PHYS 1100 LESSON 5 FOR
TUESDAY, AUGUST 10, 2010
I Introduction
II Logistics
III Return of tests, labs, papers; collection of overdue assignments
IV Running Grades
V Review of Lesson 4: Newton's Laws of Motion
A. Objects at rest, stay at rest; objects in motion stay in motion; UNLESS acted upon by an Fext.
B. F = m a à ties into the Universal Law of Gravity
C. Every force has an equal and opposite force, or, F1 = - F2.
D. Units of force are “Newtons”, kg m/s2. = N
F = ma
v = x/t
VI Lesson 5: Forces
A. The force of friction, Ff = mN = + mmg, which means the force of friction equals a number (m) that is called the “coefficient of friction” multiplied by the equal and opposite force to gravity, N, the Normal force; or N = mg, where m is the “mass” and g is the acceleration of gravity, “9.8 m/s2”. When I put in the “plus sign, +,” that means the Normal force is positive, whereas gravity is negative (down). “Normal” here means “perpendicular to the plane of the table,” or, up.
B. Hooke's Law, Energy, Work, Momentum
1. Hooke’s Law deals with springs (like in pens, in cars as shock absorbers, in toys like the Slinky...)
F = - k x, which means that the force, which is negative, pushes (or pulls) in the opposite direction to the external force. If I stretch a spring, it will recoil back to its original shape when I let go; if I compress a spring, it will push itself out to its original shape once I stop compressing it. The “k” is called the “spring constant.” A large k means a very strong spring, like in a shock absorber. A very small k means a very weak spring, like a Slinky. “x” means how far the spring is compressed in, or, stretched out.
In the Hooke's Law lab, we will use F = - k y instead, since we will be stretching a string downward in the negative y-direction. The force that we will apply will be a weight which will have the force of = m g. So, in the lab,
- k y = m g. We will measure “y” with a ruler, we will know the mass, and we know gravity.
2. Forms of Energy include, but are not restricted to:
· Heat
· Light
· Mechanical
· Acoustical, etc.
3. Kinetic energy (K.E.) is in motion = ½ mv2, the mass must have a speed or velocity. If it is not moving, NO K.E.
Potential energy (P.E.) is stored energy. For storing potential energy when working against the force of gravity, P.E. = mgh = (mass)(acceleration)(height above starting point).
An Example: take a 5-kg cat and carry him up a hill about 100 meters. You will have stored this much P.E. in the cat: mgh = (5kg)(- 9.8 m/s2)(100 m) ~ 5000 Joules.
If you then release (drop) the cat over the edge of a cliff that is 100 meters above where you started, 5000 Joules will be released and turn into K.E. as the cat is falling, until he impacts with the ground. The instant just before impact all the 5000 Joules have changed from P.E. to K.E. After impact, the energy then is transferred into heat (thermal energy), sound (acoustical energy), destruction (mechanical energy), and so forth.
4. The only good thing about energy is to use it to do work for us. However, nothing is 100% efficient, so the amount of work done divided by the available energy to do that work is a ratio: Work / Energy < 1, which means, less than 100%.
If I were to use a force, F, and push a mass, m, a distance, d, then the amount of work that I would do on that object would be the force times the distance moved, or F x d = energy, units 1.0 N-m == Joule
5. In chemistry, calories are used as the main concept of energy, and 4.186 calorie = 1 joule.
However, 1,000 calories, or, 1 kcal is what is called a “food calorie” or, a Calorie with a capital “C”. 1.0 Calorie = kcal.
The stuff below was NOT covered, but will be:
6. Momentum = energy of motion
p = m v
Dp = 0 in a closed system
VII Laboratory Exercise 5: Potential and Kinetic Energy
VIII HWK Assignment 2: 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292.due 8/12
IX Essay 2: Sir Isaac Newton, due 8/12
Tuesday, August 10, 2010
LAB 5
Physics Lab 5 Tuesday, August 10, 2010 Name __________________
I Title: Acceleration of Gravity
II Purpose: To determine the acceleration of gravity using simple equipment.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Weighted Ring Stand Optional
IV Procedure
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Have that same person, or another, attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “y”) as accurately as possible.
5. Use the Data Table below; one column is for the number of the trials; the other for the time (in seconds) for each cycle.
6. Have one of the lab partners pull the pendulum back, to about an angle of θ = 45° (but no further) as seen in the diagram to the right
6. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
7. Allow Bob to swing out and come back to where it was released. Stop the watch. That is one cycle. Record the time in the table. Reset the watch and get ready to repeat.
8. Have the same CroMagnon Repeat steps #6 - #8, nine more times, and place your data in the table. You should have a total of ten trials.
9. Find the average period of oscillation of the ten trials. This means, add up all the numbers in the second column, then divide by 10. Record.
10. Use the data to find the acceleration of Earth’s gravity, g, by re-writing Galileo’s Period-Length equation so that the acceleration, g, is all alone on the left side.
-MORE-
Physics Lab 5 Tuesday, August 1, 2010 page 2 Name _____________
P = 2 p [(y / g)] ½
P is the period in seconds, y is the length in meters, and g is what we want to find. It will be in meters per square second. This equation reads “Period equals two times pi times the square root of (y/g).”
Let’s continue to re-write this relationship, in order for us to get, g all alone:
[P / 2 p ] = [(y / g)] ½
[P / 2 p ]2 = (y / g)
g = y [2 p / P] 2
Remember that “y” is the length of the string (about 1.0 meter, but make sure it’s exactly measured), and “P” is the average period of time that you found by combining the 10 periods above. And, of course, p = 3.14.
V Data & Calculations:
1. Exact length of string, in meters: _______________
Table of Data
Trial # Time (seconds) Trial # Time (seconds)
1 6
2 7
3 8
4 9
5 10
Average Period of Oscillation of these ten trials (seconds)
P = __________________
YOUR acceleration of gravity, in m/s2:
g = y [2 p / P] 2 = ______________________
-MORE-
Physics Lab 5 Tuesday, August 10, 2010 page 3 Name _____________
The true value of acceleration is g0 = - 9.8 m/s2.
VI Results:
The purpose of determining the acceleration of gravity using the equipment and procedures above was or was not achieved due to: (explain in detail)
VII Error Analysis:
A. Qualitative Error:
Personal: (what did you or your partner do to screw up?)
Systematic: (what external factors happened that you could not control, e.g., broken equipment, doing the lab in a hurricane, etc.)
Random: There is always random error, unless one does multiple trials. Since we did 10 trials and took an average, there is NO random error in this lab.
B. Quantitative Error:
Find the quantitative error for this experiment: (% error). Find this by using the error analysis formula:
|[True Answer – Your Answer]| / [True Answer] x 100% = ________%
Remember, the true value of acceleration is g0 = - 9.8 m/s2.
VIII Questions
Information: The acceleration of gravity, g, for any planet, is equal to: g = GM / R2 where G is the universal constant of Gravity = 6.67 x 10-11 Nm2/kg2; M is the mass of any planet given, and R is the radius of any planet given. You are dividing GM by the square of R = R2.
1. The mass of Mars is M = 6.4 x 1023 kg; the radius of Mars is R = 3.4 x 106 meters. Find the acceleration of gravity of Mars, g♂.
2. The mass of Jupiter is M = 1.9 x 1027 kg ; the radius of Jupiter is R = 7.13 x 107 m . Find the acceleration of gravity of Jupiter, g♃.
END
I Title: Acceleration of Gravity
II Purpose: To determine the acceleration of gravity using simple equipment.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Weighted Ring Stand Optional
IV Procedure
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Have that same person, or another, attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “y”) as accurately as possible.
5. Use the Data Table below; one column is for the number of the trials; the other for the time (in seconds) for each cycle.
6. Have one of the lab partners pull the pendulum back, to about an angle of θ = 45° (but no further) as seen in the diagram to the right
6. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
7. Allow Bob to swing out and come back to where it was released. Stop the watch. That is one cycle. Record the time in the table. Reset the watch and get ready to repeat.
8. Have the same CroMagnon Repeat steps #6 - #8, nine more times, and place your data in the table. You should have a total of ten trials.
9. Find the average period of oscillation of the ten trials. This means, add up all the numbers in the second column, then divide by 10. Record.
10. Use the data to find the acceleration of Earth’s gravity, g, by re-writing Galileo’s Period-Length equation so that the acceleration, g, is all alone on the left side.
-MORE-
Physics Lab 5 Tuesday, August 1, 2010 page 2 Name _____________
P = 2 p [(y / g)] ½
P is the period in seconds, y is the length in meters, and g is what we want to find. It will be in meters per square second. This equation reads “Period equals two times pi times the square root of (y/g).”
Let’s continue to re-write this relationship, in order for us to get, g all alone:
[P / 2 p ] = [(y / g)] ½
[P / 2 p ]2 = (y / g)
g = y [2 p / P] 2
Remember that “y” is the length of the string (about 1.0 meter, but make sure it’s exactly measured), and “P” is the average period of time that you found by combining the 10 periods above. And, of course, p = 3.14.
V Data & Calculations:
1. Exact length of string, in meters: _______________
Table of Data
Trial # Time (seconds) Trial # Time (seconds)
1 6
2 7
3 8
4 9
5 10
Average Period of Oscillation of these ten trials (seconds)
P = __________________
YOUR acceleration of gravity, in m/s2:
g = y [2 p / P] 2 = ______________________
-MORE-
Physics Lab 5 Tuesday, August 10, 2010 page 3 Name _____________
The true value of acceleration is g0 = - 9.8 m/s2.
VI Results:
The purpose of determining the acceleration of gravity using the equipment and procedures above was or was not achieved due to: (explain in detail)
VII Error Analysis:
A. Qualitative Error:
Personal: (what did you or your partner do to screw up?)
Systematic: (what external factors happened that you could not control, e.g., broken equipment, doing the lab in a hurricane, etc.)
Random: There is always random error, unless one does multiple trials. Since we did 10 trials and took an average, there is NO random error in this lab.
B. Quantitative Error:
Find the quantitative error for this experiment: (% error). Find this by using the error analysis formula:
|[True Answer – Your Answer]| / [True Answer] x 100% = ________%
Remember, the true value of acceleration is g0 = - 9.8 m/s2.
VIII Questions
Information: The acceleration of gravity, g, for any planet, is equal to: g = GM / R2 where G is the universal constant of Gravity = 6.67 x 10-11 Nm2/kg2; M is the mass of any planet given, and R is the radius of any planet given. You are dividing GM by the square of R = R2.
1. The mass of Mars is M = 6.4 x 1023 kg; the radius of Mars is R = 3.4 x 106 meters. Find the acceleration of gravity of Mars, g♂.
2. The mass of Jupiter is M = 1.9 x 1027 kg ; the radius of Jupiter is R = 7.13 x 107 m . Find the acceleration of gravity of Jupiter, g♃.
END
Monday, August 9, 2010
Solution Set 2
Chapter 5: Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143
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?
Solution: At t=0, v= 0 for the mass. At some other time, t, v=? A force is applied, F. We know that m v = F t, so v = F t/m = (F/m) t = a t.
3. 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?
Solution: 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? m, F, t; x?
F = m a, then a = F/m = 10.1/12.3 = 0.82 m/s2.
V = Dx/Dt = x/t
Vi = 0
Vf = ?
Dv/Dt = a; Vf = a t = (0.82 m/s2)(2.50 s) = 2.05 m/s.= x/t
2.5(2.05) = x = 5.125 m = 5.13 m
9. In baseball a pitcher can accelerate a 0.15-kg ball from rest to 98 mi//h in a distance of 1.7 meters.
a. Find the average force exerted on the ball during the pitch.
b. If the mass of the ball is increased is the force required of the pitcher increased, decreased, or unchanged? Explain.
Solution: m, v, x; F?
98 mi/h ~ 160 km/h = 160,000 m/3600s = 44.4 m/s
F = m a = (0.15 kg)(1162 m/s2) = 174 N.
a = v/t = 98/t; t = v/a = x/v
v = x/t; t = x/v; v2 = a x
a = v2/x = (44.4)2 / 1.7 m = 1975/1.7 = 1162 m/s2.
13. A drag racer crosses the finish line doing 202 mph and promptly deploys a drag chute.
a. What force must the drag chute exert on the 891-kg car to slow it to 45.0 mph over a distance of 185 meters.
b. Describe how you came to that solution.
Solution: vi, m, vf, x, F? Vi = 202 mph ~ 320 km/h = 320,000 m/h = 320,000/3600 m/s = 88.9 m/s, and 45 mph ~ 72 km/h = 72,000 m/h = 72,000 m/3600 sec = 20 m/s.
We know that F = m a, and we have m, but must find “a.” Using vf2 – vi2 = 2 a x, we see that a = (vf2 – vi2)/2x = [(20)2 – (88.9)2]/(2)(185) = [400-7901]/(370) = - 20.27 m/s2. Thus, F = (891)(20.27 m/s2) = - 18064 N = - 1.8 x 10^3 N.
19. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They pace 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?
c. If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration?
Solution:
a. same
b. more
c. mp ap = mc ac ; mc ac / mp = ap;
(mc/mp) ac = (19/71) 2.6 m/s2 =
0.2676 (2.6 m/s2) = 0.7 m/s2.
25. 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)?
Solution: Without the tractor, the trailer would accelerate down at a = g sin (18°) = 9.8 (0.309) = 3.03 m/s2, so that means the tractor must be pulling up with an equal and opposite force, F = m a = (3700 kg)(3.03 m/s2) = 11,211 N. = 1.1 x 10^4 N. The net force is zero, as the velocity (speed) is constant.
29. A hockey puck is acted upon by one or more forces as shown in Figure 5-24 on page 142. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, starting with the smallest. Indicate ties where appropriate.
Solution: Later
31. Before practicing his routine on the rings, a 67-kg gymnast stands motionless with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.
a. if the force exerted by the rings on each arm has a magnitude of 290 N and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?
b. If the angle that his arms make with the horizontal is great than 24°, and everything else remain the same, is the force exerted by the floor on his feet greater, less, or the same as the value found in (a)? Explain.
Solution: Each arm has a y-component of (290 N)(Cos 24) = 264.93 N, so two arms has ~530 N. The 67-kg gymnast would normally have m g = (67)(9.8) = 657 N, so the difference is (657 - 530) = 127 N.
39. At the bow of ship on a stormy sea a crewman conducts an experiment by standing on a bathroom scale. In calm water, the scale reads 182 lb. During he storm the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Solution:
Chapter 6: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184
1. You push two identical bricks across a tabletop with constant speed, v, as shown figure 6-16 on page 178 in the book. The the first case, you place the bricks end to end. In the second case you stack the bricks one on top of the other.
a. Is the force of kinetic friction in the first case greater, less, or equal to the force of Kinetic friction in case 2?
b. Choose the best explanation
I. The normal force in case #2 is larger, and hence the bricks press down more firmly against the table.
II. The normal force is the same on both cases, and friction is independent of surface area.
III. The first case has more surface area in contact with the table top and this leads to more friction.
Solution:
3. 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?
Solution:
9. A tie of uniform width is laid out ion a table with a fraction of its length hanging over the edge. Initially, the tie is at rest.
a. if the friction hanging from the table is increased, the tie eventually slides to the ground. Explain.
b. what is the coefficient of static friction between the tie and table if the tie begins to slide when 1/4th of its length hangs over the edge?
Solution:
13. 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.
Solution:
19. 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?
Solution:
25. If the 52-N backpack in figure 6-19 on page 179 in the book begins to slide when the spring stretches by 2.50 cm, what is the coefficient of static friction between the backpack and the table? Let k = 150 N/m.
Solution:
29. Your friend's 13.6-g graduation tassel hangs on a string from his rear-view mirror.
a. When is accelerates from a stop, the tassel deflects backward toward the rear of the car. Explain.
b. if the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
Solution:
31. A picture hangs on a wall suspended by two strings, as shown in figure 6-21 on page 180 in the book. The tension in String 1 (on the left) is 1.7 Newtons.
a. Is the tension in String 2 greater, lesser, or the same as in String 1? Explain.
b. Calculate the tension in String 2 to verify
c. Find the weight of the picture in Newtons
Solution:
39. 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.
Solution:
53. 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?
Solution:
63. At what speed must you drive over the hump in the road as seen in figure 6-35 on page 183 in the book if your passengers are going to experience the phenomenon of weightlessness? The radius of curvature of the hump is 35 meters.
Solution:
65. If you weigh yourself on a “bathroom scale” at the equator, is it higher, lower, or the same as it would be a the North Pole? Explain.
Solution:
Chapter 7: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213
1. 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.
Solution:
3. A plumb Bob pendulum is seen in figure 7-14 on page 211 in the book.
a. Is the work done on Bob by Earth's gravity positive, negative, or zero? Explain.
Solution:
9. A towing rope, parallel to the water pulls a water skier directly behind the boat with constant velocity for a distance of 65 meters before the skier falls. The rope's tension is 120 N.
a. Is the work done on the skier positive, negative, or zero? Explain.
Solution:
13. 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.
Solution:
19. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution:
25. 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?
Solution:
29. 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.
Solution:
31. A block of mass, m, and speed, v, collides with a spring, compressing it a distance of Dx. What is the compression of the spring if the force constant of the spring is increase by 4 times?
39. 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.
Solution:
53. A certain car can accelerate from rest to a speed, v, in “t” seconds. If the power output of the car remains constant,
a. How long does it take for the car to accelerate from “v” to “2v”?
b. How fast is the car moving at a time of “2t”?
Solution:
Chapter 8: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248
1. The work done by a conservative force is indicated in figure 8-14 on page 244 in the book, is for a variety of different paths connected to the point A and B. What is the work done by this force on path 1 and on path 2?
Solution:
3. 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.
Solution:
9. As an Acapulco cliff diver drops to the water from a height of 46 meters, his gravitational potential energy decreases by 25,000 Joules. Find his weight in Newtons.
13. A vertical spring stores energy (0.962 J) as spring potential energy. When a 3.5-kg mass is suspended from it.
a. by what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?
b. Verify it.
Solution:
19. Suppose the situation descrige in Conceptual Checkpoint 8-2 (page 232 in the book) is repeated on the fictional planet Epsilon, where the acceleration due to its gravity is less than it is on Earth.
a. Would the height of a hill on Epsilon that causes a reduction in speed from 4.0 m/s to 0.0 m/s be greater, lesser, or equal to that on Earth. Explain.
b. Re-consider this Epsilonian hill. If the initial speed at the bottom of the hill is 5.0 m/s will the final speed at the top of the hill be greater, less, or equal to 3.0- m/s? Explain.
Solution:
25. At a water park, a swimmer uses a water slide to enter the main swimming pool. If the swimmer starts at the top of the slide with an initial velocity of 0.840 m/s, find the swimmer's speed at the bottom of the slide. Assume that the height of the slide is 2.31 meters and that the slide has no friction.
Solution:
29. An apple of mass 0.21 kg falls of an apple tree and lands on the ground, 4.0 meters below. Determine the apple's kinetic energy (KE), gravitational potential energy (PE), and total mechanical energy of the system (E) when the apple's height (y) is
a. 4.0 m
b. 3.0 m
c. 2.0 m
d. 1.0 m
e. 0.0 m
Ignore air friction. Also the ground is at y = 0.0 meters
Solution:
31. A rock of mass 0.26 kg is thrown “straight up” from a cliff that is 32 meters above the ground level (where y = 0.0 m). The rock rises, stops, and then is in free fall all the way to the bottom of the cliff, where its final speed, just before impact, is 29.0 m/s. Ignore air friction. For the sake of this problem, assume that the rock is thrown up from slightly beyond the cliff's edge (or else it would just hit the top of the cliff).
a. Find the initial speed of the rock (upward)
b. The maximum height of the rock (y) relative to the base of the cliff.
Solution:
39. When the Space Shuttle re-enters the Earth's atmosphere, its protective tiles get really hot.,
a. Is the mechanical energy of the Shuttle-Earth system when the Shuttle lands, greater, less, or equal to when it is in orbit?
b. Choose the best explanation from the three choices below.
I Dropping out of orbit increases the mechanical energy of the Shuttle
II Gravity is a conservative force
III A portion of the mechanical energy has been converted to heat energy.
Solution:
53. A block of mass 1.80 kg slides along a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is mk = 0.560, what is the force constant of the spring (k)?
Solution: skip this one
Chapter 9: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292
1. Exercise 9-1 on page 255 states: “An 1180-kg car drives along a city street at 30.0 miles per hour = 13.4 m/s.
(a) “What is the magnitude of the car's momentum?
(b) “A major league pitcher can give a 0.142-kg baseball a speed of 101 miles per hour (45.1 m/s). Find the magnitude of the baseball's momentum.”
The answer to (a) is pcar = 15,800 kg-m/s, where “p” stands for momentum, i.e.,
p = m v.
In this first question, it asks us, “What speed must the baseball go if its momentum were to equal that of the car?” In other words, how fast must the ball go if it, too, has a momentum of p = 15,800 kg-m/s? They want the answer in miles per hour, probably because you are still converting back and forth and they don't want you to forget. (“They” refers to the authors).
pcar = 15,800 kg-m/s and pball = (0.142 kg)(vball), so now we need to find vball.
vball = (15,800 kg-m/s) / (0.142 kg) = 111,268 m/s = 111 km/sec = 69.54 miles/sec = 250,352 mi/hour = 2.5 x 105 mi/hr.
3. A 26.2-kg dog is running northward at 2.7 m/s (it's a vector!), while a 5.30-kg cat is running eastward at 3.04 m/s (another vector!). Their 74.0-kg owner has the same momentum s the two pets taken together. Find the direction and magnitude of the owner's velocity.
Solution: Dog is 26.2 kg, 2.7 m/s @90°, cat 5.3 kg, 3.04 m/s @ 0°. Homosapien, 74 kg.
p1 = m1v1 = (26.2)(2.7) = 70.7 kgm/s @ 90°
p2 = m2v2 = (5.3)(3.04) = 16.1 kgm/s @ 0°.
pt2 = (70.7)2 + (16.1)2 = 4998.5 + 259.6 = 5258
pt = (5258)½ = 72.5 m/s;
16.1 = 72.5 cos a
cos a = (16.1)/(72.5) = 0.222
a = cos-1 (0.222) = 77°
p = 72.5 kgm/s @ 77°.
v = p/m = (72.5 kgm/s)/(74.0 kg) = 0.98 m/s = 9.8 x 10-1 m/s @ 77°.
9. A net force of F = 200 N acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble.
(a) is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum over the same time period?
(b) Choose the best explanation from among the 3 following:
I. The large mass of the boulder gives it the greater momentum
II. The force causes a much greater speed in the 100-g pebble, resulting in more momentum.
III. Equal force means equal change in momentum for a given time.
Solution: Momentum, p = m v, but it is equivalent to F Dt.
(a) for the boulder, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
for the pebble, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
Thus, the same
(b) III
13. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1,250.00 N. Assume that the player's food is in contact with the ball for 5.95 x 10-3 seconds.
Solution: Impulse is force multiplied by the small incremental change in time, I = F Dt, so in this case, I = (1250.00 N)(5.95 x 10-3 s) = 74337.5 10-3 = 74.3375 = 7.43 x 101 N-s.
19. A 0.14-kg baseball moves toward home plate with a velocity (a vector) of v = (- 36 m/s)x. After striking the bat, the ball moves vertically upward with a velocity of v = (+18 m/s)ŷ
(a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume contact between them lasted a time of Dt = 1.5 milliseconds.
(b) How would your answer to part (a) change if the mass of the ball were doubled?
(c) How would your answer to part (a) change if the mass of the bat were doubled?
Solution: Remember that I = F Dt = D(mv) = mDv. (The mass does not change).
(a) I = (0.14 kg)(- 36 – 18) = (0.14 kg)(– 54) = - 7.5 N-s.
(b) It would double.
(c) no change.
25. A 92-kg astronaut and a 1200-kg satellite are at rest relative to a Space Shuttle near by. The astronaut pushes on the satellite, give it a speed of 0.14 m/s directly away from the Shuttle. Then, 7.5 seconds later, the astronaut comes into contact with the Shuttle. What was the initial distance from the Shuttle to the Astronaut?
Solution: In this case, there is no momentum in the system, so after the astronaut pushes the satellite, the net momentum is still zero. So, m1v1 + m2v2 = 0. But m1 = 92 kg, v1 = ?; m2 = 1200 kg, v2 = - 0.14 m/s. So, let's set up a ratio type thing:
m1v1 + m2v2 = 0
m1v1 = - m2v2
v1 = - [(m2) / (m1)] v2 = - [(1200 kg) / (92 kg)] (- 0.14 m/s) = + 1.83 m/s. If the astronaut traveled 1.83 m/s for 7.5 seconds, then he traveled a distance of x = (1.83)(7.5)= 13.7 m, or 1.37 x 101 meters.
29. In Example 9-6 on page 271, a 950-kg car moving at v = 16 m/s east crashes into a 1300-kg minivan going v = 21 m/s north; they stick together and the wreckage moves off towards the general direction of northeast.
Now, let's change some stuff …. suppose the car in Example 9-6 has an initial speed of 20.0 m/s (not 16.0 m/s) and that the direction of the wreckage after the collision is 40.0° above the x-axis (counter-clockwise from the + x-axis). Find the initial speed of the minivan and the final speed of the combined wreckage.
Solution: In a “closed” system, which this is, we don't have any outside sources of speed, momentum, force, or acceleration. So, the momentum of both vehicles before the impact must equal the momentum of both vehicles (stuck together) after the collision.
pi = pc + pv = mcvc + mvvv = pf = (mc + mv)vf .And we are looking for vv.
mcvc = (950 kg)(20 m/s) = 19,000 kg-m/sec @ 0°. This is the initial momentum for the car.
mvvv = (1300 kg)(21 m/s) = 27,300 kg-m/sec @ 90°. This is the initial momentum for the van.
As the initial momenta are at right angles, then the net original momentum would be the vector addition of pc + pv = 19,000 @ 0° + 27,300 @ 90°. In this case, using Pythagoras, a = 19,000 and b = 27,300, so we need c, which we get from the Pythagorean relationship: a2 + b2 = c2, or, (1.9 x 104)2 + (2.73 x 104)2 = (c)2 ; = (3.61 x 108) + (7.45 x 108) = 11.06 x 108. Therefore, c = √(11.06 x 108) = 3.33 x 104 kg-m/s. But what is the angle, θ?
Tanθ = (27300)/(19000) = (27.3)/(19) = 1.437; so the angle is: θ = tan-1 (1.437) = 55.17°.
So, in the end, the final momentum of the wreckage is 3.33 x 104 kg-m/s @ 55.17°
However, what if the angle is 40.0°? Then that means that Tanθ = tan 40.0° = 0.839. So, pv/pc = 0.839; and pc is still 19,000 kg-m/s @ 0°. Thus, pv/pc = (pv)/(19,000) = 0.839. Or, pv = (19000)(0.839) = 15,943 kg-m/s. Since the mass of the van is 1300 kg, then the initial velocity of the mini van is: pv = mvvv = (1300 kg)(vv) = 15,943 kg-m/s, or, vv = (15,943 kg-m/s) / (1300 kg) = 12.26 m/s.
31. Exercise 9-2 on page 268 states: “A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. If the vehicles stick together after the collision, what's their speed immediately after the collision?”
The answer is vf = 5.0 m/s.
In this question, it asks us,(a) is the final KE of the car+truck greater than, less than, or equal to the sum of the initial KE's of the car and the truck separately? In other words, is KEf > KE1 + KE2, is KEf < KE1 + KE2, or is KEf = KE1 + KE2. Explain.
(b) Verify your answer to part (a) by calculating the initial and final KE's of the system.
Solution: Okay, the relationship for kinetic energy is: KE = ½ m v2.
(a) The KE of the car, or KE1 = ½ m1 v12 = (0.5)(1200)(2.5)2 = 3,750 Joules.
The KE of the truck, or KE2 = ½ m2 v22 = (0.5)(2600)(6.2)2 = 49,972 Joules.
So, KE1 + KE2 = 3750 Joules + 49,972 Joules = 53,722 Joules. But KEf = ½ (m1 + m2)vf2 = (0.5)(3800 kg)(5.0 m/s)2 = 47,500 Joules. Since 53,722 Joules > 47,500 Joules, then KEf < KE1 + KE2 or KE1 + KE2 > KEf.
(b) we did this already in part (a)
39. A charging bull elephant with a mass of M = 5240 kg comes directly towards you at ve = 4.55 m/s. In a panic, you toss a rubber ball of mass, m = 0.150 kg , at the elephant, with a speed of vb = 7.81 m/s.
(a) When the ball bounces back towards you, what is its speed?
(b) How do you account for the fact that the ball's KE has increased?
Solution:
(a) 4.55 + 7.81 = 12.36 m/s.
(b) The hefalump gave it more energy.
53. Three uniform meter sticks, each of mass, m, are placed on the floor as follows: stick 1 lies along the y-axis from (0,0) to (0,1); stick 2 lies along the x-axis from (0,0) to (1,0). Stick 3 lies along the x-axis from (1,0) to (2,0).
(a) Find the location of the center of mass of the meter sticks.
(b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?
Solution: This is the same as if we had a wooden stick of 1 meter whose center of mass is at 0.5 meters in the y-direction, and a wooden stick of 2 meters, whose center of mass is at 1.0 meters in the x-direction.
(a) So the center of mass is some where between (0, 0.5) and (1.0, 0). And in fact, it's at
(1.0, 0.5).
(b) no change
END
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?
Solution: At t=0, v= 0 for the mass. At some other time, t, v=? A force is applied, F. We know that m v = F t, so v = F t/m = (F/m) t = a t.
3. 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?
Solution: 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? m, F, t; x?
F = m a, then a = F/m = 10.1/12.3 = 0.82 m/s2.
V = Dx/Dt = x/t
Vi = 0
Vf = ?
Dv/Dt = a; Vf = a t = (0.82 m/s2)(2.50 s) = 2.05 m/s.= x/t
2.5(2.05) = x = 5.125 m = 5.13 m
9. In baseball a pitcher can accelerate a 0.15-kg ball from rest to 98 mi//h in a distance of 1.7 meters.
a. Find the average force exerted on the ball during the pitch.
b. If the mass of the ball is increased is the force required of the pitcher increased, decreased, or unchanged? Explain.
Solution: m, v, x; F?
98 mi/h ~ 160 km/h = 160,000 m/3600s = 44.4 m/s
F = m a = (0.15 kg)(1162 m/s2) = 174 N.
a = v/t = 98/t; t = v/a = x/v
v = x/t; t = x/v; v2 = a x
a = v2/x = (44.4)2 / 1.7 m = 1975/1.7 = 1162 m/s2.
13. A drag racer crosses the finish line doing 202 mph and promptly deploys a drag chute.
a. What force must the drag chute exert on the 891-kg car to slow it to 45.0 mph over a distance of 185 meters.
b. Describe how you came to that solution.
Solution: vi, m, vf, x, F? Vi = 202 mph ~ 320 km/h = 320,000 m/h = 320,000/3600 m/s = 88.9 m/s, and 45 mph ~ 72 km/h = 72,000 m/h = 72,000 m/3600 sec = 20 m/s.
We know that F = m a, and we have m, but must find “a.” Using vf2 – vi2 = 2 a x, we see that a = (vf2 – vi2)/2x = [(20)2 – (88.9)2]/(2)(185) = [400-7901]/(370) = - 20.27 m/s2. Thus, F = (891)(20.27 m/s2) = - 18064 N = - 1.8 x 10^3 N.
19. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They pace 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?
c. If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration?
Solution:
a. same
b. more
c. mp ap = mc ac ; mc ac / mp = ap;
(mc/mp) ac = (19/71) 2.6 m/s2 =
0.2676 (2.6 m/s2) = 0.7 m/s2.
25. 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)?
Solution: Without the tractor, the trailer would accelerate down at a = g sin (18°) = 9.8 (0.309) = 3.03 m/s2, so that means the tractor must be pulling up with an equal and opposite force, F = m a = (3700 kg)(3.03 m/s2) = 11,211 N. = 1.1 x 10^4 N. The net force is zero, as the velocity (speed) is constant.
29. A hockey puck is acted upon by one or more forces as shown in Figure 5-24 on page 142. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, starting with the smallest. Indicate ties where appropriate.
Solution: Later
31. Before practicing his routine on the rings, a 67-kg gymnast stands motionless with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.
a. if the force exerted by the rings on each arm has a magnitude of 290 N and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?
b. If the angle that his arms make with the horizontal is great than 24°, and everything else remain the same, is the force exerted by the floor on his feet greater, less, or the same as the value found in (a)? Explain.
Solution: Each arm has a y-component of (290 N)(Cos 24) = 264.93 N, so two arms has ~530 N. The 67-kg gymnast would normally have m g = (67)(9.8) = 657 N, so the difference is (657 - 530) = 127 N.
39. At the bow of ship on a stormy sea a crewman conducts an experiment by standing on a bathroom scale. In calm water, the scale reads 182 lb. During he storm the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Solution:
Chapter 6: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184
1. You push two identical bricks across a tabletop with constant speed, v, as shown figure 6-16 on page 178 in the book. The the first case, you place the bricks end to end. In the second case you stack the bricks one on top of the other.
a. Is the force of kinetic friction in the first case greater, less, or equal to the force of Kinetic friction in case 2?
b. Choose the best explanation
I. The normal force in case #2 is larger, and hence the bricks press down more firmly against the table.
II. The normal force is the same on both cases, and friction is independent of surface area.
III. The first case has more surface area in contact with the table top and this leads to more friction.
Solution:
3. 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?
Solution:
9. A tie of uniform width is laid out ion a table with a fraction of its length hanging over the edge. Initially, the tie is at rest.
a. if the friction hanging from the table is increased, the tie eventually slides to the ground. Explain.
b. what is the coefficient of static friction between the tie and table if the tie begins to slide when 1/4th of its length hangs over the edge?
Solution:
13. 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.
Solution:
19. 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?
Solution:
25. If the 52-N backpack in figure 6-19 on page 179 in the book begins to slide when the spring stretches by 2.50 cm, what is the coefficient of static friction between the backpack and the table? Let k = 150 N/m.
Solution:
29. Your friend's 13.6-g graduation tassel hangs on a string from his rear-view mirror.
a. When is accelerates from a stop, the tassel deflects backward toward the rear of the car. Explain.
b. if the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
Solution:
31. A picture hangs on a wall suspended by two strings, as shown in figure 6-21 on page 180 in the book. The tension in String 1 (on the left) is 1.7 Newtons.
a. Is the tension in String 2 greater, lesser, or the same as in String 1? Explain.
b. Calculate the tension in String 2 to verify
c. Find the weight of the picture in Newtons
Solution:
39. 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.
Solution:
53. 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?
Solution:
63. At what speed must you drive over the hump in the road as seen in figure 6-35 on page 183 in the book if your passengers are going to experience the phenomenon of weightlessness? The radius of curvature of the hump is 35 meters.
Solution:
65. If you weigh yourself on a “bathroom scale” at the equator, is it higher, lower, or the same as it would be a the North Pole? Explain.
Solution:
Chapter 7: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213
1. 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.
Solution:
3. A plumb Bob pendulum is seen in figure 7-14 on page 211 in the book.
a. Is the work done on Bob by Earth's gravity positive, negative, or zero? Explain.
Solution:
9. A towing rope, parallel to the water pulls a water skier directly behind the boat with constant velocity for a distance of 65 meters before the skier falls. The rope's tension is 120 N.
a. Is the work done on the skier positive, negative, or zero? Explain.
Solution:
13. 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.
Solution:
19. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution:
25. 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?
Solution:
29. 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.
Solution:
31. A block of mass, m, and speed, v, collides with a spring, compressing it a distance of Dx. What is the compression of the spring if the force constant of the spring is increase by 4 times?
39. 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.
Solution:
53. A certain car can accelerate from rest to a speed, v, in “t” seconds. If the power output of the car remains constant,
a. How long does it take for the car to accelerate from “v” to “2v”?
b. How fast is the car moving at a time of “2t”?
Solution:
Chapter 8: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248
1. The work done by a conservative force is indicated in figure 8-14 on page 244 in the book, is for a variety of different paths connected to the point A and B. What is the work done by this force on path 1 and on path 2?
Solution:
3. 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.
Solution:
9. As an Acapulco cliff diver drops to the water from a height of 46 meters, his gravitational potential energy decreases by 25,000 Joules. Find his weight in Newtons.
13. A vertical spring stores energy (0.962 J) as spring potential energy. When a 3.5-kg mass is suspended from it.
a. by what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?
b. Verify it.
Solution:
19. Suppose the situation descrige in Conceptual Checkpoint 8-2 (page 232 in the book) is repeated on the fictional planet Epsilon, where the acceleration due to its gravity is less than it is on Earth.
a. Would the height of a hill on Epsilon that causes a reduction in speed from 4.0 m/s to 0.0 m/s be greater, lesser, or equal to that on Earth. Explain.
b. Re-consider this Epsilonian hill. If the initial speed at the bottom of the hill is 5.0 m/s will the final speed at the top of the hill be greater, less, or equal to 3.0- m/s? Explain.
Solution:
25. At a water park, a swimmer uses a water slide to enter the main swimming pool. If the swimmer starts at the top of the slide with an initial velocity of 0.840 m/s, find the swimmer's speed at the bottom of the slide. Assume that the height of the slide is 2.31 meters and that the slide has no friction.
Solution:
29. An apple of mass 0.21 kg falls of an apple tree and lands on the ground, 4.0 meters below. Determine the apple's kinetic energy (KE), gravitational potential energy (PE), and total mechanical energy of the system (E) when the apple's height (y) is
a. 4.0 m
b. 3.0 m
c. 2.0 m
d. 1.0 m
e. 0.0 m
Ignore air friction. Also the ground is at y = 0.0 meters
Solution:
31. A rock of mass 0.26 kg is thrown “straight up” from a cliff that is 32 meters above the ground level (where y = 0.0 m). The rock rises, stops, and then is in free fall all the way to the bottom of the cliff, where its final speed, just before impact, is 29.0 m/s. Ignore air friction. For the sake of this problem, assume that the rock is thrown up from slightly beyond the cliff's edge (or else it would just hit the top of the cliff).
a. Find the initial speed of the rock (upward)
b. The maximum height of the rock (y) relative to the base of the cliff.
Solution:
39. When the Space Shuttle re-enters the Earth's atmosphere, its protective tiles get really hot.,
a. Is the mechanical energy of the Shuttle-Earth system when the Shuttle lands, greater, less, or equal to when it is in orbit?
b. Choose the best explanation from the three choices below.
I Dropping out of orbit increases the mechanical energy of the Shuttle
II Gravity is a conservative force
III A portion of the mechanical energy has been converted to heat energy.
Solution:
53. A block of mass 1.80 kg slides along a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is mk = 0.560, what is the force constant of the spring (k)?
Solution: skip this one
Chapter 9: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292
1. Exercise 9-1 on page 255 states: “An 1180-kg car drives along a city street at 30.0 miles per hour = 13.4 m/s.
(a) “What is the magnitude of the car's momentum?
(b) “A major league pitcher can give a 0.142-kg baseball a speed of 101 miles per hour (45.1 m/s). Find the magnitude of the baseball's momentum.”
The answer to (a) is pcar = 15,800 kg-m/s, where “p” stands for momentum, i.e.,
p = m v.
In this first question, it asks us, “What speed must the baseball go if its momentum were to equal that of the car?” In other words, how fast must the ball go if it, too, has a momentum of p = 15,800 kg-m/s? They want the answer in miles per hour, probably because you are still converting back and forth and they don't want you to forget. (“They” refers to the authors).
pcar = 15,800 kg-m/s and pball = (0.142 kg)(vball), so now we need to find vball.
vball = (15,800 kg-m/s) / (0.142 kg) = 111,268 m/s = 111 km/sec = 69.54 miles/sec = 250,352 mi/hour = 2.5 x 105 mi/hr.
3. A 26.2-kg dog is running northward at 2.7 m/s (it's a vector!), while a 5.30-kg cat is running eastward at 3.04 m/s (another vector!). Their 74.0-kg owner has the same momentum s the two pets taken together. Find the direction and magnitude of the owner's velocity.
Solution: Dog is 26.2 kg, 2.7 m/s @90°, cat 5.3 kg, 3.04 m/s @ 0°. Homosapien, 74 kg.
p1 = m1v1 = (26.2)(2.7) = 70.7 kgm/s @ 90°
p2 = m2v2 = (5.3)(3.04) = 16.1 kgm/s @ 0°.
pt2 = (70.7)2 + (16.1)2 = 4998.5 + 259.6 = 5258
pt = (5258)½ = 72.5 m/s;
16.1 = 72.5 cos a
cos a = (16.1)/(72.5) = 0.222
a = cos-1 (0.222) = 77°
p = 72.5 kgm/s @ 77°.
v = p/m = (72.5 kgm/s)/(74.0 kg) = 0.98 m/s = 9.8 x 10-1 m/s @ 77°.
9. A net force of F = 200 N acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble.
(a) is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum over the same time period?
(b) Choose the best explanation from among the 3 following:
I. The large mass of the boulder gives it the greater momentum
II. The force causes a much greater speed in the 100-g pebble, resulting in more momentum.
III. Equal force means equal change in momentum for a given time.
Solution: Momentum, p = m v, but it is equivalent to F Dt.
(a) for the boulder, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
for the pebble, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
Thus, the same
(b) III
13. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1,250.00 N. Assume that the player's food is in contact with the ball for 5.95 x 10-3 seconds.
Solution: Impulse is force multiplied by the small incremental change in time, I = F Dt, so in this case, I = (1250.00 N)(5.95 x 10-3 s) = 74337.5 10-3 = 74.3375 = 7.43 x 101 N-s.
19. A 0.14-kg baseball moves toward home plate with a velocity (a vector) of v = (- 36 m/s)x. After striking the bat, the ball moves vertically upward with a velocity of v = (+18 m/s)ŷ
(a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume contact between them lasted a time of Dt = 1.5 milliseconds.
(b) How would your answer to part (a) change if the mass of the ball were doubled?
(c) How would your answer to part (a) change if the mass of the bat were doubled?
Solution: Remember that I = F Dt = D(mv) = mDv. (The mass does not change).
(a) I = (0.14 kg)(- 36 – 18) = (0.14 kg)(– 54) = - 7.5 N-s.
(b) It would double.
(c) no change.
25. A 92-kg astronaut and a 1200-kg satellite are at rest relative to a Space Shuttle near by. The astronaut pushes on the satellite, give it a speed of 0.14 m/s directly away from the Shuttle. Then, 7.5 seconds later, the astronaut comes into contact with the Shuttle. What was the initial distance from the Shuttle to the Astronaut?
Solution: In this case, there is no momentum in the system, so after the astronaut pushes the satellite, the net momentum is still zero. So, m1v1 + m2v2 = 0. But m1 = 92 kg, v1 = ?; m2 = 1200 kg, v2 = - 0.14 m/s. So, let's set up a ratio type thing:
m1v1 + m2v2 = 0
m1v1 = - m2v2
v1 = - [(m2) / (m1)] v2 = - [(1200 kg) / (92 kg)] (- 0.14 m/s) = + 1.83 m/s. If the astronaut traveled 1.83 m/s for 7.5 seconds, then he traveled a distance of x = (1.83)(7.5)= 13.7 m, or 1.37 x 101 meters.
29. In Example 9-6 on page 271, a 950-kg car moving at v = 16 m/s east crashes into a 1300-kg minivan going v = 21 m/s north; they stick together and the wreckage moves off towards the general direction of northeast.
Now, let's change some stuff …. suppose the car in Example 9-6 has an initial speed of 20.0 m/s (not 16.0 m/s) and that the direction of the wreckage after the collision is 40.0° above the x-axis (counter-clockwise from the + x-axis). Find the initial speed of the minivan and the final speed of the combined wreckage.
Solution: In a “closed” system, which this is, we don't have any outside sources of speed, momentum, force, or acceleration. So, the momentum of both vehicles before the impact must equal the momentum of both vehicles (stuck together) after the collision.
pi = pc + pv = mcvc + mvvv = pf = (mc + mv)vf .And we are looking for vv.
mcvc = (950 kg)(20 m/s) = 19,000 kg-m/sec @ 0°. This is the initial momentum for the car.
mvvv = (1300 kg)(21 m/s) = 27,300 kg-m/sec @ 90°. This is the initial momentum for the van.
As the initial momenta are at right angles, then the net original momentum would be the vector addition of pc + pv = 19,000 @ 0° + 27,300 @ 90°. In this case, using Pythagoras, a = 19,000 and b = 27,300, so we need c, which we get from the Pythagorean relationship: a2 + b2 = c2, or, (1.9 x 104)2 + (2.73 x 104)2 = (c)2 ; = (3.61 x 108) + (7.45 x 108) = 11.06 x 108. Therefore, c = √(11.06 x 108) = 3.33 x 104 kg-m/s. But what is the angle, θ?
Tanθ = (27300)/(19000) = (27.3)/(19) = 1.437; so the angle is: θ = tan-1 (1.437) = 55.17°.
So, in the end, the final momentum of the wreckage is 3.33 x 104 kg-m/s @ 55.17°
However, what if the angle is 40.0°? Then that means that Tanθ = tan 40.0° = 0.839. So, pv/pc = 0.839; and pc is still 19,000 kg-m/s @ 0°. Thus, pv/pc = (pv)/(19,000) = 0.839. Or, pv = (19000)(0.839) = 15,943 kg-m/s. Since the mass of the van is 1300 kg, then the initial velocity of the mini van is: pv = mvvv = (1300 kg)(vv) = 15,943 kg-m/s, or, vv = (15,943 kg-m/s) / (1300 kg) = 12.26 m/s.
31. Exercise 9-2 on page 268 states: “A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. If the vehicles stick together after the collision, what's their speed immediately after the collision?”
The answer is vf = 5.0 m/s.
In this question, it asks us,(a) is the final KE of the car+truck greater than, less than, or equal to the sum of the initial KE's of the car and the truck separately? In other words, is KEf > KE1 + KE2, is KEf < KE1 + KE2, or is KEf = KE1 + KE2. Explain.
(b) Verify your answer to part (a) by calculating the initial and final KE's of the system.
Solution: Okay, the relationship for kinetic energy is: KE = ½ m v2.
(a) The KE of the car, or KE1 = ½ m1 v12 = (0.5)(1200)(2.5)2 = 3,750 Joules.
The KE of the truck, or KE2 = ½ m2 v22 = (0.5)(2600)(6.2)2 = 49,972 Joules.
So, KE1 + KE2 = 3750 Joules + 49,972 Joules = 53,722 Joules. But KEf = ½ (m1 + m2)vf2 = (0.5)(3800 kg)(5.0 m/s)2 = 47,500 Joules. Since 53,722 Joules > 47,500 Joules, then KEf < KE1 + KE2 or KE1 + KE2 > KEf.
(b) we did this already in part (a)
39. A charging bull elephant with a mass of M = 5240 kg comes directly towards you at ve = 4.55 m/s. In a panic, you toss a rubber ball of mass, m = 0.150 kg , at the elephant, with a speed of vb = 7.81 m/s.
(a) When the ball bounces back towards you, what is its speed?
(b) How do you account for the fact that the ball's KE has increased?
Solution:
(a) 4.55 + 7.81 = 12.36 m/s.
(b) The hefalump gave it more energy.
53. Three uniform meter sticks, each of mass, m, are placed on the floor as follows: stick 1 lies along the y-axis from (0,0) to (0,1); stick 2 lies along the x-axis from (0,0) to (1,0). Stick 3 lies along the x-axis from (1,0) to (2,0).
(a) Find the location of the center of mass of the meter sticks.
(b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?
Solution: This is the same as if we had a wooden stick of 1 meter whose center of mass is at 0.5 meters in the y-direction, and a wooden stick of 2 meters, whose center of mass is at 1.0 meters in the x-direction.
(a) So the center of mass is some where between (0, 0.5) and (1.0, 0). And in fact, it's at
(1.0, 0.5).
(b) no change
END
Sunday, August 8, 2010
LAB EXERCISE 4
PhysicsLab4, Monday, August 9, 2010 NAME __________________
Dr Dave Menke, Instructor
I Title: Centripetal Acceleration
II Purpose: To study centripetal acceleration and have fun
Theory: Planets, like Earth, travel around the Sun similar to how a weight on a string travels in a circular path if you swing it around. For the Sun and the planets, there is no “string,” but the force is Gravity. The planets are like weights. Each planet has a velocity or speed and an acceleration. You will notice that the force in this lab is a central force, so that the acceleration is a central one, i.e., ac.
III Equipment
- White String
- Metal weight (washer, nut, whatever)
- Wooden metric ruler
- stopwatch
- Scissors
IV Procedure
1. Select a weight
2. Obtain approximately a 1.0-meter length of string
3. Attach the weight to one end of the string, just like before
4. Suspend the (string + weight) by holding the top of the string tightly at one end. This is your friend, “plumb bob.” He’s related to Sponge Bob, Bob the Builder, and 97.5 Bob FM. He’s the same guy as last time,
5. Measure exactly the length, l, of Bob (from your fingers to the middle of the weight) in meters.
6. Leave the classroom and to find an open area reasonably clear of muggles*.
7. Have one lab partner to practice - carefully - swinging Bob in a circle until he/she/it has achieved a relative constant velocity. Don’t hit anyone. Some students swing it overhead, like a lasso. It is not likely that you will hit anyone who is walking on the ceiling.
8. Have another lab partner practice using the stop watch.
9. When ready, have the swinging partner (SP) begin swinging Bob in circles at a constant rate. When ready, have the stopwatcher lab partner (SWLP) click the stop watch and count 10 cycles, then have the SWLP stop the watch. Record. The SP can keep swinging or not. Personal preference.
10. Repeat this three times to get an average amount of time for each 10-cycle period. Record. Now stop the SP if he/she/it hasn’t already.
*muggle (1) common, ordinary, ignorant person; (2) someone with NO magical powers – from the Harry Potter series of books; (3) a marijuana “joint” – from the 1920’s New Orleans
More…
Lab 3, page 2, August 9 Name _____
11. Return to the classroom, and encourage the SP and SWLP to join you. Put away your toys, and write up your report.
12. Divide your average cycle time by 10 to get the period, P, of one cycle. Record.
13. Find the circumference, c, of the orbital path. Do this by multiplying Bob’s length that you found in #5, l, by the number 2 pi or 2p = 2(3.14). Record.
14. Calculate the average linear velocity, v, of the mass. Do this by dividing the circumference that you found in #13 by the period (time) that you found in #12. Record.
15. Calculate the mean centripetal acceleration, ac, of the mass. Do this by squaring the velocity, v2 (multiply it by itself) that you found in #14 and dividing that by the length of the string that you find in #5, l. Record.
V Data & Calculations (This is where you put your data)
1. Bob’s length, l, in meters: _____________________
2. Trials and Times
TRIAL NUMBER of 10 Cycles TIME IN SECONDS of each 10 Cycles
1
2
3
AVE
3. Period of one cycle, P (divide the average of 10 cycles by 10) ______ s
4. The circumference of the orbital path, 2 p l = ______________ m
5. The average linear velocity of the mass, v = _______________m/s
6. The mean centripetal acceleration of the mass, ac = __________m/s2
VI Results
“The purpose of the lab was to go Bob-Bob-Bobbin’ along.” No, for “reals” it was to study orbital revolutions and have fun, and it (was, was not) [circle one] achieved because …
More…
Lab 3, page 3, August 9 Name _____
VII Error Analysis
A. Quantitative Error – NA
B. Qualitative Error:
1. Personal
2. Systematic
3. Random
VIII Questions
1. Find the circumference of Earth’s orbit around Sun (in meters) if Bob’s length, l, (the radius of Earth’s orbit) is 150,000,000 km, just like you did in Procedure #13 above.
2. Find the period of the Earth’s orbit (in seconds). Do this by multiplying the number of seconds in a day, 86,400, by the number of days in a year, 365.
3. Find the linear velocity of Earth (in m/s). Do this by dividing what you found in Question #1 with what you found in Question #2.
4. Find the centripetal acceleration of Earth around the Sun (in m/s2). Do this by squaring the velocity that you found in Question #3 and then dividing it with the radius of Earth’s orbit, 150,000,000 km.
5. There is no number 5.
Dr Dave Menke, Instructor
I Title: Centripetal Acceleration
II Purpose: To study centripetal acceleration and have fun
Theory: Planets, like Earth, travel around the Sun similar to how a weight on a string travels in a circular path if you swing it around. For the Sun and the planets, there is no “string,” but the force is Gravity. The planets are like weights. Each planet has a velocity or speed and an acceleration. You will notice that the force in this lab is a central force, so that the acceleration is a central one, i.e., ac.
III Equipment
- White String
- Metal weight (washer, nut, whatever)
- Wooden metric ruler
- stopwatch
- Scissors
IV Procedure
1. Select a weight
2. Obtain approximately a 1.0-meter length of string
3. Attach the weight to one end of the string, just like before
4. Suspend the (string + weight) by holding the top of the string tightly at one end. This is your friend, “plumb bob.” He’s related to Sponge Bob, Bob the Builder, and 97.5 Bob FM. He’s the same guy as last time,
5. Measure exactly the length, l, of Bob (from your fingers to the middle of the weight) in meters.
6. Leave the classroom and to find an open area reasonably clear of muggles*.
7. Have one lab partner to practice - carefully - swinging Bob in a circle until he/she/it has achieved a relative constant velocity. Don’t hit anyone. Some students swing it overhead, like a lasso. It is not likely that you will hit anyone who is walking on the ceiling.
8. Have another lab partner practice using the stop watch.
9. When ready, have the swinging partner (SP) begin swinging Bob in circles at a constant rate. When ready, have the stopwatcher lab partner (SWLP) click the stop watch and count 10 cycles, then have the SWLP stop the watch. Record. The SP can keep swinging or not. Personal preference.
10. Repeat this three times to get an average amount of time for each 10-cycle period. Record. Now stop the SP if he/she/it hasn’t already.
*muggle (1) common, ordinary, ignorant person; (2) someone with NO magical powers – from the Harry Potter series of books; (3) a marijuana “joint” – from the 1920’s New Orleans
More…
Lab 3, page 2, August 9 Name _____
11. Return to the classroom, and encourage the SP and SWLP to join you. Put away your toys, and write up your report.
12. Divide your average cycle time by 10 to get the period, P, of one cycle. Record.
13. Find the circumference, c, of the orbital path. Do this by multiplying Bob’s length that you found in #5, l, by the number 2 pi or 2p = 2(3.14). Record.
14. Calculate the average linear velocity, v, of the mass. Do this by dividing the circumference that you found in #13 by the period (time) that you found in #12. Record.
15. Calculate the mean centripetal acceleration, ac, of the mass. Do this by squaring the velocity, v2 (multiply it by itself) that you found in #14 and dividing that by the length of the string that you find in #5, l. Record.
V Data & Calculations (This is where you put your data)
1. Bob’s length, l, in meters: _____________________
2. Trials and Times
TRIAL NUMBER of 10 Cycles TIME IN SECONDS of each 10 Cycles
1
2
3
AVE
3. Period of one cycle, P (divide the average of 10 cycles by 10) ______ s
4. The circumference of the orbital path, 2 p l = ______________ m
5. The average linear velocity of the mass, v = _______________m/s
6. The mean centripetal acceleration of the mass, ac = __________m/s2
VI Results
“The purpose of the lab was to go Bob-Bob-Bobbin’ along.” No, for “reals” it was to study orbital revolutions and have fun, and it (was, was not) [circle one] achieved because …
More…
Lab 3, page 3, August 9 Name _____
VII Error Analysis
A. Quantitative Error – NA
B. Qualitative Error:
1. Personal
2. Systematic
3. Random
VIII Questions
1. Find the circumference of Earth’s orbit around Sun (in meters) if Bob’s length, l, (the radius of Earth’s orbit) is 150,000,000 km, just like you did in Procedure #13 above.
2. Find the period of the Earth’s orbit (in seconds). Do this by multiplying the number of seconds in a day, 86,400, by the number of days in a year, 365.
3. Find the linear velocity of Earth (in m/s). Do this by dividing what you found in Question #1 with what you found in Question #2.
4. Find the centripetal acceleration of Earth around the Sun (in m/s2). Do this by squaring the velocity that you found in Question #3 and then dividing it with the radius of Earth’s orbit, 150,000,000 km.
5. There is no number 5.
HOMEWORK SET 2
Chapter 5: Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143
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?
3. 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? m, F, t; x?
9. In baseball a pitcher can accelerate a 0.15-kg ball from rest to 98 mi/h in a distance of 1.7 meters. m, v, x; F?
a. Find the average force exerted on the ball during the pitch.
b. If the mass of the ball is increased is the force required of the pitcher increased, decreased, or unchanged? Explain.
13. A drag racer crosses the finish line doing 202 mph and promptly deploys a drag chute.
a. What force must the drag chute exert on the 891-kg car to slow it to 45.0 mph over a distance of 185 meters?
b. Describe how you came to that solution.
19. 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?
c. If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration?
25. 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).
29. A hockey puck is acted upon by one or more forces as shown in Figure 5-24 on page 142. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, staring with the smallest. Indicate ties where appropriate.
31. Before practicing his routine on the rings, a 67-kg gymnast stands motionless with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.
a. if the force exerted by the rings on each arm has a magnitude of 290 N and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?
b. If the angle that his arms make with the horizontal is great than 24°, and everything else remain the same, is the force exerted by the floor on his feet greater, less, or the same as the value found in (a)? Explain.
39. At the bow of ship on a stormy sea a crewman conducts an experiment by standing on a bathroom scale. In calm water, the scale reads 182 lb. During the storm the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Chapter 6: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184
1. You push two identical bricks across a tabletop with constant speed, v, as shown figure 6-16 on page 178 in the book. In the first case, you place the bricks end to end. In the second case you stack the bricks one on top of the other.
a. Is the force of kinetic friction in the first case greater, less, or equal to the force of Kinetic friction in case 2?
b. Choose the best explanation
I. The normal force in case #2 is larger, and hence the bricks press down more firmly against the table.
II. The normal force is the same on both cases, and friction is independent of surface area.
III. The first case has more surface area in contact with the table top and this leads to more friction.
3. 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 ground is 0.46, how far does the player slide before coming to rest?
9. A tie of uniform width is laid out on a table with a fraction of its length hanging over the edge. Initially, the tie is at rest.
a. if the friction hanging from the table is increased, the tie eventually slides to the ground. Explain.
b. what is the coefficient of static friction between the tie and table if the tie begins to slide when 1/4th of its length hangs over the edge?
13. 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.
19. 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?
25. If the 52-N backpack in figure 6-19 on page 179 in the book begins to slide when the spring stretches by 2.50 cm, what is the coefficient of static friction between the backpack and the table? Let k = 150 N/m.
29. Your friend's 13.6-g graduation tassel hangs on a string from his rear-view mirror.
a. When is accelerates from a stop, the tassel deflects backward toward the rear of the car. Explain.
b. if the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
31. A picture hangs on a wall suspended by two strings, as shown in figure 6-21 on page 180 in the book. The tension in String 1 (on the left) is 1.7 Newtons.
a. Is the tension in String 2 greater, lesser, or the same as in String 1? Explain.
b. Calculate the tension in String 2 to verify
c. Find the weight of the picture in Newtons
39. 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.
53. A car is driven with a constant speed around a circular track. Answer each of these following questions 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?
63. At what speed must you drive over the hump in the road as seen in figure 6-35 on page 183 in the book if your passengers are going to experience the phenomenon of weightlessness? The radius of curvature of the hump is 35 meters.
65. If you weigh yourself on a “bathroom scale” at the equator, is it higher, lower, or the same as it would be at the North Pole? Explain.
Chapter 7: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213
1. 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.
3. A plumb Bob pendulum is seen in figure 7-14 on page 211 in the book.
a. Is the work done on Bob by Earth's gravity positive, negative, or zero? Explain.
9. A towing rope, parallel to the water pulls a water skier directly behind the boat with constant velocity for a distance of 65 meters before the skier falls. The rope's tension is 120 N.
a. Is the work done on the skier positive, negative, or zero? Explain.
13. 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.
19. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
25. 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?
29. 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.
31. A block of mass, m, and speed, v, collides with a spring, compressing it a distance of Dx. What is the compression of the spring if the force constant of the spring is increase by 4 times?
39. 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.
53. A certain car can accelerate from rest to a speed, v, in “t” seconds. If the power output of the car remains constant,
a. How long does it take for the car to accelerate from “v” to “2v”?
b. How fast is the car moving at a time of “2t”?
Chapter 8: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248
1. The work done by a conservative force is indicated in figure 8-14 on page 244 in the book, is for a variety of different paths connected to the point A and B. What is the work done by this force on path 1 and on path 2?
3. 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.
9. As an Acapulco cliff diver drops to the water from a height of 46 meters, his gravitational potential energy decreases by 25,000 Joules. Find his weight in Newtons.
13. A vertical spring stores energy (0.962 J) as spring potential energy. When a 3.5-kg mass is suspended from it.
a. by what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?
b. Verify it.
19. Suppose the situation describe in Conceptual Checkpoint 8-2 (page 232 in the book) is repeated on the fictional planet Epsilon, where the acceleration due to its gravity is less than it is on Earth.
a. Would the height of a hill on Epsilon that causes a reduction in speed from 4.0 m/s to 0.0 m/s be greater, lesser, or equal to that on Earth? Explain.
b. Re-consider this Epsilonian hill. If the initial speed at the bottom of the hill is 5.0 m/s will the final speed at the top of the hill be greater, less, or equal to 3.0- m/s? Explain.
25. At a water park, a swimmer uses a water slide to enter the main swimming pool. If the swimmer starts at the top of the slide with an initial velocity of 0.840 m/s, find the swimmer's speed at the bottom of the slide. Assume that the height of the slide is 2.31 meters and that the slide has no friction.
29. An apple of mass 0.21 kg falls of an apple tree and lands on the ground, 4.0 meters below. Determine the apple's kinetic energy (KE), gravitational potential energy (PE), and total mechanical energy of the system (E) when the apple's height (y) is
a. 4.0 m
b. 3.0 m
c. 2.0 m
d. 1.0 m
e. 0.0 m
Ignore air friction. Also the ground is at y = 0.0 meters
31. A rock of mass 0.26 kg is thrown “straight up” from a cliff that is 32 meters above the ground level (where y = 0.0 m). The rock rises, stops, and then is in free fall all the way to the bottom of the cliff, where its final speed, just before impact, is 29.0 m/s. Ignore air friction. For the sake of this problem, assume that the rock is thrown up from slightly beyond the cliff's edge (or else it would just hit the top of the cliff).
a. Find the initial speed of the rock (upward)
b. The maximum height of the rock (y) relative to the base of the cliff.
39. When the Space Shuttle re-enters the Earth's atmosphere, its protective tiles get really hot.
a. Is the mechanical energy of the Shuttle-Earth system when the Shuttle lands, greater, less, or equal to when it is in orbit?
b. Choose the best explanation from the three choices below.
I Dropping out of orbit increases the mechanical energy of the Shuttle
II Gravity is a conservative force
III A portion of the mechanical energy has been converted to heat energy.
53. A block of mass 1.80 kg slides along a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is mk = 0.560, what is the force constant of the spring (k)?
Chapter 9: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292
1. Exercise 9-1 on page 255 states: “An 1180-kg car drives along a city street at 30.0 miles per hour = 13.4 m/s.
(a) “What is the magnitude of the car's momentum?
(b) “A major league pitcher can give a 0.142-kg baseball a speed of 101 miles per hour (45.1 m/s). Find the magnitude of the baseball's momentum.”
The answer to (a) is pcar = 15,800 kg-m/s, where “p” stands for momentum, i.e., p = m v.
In this first question, it asks us, “What speed must the baseball go if its momentum were to equal that of the car?” In other words, how fast must the ball go if it, too, has a momentum of p = 15,800 kg-m/s? They want the answer in miles per hour, probably because you are still converting back and forth and they don't want you to forget. (“They” refers to the authors).
3. A 26.2-kg dog is running northward at 2.7 m/s (it's a vector!), while a 5.30-kg cat is running eastward at 3.04 m/s (another vector!). Their 74.0-kg owner has the same momentum s the two pets taken together. Find the direction and magnitude of the owner's velocity.
9. A net force of F = 200 N acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble.
(a) is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum over the same time period?
(b) Choose the best explanation from among the 3 following:
I. The large mass of the boulder gives it the greater momentum
II. The force causes a much greater speed in the 100-g pebble, resulting in more momentum.
III. Equal force means equal change in momentum for a given time.
13. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1,250.00 N. Assume that the player's food is in contact with the ball for 5.95 x 10-3 seconds.
19. A 0.14-kg baseball moves toward home plate with a velocity (a vector) of v = (- 36 m/s)x. After striking the bat, the ball moves vertically upward with a velocity of v = (+18 m/s)ŷ
(a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume contact between them lasted a time of Dt = 1.5 milliseconds.
(b) How would your answer to part (a) change if the mass of the ball were doubled?
(c) How would your answer to part (a) change if the mass of the bat were doubled?
25. A 92-kg astronaut and a 1200-kg satellite are at rest relative to a Space Shuttle near by. The astronaut pushes on the satellite, give it a speed of 0.14 m/s directly away from the Shuttle. Then, 7.5 seconds later, the astronaut comes into contact with the Shuttle. What was the initial distance from the Shuttle to the Astronaut?
29. In Example 9-6 on page 271, a 950-kg car moving at v = 16 m/s east crashes into a 1300-kg minivan going v = 21 m/s north; they stick together and the wreckage moves off towards the general direction of northeast. Now, let's change some stuff …. suppose the car in Example 9-6 has an initial speed of 20.0 m/s (not 16.0 m/s) and that the direction of the wreckage after the collision is 40.0° above the x-axis (counter-clockwise from the + x-axis). Find the initial speed of the minivan and the final speed of the combined wreckage.
31. Exercise 9-2 on page 268 states: “A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. If the vehicles stick together after the collision, what's their speed immediately after the collision?”
The answer is vf = 5.0 m/s.
In this question, it asks us,(a) is the final KE of the car+truck greater than, less than, or equal to the sum of the initial KE's of the car and the truck separately? In other words, is KEf > KE1 + KE2, is KEf < KE1 + KE2, or is KEf = KE1 + KE2. Explain.
(b) Verify your answer to part (a) by calculating the initial and final KE's of the system.
39. A charging bull elephant with a mass of M = 5240 kg comes directly towards you at ve = 4.55 m/s. In a panic, you toss a rubber ball of mass, m = 0.150 kg , at the elephant, with a speed of vb = 7.81 m/s.
(a) When the ball bounces back towards you, what is its speed?
(b) How do you account for the fact that the ball's KE has increased?
53. Three uniform meter sticks, each of mass, m, are placed on the floor as follows: stick 1 lies along the y-axis from (0,0) to (0,1); stick 2 lies along the x-axis from (0,0) to (1,0). Stick 3 lies along the x-axis from (1,0) to (2,0).
(a) Find the location of the center of mass of the meter sticks.
(b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?
END
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?
3. 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? m, F, t; x?
9. In baseball a pitcher can accelerate a 0.15-kg ball from rest to 98 mi/h in a distance of 1.7 meters. m, v, x; F?
a. Find the average force exerted on the ball during the pitch.
b. If the mass of the ball is increased is the force required of the pitcher increased, decreased, or unchanged? Explain.
13. A drag racer crosses the finish line doing 202 mph and promptly deploys a drag chute.
a. What force must the drag chute exert on the 891-kg car to slow it to 45.0 mph over a distance of 185 meters?
b. Describe how you came to that solution.
19. 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?
c. If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration?
25. 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).
29. A hockey puck is acted upon by one or more forces as shown in Figure 5-24 on page 142. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, staring with the smallest. Indicate ties where appropriate.
31. Before practicing his routine on the rings, a 67-kg gymnast stands motionless with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.
a. if the force exerted by the rings on each arm has a magnitude of 290 N and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?
b. If the angle that his arms make with the horizontal is great than 24°, and everything else remain the same, is the force exerted by the floor on his feet greater, less, or the same as the value found in (a)? Explain.
39. At the bow of ship on a stormy sea a crewman conducts an experiment by standing on a bathroom scale. In calm water, the scale reads 182 lb. During the storm the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Chapter 6: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184
1. You push two identical bricks across a tabletop with constant speed, v, as shown figure 6-16 on page 178 in the book. In the first case, you place the bricks end to end. In the second case you stack the bricks one on top of the other.
a. Is the force of kinetic friction in the first case greater, less, or equal to the force of Kinetic friction in case 2?
b. Choose the best explanation
I. The normal force in case #2 is larger, and hence the bricks press down more firmly against the table.
II. The normal force is the same on both cases, and friction is independent of surface area.
III. The first case has more surface area in contact with the table top and this leads to more friction.
3. 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 ground is 0.46, how far does the player slide before coming to rest?
9. A tie of uniform width is laid out on a table with a fraction of its length hanging over the edge. Initially, the tie is at rest.
a. if the friction hanging from the table is increased, the tie eventually slides to the ground. Explain.
b. what is the coefficient of static friction between the tie and table if the tie begins to slide when 1/4th of its length hangs over the edge?
13. 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.
19. 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?
25. If the 52-N backpack in figure 6-19 on page 179 in the book begins to slide when the spring stretches by 2.50 cm, what is the coefficient of static friction between the backpack and the table? Let k = 150 N/m.
29. Your friend's 13.6-g graduation tassel hangs on a string from his rear-view mirror.
a. When is accelerates from a stop, the tassel deflects backward toward the rear of the car. Explain.
b. if the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
31. A picture hangs on a wall suspended by two strings, as shown in figure 6-21 on page 180 in the book. The tension in String 1 (on the left) is 1.7 Newtons.
a. Is the tension in String 2 greater, lesser, or the same as in String 1? Explain.
b. Calculate the tension in String 2 to verify
c. Find the weight of the picture in Newtons
39. 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.
53. A car is driven with a constant speed around a circular track. Answer each of these following questions 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?
63. At what speed must you drive over the hump in the road as seen in figure 6-35 on page 183 in the book if your passengers are going to experience the phenomenon of weightlessness? The radius of curvature of the hump is 35 meters.
65. If you weigh yourself on a “bathroom scale” at the equator, is it higher, lower, or the same as it would be at the North Pole? Explain.
Chapter 7: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213
1. 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.
3. A plumb Bob pendulum is seen in figure 7-14 on page 211 in the book.
a. Is the work done on Bob by Earth's gravity positive, negative, or zero? Explain.
9. A towing rope, parallel to the water pulls a water skier directly behind the boat with constant velocity for a distance of 65 meters before the skier falls. The rope's tension is 120 N.
a. Is the work done on the skier positive, negative, or zero? Explain.
13. 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.
19. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
25. 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?
29. 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.
31. A block of mass, m, and speed, v, collides with a spring, compressing it a distance of Dx. What is the compression of the spring if the force constant of the spring is increase by 4 times?
39. 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.
53. A certain car can accelerate from rest to a speed, v, in “t” seconds. If the power output of the car remains constant,
a. How long does it take for the car to accelerate from “v” to “2v”?
b. How fast is the car moving at a time of “2t”?
Chapter 8: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248
1. The work done by a conservative force is indicated in figure 8-14 on page 244 in the book, is for a variety of different paths connected to the point A and B. What is the work done by this force on path 1 and on path 2?
3. 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.
9. As an Acapulco cliff diver drops to the water from a height of 46 meters, his gravitational potential energy decreases by 25,000 Joules. Find his weight in Newtons.
13. A vertical spring stores energy (0.962 J) as spring potential energy. When a 3.5-kg mass is suspended from it.
a. by what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?
b. Verify it.
19. Suppose the situation describe in Conceptual Checkpoint 8-2 (page 232 in the book) is repeated on the fictional planet Epsilon, where the acceleration due to its gravity is less than it is on Earth.
a. Would the height of a hill on Epsilon that causes a reduction in speed from 4.0 m/s to 0.0 m/s be greater, lesser, or equal to that on Earth? Explain.
b. Re-consider this Epsilonian hill. If the initial speed at the bottom of the hill is 5.0 m/s will the final speed at the top of the hill be greater, less, or equal to 3.0- m/s? Explain.
25. At a water park, a swimmer uses a water slide to enter the main swimming pool. If the swimmer starts at the top of the slide with an initial velocity of 0.840 m/s, find the swimmer's speed at the bottom of the slide. Assume that the height of the slide is 2.31 meters and that the slide has no friction.
29. An apple of mass 0.21 kg falls of an apple tree and lands on the ground, 4.0 meters below. Determine the apple's kinetic energy (KE), gravitational potential energy (PE), and total mechanical energy of the system (E) when the apple's height (y) is
a. 4.0 m
b. 3.0 m
c. 2.0 m
d. 1.0 m
e. 0.0 m
Ignore air friction. Also the ground is at y = 0.0 meters
31. A rock of mass 0.26 kg is thrown “straight up” from a cliff that is 32 meters above the ground level (where y = 0.0 m). The rock rises, stops, and then is in free fall all the way to the bottom of the cliff, where its final speed, just before impact, is 29.0 m/s. Ignore air friction. For the sake of this problem, assume that the rock is thrown up from slightly beyond the cliff's edge (or else it would just hit the top of the cliff).
a. Find the initial speed of the rock (upward)
b. The maximum height of the rock (y) relative to the base of the cliff.
39. When the Space Shuttle re-enters the Earth's atmosphere, its protective tiles get really hot.
a. Is the mechanical energy of the Shuttle-Earth system when the Shuttle lands, greater, less, or equal to when it is in orbit?
b. Choose the best explanation from the three choices below.
I Dropping out of orbit increases the mechanical energy of the Shuttle
II Gravity is a conservative force
III A portion of the mechanical energy has been converted to heat energy.
53. A block of mass 1.80 kg slides along a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is mk = 0.560, what is the force constant of the spring (k)?
Chapter 9: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292
1. Exercise 9-1 on page 255 states: “An 1180-kg car drives along a city street at 30.0 miles per hour = 13.4 m/s.
(a) “What is the magnitude of the car's momentum?
(b) “A major league pitcher can give a 0.142-kg baseball a speed of 101 miles per hour (45.1 m/s). Find the magnitude of the baseball's momentum.”
The answer to (a) is pcar = 15,800 kg-m/s, where “p” stands for momentum, i.e., p = m v.
In this first question, it asks us, “What speed must the baseball go if its momentum were to equal that of the car?” In other words, how fast must the ball go if it, too, has a momentum of p = 15,800 kg-m/s? They want the answer in miles per hour, probably because you are still converting back and forth and they don't want you to forget. (“They” refers to the authors).
3. A 26.2-kg dog is running northward at 2.7 m/s (it's a vector!), while a 5.30-kg cat is running eastward at 3.04 m/s (another vector!). Their 74.0-kg owner has the same momentum s the two pets taken together. Find the direction and magnitude of the owner's velocity.
9. A net force of F = 200 N acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble.
(a) is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum over the same time period?
(b) Choose the best explanation from among the 3 following:
I. The large mass of the boulder gives it the greater momentum
II. The force causes a much greater speed in the 100-g pebble, resulting in more momentum.
III. Equal force means equal change in momentum for a given time.
13. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1,250.00 N. Assume that the player's food is in contact with the ball for 5.95 x 10-3 seconds.
19. A 0.14-kg baseball moves toward home plate with a velocity (a vector) of v = (- 36 m/s)x. After striking the bat, the ball moves vertically upward with a velocity of v = (+18 m/s)ŷ
(a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume contact between them lasted a time of Dt = 1.5 milliseconds.
(b) How would your answer to part (a) change if the mass of the ball were doubled?
(c) How would your answer to part (a) change if the mass of the bat were doubled?
25. A 92-kg astronaut and a 1200-kg satellite are at rest relative to a Space Shuttle near by. The astronaut pushes on the satellite, give it a speed of 0.14 m/s directly away from the Shuttle. Then, 7.5 seconds later, the astronaut comes into contact with the Shuttle. What was the initial distance from the Shuttle to the Astronaut?
29. In Example 9-6 on page 271, a 950-kg car moving at v = 16 m/s east crashes into a 1300-kg minivan going v = 21 m/s north; they stick together and the wreckage moves off towards the general direction of northeast. Now, let's change some stuff …. suppose the car in Example 9-6 has an initial speed of 20.0 m/s (not 16.0 m/s) and that the direction of the wreckage after the collision is 40.0° above the x-axis (counter-clockwise from the + x-axis). Find the initial speed of the minivan and the final speed of the combined wreckage.
31. Exercise 9-2 on page 268 states: “A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. If the vehicles stick together after the collision, what's their speed immediately after the collision?”
The answer is vf = 5.0 m/s.
In this question, it asks us,(a) is the final KE of the car+truck greater than, less than, or equal to the sum of the initial KE's of the car and the truck separately? In other words, is KEf > KE1 + KE2, is KEf < KE1 + KE2, or is KEf = KE1 + KE2. Explain.
(b) Verify your answer to part (a) by calculating the initial and final KE's of the system.
39. A charging bull elephant with a mass of M = 5240 kg comes directly towards you at ve = 4.55 m/s. In a panic, you toss a rubber ball of mass, m = 0.150 kg , at the elephant, with a speed of vb = 7.81 m/s.
(a) When the ball bounces back towards you, what is its speed?
(b) How do you account for the fact that the ball's KE has increased?
53. Three uniform meter sticks, each of mass, m, are placed on the floor as follows: stick 1 lies along the y-axis from (0,0) to (0,1); stick 2 lies along the x-axis from (0,0) to (1,0). Stick 3 lies along the x-axis from (1,0) to (2,0).
(a) Find the location of the center of mass of the meter sticks.
(b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?
END
Thursday, August 5, 2010
LESSON 3
PHYS 1100 LESSON 3 FOR
THURSDAY, AUGUST 05, 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
Vectors
Virtual arrow: magnitude (size) and direction
heads, tails
adding vectors ≠ adding algebraically
adding vectors
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 ←→
→↑ is okay; but ↑→ is not okay; and →↑ is not okay
A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:
Pythagoras (576 BC – 495 BC)
Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
Components: see above; the x-component of vector A+B is A; the y-component is B.
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
Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:
Projectile Motion
Range, height, angle (above)
Lesson 3: Circular motion
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).
v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2
So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r
And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
However, 2pn = w in rad/sec, thus a = w2r
If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sin θ = 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°.
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 Lesson 3: Newton's Laws of Motion
Objects at rest, stay at rest; objects in motion stay in motion; UNLESS acted upon by an Fext.
F = m a -à Universal Law of Gravity
Every force has an equal and opposite force, or, F1 = - F2.
Units of force are “Newtons”, kg m/s2. = N
F = ma
V = x/t
V Laboratory Exercise 3: Circular Motion
Centrifugal force is a reaction to a real force. Centripetal
VI HWK Assignment 2: 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292.due 8/12
VII Essay 2: Sir Isaac Newton, due 8/12
THURSDAY, AUGUST 05, 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
Vectors
Virtual arrow: magnitude (size) and direction
heads, tails
adding vectors ≠ adding algebraically
adding vectors
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 ←→
→↑ is okay; but ↑→ is not okay; and →↑ is not okay
A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:
Pythagoras (576 BC – 495 BC)
Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
Components: see above; the x-component of vector A+B is A; the y-component is B.
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
Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:
Projectile Motion
Range, height, angle (above)
Lesson 3: Circular motion
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).
v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2
So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r
And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
However, 2pn = w in rad/sec, thus a = w2r
If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sin θ = 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°.
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 Lesson 3: Newton's Laws of Motion
Objects at rest, stay at rest; objects in motion stay in motion; UNLESS acted upon by an Fext.
F = m a -à Universal Law of Gravity
Every force has an equal and opposite force, or, F1 = - F2.
Units of force are “Newtons”, kg m/s2. = N
F = ma
V = x/t
V Laboratory Exercise 3: Circular Motion
Centrifugal force is a reaction to a real force. Centripetal
VI HWK Assignment 2: 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292.due 8/12
VII Essay 2: Sir Isaac Newton, due 8/12
Tuesday, August 3, 2010
LAB 3 FOR THURSDAY AUGUST 5
PhysicsLab3 Thursday, August 5, 2010 Name __________________
Dr Dave Menke, Instructor
I. Title: Linear Motion
II. Purpose: To observe objects moving at a constant speed. Graph the relationships; interpret the graphs
III. Equipment
A rolling object (on wheels or a ball)
Brick or block
Graph paper, pencil, ruler
Masking Tape
Metric ruler
Timing Chronometer (Stop Watch or similar)
IV. Procedure
1. Find a clear, flat surface a few meters long.
2. Using masking tape, mark a starting point, known as “The Starting Point.”
3. Place rolling object on starting point.
4. Have the lab partner practice pushing the object with a consistent force to get the same initial speed each time.
5. After practicing to get a consistent speed, push the object, and timing device simultaneously (same person).
6. At the 2.0-second point, shout “2-second point!” while the second lab partner notes the displacement of the object and marks it with tape. Do this 4 more times. The ball should pass the same point every time, or very close to it. Mark this point, or its average, and label it the 0.00-meter point. (This 0.0 meter point is NOT “The Starting Point.”)
7. Now you are ready. Give the stopwatch to the second lab partner. The first lab partner will then place the object at the starting point, and push it to go. When it crosses the 0.00-meter point, the second lab partner will start the timing device. After 10 seconds, the 2nd lab partner will shout “10-second point!” while a third lab partner notes the displacement of the object and marks it with tape. Have a lab partner note the distance traveled from 0.00-meters. Repeat this 8 more times: one for 9 sec, 8, 7, 6, 5, 4, 3, and 2 seconds. Record the distances v. times in a table. Measure the exact lengths with the meter stick - from the 0.00 point.
9. After putting all the data in the table, graph the nine points.
V. Data & Calculations
1. Distance that the ball travels in 2.0 seconds (on average): ______________
2. The distance traveled for each second, from 10 seconds all the way to 2 seconds:
Make a table and attach
Time (seconds) Distance (m)
10 24
9 23
v1 = (x2 – x1) / (1.0 s)
v2 = (x3 – x2) / (1.0 s)
etc.
vave = Svi / n
3.Make graph of distance traveled, in meters, vs. time (in seconds).
Make graph and attach.
VI. Results
The purpose of this lab was to observe objects moving at a constant speed. Explain how well this was achieved…..
VII. Error Analysis
A. Personal
B. Systematic
C. Random
VIII. Questions
1. Did the ball speed up, slow down, or stay the same speed as it traveled?
Explain or support.
2. What is the shape of the graph you made?
3. How far did the ball travel during each 1.0-second interval?
4. Predict the position of the ball after 12.0 seconds, if you had actually done it. Support
your prediction.
Dr Dave Menke, Instructor
I. Title: Linear Motion
II. Purpose: To observe objects moving at a constant speed. Graph the relationships; interpret the graphs
III. Equipment
A rolling object (on wheels or a ball)
Brick or block
Graph paper, pencil, ruler
Masking Tape
Metric ruler
Timing Chronometer (Stop Watch or similar)
IV. Procedure
1. Find a clear, flat surface a few meters long.
2. Using masking tape, mark a starting point, known as “The Starting Point.”
3. Place rolling object on starting point.
4. Have the lab partner practice pushing the object with a consistent force to get the same initial speed each time.
5. After practicing to get a consistent speed, push the object, and timing device simultaneously (same person).
6. At the 2.0-second point, shout “2-second point!” while the second lab partner notes the displacement of the object and marks it with tape. Do this 4 more times. The ball should pass the same point every time, or very close to it. Mark this point, or its average, and label it the 0.00-meter point. (This 0.0 meter point is NOT “The Starting Point.”)
7. Now you are ready. Give the stopwatch to the second lab partner. The first lab partner will then place the object at the starting point, and push it to go. When it crosses the 0.00-meter point, the second lab partner will start the timing device. After 10 seconds, the 2nd lab partner will shout “10-second point!” while a third lab partner notes the displacement of the object and marks it with tape. Have a lab partner note the distance traveled from 0.00-meters. Repeat this 8 more times: one for 9 sec, 8, 7, 6, 5, 4, 3, and 2 seconds. Record the distances v. times in a table. Measure the exact lengths with the meter stick - from the 0.00 point.
9. After putting all the data in the table, graph the nine points.
V. Data & Calculations
1. Distance that the ball travels in 2.0 seconds (on average): ______________
2. The distance traveled for each second, from 10 seconds all the way to 2 seconds:
Make a table and attach
Time (seconds) Distance (m)
10 24
9 23
v1 = (x2 – x1) / (1.0 s)
v2 = (x3 – x2) / (1.0 s)
etc.
vave = Svi / n
3.Make graph of distance traveled, in meters, vs. time (in seconds).
Make graph and attach.
VI. Results
The purpose of this lab was to observe objects moving at a constant speed. Explain how well this was achieved…..
VII. Error Analysis
A. Personal
B. Systematic
C. Random
VIII. Questions
1. Did the ball speed up, slow down, or stay the same speed as it traveled?
Explain or support.
2. What is the shape of the graph you made?
3. How far did the ball travel during each 1.0-second interval?
4. Predict the position of the ball after 12.0 seconds, if you had actually done it. Support
your prediction.
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