Physics 2A
Exam 1 Rubric:
Question 1:
Q (4.0): Perfect Solution: You can tell the net force on an object is
zero when the acceleration of the object is equal to zero. This
means that the object can either be at rest or moving with a
constant velocity. (6 students)
W (3.7): Stated that the object is at rest and that all of the forces acting
on the object cancel out. (Failed to state that the object could have a
constant velocity.) I also placed students in this category that stated that
the object can’t be accelerating (correct), but can’t move (incorrect). (9
students)
E (3.3): Stated that the object needed to be at rest. (20 students)
R (2.0): Stated that the forces needed to cancel for the object’s net force
to equal zero (9 students)
T (1.5): Some progress made. (6 students)
Question 2:
Q (4.0): Perfect Solution: part a) I do not agree with the student because
this equation gives the initial velocity in the y direction, not the overall
height of the ball. part b) I disagree with the student because the
velocity in the y direction is zero at maximum height, not the velocity in
the x direction. (8 students)
A (4.0): Just as in “Q” however, stated that they agreed with the student
in ‘part a,’ and then went on to indicate in some form that they
understood that vosinθ = voy and not hmax. Notice that this is not
technically correct (because you said you agreed with the student), but
you indicated you understood why the student was incorrect, so you
receive full credit. (3 students)
R (3.7): Just as in “Q,” although ‘part a’ needs a little more justification. I
am not convinced you understand why the student is incorrect. (4
students)
S (2.5): Part a is correct, but in part b it is stated that velocity is zero in
the x-direction at max height. (It is the y-direction velocity that is zero
at max height.) (8 students)
X (2.5): Part b is correct, but in part a, it’s stated that the student is
correct because the height can be found in the way that the student
indicates. (7 students)
C (2.0) : Either part a or part b is correct (see “Q” to see which is
correct), but the other part uses flawed logic different than from what is
written in “S” or “T”. (4 students)
W (1.0): Both parts a and b are incorrect. In part a, students draw a
picture and/or state that the height is equal to the sin component of the
initial velocity. In part b, students state that at maximum height vx or v =
0. (3 students)
P (0.5): Both parts a and b are incorrect. Flawed logic is used to either
disagree or agree with the student. (13 students)
Question 3:
Q (4.0): Perfect: Part a: Both balls would hit the ground at the same time
because they have the same initial velocity in the y direction (zero), and
each only has the force of gravity acting on them. Therefore it will take
them both the same amount of time to fall to the ground. Part b: The ball
that was thrown horizontally will have the greater speed overall
because while it’s final velocity in the y direction will be equal to the ball
that is dropped, it will also have a velocity in the x direction. The sum of
the velocities in the x and the y direction is equal to the total velocity. (9
students)
L (3.7): Similar to “Q” but, in part a, neglected to state that both balls
start with same velocity in the y direction. Part b is correct (as in “Q”). (4
students)
V (2.5): In part a it’s stated (in some form) that the dropped ball hits the
ground first due to the fact that the velocity in the x direction for the
thrown ball somehow keeps it in the air longer. Part b is correct (as in
“Q”). (5 students)
Y (2.5): Part a is as in “Q” or “L”, but in part b, students argue that the
velocity the horizontally thrown ball is greater because the velocity =d/t
and so since it traveled further, it’s velocity must be greater. This is not
complete because it does not take into account the y direction. (4
students)
G (2.0): Part a is incorrect, the students either make the same mistake as
in “V” or “L”, and in part b they then use a force argument to account for
the thrown ball to have a greater final velocity. OR error in b is that it’s
not clear which velocity they are talking about (final, or initial, x or y).
(13 students)
D (1.3): In part a, it is incorrectly argued that the dropped ball will hit
first because it has a shorter path to follow. Part b is incorrect for
various reasons. (6 students)
H (0.7): Both a and b are incorrect. Flawed logic is used to make
decision. (7 students)
C (0.5): part a is correct, part b is blank. (1 student)
Z (0.0): Blank (1 student)
Question 4:
Q (4.0): Perfect:
a)
Δd = v0 t + ½ a t2
4 = v0 (1.5) + 1/2 (-9.8)(1.5)2
v0 = 10 m/s
b)
vf = v0 + at
vf = 10 + (-9.8)(1.5)
vf = -4.7m/s
c)
Picture 1, where the measuring tape has reached is maximum and
then is caught on its way back down, is correct. This is because
the velocity is found in part b is negative and therefore it must be
traveling downwards. (11 students)
V (4.0): Identical to “Q” but used vf2 = v02 + 2 a Δd instead of vf = v0 + at.
The problem with this is that because the velocities are squared, you lose all
information regarding the direction of the velocity. vf could be positive or
negative, and you have to choose. You all (incorrectly) chose positive, and
then therefore accordingly chose picture 2 and argued that the velocity was
positive so the measuring tape must be on it’s way up (which is consistant
with your answer from part b). Since we did not discuss this in class and it
did not come up in the homework, this error is worth the same as a perfect
score. (2 students)
R (3.7): Like Q or V, but minor math error such as missed negative sign or
forgot to square something. In these cases, equation is initially written out
correctly, and it is obvious the student made an error only when they got to
the algebra portion. Picture 1 is selected with correct reasoning. (1 student)
H (3.4): Part a correct. Part b has error identical to error in V, or made small
math error using equations from a, resulting in a positive final velocity. In
part c, does not recognize that the sign of the velocity indicates the direction
it is traveling when caught. (8 students)
N (3.3): Math error in part a propagates through part b. In part c, recognizes
that the sign of the velocity indicates the direction it is traveling when
caught. (2 students)
M (2.7): Math error in part a propagates through part b. In part c, does not
recognize that the sign of the velocity indicates the direction it is traveling
when caught. (3 students)
K (2.3): Uses equation for average velocity to find a velocity of 2.7m/s
(which is not the initial velocity) then incorrectly applies it as an initial
velocity in part b. Picture is correctly selected with respect to the sign of the
velocity found. (2 students)
S (2.0): Sets either v, a or change in d equal to zero when this is not the case.
Part c does not reference the direction of any of the velocities found above.
(5 students)
J (1.7): Like “K”, uses equation for average velocity to find a velocity of
2.7m/s (which is not the initial velocity) then incorrectly applies it as an
initial velocity in part b. Part c does not reference the direction of any of the
velocities found above. (2 students)
G (1.5): Like “K” uses equation for average velocity and or average
acceleration. Then either stops or applies it incorrectly. Part c does not
reference the direction of any of the velocities found above. (8 students)
L (0.7): Little progress made. (3 students)
Z (0.5): Little progress made and part c blank. (2 students)
Question 5:
Q (4.5): perfect solution (note that this problem is worth extra if you got it
correct because it was so difficult, the rest of the categories are bumped up
too):
Change meters to km  100m = 0.1km
do
df
vo
vf
a
t
My car 0
65km/h 65 km/h 0
Truck
100m
30 km/h 30 km/h 0
I have defined my interval to start from when the bumpers are 100m apart
until my car passes the truck. The variables of time and d final should be the
same for both vehicles because this is the time and place where the passing
happens.
My car:
Δd = v0 t + ½ a t2
df – 0 = (65km/h) t + 0 (because acceleration=0)
df = (65km/h) t
The Truck:
Δd = v0 t + ½ a t2
df – 0.1km = (30 km/h) t
df = (30 km/h) t + 0.1km
We are looking for time so we can set the two equations equal to each other
and solve for time.
(65km/h) t = (30 km/h) t + 0.1km
t=2.86x10-3 h OR after unit conversion: t = 10.3s
(2 students)
A (4.0): The chart is filled in incorrectly (most common error is that di and df
are conflicting or not included). The problem is solved by using the
difference in velocities instead of setting the two equations equal to each
other. (5 students)
V (3.7): The chart is filled out correctly, except that the final velocities are
missing. The problem does not explicitly say that the vehicles each remain
at constant velocities, (although this was the implied) and so therefore
students with only this specific omission have been given the benefit of the
doubt. They also continued to work through the problem despite not having
enough information. (3 students)
R (3.7): Parts a through d are completely correct, meaning the problem is set
up correctly. Error is from algebra errors, unit conversions, or just stopped
solving right before the end. (5 students)
S (3.5): Significant progress is made. Error results from misconception with
starting and ending distances of the two vehicles, but otherwise process is
correct and complete. Must have written two equations, one for truck and
one for car, and solved for a time. May also have minor errors listed in “R.”
(6 students)
D (2.3): Error is in regard to initial and final distances. The fact that the
change in distance is different for each vehicle, but that the final distance for
each vehicle is the same, is not indicated. Either only wrote an equation for
one of the vehicles, or did not finish using equations. (To be in this category,
students must have made fewer than three of the omissions listed in “E.”)
(6 students)
M (2.0): Made same errors in “D” but also neglected to indicate final
velocities. Followed through to incorrectly find a final time. (Also made
other errors as in “R”, but made fewer than three of the omissions listed in
“E.”) (3 students)
G (1.0): Acceleration used is either 9.8, or has units of velocity. These are
both major conceptual errors. (To be in this category, students must have
made fewer than three of the omissions listed in “E.”) (9 students)
E (0.5): There are at least 3 of the 4 omissions mentioned here: The chart is
incomplete, the explanation (part c) is missing, there are no initial equations
written in variable form to solve for time, there is no numerical calculation
present (other than unit conversions). (11 students)
Question 6:
Q (4.0): Perfect solution:
|A| = 75m
|B| = 80m
Ax = 75 cos (20) = - 70.48m
Ay = 75 sin (20) = 25.65m
Bx = 0
By = -80m
Rx = Ax + Bx = -70.48 + 0 = -70.48m
Ry = Ay + By = 25.65 – 80 = -54.35m
Rx = 70.48m
Ry = -54.35m
Rx2 + Ry2 = |R|2
(-70.48) 2 + (-54.35) 2 = |R|2
|R| = 89m
Now we know that Dr. Paul’s displacement is 89m from where she started, we
now need to determine the direction she needs to travel back to the start.
tanΦ = o/a = 70.48/54.35
θ = 37.63
This is saying that the angle of R is ~38° below the negative x axis. Which means Dr.
Paul originally traveled 38° south of west. In order to get back where she was, she
needs to travel 38° north of east.
θ
Originally travelled
θ
To go back
(6 students)
R (3.7): Correctly found the distance I was away from my original position, also
correctly found the angle that I needed to travel at, but incorrectly stated with what
which direction the angle was respect to. Common errors included drawing the first
vector in the eastward direction, and determining the direction I traveled and
stating it was the direction I needed to travel back. (7 students)
S (3.3): Correctly found the magnitude of R, but then mixed up opposite and
adjacent sides which resulted in an incorrect angle. OR incorrectly assiged the x and
y components leading to an incorrect magnitude for R but carryied that mistake
through the rest of the problem correctly. (5 students)
X (2.5): Correctly finds the components of the first vector, makes a conceptual error
in summing the vector components directly instead of using the Pythagorean
theorem. The angle of this (incorrect) resultant is then found. (2 students)
Y (1.7): Correctly finds the components of the first vector, but then makes several
errors from there. (7 students).
P (1.3): Used Pythagorean theorem on the two vectors that given in the question.
These vectors do not make a right triangle so you can’t do this. OR made a right
triangle and used trig equations. (13 students)
L (1.0): Summed x and y components as they were scalars. You can only sum like
components (x’s with x’s and y’s with y’s). (4 students)
B (0.5): Little progress shown. (6 students)
Problem 7:
Q (4.0) perfect solution
a)
Fnormal
Fpull or tension
Fgravity, mg, or weight
b) ΣF = Fn + Fmg + Fpull = ma
c) ΣFy = Fny + Fmgy + |Fpull| sin(30) = ma
ΣFy = Fny + mg + 10 sin(30) = 0  equals zero because not accelerating in y direction
ΣFx = Fnx + Fmgx + |Fpull| cos(30) = ma
ΣFx = 0 + 0 + 10 cos(30) = ma
d) from above: 10 cos(30) = ma
need to find mass:
weight = mg
350N = m(9.8m/s2)
m = 35.71kg
10 cos(30) = ma
10 cos(30) = (35.71)a
a= 0.24m/s2
(zero students!)
R (3.9): Very close to a perfect solution, although made one of the following errors:
failed to include the y component of the pull in the sum of the y forces, wrote that
the sum of the x forces equaled to zero (even though they never used this), did not
explicitly write out the equation for the sum of all forces, did not set the sum of the
total forces equal to ma, did not explicitly indicate all forces equal to zero, did not
include y component of pull in sum of y forces, made small math error. (7 students)
S (3.7): Drew the force diagram correctly, wrote correct sum for forces. Found mass
of cousin and sled correctly. However when calculating for the final acceleration
used 10N instead of the x component of the pull. (May have had one of errors listed
in “R” as well.) (5 students)
P (3.0): One conceptual error. Used weight as mass, or used total y force of pull
instead of component. (2 students)
W (2.9): Drew force diagram wrong, has wrong equations for total force and force in
each component, but realizes the force of the pull in the x direction is responsible for
the acceleration and solves correctly for it. (2 students)
J (2.5): Draws incorrect force diagram and makes one conceptual error (ex: finds the
y component of force instead of the x, or after correctly finding x component of force
sums it to the y to find the total acceleration. (2 students)
C (1.7): Correct force diagram, but more than one conceptual error (adding in
acceleration due to gravity was common) or incomplete. (5 students)
K (1.3): Notes that ΣFx = FT (it is actually equal to FTx ) and then solves using FT =
ma, and (incorrectly) subbing in 350N for the mass. (6 students)
V (0.8): Incorrect Force diagram and multiple conceptual errors. (10 students)
Y (0.6): Draws incorrect force diagram, writes sum of forces, and then either does
not plug in variables or plugs in only two forces listed in the problem. (3 students)
F (0.5): Drew force diagram (incorrectly) then stopped. (4 students)
Z (0.0): Blank or essentially blank. (3 students)