Training Exam 2 S14
__________
your name
Each problem is of equal value.
You can skip one problem.
If you work all problems, we will count the best five grades.
Tips for better exam grades:
Read all problems right away and ask questions as early as possible.
Make sure that you give at least a basic relevant equation or figure for each
sub-problem.
Make use of the entire exam time.
Show your work for full credit. The answer ‘42’ only earns you any credit IF
‘42’ is the right answer. We reserve points for ‘steps in between’, figures,
units, etc. If all you give us is a numeric answer you may receive only a C
grade for the problem and in some cases even less.
No credit given for illegible handwriting or flawed logic in an argument.
All multiple choice questions may have more than one correct answer. For
full credit, you need to mark all correct answers and mark no incorrect
answer.
1. Energy
A wooden block with mass 1.5[kg] is placed against a compressed spring at the bottom of
an incline of slope 30° (point A). When the spring is released, it projects the block up the
incline. At point B, 6[m] up the incline from A, the block is moving up at 7[m/s]. k = 0.5
between block and incline. The mass of the spring is negligible.
Calculate the amount of potential energy that was initially stored in the spring.
2. Energy II
2. Work and Energy – use Energy Methods!
A tractor pulls an object down a hill of constant downward slope of 30° at constant velocity. The
weight force of the object is 2000[N]. The chain by which the tractor pulls the object is 30° above
the slope. The object that is being pulled experiences friction: k= 0.1.
a) Is the weight force of the object equal in magnitude but opposite in direction, or just equal in
magnitude, or just equal in direction, or neither with regard to the normal force of the object?
Explain!
b) How much work is done in pulling the object 10[m]?
3. Momentum I
Three objects A, B, and C are approaching the origin as they slide on a
frictionless horizontal air table. The initial positions are given in figure 1.
The masses of the objects are mA= 2[kg], mB= 4[kg], mC= 4[kg].
All objects arrive at the origin at the same time, 2.5 [s], and stick together
after the collision.
Figure 1
a) Where is the center of mass of the system of the three objects relative to
the origin when the objects are all at the initial positions?
b) What is the velocity of the center of mass?
c) What is the direction and magnitude of the velocity vector of the
compound object after the collision?
4. Momentum Conservation II
Two frictionless gliders on an air track undergo an elastic collision. The gliders
have different masses m1 and m2. Glider m1 moves initially to the right with
4[m/s] and glider m2 moves initially to the left with 6 [m/s].
a) If m1 has half the mass of m2 what will be the final velocity of each glider?
5. Rotational Motion I
Consider a modified Atwood Machine with torque (the pulley
possesses mass) – figure 2.
The rope is weightless and the contact between axle and pulley is
frictionless. The mass on the table does not experience friction.
mA = 20[kg] , mB= 12[kg], mP=10[kg]
Hint: Ignore air drag. The no slip condition applies to all motions.
Figure 2: The system.
Figure 3: The pulley.
a) The pulley has the shape of a thin walled hollow cylinder filled in part by
a solid cylinder and being supported and connected by 4 spokes, which
are slender rods (see figure 3). The radius of the inner cylinder is R, the
radius of the outer cylinder is 1.5 R. The mass of the inner cylinder is 6
[kg], the mass of the outer cylinder is 2 [kg], and the mass of each of the
four spokes is 0.5[kg].
Find the moment of inertia of the pulley as function of R.
b) Find the acceleration of the system. Use force methods and Newton’s
laws/ torque equivalent of Newton’s Laws. Show your work.
Hint: If you could not determine the answer to part a), use here the
general dependence of any moment of inertia on mass and radius so
that you can simplify the final formulas and leave an unspecified
constant, c, in the formula for I that carries the undetermined details of
the result of a) in it.
Use /g/ = 10 [m/s2]
c) Find the velocity of the system after the hanging mass has descended by
1[m] from the position it had at rest for the case where R= 0.4[m]. This
time, use energy methods. Show your work.
Use /g/ = 10 [m/s2]
6. Torque
All rods are 4[m] long. Places where forces act are in some cases indicated by the
distance measurements in the figures.
For the 8 situations, (a) to (h) in the figure, rank the torque magnitudes.
Hint: Count clockwise as positive and include sign into the ranking decision.
6. Rotational Motion II
Consider the situation in the figure and solve with energy methods. The mass on
the incline does not experience friction.
The mass on the incline is a solid sphere, which is connected to the string via a
yoke so that it can roll freely. There are neither air drag nor slip. The sphere has a
mass of 2M and a radius of 2R and the weight has a mass of M. Find the
acceleration a. Take v0 =0.
Master Equations – Physics 1210
One-dimensional motion with constant acceleration:

𝑥 = 𝑥0 + 𝑣0𝑥 𝑡 + 12𝑎𝑥 𝑡 2 find the other forms of master equation 1 by
(a) building the derivative of the equation
(b) solving the new equation for t and substituting it back into the master equation, and
(c) using the equation for average velocity times time
Two-dimensional motion with constant acceleration:
𝑥 =
𝑣0 𝑐𝑜𝑠𝛼 𝑡
𝑦 = 𝑣0 𝑠𝑖𝑛𝛼 𝑡 −
1
𝑎 𝑡2
2 𝑦
find the related velocities by building the derivatives of the equations
Newton’s Laws
Σ𝐹⃗ = 0,
Σ𝐹⃗ = 𝑚 𝑎⃗, 𝐹⃗𝐴→𝐵 = −𝐹⃗𝐵→𝐴 find the related component equations by replacing all
relevant properties by their component values
Energy
𝑊 =
1
1
𝐹⃗ ∙ 𝑠⃗𝐾 = 2 𝑚𝑣 2 𝑈𝑔𝑟𝑎𝑣 = 𝑚𝑔ℎ𝑈𝑒𝑙 = 2 𝑘𝑥 2 
energy conservation 𝐸1 = 𝐸2 , 𝐸 = ∆𝐾 + ∆𝑈 + 𝑊𝑜𝑡ℎ𝑒𝑟
Momentum
𝑝⃗ = 𝑚 𝑣⃗momentum conservation for a system: P1,sys = P2,sys
𝑟⃗𝐶𝑀 =
Σ𝑟⃗𝑖 𝑚𝑖
Σ𝑚𝑖
position of the center of mass
find the velocity equation for the center of mass by building the derivative
𝑚𝐴 𝑣⃗𝐴1 + 𝑚𝐵 𝑣⃗𝐵1 = (𝑚𝐴 + 𝑚𝐵 ) ∙ 𝑣⃗2 for one type of collisions, and
𝑣⃗𝐴𝑥 =
𝑚𝐴 −𝑚𝐵
𝑚𝐴 +𝑚𝐵
∙ 𝑣⃗𝑥 𝑎𝑛𝑑 𝑣⃗𝐵𝑥 =
2𝑚𝐴
𝑚𝐴 +𝑚𝐵
∙ 𝑣⃗𝑥 for a collision of two objects, A and B, where 𝑣⃗𝑥 is
the relative velocity of the incoming object.
𝑠 = 𝑟𝜃, 𝑥 → 𝜃, 𝑣 → 𝜔𝑧 , 𝑎 → 𝛼𝑧 , 𝑚 → 𝐼, 𝐹 → 𝜏𝑧 , 𝑝 → 𝐿
the translation key to apply the equations of linear motion to rotational motion.
𝑣 = 𝑟 𝜔𝑧 , 𝑎𝑡𝑎𝑛 = 𝑟𝛼𝑧,
𝐾𝑟𝑜𝑡 =
1
𝐼
2
𝑎𝑟𝑎𝑑 = 𝑟 𝜔𝑧2 rotational kinematics
𝜔𝑧2 , 𝐼 = Σ𝑚𝑖 𝑟𝑖2 , 𝐼 = ∫ 𝑟 2 𝑑𝑚 , 𝐼𝑃 = 𝐼𝐶𝑀 ∙ 𝑑2 rotational dynamics
𝜏⃗𝑧 = 𝑟⃗ × 𝐹⃗ , Σ𝜏𝑧 = 𝐼 𝛼𝑧
Lego formulas:
2
𝑀𝑅2
Solid sphere
𝐼=
Solid cylinder
𝐼=
Thin hollow cylinder
𝐼 = 𝑀𝑅2
5
1
2
𝑀𝑅2
Thin rod spinning at center (short spoke) 𝐼 =
Thin rod spinning at end (long spoke) 𝐼 =
1
12
1
3
𝑀𝑅2
𝑀𝑅 2
Need to take your mind of the exam for a second? Check this out:
If you want to be a physicist, you must do three things -- First, study
mathematics, second, study more mathematics, and third, do the same.
-- Arnold Sommerfeld (German Physicist, 1868-1951)
What Einstein said wasn't all that stupid.
-- Wolfgang Pauli as a student, after hearing Einstein, 20 years his
senior give a lecture.
Physics is becoming so unbelievably complex that it is taking longer and
longer to train a physicist. It is taking so long, in fact, to train a
physicist to the place where he understands the nature of physical problems
that he is already too old to solve them.
-- Eugene Wigner
The Feynman Problem Solving Algorithm:
1) Write down the problem.
2) Think very hard.
3) Write down the solution.
"Physics is like sex: sure, it may give some practical results, but that's not why we do it." R. Feynman
About eight years ago, when I was studying in high school in
India, my Chemistry professor was trying to explain the "screening
effect" of electrons (a phenomenon that makes metals bind their
electrons less losely then other elements, resulting in conductivity).
He tried to give an analogy, using earth and moon.
He said, "Imagine if their was another moon orbiting earth, then the
pull that our true moon faces will be smaller." I was puzzled and
declared that it is not possible. To which he further explained," Well
it's like this. The earth now has to pull two moons instead of one
hence it has to divide its force among the too, hence its pull on the
moon will be halved."
At this point I argued that all the artificial satellites in
the sky must face lesser pull by earth when ever a new satellite is
launched. " That's true," he said,"and that's why the cost of
launching satellites is going up these days....
Physics is not a religion. If it were, we'd have a much easier time
raising money. - Leon Lederman
"Heavier-than-air flying machines are impossible."
Lord Kelvin, president, Royal Society, 1895.