Honors Physics Assessment 5th Week
Standard: 1D Velocity and Acceleration (absolutely the last go around in class).
1. A ball is thrown up at 50 m/s.
a. How long does it stay in the air (assume g = -10 m/s/s).
It takes 5 seconds to slow down to 0 m/s then the next 5 seconds are required to return to earth. 10 seconds.
b. How high does it go?
Average speed on the way up is 25 m/s. It spends 5 seconds moving up for 5 ¥ 25 = 125 m.
2. A car is traveling 30.0 m/s when it begins to accelerate at uniform rate (in physics weÕll say the slowing down is acceleration as well as speeding up is – itÕs just that slowing down gets a negative value). 4.0 seconds after it begins accelerating it is traveling 20.0 m/s.
a. What is its rate of acceleration?
a = Æv/t = -10/4 = -2.5 m/s/s
b. How far does it travel during those 5.0 4.0
seconds?
average velocity is 25 m/s. For 4.0 seconds: 4 ¥ 25 = 100 m
Standard 2c Simple Work and Energy
1. A 5000 kg truck skids to a stop at a work site. If it exerts* 25,000 N of stopping force, how much distance will be required to stop the truck from a speed of 40 m/s?
W = ÆE = .5¥m¥v2 = .5 ¥ 5000 ¥ 40¥40 = 4,000,000 J
W = F¥d so d = WÖF = 4,000,000 J Ö 25,000 N = 160 m
*Really the truck exerts force on the road and the road exerts 25,000 N against the truck. Here the 25,000 N force is consistent with a weight of mg = 5000 kg ¥ 10 N/kg = 50,000 N then times µ = .5 = 25,000 N of stopping force.
2. A 50 kg (or 80 kg) circus acrobat dives into a bucket from 30.0 meters above the ground.
a. How much gravitational potential energy does he have before he jumps off his high platform?
Ep = mgh = 50¥10¥30 = 15,000 J (80¥10¥30 = 24,000 J)
b. How much gravitational potential energy does he have when he is 20.0 meters above the ground?
50¥10¥20 = 10,000 J (80¥10¥20 = 16000 J)
c. How much kinetic energy does he have when he is 10.0 meters above the ground?
ÆEk = -ÆEp= mgÆh = 50¥10¥20 = 10,000 J (80¥10¥20 = 16,000 J)
d. How fast is he falling when he is 10.0 meters above the ground?
v = sqrt(2E/m) = sqrt(2¥10,000/50)= 20 m/s sqrt(2¥16,000/80) = 20 m/s
Standard: 3c Measurement and Analysis Name _________________________________Hour_______________________
There are eight stations in the room. Each has one or more physics books stacked under one end. Note how many books are stacked at your station:
Station Letter _______Number of books: ________ Position of the photogate ________________ cm.
Color of books: circle one: red yellow/blue white
Make sure the higher end of the track is flush with the back edge of the top book (like last week). Release the cart so that it rolls through the photogate. Place the back end of the track at the back edge of the stack of books. Obtain a LabQuest from the supply stand. Obtain one cart of your choice (ALL CARTS HAVE MASS = 0.50 kg). The box of accessories is nearby, choose whatever accessories you desire to add to the cart. Using the LabQuest and photogate, determine the momentum of the cart as it passes down the track and through the photogate. Show ALL data used and all necessary calculations in this space. Label data and results.
Varies, of course, from station to station:
Very few asked from where to release the cart. I did just judge based on the ÒformÓ of your answers, but for those whoÕve been paying attention, it should be obvious that releasing it from one spot rather than another will influence the speed.
DonÕt think: divide the distance between the first rod and the second rod by the time from the first row of gate data to the fourth row of gate data. All of that can change with your equipment. Think about the equipment and how it is used. DonÕt count on me to tell you how it works in a particular situation.
I needed to see the following:
distance (probabaly in cm first) in meters – it might just have been the diameter of one single rod.
time (probably the spread from the first to fourth row of the data table, but if you used the block or a single rod it could be from the first to the second row).
velocity = distanceÖtime
momentum = mass¥velocity
units of momentum: kg¥m/s