Download Final Exam Problems - Engineering Physics I | PHY 303K and more Exams Physics in PDF only on Docsity!
This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points One particle, of mass m , moves with a speed v in the x-direction, and another particle, of mass 2 m , moves with a speed v/2 in the y-direction. What is the velocity of the center of mass of these two particles?
v 2
ˆx +
v 2
ˆy
v 3
ˆx +
v 2
ˆy
v 2
ˆx +
v 3
ˆy
- v xˆ +
v 2
yˆ
v 3
ˆx +
v 3
ˆy correct
Explanation:
Let : m 1 = m , ~v 1 = v x ,ˆ m 2 = 2 m , and ~v 2 =
v 2
ˆy.
The velocity of the center of mass of the two particles is
V^ ~CM = m^1 ~v^1 +^ m^2 ~v^2 m 1 + m 2
=
m v ˆx + 2 m (v/2) ˆy m + 2 m
=
v 3
ˆx +
v 3
ˆy.
002 10.0 points How fast must a roller coaster car go through a circular dip for you to feel three times as “heavy” as usual, due to the upward force of the seat on your bottom being three times as large as usual? The center of the car moves along a circular arc of radius R as in the figure below.
b (^) b b b b b
~v
R
- |~v| =
2 g R correct
- |~v| =
g R 2
- |~v| = 3g R
- |~v| =
g R
- |~v| = 3
gR
- |~v| = 2g R
- |~v| = 2
g R
- |~v| =
g R
- |~v| =
3 gR
- |~v| = g R
Explanation: At the bottom of a hill (or a “dip”), the net force on the rider is upward. The force by the seat on the rider is also upward, and in this case has a magnitude of 3m g. The net force on the rider is
Fnet = Fseat + Fgrav
〈 0 ,
m |~v|^2 R
, 0 〉 = 〈 0 , 3 m g, 0 〉 + 〈 0 , −m g, 0 〉
m |~v|^2 R
= 2m g
⇒ |~v| =
2 g R.
003 10.0 points Two blocks, one of mass M and the other of mass m, on a frictionless horizontal surface
are connected by a massless string. A horizon- tal force F pulls the mass m in the direction away from mass M. What is the tension in the string?
M F
M + 2m
- m F.
- (m + M ) F.
- M F.
2 m F M + m
m F 2 M + m
2 M F
M + 2m
2 M F
M + m
m F M + m
M F
M + m
. correct
Explanation: The force F accelerates the system of mass m+M , so F = (m+M ) a. Applying Newton’s second law to mass M ,
M a = T T =
M F
M + m
004 10.0 points A ball moves in the direction of the arrow labeled d in the following diagram. The ball is struck by a stick that briefly exerts a force on the ball in the direction of the arrow labeled g. Which arrow best describes the direction of ∆~p, the change in the ball’s momentum?
a b
c d e
f
g
h
- h
- e
- d
- f
- c
- a
- g correct
- b
Explanation: Recall the definition of impulse:
Impulse = F~net∆t = ∆~p.
Therefore, whatever direction the net force points in, that will be the direction of the change in the ball’s momentum. We are told that the force is in the direction of g, so that is the correct answer.
005 10.0 points
A uniform rod, supported and pivoted at its midpoint, but initially at rest, has a mass 2 m and a length ℓ. A piece of clay with mass m and velocity v 0 hits one end of the rod, gets stuck and causes the clay-rod system to spin about the pivot point O at the center of the rod in a horizontal plane. Viewed from above the scheme is
y ˆ
ˆz^ ˆx
In which direction is the net torque vector ~τ about the hinge due to the force applied to the door handle?
- ˆz
- xˆ
- Insufficient information is given.
- −yˆ
- ˆy correct
- −xˆ
- −ˆz
Explanation:
~τ = r × F
r ~τ
F ~r points in the ˆz direction from the hinge to the spot where F~ acts, so by the right hand rule, ~r × F~ points in the direction ˆy, along the hinge line.
008 10.0 points Object A with mass 4 kg moves at an initial speed of 2 m/s along a frictionless horizontal plane. A horizontal force F is applied oppo- site to the direction of the motion and brings object A to a stop in a distance ∆xA. Object B with mass 2 kg moves at an initial speed of 4 m/s along the frictionless horizontal plane. The same horizontal force F is applied oppo- site to the direction of the motion of B and brings object B to a stop in a distance ∆xB. What is the relation between ∆xA and ∆xB?
- ∆xA=8∆xB
- ∆xB =8∆xA
- ∆xB =6∆xA
- ∆xA=6∆xB
- ∆xA=2∆xB
- ∆xB =2∆xA correct
- ∆xA=4∆xB
- ∆xA=∆xB
- ∆xB =4∆xA Explanation: Consider each ball and the earth as three separate systems. Using the Energy Principle with the rest mass not changing, we know ∆K = W. The displacement is only in the x-direction and the only force is F , which is opposite to the direction of motion so
WA = −F ∆xA
and WB = −F ∆xB For both blocks, Kf = 0 so the Energy Principle says:
F ∆xA = 0 −
mAv A^2
and F ∆xB = 0 −
mB v^2 B
Taking the ratio of the two equations gives:
(F ∆xA)/(F ∆xB) = (mAv^2 A)/(mB v^2 B ) (∆xA)/(∆xB) = (mAv^2 A)/(mB v^2 B ) = [4kg(2m/s)^2 ]/[2kg(4m/s)^2 ] = 1/2 so ∆xB = 2∆xA
009 10.0 points
Two spheres, A and B, have the same mass and radius. However, sphere B is made of a dense core and a less dense shell around it. How does the moment of inertia of sphere A about its center of mass compare to the moment of inertia of sphere B about its center of mass?
Ia. IA > IB Ib. IA < IB Ic. IA = IB
If the two spheres are rolled down an incline from the same height simultaneously,
IIa. sphere A reaches the bottom first. IIb. sphere B reaches the bottom first. IIc. spheres A and B reach the bottom simul- taneously.
Choose the correct pair of statements.
- Ia,IIc
- Ib,IIa
- Ic,IIa
- Ic,IIb
- Ia,IIb correct
- Ic,IIc
- Ib,IIb
- Ib,IIc
- Ia,IIa
Explanation:
Since the two spheres share the same mass and radius, a comparison of their moment of inertias depends on how their mass is dis- tributed relative to an axis through the center of mass. Sphere B has a dense core, implying that its mass is on average closer to the axis than A’s mass. Thus, A has a greater moment of inertia than B. Let β be the unitless parameter in the mo- ment of inertia formula I = βM R^2. Sphere B has a smaller β from the preceding argument. As a consequence of energy conservation, it can be shown that
vCM =
2 gh 1 + β
Therefore, B reaches the bottom first.
010 10.0 points A simple harmonic oscillator is described by the function
x(t) = (6) cos(π t + 2).
where the amplitude is measured in cm and the frequency is measured in s−^1. What is the acceleration of this oscillator at t = 0?
- 2 cm/s^2
- − 18 .8 cm/s^2
- 24 .6 cm/s^2 correct
- 18 .8 cm/s^2
- 59 .2 cm/s^2
- Zero
- 6 cm/s^2
- − 24 .6 cm/s^2
- − 59 .2 cm/s^2
- −6 cm/s^2 Explanation:
Ephoton = |∆E 1 , 3 | = |E 1 − E 3 | = |−8 eV + 2.6 eV| = 5.4 eV Ephoton = |∆E 2 , 3 | = |E 2 − E 3 | = |− 5 .2 eV + 2.6 eV| = 2.6 eV
013 10.0 points
q 1 q 2 Ω 1 Ω 2 Ω 1 Ω 2 0 4 1 5985 5985 1 3 18 1140 20 , 520 2 2 171 171 29 , 241 3 1 1140 18 20 , 520 4 0 5985 1 5985
Two blocks contain 18 oscillators each and share a total of 4 quanta. Above is a table showing the number of microstates in block 1 (Ω 1 ), the number of microstates in block 2 (Ω 2 ), and the total number of microstates (Ω 1 Ω 2 ) for each possible pair of q 1 and q 2. If you look at one particular oscillator once every minute, approximately how long would it take (on average) to observe this oscillator having all 4 quanta?
- 2 billion years
- 2 days
- 2 minutes
- A few seconds
- 2 million years
- 2 months correct
- Longer than the age of the universe
- 2 hours
- 2 thousand years
- 2 years
Explanation: The total of the last column is 82,251. Thus, the probability of finding the system in one particular state is 1/ 82 ,251, and the system would have to be checked 82,251 times (on average) to observe this result, taking 82 ,251 minutes, or about 60 days:
(82,251 m)
1 day 24 · 60 m
≈ 57 .12 days.
An alternate solution is to consider the two blocks as one large block. Then, the mi- crostates formula gives
with the time calculation proceeding as be- fore.
014 10.0 points Find the approximate kinetic energy of a circular wheel of radius r and mass M that is rotating about its center at 2 cycles/s. As- sume the wheel’s mass is concentrated at the rim and the mass of the wheel’s spokes is neg- ligible.
r
- 40 M r^2
- 6 M r
- 80 M r^2 correct
- 4 M r^2
- 2 M r Explanation: The moment of inertia of a thin ring or hollow cylinder about its axis is
I = M r^2.
The angular speed of the wheel in radians is
ω =
2 cycles s
2 π rad cycle
= 4 π rad/s
The wheel is only rotating, not translating, so its kinetic energy is
K = Krot =
I ω^2
=
M r^2
(4 π)^2 = 78. 9568 M r^2.
015 10.0 points A certain laser outputs pure red light (photon energy 1.8 eV) with power 600 mW (milli- Watts = 0.6 Watts). How many photons per second does this laser emit?
- 3.03819e+
- 1.90972e+
- 2.86458e+
- 2.60417e+
- 2.77778e+
- 2.17014e+
- 2.69097e+
- 2.08333e+
- 3.125e+
- 1.99653e+
Correct answer: 2. 08333 × 1018. Explanation: Convert eV to Joules:
(1.8 eV)
1. 6 × 10 −^19 J
eV
≈ 2. 88 × 10 −^19 J.
Then, number of photons emitted per sec- ond is simply
0 .6 W
2. 88 × 10 −^19 J
≈ 2. 08333 × 1018
016 10.0 points Alcohol is a liquid less dense than water, and mercury is a liquid denser than water. Given this information, any object that will float on water
- may float on mercury and will float on alcohol.
- may sink in mercury and will sink in alcohol.
- will float on mercury and may float on alcohol. correct
- will sink in mercury and may sink in alcohol.
- will sink in mercury and alcohol.
- will float on mercury and alcohol.
Explanation: An object floats when the buoyancy force balances the object’s weight. The buoyancy force is equal to the weight of the fluid dis- placed by the object. A liquid with higher density than water has a buoyancy force equal to or greater than that of water. So, any object that will float on water will float on mercury. A liquid with lower density than water has a buoyancy force equal to or less than that of water. If the buoyancy force is equal to that of water, the object will float lower in the liquid. But, if the buoyancy force is less than that of water, the object will sink. So, any object that will float on water may or may not float on alcohol.
017 10.0 points A conservative force has the potential energy function U (x), shown by the graph. A particle moving in one dimension under the influence of this force has kinetic energy 1.0 Joule when it is at position x 1.
Which statement is correct about the mo- tion?
- The particle slows down, comes momen- tarily to a rest, then returns with increasing speed to its starting point. correct
- The particle moves at constant speed, but not at constant velocity.
- The particle moves along a portion of a circle, as the sketch shows directly.
- The particle follows the path of a projec- tile, like a thrown baseball.
- The particle at first speeds up, then slows down, then speeds up again.
Explanation: Since vx is the slope of x(t), the particle slows down, comes to momentary rest, and then reverses direction, increasing speed as it returns to its starting point.
020 10.0 points Static friction 0.49 between a 0.3 kg block and a 4.3 kg cart. There is no kinetic friction between the cart and the horizontal surface. The acceleration of gravity is 9.8 m/s^2.
0 .3 kg F
4 .3 kg μ
9..^49
8 m
/s
2
What minimum force F must be exerted on the 4.3 kg cart in order for the 0.3 kg block not to fall?
- F = 68 N
- F = 48 N
- F = 92 N correct
4. F = 42 N
5. F = 49 N
6. F = 90 N
7. F = 46 N
8. F = 82 N
9. F = 91 N
10. F = 84 N
Explanation:
Let : mA = 4.3 kg , Cart A mB = 0.3 kg , Block B μAB = 0. 49 , between A and B μk = 0 , horizontal surface, and g = 9.8 m/s^2.
mA = 4.3 kg (^) m B = 0 .3 kg
F NBA NAB
mA g mB g
fμ
fμ
The condition that block B not fall implies that its vertical acceleration is zero. Applying Newton’s second law for B in the horizontal and vertical directions yields ∑ Fx = N = mB ax (1) ∑ Fy = μAB N − mB g = 0 , (2)
where N is the normal force exerted on block B by cart A. Applying Newton’s law on cart A in the horizontal direction we have
F − N = mA ax.
Solving for F ,
F = N + mA ax. (3)
The normal force N and the acceleration ax may be determined from the first two equa- tions. From Eq. 2 we find
N =
mB g μAB
From Eq. 1 and 2, and solving for ax yields
ax =
N
mB
g μAB
Since we now have expressions for N and ax, we can find the force F required so that block B does not fall. From Eq. 3 again
F = N + mA ax = mB
g μAB
g μAB
= (4.3 kg + 0.3 kg)
9 .8 m/s^2
- 49 = 92 N.
021 10.0 points
Moving objects left traces labeled I-VI in the figure above. The dots were deposited at equal time intervals (for example, one dot each second.) In each case the object starts from the square. Which trajectories show ev- idence that the moving object was interacting with another object somewhere?
- II,IV,V,VI
- II,III,IV,V,VI correct
- I,II,III,IV,V
- I,III,IV,V,VI
- II,III,IV,V
- III,IV,V,VI
- I,II,III,IV,V,VI
- I,II,IV,V
- I,II,III,VI
- II,III,IV,VI
Explanation: According to Newton’s First Law, an ob- ject will travel at constant velocity except to the extent that it interacts with other objects. The only path that is constant velocity (con- stant speed and straight line motion) is path I; thus, all of the others exhibit evidence of
m
D
μk
F
θ
Calculate the work done by the friction f.
- f = −μk (m g cos θ + F sin θ) D correct
- f = μk m g sin θ D
- f = −μk m g cos θ D
- f = μk (m g sin θ − F cos θ) D
- f = −μk (m g sin θ + F cos θ) D
- f = μk (m g cos θ + F sin θ) D
- f = μk m g cos θ D
- f = −μk m g sin θ D
- f = −μk (m g sin θ − F sin θ) D
- f = −μk (m g cos θ − F cos θ) D
Explanation: The force of friction has a magnitude Ff riction = μk N. Since it is in the direc- tion opposite to the motion, we get
Wf riction = −Ff rictionD = −μk N D. = −μk (m g cos θ + F sin θ) D
025 10.0 points A circular disk with moment of inertia 1 2
m R^2 , mass m = 8 kg and radius R = 0.4 m
is mounted at its center, about which it can rotate freely. A light cord wrapped around it supports a weight m g. Assume acceleration due to gravity is 9.8 m/s^2.
R
I
ω
m
T
g
Find the total kinetic energy of the system when the weight is moving at a speed v = 2 m/s.
Correct answer: 24 J.
Explanation: v = R ω , so the total kinetic energy is
Ktot = Km + Krot
m v^2 +
I ω^2
m v^2 +
m R^2
v R
m v^2
=
(8 kg)(2 m/s)^2
= 24 J
026 10.0 points
Blocks A and B have 300 and 200 oscilla- tors, respectively. The blocks are placed in contact with each other in an insulated en- vironment. At the instant the blocks touch, block A has 10 quanta of energy while block B has 90 quanta of energy. What will happen some time after contact?
I. The entropy of block A will increase. II. The entropy of block B will increase. III. The entropy of the two block system will increase.
- III only
- II,III only
- I only
- I,II,III
- None
- I,III only correct
- II only
- I,II only
Explanation: Since block A has more oscillators than block B, A will have more quanta than B at thermal equilibrium. This situation is similar to the left side of the graph in text figure 12.26. As block B transfers quanta to block A, block B’s entropy decreases while block A’s entropy increases. But block A gains more entropy than block B loses, so the system increases in entropy overall, as required by the second law of thermodynamics.
027 10.0 points
A newly discovered planet has twice the mass of the Earth, but the acceleration due to grav- ity on the new planet’s surface is exactly the same as the acceleration due to gravity on the Earth’s surface. What is the radius Rp of the new planet in terms of the radius R of Earth?
- Rp =
R
- Rp = 4 R
- Rp = 2 R
- Rp =
2 R correct
- Rp =
R
Explanation: From Newton’s second law and the law of universal gravitation, the gravitational force near the surface is
Fg = m g = G
M m r^2 g =
G M
r^2
Mp = 2 Me and gp = ge, so
G Me R^2
G Mp R^2 p
2 G Me R^2 p 1 R^2
R^2 p Rp =
2 R.
028 10.0 points A sample of an ideal gas is in a tank of con- stant volume. The sample absorbs heat en- ergy so that its temperature changes from 206 K to 412 K. If v 1 is the average speed of the gas molecules before the absorption of heat and v 2 their average speed after the absorption of heat, what is the ratio
v 2 v 1
? Hint: the aver- age kinetic energy of a molecule is given by E =
kB T.
- II, IV and V.
- V and VI.
- I, III, and IV.
- All of the above.
- III, IV, V
- II, III and IV.
- III and V.
Explanation: I. is true because using graphic subtraction of vectors, B~ − ~R is a vector that points from the tip of ~R to the tip of B~, which is D~. II. is false. ~A − B~ is the vector −P~ III. is true because when using graphic ad- dition of vectors, we put vectors ~A and B~ tip to tail. ~R points from the tail of A~ to the tip of B~. IV. is true. We can use P~ = B~ − A~ and R~ = B^ ~ + ~A so that P~ + ~R = B~ − ~A + B~ + A~ = 2 B~. V. is false using graphic vector subtraction. IV. is true because ~A = −D~ and ~B = −C~ so adding them together gives ~A + B~ + C~ + D~ = −D~ − C~ + C~ + D~ = ~ 0 Therefore, the correct answer is 1.
