Exam 7 review
For
each class, I will select six questions from the Webassign review set and
modify them slightly, as in changing the diagram or graph slightly or changing
the numbers. Each question will have
four steps:
1.
Entering the principle most applicable to the situation
2.
Entering the system to which the principle is applied
3.
Choosing the correct answer
4.
Tying all three together with the explanation.
These
are the principles tested, which will also be provided at the beginning of the
test:
Net external force equals mass times
acceleration
Net external work equals change in system
energy
Net external
impulse equals change in system momentum
Net external force equals system mass times center of mass acceleration
Net external torque equals rotational
inertia times angular acceleration
Net external rotational impulse equals
change in system angular momentum
Though
not always true, here is some guidance about when a principle applies:
∑F
= ma 
To
determine the acceleration of an object or to determine an unknown force on
an object. 
∑W
= ΔE 
To
determine the speed or height of an object after some change has occurred. 
∑J
= Δp 
When
objects collide or fly apart (as in an explosion). 
∑F
= (m_{sys})(a_{cm}) 
When
internal forces rearrange a system; these questions often directly refer to
the center of mass concept. 
∑τ = Iα 
When
outside forces are applied to a bar or disk.
Sometimes they keeps the bar or disk stationary and sometimes they
cause an angular acceleration. 
∑J
= ΔL 
Conservation
of angular momentum questions when spinning systems rearrange themselves or
when a collision leads to rotational motion. 
And
some general guidance about what belongs in a system; again, not necessarily
true in every situation:
∑F
= ma 
Whatever
has mass and is accelerating. If a
block slides down a stationary ramp, the ramp is not accelerating, so it
doesn’t belong in the system. 
∑W
= ΔE 
It’s
often most convenient to put everything in the system so that the net
external work is zero. 
∑J
= Δp 
Whatever
has mass and is moving in the initial and/or final stage of motion. 
∑F
= (m_{sys})(a_{cm}) 
Whatever
has mass and is changing in position. 
∑τ = Iα 
Whatever
is spinning or potentially spinning. 
∑J
= ΔL 
Whatever
is spinning and/or objects moving with linear momentum that influence the
rotation of a rotating object. 
Before
I show some examples, I should mention a few common errors I’ve noticed in
Webassign explanations.
1.
Stating that an object moves with a constant velocity or that the center of
mass of a system moves with a constant velocity “because there are no external
forces on the system”. The idea that
there are no forces on the object or
system is really only approximately true in deep space. Everywhere else, it’s only correct to say,
“because the net external force is zero”.
For instance, a puck sliding on ideal ice coasts at a constant velocity
not “because there are no outside forces” but because “the net external force
(consisting of gravity and the normal force) add to zero.”
2.
Not being careful about defining systems when using conservation of
energy. For example, if spring energy is
in your equations, the spring must be in the system. If gravitational potential energy is in your
equations, the Earth must be in the system.
3.
Not being careful about defining systems when using Newton’s second law. For example, when a block slides along a
frictive surface and you are calculating the acceleration of it slowing down,
the block is the system. Just because
the floor is involved in the situation, does not mean the floor is in the
system.
Below
are some example questions. Each has the
original Webassign question, an example of how I would change it slightly, and then
an example of what a good test answer would be.
Original
question:

Modified
question:
The
coefficient of friction between a 120kg crate and the floor is 0.60. How hard must the crate be pushed so that
it slides to the right with an acceleration of 3m/s^{2}? a.
1200N b.
720N c.
360N d.
1560N e.
1080N 
Good
test answer:
1.
Entering the principle most applicable to the situation Net external force equals mass times
acceleration 2.
Entering the system to which the principle is applied The 120kg crate 3.
Choosing the correct answer e 4.
Tying all three together with the explanation. The sum of the force of gravity and the normal force are zero, so the net external force on the crate is equal to the force of the push plus the force of kinetic friction. Kinetic friction magnitude is equal to the normal force of 1200N times the coefficient of 0.60 and is negative if the crate is sliding rightward and kinetic friction opposes the sliding, so kinetic friction force is 720N. Net external force equals mass times acceleration then becomes F_{push} + (720N) = (120kg)(3m/s^{2}) and F_{push}= 1080N. 
Example
grading rubric:
Choosing
the correct principle 
1 pt 
Choosing
the correct system 
1 pt 
Choosing
the correct answer 
1 pt 
Stating
that the normal force is 1200N 
1 pt 
Stating
that the force of kinetic friction is –720N 
1 pt 
Stating
that the push force plus the force of kinetic friction equals the mass of the
crate times the acceleration of the crate 
2
pts 
With only six questions, you should have enough time
to write an overlythorough explanation for each answer.
Original
question:

Modified
question:

Good
test answer:
1.
Entering the principle most applicable to the situation Net external work equals change in
system energy 2.
Entering the system to which the principle is applied The block 3.
Choosing the correct answer d 4.
Tying all three together with the explanation. The only force inputting work to the system of the block is the force of the spring scale. This work input then appears as the kinetic energy of the block, ½mv^{2}. Because work is the area under a force versus position (or distance) graph, the option with the greatest area (d) has the greatest work input to the block. This implies the greatest kinetic energy for (d) and thus the greatest speed, v. 
Original
question:

Modified
question:
As shown above, a
disc of mass m is moving
horizontally to the right with a speed v
on a table with negligible friction when it collides with a second disc of
mass 2m. The second disc is moving horizontally to
the right with speed v/2 at the moment
of impact. After colliding, the
smaller disk rebounds with a velocity of –v/2. What is the final velocity of the larger
disk? (A) v/2 (B) 5v/4 (C) v (D) 2v (E) 3v/4 
Good
test answer:
1.
Entering the principle most applicable to the situation Net external impulse equals change in
system momentum 2.
Entering the system to which the principle is applied The small disk and the large disk 3.
Choosing the correct answer b 4.
Tying all three together with the explanation. The net external force on the system is
zero, so the net external impulse input to the system is zero. Therefore, there is no change in the
system’s momentum. The initial momentum
(p = mv) equals (m)(v)
+ (2m)(v/2) = 2mv and this must equal the final momentum. If 2mv = (m)(v)
+ (2m)(v_{f}), then v_{f}
= 5v/4. 
Original
question:

Modified
question:
A
student stands at the center of a raft floating in a pool with equally spaced
marks along the bottom. The student
and the raft have the same mass. The
student walks to the right end of the raft.
Which of the following best shows the final locations of the raft and
student relative to the marks at the bottom of the pool? Assume that there is no drag force between
the raft and water. 
Good
test answer:
1.
Entering the principle most applicable to the situation Net external force equals system mass
times center of mass acceleration 2.
Entering the system to which the principle is applied The raft and the person 3.
Choosing the correct answer a 4.
Tying all three together with the explanation. The net external force on the system is
zero because the downward force of gravity on the person and raft plus the
upwards force of the water on the person and raft add to zero. Therefore, the center of mass acceleration
of the person/raft system is zero. The
initial velocity of the center of mass is zero, so it remains zero as the
person walks. If the center of mass
velocity is always zero, then the center of mass does not move. The initial center of mass (taking the left
side as the origin and each mark as 1 equals (3)(m)
+ (3)(m) / 2m = 3. After the person
walks, the center of mass must also end at this position of 3. This is only true for option (a) where (m)(2.5) + (m)(3.5) / (2m) = 3. 
Original
question:

Modified
question:

Good
test answer:
1.
Entering the principle most applicable to the situation Net external torque equals rotational
inertia times angular acceleration 2.
Entering the system to which the principle is applied The light rod and two attached masses 3.
Choosing the correct answer D 4.
Tying all three together with the explanation. The net external torque on the system is
the torque of gravity on the left mass, M_{0}gl, plus the torque of
gravity on the right mass, 6M_{0}gl, which is 5M_{0}gl. The net rotational inertia of the system is
the rotational inertia of the left mass, M_{0}l^{2}, plus the
rotational inertia of the right mass, 12M_{0}l^{2}, which is
13 M_{0}l^{2}. The
angular acceleration is then equal to the net external torque divided by the
net rotational inertia, α =
which has a magnitude
of . 
Original
question:

Modified
question:

Good
test answer:
1.
Entering the principle most applicable to the situation Net external rotational impulse equals change
in system angular momentum 2.
Entering the system to which the principle is applied The bar and the puck 3.
Choosing the correct answer v = 4.
Tying all three together with the explanation. There
is no external torque on the bar/puck system, so the change in angular
momentum is zero and angular momentum is conserved. L_{bar} +
0 = 0 + L_{puck} if the puck begins at rest
and the bar ends at rest. Iω = Lmv so v = which has a magnitude of v = . 