab Exercise #4: Centripetal Motion
d. Lesson 5: Energy, Momentum
e. Homework Assignment 2: Handout. Turn in Thursday, 6/10
f. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
3. Wednesday
a. Running Grades
b. Review of Lesson 5: Energy, Momentum
c. Lab Exercise 5: Finding Gravity
d. Lesson 6: Momentum & Impulse
e. Homework Assignment 2: Handout. Turn in Thursday, 6/10
f. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
4. Thursday
a. Turn in Homework Assignment 2
b. Turn in Essay 2
c. Review Lesson 6: Momentum & Impulse
d. Test 2 on Lessons 3, 4, 5, 6
e. Mind Game 2
f. Lesson 7:
g. Essay 3: James Prescott Joule, Due 6/17
h. Homework Assignment 3: Handout. Turn in Thursday, 6/17
5. Friday: No School
Week 3, June 14-18
1. Monday
a. Return of papers
b. Running Grades
c. Review of Lesson 6 Energy, Work, Power
d. Lesson 8: Simple Harmonic Motion
e. Lab Exercise #6 Hooke’s Law
f. Lab Exercise #7: Simple Harmonic Motion
g. Essay 3: James Prescott Joule, Due 6/17
h. Homework Assignment 3: Handout. Turn in Thursday, 6/17
2. Tuesday
a. Review of Lesson 7: Simple Harmonic Motion
b. Running Grades
c. Lesson 9
d. Lab Exercise #8: Momentum
e. Essay 3: James Prescott Joule, Due 6/17
f. Homework Assignment 3: Handout. Turn in Thursday, 6/17
3. Wednesday
a. Lesson 10:
b. Lab Exercise #9: KE, PE
c. Essay 3: James Prescott Joule, Due 6/17
d. Homework Assignment 3: Handout. Turn in Thursday, 6/17
4. Thursday
a. Turn in Homework Assignment 3
b. Turn in Essay 3
c. Test 3 on Lessons 7, 8, 9, 10
d. Mind Game 3
e. Review Lesson 7:
f. Homework Assignment 4: Handout. Turn in Thursday, 6/24
g. Essay 4: Albert Einstein, due 6/24
h. Lesson 11
5. Friday: No School
Week 4, June 21-25
1. Monday:
a. Return of papers
b. Running grades
c. Questions
d. Review Lesson 9
e. Lesson 12:
f. Lab Exercise #10: Archimedes Principle
g. Homework Assignment 4: Handout. Turn in Thursday, 6/24
h. Essay 4: Albert Einstein, due 6/24
2. Tuesday
a. Return of papers
b. Running grades
c. Questions
d. Review Lesson 10
e. Lesson 13:
f. Lab Exercise 11: Waves on a string
g. Homework Assignment 4: Handout. Turn in Thursday, 6/24
h. Essay 4: Albert Einstein, due 6/24
3. Wednesday
a. Return of papers
b. Running grades
c. Review Lesson 13
d. Lesson 14
e. Homework Assignment 4: Handout. Turn in Thursday, 6/24
f. Essay 4: Albert Einstein, due 6/24
4. Thursday
a. Turn in Homework Assignment 4
b. Turn in Essay 4
c. Test 4 on Lessons 11, 12, 13, 14
d. Mind Game 4
e. Review Lesson 7:
f. Homework Set 5: Hand out, due Thursday, 7/1
g. Essay 5: Enrico Fermi, due Thursday 7/1
h. Lesson 15
5. Friday: No School
Week 5, June 28- July 2
1. Monday:
a. Return of papers
b. Running grades
c. Review Lesson 9,
d. Lesson 16:
e. Lab Exercise #10: Waves on a string
f. Homework Set 5: Hand out, due Thursday, 7/1
g. Essay 5: Enrico Fermi, due Thursday 7/1
2. Tuesday
a. Return of papers
b. Running grades
c. Review Lesson 10
d. Lab exercise 11:
e. Lesson 17:
f. Homework Set 5: Hand out, due Thursday, 7/1
g. Essay 5: Enrico Fermi, due Thursday 7/1
3. Wednesday
a. Return of papers
b. Running grades
c. Lab Exercise 12:
d. Lesson 18
e. Homework Set 5: Hand out, due Thursday, 7/1
f. Essay 5: Enrico Fermi, due Thursday 7/1
4. Thursday
a. Turn in Homework Assignment 5
b. Turn in Essay 5
c. Test 5 on Lessons 15, 16, 17, 18
d. Mind Game 5
5. Friday: No School
Week 6: July 5 - 9
1. Monday, July 5: No School
2. Tuesday, July 6: Banquet
END
Weekly Workload
Week 1, May 31 – June 4, 2010
1. Monday: No School
2. Tuesday
a. Introduction
b. Syllabus
c. Lesson 1: History of Physics, Kinematics
d. Lab Exercise #1: Density of Water
e. Homework Assignment 1: Hand out. Turn in Thursday, 6/3
f. Essay 1: Aristotle, due Thursday, 6/3
3. Wednesday
a. Review of Lesson 1
b. Lesson 2: Vector Physics
c. Lab Exercise #2: Measurement
d. Homework Assignment 1: Hand out. Turn in Thursday, 6/3
e. Essay 1: Aristotle, due Thursday, 6/3
4. Thursday
a. Turn in Homework Assignment 1
b. Turn in Essay 1 on Aristotle
c. Test 1 on Lessons 1 & 2
d. Mind Game 1
e. Lesson 3: Newton’s Laws
f. Homework Assignment 2: Handout. Turn in Thursday, 6/10
g. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
5. Friday: No School
Week 2, June 7-11
1. Monday –
a. Return of papers
b. Running Grades
c. Review of Lesson 3: Newton's Laws
d. Lab Exercise #3: Linear motion
e. Lesson 4: Circular Motion
f. Homework Assignment 2: Handout. Turn in Thursday, 6/10
g. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
2. Tuesday
a. Running Grades
b. Review of Lesson 4: Circular Motion
c. Lab Exercise #4: Centripetal Motion
d. Lesson 5: Energy, Momentum
e. Homework Assignment 2: Handout. Turn in Thursday, 6/10
f. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
3. Wednesday
a. Running Grades
b. Review of Lesson 5: Energy, Momentum
c. Lab Exercise 5: Finding Gravity
d. Lesson 6: Momentum & Impulse
e. Homework Assignment 2: Handout. Turn in Thursday, 6/10
f. Essay 2: Isaac Newton. Turn in Thursday, 6/10.
4. Thursday
a. Turn in Homework Assignment 2
b. Turn in Essay 2
c. Review Lesson 6: Momentum & Impulse
d. Test 2 on Lessons 3, 4, 5, 6
e. Mind Game 2
f. Lesson 7:
g. Essay 3: James Prescott Joule, Due 6/17
h. Homework Assignment 3: Handout. Turn in Thursday, 6/17
5. Friday: No School
Week 3, June 14-18
1. Monday
a. Return of papers
b. Running Grades
c. Review of Lesson 6 Energy, Work, Power
d. Lesson 8: Simple Harmonic Motion
e. Lab Exercise #6 Hooke’s Law
f. Lab Exercise #7: Simple Harmonic Motion
g. Essay 3: James Prescott Joule, Due 6/17
h. Homework Assignment 3: Handout. Turn in Thursday, 6/17
2. Tuesday
a. Review of Lesson 7: Simple Harmonic Motion
b. Running Grades
c. Lesson 9
d. Lab Exercise #8: Momentum
e. Essay 3: James Prescott Joule, Due 6/17
f. Homework Assignment 3: Handout. Turn in Thursday, 6/17
3. Wednesday
a. Lesson 10:
b. Lab Exercise #9: KE, PE
c. Essay 3: James Prescott Joule, Due 6/17
d. Homework Assignment 3: Handout. Turn in Thursday, 6/17
4. Thursday
a. Turn in Homework Assignment 3
b. Turn in Essay 3
c. Test 3 on Lessons 7, 8, 9, 10
d. Mind Game 3
e. Review Lesson 7:
f. Homework Assignment 4: Handout. Turn in Thursday, 6/24
g. Essay 4: Albert Einstein, due 6/24
h. Lesson 11
5. Friday: No School
Week 4, June 21-25
1. Monday:
a. Return of papers
b. Running grades
c. Questions
d. Review Lesson 9
e. Lesson 12:
f. Lab Exercise #10: Archimedes Principle
g. Homework Assignment 4: Handout. Turn in Thursday, 6/24
h. Essay 4: Albert Einstein, due 6/24
2. Tuesday
a. Return of papers
b. Running grades
c. Questions
d. Review Lesson 10
e. Lesson 13:
f. Lab Exercise 11: Waves on a string
g. Homework Assignment 4: Handout. Turn in Thursday, 6/24
h. Essay 4: Albert Einstein, due 6/24
3. Wednesday
a. Return of papers
b. Running grades
c. Review Lesson 13
d. Lesson 14
e. Homework Assignment 4: Handout. Turn in Thursday, 6/24
f. Essay 4: Albert Einstein, due 6/24
4. Thursday
a. Turn in Homework Assignment 4
b. Turn in Essay 4
c. Test 4 on Lessons 11, 12, 13, 14
d. Mind Game 4
e. Review Lesson 7:
f. Homework Set 5: Hand out, due Thursday, 7/1
g. Essay 5: Enrico Fermi, due Thursday 7/1
h. Lesson 15
5. Friday: No School
Week 5, June 28- July 2
1. Monday:
a. Return of papers
b. Running grades
c. Review Lesson 9,
d. Lesson 16:
e. Lab Exercise #10: Waves on a string
f. Homework Set 5: Hand out, due Thursday, 7/1
g. Essay 5: Enrico Fermi, due Thursday 7/1
2. Tuesday
a. Return of papers
b. Running grades
c. Review Lesson 10
d. Lab exercise 11:
e. Lesson 17:
f. Homework Set 5: Hand out, due Thursday, 7/1
g. Essay 5: Enrico Fermi, due Thursday 7/1
3. Wednesday
a. Return of papers
b. Running grades
c. Lab Exercise 12:
d. Lesson 18
e. Homework Set 5: Hand out, due Thursday, 7/1
f. Essay 5: Enrico Fermi, due Thursday 7/1
4. Thursday
a. Turn in Homework Assignment 5
b. Turn in Essay 5
c. Test 5 on Lessons 15, 16, 17, 18
d. Mind Game 5
5. Friday: No School
Week 6: July 5 - 9
1. Monday, July 5: No School
2. Tuesday, July 6: Banquet
END
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