AP50 Fall 2016
Problem Set 1Solutions 1) Food for Thought: In an effort to cover all four years of her private university tuition, books, and living expenses, Penny, friends of Leonard and Sheldon, decides to collect soda cans from people’s trash and redeem deposits. Will Penny need to take out a loan for her college education? I: Getting Started: We want to know whether or not Penny can make enough money by collecting soda cans and redeeming the deposits to pay for her university or whether she will need to take out a loan to cover her expenses. First we need to decide on the average cost of a private university per year. Next we need to determine the number of cans that Penny would need to collect per year to cover her college education. Once we know that we can determine how many cans she would need to collect per day as well as how many cans she would need to collect per minute. With this calculation completed we will be able to decide whether realistically she can collect enough cans to pay for her education or whether she should take out a loan. II: Devise Plan: Step 1: First we determine the average cost of private university tuition, books, and living expenses per year. Step 2: Next we determine the number of cans that Penny would need to collect per year to cover all of her expenses. Step 3: We then determine the number of cans that she would need to collect per day and per minute. Step 4: We then determine how many hours per day Penny would realistically be able to collect cans Step 5: Finally, we decide whether a person would actually be able to collect the number of cans we calculate or whether Penny should take out a loan. III: Execute Plan: Step 1: Determine the average cost of private university tuition, books, and living expenses per year. How do you figure out this cost? You could look up the average cost of tuition at a private university per year and add that to an estimate for how much books cost per year and an estimate for how much living expenses cost per year. Assume average cost per year of college (private) education is $50,000/year. AP50 Fall 2016
Step 2: Determine the number of cans that Penny would need to collect per year to cover all of her expenses. We know that the deposit she will get for each can is $0.05. Number of cans needed per year = $"#,!!!
$".!"
= 1,000,000 cans Step 3: Determine the approximate number of cans that she would need to collect per day. # of cans needed per day = !,!!!,!!! !"#$
! !"#$ ! !"#$
x !"# !"#$≅3,000 cans/day Step 4: Determine how many hours per day Penny would realistically be able to collect cans and how many cans per hour she would need. Since Penny needs to study, sleep, exercise, eat, and socialize, only a portion of her day can be devoted to collecting and redeeming soda cans, limited dorm space notwithstanding! We can therefore estimate that Penny would allot 4 hours/day for collecting/ redeeming soda cans. # of cans needed per minute = !,!!! !"#$
! !"#
! !"#
! !"#$
x ! !"#$% x !" !"#$%&' = 12 cans/minute Step 5: Decide whether a person would actually be able to collect the number of cans we calculate or whether Penny should take out a loan. Even if Penny allots 4 hours/day collecting/redeeming soda cans, 1/6 of each day is used and she would need to collect 12 cans/minute! Penny should take out a loan. IV: Reflection: Is there any other way that Penny would be able to collect cans to earn enough money for university? If, by chance, Penny collected cans for 8 years prior to college at a rate of 4 hours/day, Penny would still have to maintain an hourly rate of 4 cans every minute (both during the 8 years and the 4 years of college). Since Penny cannot support her private college education by collecting/redeeming soda cans, she will need to apply for a loan and/or scholarship/fellowship. AP50 Fall 2016
2) Road Rally Race: In a road rally race, you are told to drive half the trip at 25 m/s and half the trip at 35 m/s. It’s not clear from the directions whether this means to drive half the time at each speed or drive half the distance at each speed. Which would yield the shorter travel time for the entire trip? I: Getting Started: We want to know whether it is faster to drive a certain distance given the two speeds when you drive half the time at each of the two speeds given or if you drive half the distance at each of the two speeds. As we are not given the length of the trip we are going to need to write expressions for both of the two options (distance and time) and then compare to figure out which one would be shorter. II: Devise Plan: Step 1: First we want to determine the average speed if you were to drive half of the race at 25 m/s and half the race at 35 m/s. Step 2: Next we want to determine the average speed if you were to drive half of the distance of the race at 25 m/s and half of the race at 35 m/s. Step 3: Lastly we want to compare the two average speeds and determine which one is smaller. III: Execute Plan: Step 1: Determine the average speed if you were to travel half of the time at 25 m/s and half of the time at 35 m/s. For this we just need to take the average of the two velocities since they will be driven for the same amounts of time. Average speed = !
!
!" !!"
!
!
!
= 30 m/s Step 2: Determine the average speed if you were to travel half of the distance at 25 m/s and half of the distance at 35 m/s. For this we assume that the total distance that is traveled is L. We can then say that the distance travelled when driving at 25 m/s is equal to the distance travelled when driving at 35 m/s which with our notation is L/2. !!"#$%&' !"#$%&'(
Since we know that velocity = !"#$ we can rearrange this to time = !"#$%&'( . AP50 Fall 2016
The total time that it will take to travel a distance L will be equal to the time it takes to drive L/2 at 25 m/s plus the time it takes to drive L/2 at 35 m/s. !
!
+ !∗!" !/! !∗!" !/!
Total time when driving half distance at different speeds = 7L
5L
Total time = 350 m/s + 350 m/s Total time =
12L
350 m/s
To find the average speed we just need to realize that time = distance/velocity. Therefore from the expression above we can say that the average speed is equal to: Average speed = !"# !/!
!"
~ 29 m/s Since the average velocity when travelling half of the time at 25 m/s and half the time at 35 m/s is larger than that when travelling at those speeds for equal distances, the shorter travel time for the entire trip would occur if you travelled half the time at 25 m/s and half of the time at 35 m/s. IV: Reflection: This is a good example of when we don’t have all of the values to calculate the exact value for a given quantity, such as the time it will take in the two different situations, but we can still compare two situations by writing expressions and determining a different quantity (in this case the average velocity) that aides in determining the final answer. One way that we can see if our answer makes sense is to pick a distance for the race, say 100m and see if we find that our answer is correct. Average speed traveling half the time at 25m/s and half the time at 35 m/s: Average speed = !
!
!" !!"
!
!
!
= 30 m/s This means the travel time for travelling in this way is: !"#$%&'(
!""#
travel time = !"#$%&'( = !" !/! = 3.3 seconds Tarvel time if traveling half the distance (50m) at 25m/s and half the distance (50m) at 35m/s: !"#$%&'(
!"#
!"!
Tarvel time = !"#$%&'( = !" !/! + !" !/! = 3.4 seconds AP50 Fall 2016
You will find for this given distance (100m) it does take less time to travel half the time at 25 m/s and half the time at 35 m/s. You should take note however that while this is a good first check to see if an answer makes sense picking a certain distance does not necessarily mean that our answer is correct for all values but it does help us make sure we are on the right track. It is also often helpful to check extreme cases. For example, suppose in this case the smaller velocity was zero and the larger velocity was 50 m/s. Then, going half the time at 0 m/s and half the time at 50 m/s means an average of 25 m/s. However, going at 0 m/s means you will never get to the halfway distance, so the time is infinity – much higher than going half the time at the given speeds. AP50 Fall 2016
3) Acceleration Down a Slide: You and a friend ride what are billed as the “world’s longest slides” at a county fair. Your slide is 100m long, and your trip takes 10 s,. Your friend chooses a taller, 150-­‐m-­‐long slide made from the same material as yours and with the same angle of incline. (a) What is the magnitude of your acceleration down the slide? (b) What is the magnitude of your friend’s acceleration down the slide? (c) How many seconds does it take your friend to get to the bottom? (d) What is your speed when you hit the bottom? (e) What is your friend’s speed when she reaches the bottom? I: Getting Started: We want to know the magnitude of your acceleration down the slide. Consider the forces that are acting on you as you travel down the slide. We know that: x = 0.5*a*t2 as discussed in the textbook. In part b we want to know the magnitude of your friend’s acceleration where the same forces are at play. In part c we are asked to determine how long it takes for your friend to reach the bottom of the slide given the length of the slide and the acceleration calculated in the previous part. In parts d and e we are asked to calculate your speed and your friend’s speed respectively at the bottom of the slide. We know that a =v/t and can use this to get to the answer. II: Devise Plan: Step 1: First we determine the magnitude of your acceleration down the slide given the length of the slide and the time it takes for you to reach the bottom. Step 2: Determine the magnitude of your friend’s acceleration down the slide they went on remembering that the angle of incline of the two slides is the same. Step 3: Having determined the acceleration of your friend on the slide we can solve for the time it takes for them to reach the bottom. Step 4: Having calculated the acceleration and the time it takes for you and your friend to travel on the slide, we can calculate the final velocity for you and your friend at the bottom of the slide. III: Execute Plan: Step 1: Determine the magnitude of your acceleration down the slide. We can use the equation x = 0.5*a*t2 as discussed in the textbook. x = 0.5*a*(Δt)2 !"
a = (𝛥!)! = !∗(!""#)
(!"#)!
= 2.0 m/s2 AP50 Fall 2016
Step 2: Determine the magnitude of your friend’s acceleration down the slide they went on. The trick to this is to realize that the friend will experience the same acceleration as you. This is because the slides have the same angle of incline. Therefore, your friend’s acceleration is 2.0 m/s2. Step 3: Solve for time it takes for your friend to reach the bottom of the slide. Now that we know your friend’s acceleration and were given the length of their slide we can calculate how long it will take for them to reach the bottom. From x=0.5*a*(Δt)2, we can solve for Δt: Δt =
2x
a
= 2∗150m
2.0 m/s2
= 150 s! = 150 seconds(~12 seconds) Step 4: Calculate the final velocity for both you and your friend at the bottom of the slide. We know that velocity is the integral of acceleration and as the acceleration is constant this simplifies to simply multiplying the rate of acceleration by the time. Your velocity at bottom of the slide = 2.0 m/s2 * 10 s = 20 m/s Your friend’s velocity at the bottom of the slide = 2.0 m/s2 * 150 s = 2 150 m/s (~24 m/s) Step 5 The acceleration along the slide is the projection of g (acceleration due to gravity) along theslide : g sin α=2m/s2. Thus the incline angle is 11.5 deg IV: Reflection: Does the accelerations we calculated in the first part make sense? The accelerations we have calculated are due to gravity acting on the people as they move down the slide. As the slide is inclined, we expect the acceleration to be smaller than free fall acceleration. The rate of acceleration due to gravity at the earth’s surface is approximately 9.8m/s2. Food for thought, our calculated acceleration, which is approximately 1/5th of this value allows us to calculate the angle of incline α, The acceleration along the slide is the projection of g along the slide g sinα=2m/s2; the angle α is therefore 11.5 degrees. AP50 Fall 2016
4) Star Galactic: In an upcoming Bond film entitled “Star Galactic “, the evil villain, Sir Barron B. Bad escapes from a space station prison. The prison is located between galaxies far away from any stars. Barron steals a small space ship. He accelerates in a straight line at the ship’s maximum possible acceleration of 30 m/s2. After 12 minutes, all of the fuel is burned up and the ship coasts at a constant velocity. Meanwhile, the hero, Captain Hailey Comet, learns of the escape and she rushes off to recapture Mr. Bad. Hailey gives chase in an identical ship, which has an identical maximum acceleration, going in an identical direction. Unfortunately, Mr. Bad has a 30-­‐minute head start. As luck would have it, Barron's ship did not start with a full load of fuel. With her full load of fuel, Captain Comet can maintain maximum acceleration for 15 minutes. How long will it take Captain Comet's ship to catch up to Barron's? I: Getting Started:Barron leaves first. His spaceship travels in a straight line at a constant acceleration of 30 m/s2 for 12 minutes. Then he continues to coast at constant velocity. Dr. Comet leaves 30 minutes later traveling along the same linear route also at 30 m/s2 acceleration. She maintains this acceleration for 15 minutes and then remains at her now higher constant velocity. We want to know how long it will take for Dr. Comet to catch up to Barron. II: Devise Plan: Step 1:Find the final velocity vf,Bof Barron’s ship using vf= vi + aΔt. Step 2:Find the position of Barron’s ship, x12min, B, after he has finished his initial 12 min acceleration usingxf= xi + viΔt + ½ a(Δt)2 Step 3: Find the final velocity vf,Cof Dr. Comet’s ship after she finishes accelerating. This is att = 45min from the start of Barron’s escape (or 15 min after she started chasing him). Step 4: Now we can find the position of both Dr. Comet’s ship, x45min, C, and Barron’s ship, x45min, B, at this time. xf= xi + viΔt + ½ a(Δt)2and x = vΔt AP50 Fall 2016
Step 5: Now that we know where each ship is at and the velocity it is traveling at, we can find out what the remaining distance is between the ships and figure out how long Dr. Comet needs Δt catch up to Barron. v = Δx/ Δt III: Execute Plan: Step 1: After 12 min, Barron has finished accelerating so we can find his final velocity which will remain constant for the rest of the journey. Our acceleration is in units of m/s2, so we need to convert the time into seconds. 12 min * (60 sec/min) = 720 seconds vf ,B = vi,B+ aBΔt = 0 + (30 m/s2 )(720s) = 21,600 m/s Step 2: We can also use his initial position, velocity and constant acceleration to find the position of his ship after these 12 min. x12min, B = xi + viΔt + ½ a(Δt)2= 0 + 0 + ½ (30m/s2)(720s) = 7,776,000 m Step 3: Now we repeat a similar procedure for Dr. Comet’s ship. Let’s find her final velocity. 15 min * (60 sec/min) = 900 seconds vf ,B = vi,B+ aBΔt = 0 + (30 m/s2 )(900s) = 27,000 m/s Step 4: Next, we want to find out the relative positions of the two ships. First, let’s find where Dr. Comet’s ship is when she finishes accelerating (like we did for Barron in Step 2) x45min, C = xi + viΔt + ½ a(Δt)2= 0 + 0 + ½ (30m/s2)(900s) = 12,150,000 m Barron has been traveling at constant velocity from time t = 12 to time t=45min so we can find the new distance he has traveled as in step3. Δx = vΔt = 21,600m/s(45 min – 12min) = 21,600m/s(33min*60sec/1min) = 42,768,000m x45min, B = x12min, B + Δx= 7,776,000m + 42,768,000m = 50,544,000m Step 5: Now we know the positions of both ships and the velocity that each are traveling at. We can find out how much time is required for Comet to catch Barron by setting up a system of two equations with two unknowns using v = Δx/ Δtfor each ship. AP50 Fall 2016
IV: Evaluate result: Comet is traveling 5400m/s faster than Barron once they both reach constant velocity at t = 45min , but by that point, we found that Barron was already roughly 40,000,000 m ahead. If we divide 40,000,000m by 5400m/s we get that it would take about 123 min for Comet to make up that extra distance and catch up once neither one is accelerating anymore. If we add in the 45 min that had already passed that gives us about 168min which as a rough estimation, agrees with the 163min, 30 sec that we calculated by solving the system of equations. AP50 Fall 2016
5) Save the Trees:You want to determine whether a tall tree will fit onto your 75-­‐
m-­‐long logging trailer after you cut it down. You throw a rock straight up with a speed great enough that the rock just reaches the top of the tree, releasing the rock 2.0m above the ground. You observe that the time interval from the instant you release the rock to the moment it hits the ground is 5.0s. I.Getting Started:We know that our end goal is to know if the tree fits into our 75-­‐m trailer, so we need to figure out how tall it is. We start by sketching out the problem. We know that the rock starts at a height of 2m and with the initial velocity we throw it with. At the top of the trajectory (equal to the tree height) the velocity becomes zero as it changes direction and begins to fall back down to the ground. We know the total time up and down for the rock is 5 sec so we will use that to figure out the height of the tree. We simplify the sketch and instead draw two motion diagrams for the rock, one for the trip upward and one for the trip downward using the procedure defined in section 3.4 of the text. II.Devise Plan: AP50 Fall 2016
The motion has a constant acceleration of –g throughout the problem, so algebraically we do not need to separate the trip up and down even though we did in the motion diagram for clarity. As we designated in the motion diagram, the initial conditions of the rock (when you throw it) are: We do know that the initial velocity is upwards, so it is in the positive x-­‐direction. We choose the final conditions to be when the rock hits the ground since we know that the total time from when we release it until when it hits the ground is 5 seconds. The final conditions of the rock (when it hits the ground) are: The final velocity of the rock is unknown but we know it must be negative as the rock will be falling downwards when it hits the ground. We can use equation 3.7 to solve for the initial velocity that the rock must have had to take 5 seconds to go up and fall just under the force of gravity in 5 seconds. Once we know the initial velocity, we can use equation 3.4 to find the final velocity. After we know the final velocity we can use equation 3.13, just looking at the trip down, since we know the velocity at the top of the downward trip is zero and we just found the final velocity. III. Execute Plan: We start with equation 3.7 to find out what the initial velocity that the rock was thrown at must have been. AP50 Fall 2016
Now that we have the initial velocity, we can use eq. 3.4 to find the final velocity of the rock as it is landing on the ground. We notice that as we expected, the final velocity and initial velocity have opposite signs since one is pointing up and the other, pointing down. Finally, we now just look at the trip down where the rock is in free-­‐fall, meaning the initial velocity is zero. Using eq 3.13, This tells us that the maximum height that the rock travels to, or the height of the tree, is about 32 m so, YES, the tree will fit into our 75-­‐m-­‐long logging trailer easily. IV. Evaluate Result: Here are 3 examples of how you could possibly evaluate your result to determine if it makes sense. We do not expect you to necessarily do all of these checks, but these are just some possible ideas of things you could try. As a first example, we can check some of our calculations by going back to the original equations and plugging in the numbers we just calculated. For example, we AP50 Fall 2016
expect the time for the rock to go up will be slightly shorter than the time down and that together they will add up to 5 seconds. As we already mentioned, we also notice that the final and initial velocities have opposite signs since one is pointing up and the other, pointing down. Another way to check this would be to look only at the trajectory up. Taking the initial velocity and height we just found, we can see that eq. 3.7 is indeed satisfied. A third way is tothink of our physical intuition. It seems reasonable thatif we throw a rock about 30 meters straight up(~100 ft), we need to start with a velocity straight up of about 24 m/s (55mph.) For example, think of an in-­‐field pop-­‐up in baseball and how fast it comes off of the bat. The speed and height seem to be of the same order of magnitude, so our answer seems reasonable. Finally, we were not specifically asked to find the height of the tree; rather, we just wanted to know if it would fit in the trailer. An outside-­‐the-­‐box approachis to assume the tree height is 75m since that is the largest that would fit. Using this assumed maximum height of the tree, we can calculate how long the rock would take to rise and fall. This value would be the longest time it should take for the rock to rise and fall for a tree to still fit in the trailer. As shown in our calculation below, the rock would take almost 8 seconds to fall if the tree was 75 m tall (but in our experiment it only took 5 sec), so the tree must be shorter than 75m and would fit.