 # Sample Exam 2 PH 105-003&4 P. B. Visscher

```PH 105-003&4
Sample Exam 2
P. B. Visscher
[Note that coverage is slightly different – our exam will also cover Ch. 10,11, and 13.]
Instructions [The same as for Exam 1, you won’t need them except for reference.]
This is a closed-book exam; please put books and notebooks on the floor. Mark your
answers on the answer form given, using a No. 2 or softer pencil. Turn in this test
booklet with your green answer form – there will be a copy on the website after it is
graded so you can see which questions you got wrong.
Before turning in the answer form, be sure to write your name on it and to fill in the
appropriate circles under the letters of your name. To get your score from MyBama, you
need to also put your CWID on the test.
FORMAT: Some problems contain more than one question, and so use more than one of
the numbered lines on the answer sheet. At the end of each problem, the question line
numbers are given in braces. For example, if question says
• How many planets orbit the sun? value: {9}
This doesn’t mean there are 9 planets, it means put the numerical value of the answer on
line 9. Thus line 9 should look like this:
If the answer isn’t an integer, round to the
nearest integer. Some of the numerical answers
are not between 0 and 9, for example:
How many cents are in \$33,000? Value: {3}×10{4}
This means you should put the answer in scientific notation, 3.3×106 , round the mantissa
(3.3) to the nearest integer, and bubble 3 and 6
into lines 3 and 4:
Ignore signs in this process, i.e. bubble 3 and 6
for -3.3×10-6 as well.
To allow partial credit if you know how to do a
problem but make a numerical error, some of the problems also ask which equation to
use. Choose it from the list on the next page, and darken the circle by the appropriate
LETTER (A .. J) as in the following example:
•
A 300 kg mass is subjected to a force of magnitude 0.18 N. Its acceleration, in
m/s2, is
Formula:{8}
Value: {9}×10{10}
ANSWER: Formula A gives F = ma, so a = F/m
= 18×10-2 kg m s-2 / 3 × 102 kg = 6 ×10-4 m s-2,
so you bubble in A,6,4, and your answer sheet
looks like the figure at right:
There will be a some conventional multiple-choice problems as well, where choices (A),
(B), etc., are given and you need to bubble the letter of the best answer.
Special cases: I try to avoid answers close to a half-integer, but if you get exactly a halfinteger (and it doesn't go away when you double-check), raise your hand and show me
your work – I might have made a mistake. There are many ways to write a number in the
form (mantissa)×10exponent – for example, 91 = 91×100 = 9.1×101 = 0.91×102 =
0.091×103. To decide which to use, round each mantissa to an integer: … 91×100 ~
9×101 ~ 1×102 ~ 0×103 , . . . The first isn’t going to work because the mantissa must be
in 0…9. Of the others, choose the one with greatest accuracy, 9×101. If the answer is
exactly 0, put 0 for both the mantissa and the exponent.
To encourage you to have some familiarity with the formulas, there are a few wild cards
(indicated by “?”) in the formula list. , each wild card represents an omitted number,
letter, or phrase. You should probably begin by filling in the wild cards you know, to
make the list easier to read. Not all formulas are included, but the list should include at
least one that is needed for each problem where a formula is asked for. If two formulas
seem to be equally important, pick the most recently covered -- never bubble two answers
on the same line. You might want to tear out this page, for easier reference.
Densities in kg/m3
Aluminum
Water
Iron
2700
1000
7900
CONVERSION FACTORS
1 meter = 3.28 feet = 1.09 yards
1 km = 0.61 mi
g = 9.8 m/s
2
Constants
Formulas
(A)
F = ma
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(J)
v = v0 + a ?
P=Fv
a = v2/r
m = ? * Volume
KE + PE = constant
a = dv/dt
x2+ y2= ?2 (Pythag. Thm.)
vAC = vAB + vBC
(m1 + m2 + …) xcm =
m1x1 + m2x2 + …
Torque = F d
p = mv
Δ(KE) = work on ?
α = dω/dt
► A skateboarder starts at rest and slides frictionlessly for a distance 4.5 meters along a
ramp inclined at 45˚ to the horizontal. His velocity (in m/s) at the bottom is
Formula:{1}
Value: {2} (i.e., exponent not required here)
► A car of mass 1173 kg is moving at 15 m/s at the bottom of a hill. It coasts (with no
engine power) up the hill to a height of 8 m, where it comes to a stop. What work was
done against friction?
Formula:{3}
Value: {4}×10{5} Joules
► A block of mass 1.07 kg is at rest on a board inclined at 40˚ to the horizontal, with a
static friction coefficient of 0.9. The normal force exerted on the block by the board is
Value: {6} Newtons
► A 1500 kg Jeep on a safari has its center of mass 3 feet above the ground. A 1000 kg
elephant is then tied to its roof rack, so that the elephant’s center of mass is 8 feet above
the ground. How high above the ground is the center of mass of the Jeep + elephant, in
feet?
Formula:{7}
Value: {8}×10{9} feet
► A 0.9 kg pileated woodpecker traveling east at 5.5 m/s is struck by a 2 kg inebriated
war eagle traveling north at 3.5 m/s, and their talons become entangled. The total
momentum vector [in the form (vEast,vNorth)] of the two birds is
Formula:{10} Value: ({11},{12}) kg m/s (i.e., give 2 components, each should be < 10)
During the collision their talons become entangled. After the collision they are moving
with a common speed of
Formula:{13} Value: {14} m/s.
► A 1 kg bead slides without friction on a circular wire (radius 1 m) in a vertical plane.
It starts at the top with a very small velocity (say 0.001 m/s; it has to be nonzero or it
would never slide down). What is the acceleration of the bead at the bottom, in g’s?
(That is, give the value of a/g).
Value: {15}
What is N/mg at the bottom, i.e. the normal force in units of the weight?
Value: {16}
► How much power does a motor have to produce to lift a 1500 kg elevator at a constant
speed of 3.5 m/s?
Formula:{17}
Value: {18}×10{19} watts
If the elevator accelerates to this speed from zero in 1 second, what is the instantaneous
power produced by the motor at the end of this acceleration period (i.e. when the velocity
has just reached 3.5 m/s, but the elevator is still accelerating)?
{20}×10{21} watts
► A 0.63 kg mass hangs from a knot supported by
two light strings as shown. The tension in the
horizontal string is
{22} Newtons
And the tension in the diagonal string is
{23} Newtons
45˚
0.63 kg
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The following questions have one answer each, in the traditional way:
{24} To use Newton’s second law, you should first draw a free body diagram, showing
all of the
A) Forces exerted on a specific body.
B) Accelerations acting on a specific body.
C) Forces exerted by a specific body.
D) Bodies that are involved in the problem.
E) Conservative forces on a specific body.
{25} In an elastic collision in one dimension between particles with (x-components of)
velocities v1 and v2, a quantity that changes its sign but not its magnitude is
A) v1 + v2
B) v1
C) v1 - v2
D) ½ m1v12 + ½ m2v22
E) ½ m1v12 - ½ m2v22
F) m1v1 + m2v2
{26} The potential energy stored by a spring that obeys Hooke’s law with a spring
constant k is
C) -mgh
A) mgh
B) ½ kx2
D) v2/r
E) kx
F) - ½ kx2
{27} The impulse of a force (i.e. the momentum transferred by the force) is the integral
of
A) F dx
B) p dx
C) F dt
D) E dt
E) kx dx
F) ½ kx2
``` # 1 Spring 2015 Directions for Completing Grades 3 & 4 FSA ELA # Fabric required for the bottom color portion is top portion requires # 1. Create a model for a cooking recipe generator. Given a list of ingredients, the recipe generator should output a list of recipes starting with the simplest recipe to the most complicated recipe. # Chapter 4 Newton's First Law of Motion – Inertia Math Review Work 