Mark answers in spaces 53-75 on the answer sheet
No class Friday April 16
No class Friday April 16
PHYSICS 221
Spring 2004
EXAM 3: April 15 2004 8:00pm—9:30pm
SOLUTIONS
[53] Disk #1 of uniform density of mass 1kg and radius 2m is spinning centered on a
massless turntable with angular velocity ω=12 rad/s. Disk #2 of uniform density of mass
2kg and radius 1m is dropped on it and due to friction between the disks, the system of
two disks eventually rotate at a common angular velocity about the axis of the turntable.
What is the final angular velocity of the system?
(A) 2 rad/s
(B) 4 rad/s
Disk 2 r2=1m; M2=2kg
(C) 6 rad/s
Disk 1 r1=2m; M1=1kg
(D) 8 rad/s
(E) 10 rad/s
Massless Turntable
The moment of inertia of disk 1 is I1=(M1r1²)/2=2 kg m². The moment of inertia of disk 2
is I2=(M2r2²)/2=1 kg m² . The initial angular momentum is L1=I1ω1=(2 kg m²)(12
rad/s)=24 J s. Since the situation applies no torque to the system, the angular momentum
will be preserved so the final angular momentum will be the same. The final moment of
inertia is If=I1+I2=3 kg m². The final angular velocity is therefore ωf=Lf/ If=(24Js)/(2 kg
m²)=8rad/s.
Correct response=60%
[54] What is the x-component of the angular momentum about the origin of a particle
G
G
located at the point r = (2 ˆj ) m with mass 2kg and a velocity of v = (iˆ + 2 ˆj + 3kˆ) m / s
(A) +6 kg m²/s
(B) -6 kg m²/s
(C) +12 kg m²/s
(D) -12 kg m²/s
(E) 0 kg m²/s
Angular momentum is given by
G G
G
L = r × mv = (2kg )(2 ˆj m)(iˆ + 2 ˆj + 3kˆ)m / s = +12kg m 2 / s
Correct response=44%
The most popular wrong answer was A which I guess results from forgetting to
multiply by the mass.
[55]
Three balls are rolled down an incline ramp and roll without slipping. Ball X is a
solid ball of radius 5cm and mass 1kg. Ball Y is another solid ball of radius 10cm and
mass 0.5kg. Ball Z is a hollow ball of radius 5cm and mass 1kg. Neglecting drag, kinetic
and rolling friction, if these three balls are released simultaneously from the top of the
ramp, in which order do they arrive at the bottom?
(A) X, Y, Z
(B) Z, Y, X
(C) X tied with Y followed by Z
(D) X tied with Z followed by Y
(E) All three tied.
As these balls roll down the ramp, they should undergo uniform acceleration so that if
vf is the final velocity at the bottom of the ramp and L is the length of the ramp, the
time to get to the bottom is:
t = 2L / v f
Thus, the ordering of the objects is the objects in time is the reverse of the ordering
in terms of final velocity. The fastest final velocity is the ball which gets to the
bottom first (hopefully this intuitively obvious).
The problem is thus reduced to determining the velocity of the balls at the bottom of
the ramp.
Consider now a ball of radius r, moment of inertia I and mass m. Denote the height of
the ramp by h. The total kinetic energy which the ball will have at the bottom of the
ramp will be
K=mgh.
If a ball is rolling without slipping, ω = v / r so that the kinetic energy in terms of
velocity is
K = 12 mv 2f + 12 Iω 2 = 12 m 1 + I /(mr 2 ) v 2f
[
]
Equating the two expressions for K we find:
2 gh
1 + I /(mr 2 )
For a solid ball, regardless of mass or radius, I/(mr²)=2/5 for a hollow ball, again
regardless of mass or radius, I/(mr²)=2/3. Thus the two solid balls (XY) will tie
while the solid ball (Z) with lower vf will come in last.
vf =
Correct Response=42%
[
]
[56] Consider an object with weight 100N suspended from the ceiling by three
massless strings as shown. What is the ratio between the tension in string 1 and the
tension in string 3
(A)
(B)
(C)
(D)
(E)
T1:T3=4:5
T1:T3=5:4
T1:T3=3:5
T1:T3=5:3
T1:T3=3:4
5m
Ceiling
3m
4m
T1
T2
T3
100N
We can simplify this problem by
noticing that the strings 1 and 2
form a right triangle.
5m
Ceiling
θ1
3m
θ2
4m
G T1
F1
T2 G
The net force on the junction of
F2
the three strings must be 0 if it is
G
G G
G
F3
T3
equilibrium: F1 + F2 = − F3 . If we
take a coordinate system with the
x-axis parallel to string 2 and the y
100N
G
axis parallel to string 1, then F1 is
G
the projection of − F3 onto the y-axis. Thus, F3=T3=100N so F2=T2=(100N)sin(θ1).
Thus, T1:T3 = 1: sin(θ1). From the diagram sin(θ1)=4/5 so , T1:T3 = 4:5. Another way to
do this problem is to use a coordinate system where the y-axis is vertical and the x-axis is
G G
G
horizontal. In that case when you implement F1 + F2 = − F3 , The y component equates
the vertical components of the tension in strings 1 and 2 with the weight while the x
component tells us that the horizontal components of strings 1 and 2 cancel.
Correct response=31%
The most popular wrong answer was C which suggest to me that students are mixing up
sin and cos. A good check one can perform in this case is to imagine that we make string
2 very long then θ1 goes to 90° and the full weight is taken by string 1. It is always good
to perform a check like this if you are not sure of sin versus cos, a common problem
apparently.
[57] The horizontal beam in the figure weighs
150N and is of uniform density. Find the tension
in the 5m cable.
(A) 1200N
(B) 200N
(C) 300N
(D) 500N
(E) 625N
Hinge
Tension =?
Taking torques about the hinge, The weight will have a negative torque of -(4m)(300N).
The bar, where the weight acts on the center of mass the torque is negative and equal to –
(2m)(150N). The torque produced by the tension is positive and equal to (3/5)(4m)T.
Setting the total torque to 0 we get.
(4.00 m) (3 5)T = (4.00 m)(300 N) + (2.00 m)(150 N),
so T = 625 N.
Correct Response=53%
[58] Two uniform spheres, each with mass M and radius R , touch each another. What
is the magnitude of their gravitational force of attraction?
(A) F=GM²/(R²)
(B) F=GM²/(2R²)
(C) F=GM²/(4R²)
2R
(D) F=GM²/(8R²)
(E) F=GM²/(16R²)
The distance between the centers of the two spheres is 2R. Using Newtons law of gravity
generalized to spherically symmetric objects:
F = GMM /(2 R) 2 = GM 2 /(4 R 2 )
Correct Response=54%
[59] The mass of the Sun is about 3×105 times the mass of the Earth. The magnitude of
the gravitational force exerted by the Sun on the Earth is ________ the magnitude of the
gravitational force the exerted by the Earth on the Sun.
(A) About 9×1010 times lager then
(B) About 3×105 times lager then
(C) The same as
(D) About 3×105 times smaller then
(E) About 9×1010 times smaller then
The correct answer is C which follows from Newton’s 3rd law.
Correct response=81%
[60] Consider the three spherical masses S, T and U shown in
the diagram. S has a mass of 15kg and is located at (0m,0m,0m).
T has a mass of 50kg and is located at (0m,2m,0m). U has a mass
of 10kg and is located at (-1m,0m,0m). Assume that all the
masses are initially at rest in a region remote from any
gravitational forces aside from the forces they exert on each
other. In addition the gravitational force between these masses is
the only force they are subject to. What is the magnitude of the
initial acceleration of mass S?
40kg
2m
(A) 0.67 nm/s²
(B) 0.95 nm/s²
(C) 1.34 nm/s²
(D) 1.90 nm/s²
(E) 2.68 nm/s²
15kg
10kg
1m
By the principle of superposition, the total force on S is the sum of the forces produced
by U and T. Note that mT=4mU but rST=2rSU thus mT/rST2=mU/rSU2
G G
G
F = FSU + FST
( )
( )
2
= G (ms mU / rSU
) − iˆ + G (ms mT / rST2 ) + ˆj
2
= G (mS mU / rSU
)(−iˆ + ˆj )
The acceleration given by Newton’s second law is thus:
G G
2
a = F / mS = G (mU / rSU
)(−iˆ + ˆj )
Therefore
G
2
2
) | −iˆ + ˆj |= G 2 (mU / rSU
)
a =| a |= G (mU / rSU
= 0.95nm / s 2
Correct Response=63%
[61] Three satellites orbit the earth as shown . Satellite X orbits in a circular orbit with
radius R and period TX. Satellite Y orbits in a circular orbit with radius 2R and period TY.
Satellite Z orbits in an elliptical orbit where the distance to the center of the earth varies
between R and 2R and has period TZ. What is the ratio between the periods of the three
satellites?
Z
(A) TX:TY:TZ= 2 : 4 : 3
(B) TX:TY:TZ= 2 : 2 : 3
(C) TX:TY:TZ= 2 : 4 : 6
(D) TX:TY:TZ= 2 : 2 : 6
(E) TX:TY:TZ= 8 : 8 : 27
Y
2R
R
X
The semi-major axis of each of the orbits are aX=R aY=2R and aZ=3R/2. According to
Keplar’s third law, T ∝ a 3 / 2 therefore:
TX : TY : TZ = (a X : aY : aZ )3 / 2
= (2 : 4 : 3)3 / 2 = 8 : 64 : 27
= 8 : 8 : 27
Correct Response=44%
I am not sure why a straightforward question like this is so low. You could get the correct
answer you could just look at the circular orbits X and Y and not even worry about the
elliptical case.
[62] Planet X has a radius of 1000km and the surface gravity is 1m/s². If a projectile is
fired straight upwards from the surface of X with initial velocity 1km/s what is the
maximum height above the surface of X which the projectile achieves? (neglect air
resistance)
(A) 414 km
(B) 500 km
(C) 1000 km
(D) 1500 km
(E) 2500 km
The potential energy of the projectile of mass m is given by
U=-GmM/r
where M is the mass of planet X. The acceleration of the gravity on the surface of X is
given by
g=GmM/R²
Where R is the radius of X. We can therefore eliminate the unknown value of M and
obtain the expression for potential:
U = − gR 2 / r
The total mechanical energy of the projectile when launched from the surface of X is
E = K + U = 12 mv 2 − gRm
This will not change as the projectile moves. At the highest point, this will all be potential
energy so:
− gR 2 m / r = 12 mv 2 − gRm
 gR 
r=
R
1 2
 gR − 2 v 
where r is the farthest distance from the center of X.
Plugging in the numbers




gR
(1m / s 2 )(106 m)
r=
R
=
1000km

2
2
6
2
1
1
 gR − 2 mv 
 (1m / s )(10 m) − 2 (1000m / s ) 
= 2000km
The height of the projectile is h=r-R=1000km
Correct response=30%
The most popular wrong answer was B which is what you would get if you assumed that
the gravitational force was constant. Students who got this wrong should think carefully
about the situation where using a constant acceleration of gravity is a good approximation
and where it is not.
[63] Which of the following relationships between the acceleration, a, and the
displacement x of a particle describes a case of simple harmonic motion?
(A) 3x-a=0
(B) 3x+a=0
(C) 3xa=1
(D) 3x²+a=0
(E) 3x²-a=0
For Harmonic motion, the restoring force, and hence the acceleration, should be directed
towards the equilibrium point and be linear in the displacement from equilibrium. The
only expression which has that property is (B).
Correct Response=37%
[64] A stick of length L is suspended at one of its ends. What is the period of its small
oscillations?
L
(A) T = 2π
g
(B) T = 2π
2L
g
(C) T = 2π
L
2g
(D) T = 2π
L
3g
(E) T = 2π
2L
3g
Period=?
For this problem we use the physical pendulum formalism. The moment of inertia of the
rod about the suspension point is I=ML²/3. The distance to the suspension point is d=L/2.
The period of a physical pendulum is:
T = 2π
I
=
mgd
ML2 / 3
=
MgL / 2
2L
3g
Correct response = 19%
The most popular wrong answer was A, I assume that many students incorrectly
identified this as an ideal pendulum even though all the mass is not concentrated at the
end of the stick.
[65] A particle of mass m=2.0 kg is connected to a spring with spring constant
k=4.0N/m. If it oscillates with total mechanical energy E=16 J, what is the maximum
velocity of the mass during the oscillation? Take the zero of potential energy to be the
equilibrium point of the spring
(A)
(B)
(C)
(D)
(E)
0.5 m/s
1.0 m/s
2.0 m/s
4.0 m/s
8.0 m/s
When the kinetic energy is maximum then the potential energy is 0 so all of E is kinetic
energy. This is therefore related to the velocity by E=K=mv²/2. Thus solving
16J=(2kg)v²/2
we obtain v=4m/s.
Correct Response=75%
[66] A block of mass M=4.00kg rests on a horizontal frictionless track. It is connected
to either end of the track by two identical ideal springs with spring constant k=50.0N/m.
If the mass is displaced by a small amount, what is angular frequency of oscillation, ω?
(A) ω=3.5s−1
(B) ω=5.0s−1
(C) ω=7.1s−1
k=50.0N/m M=4.0kg k=50.0N/m
−1
(D) ω=10.0s
(E) ω=50.0 s−1
Frictionless Track
Following exercise 13.96 the effect of the
two springs is equivalent to a single spring of spring constant keff=k+k=100N/m. The
angular frequency is thus
ω=
keff
M
=
100 N / m
= 5s −1
4kg
Correct Response=50%
Most popular wrong answer was A which is what you would get if you ignored the
second spring.
[67] Four blocks of insulating material of mass m1=1kg, m2=2kg, m3=3kg, m4=4kg are
on a frictionless horizontal surface as shown on the figure below. They are evenly spaced
and the distance between the 1kg and the 4kg blocks is 10m. The 1kg block has a net
electric charge of Q1=−20µC while block 4 has a net electric charge of Q4=+50µC. The
other blocks are electrically neutral. The blocks are connected by ideal massless strings
and they are in an electric field of magnitude 3.00 × 106 N / C directed to the right as
shown below. What is the magnitude of the tension T in the string between m2 and m3.
E= 3.00 × 106 N / C
T=?
m1=1kg
Q1=−20µC
(A) T=18N
(B) T=60N
(C) T=87N
(D) T=108N
(E) T=150N
m2=2kg
Q2=0
m3=3kg
Q3=0
m4=4kg
Q4=+50µC
10m
The force between the two charged blocks is insignificant so I will neglect it. Taking the
+ve x direction to the right the force on block 1 is –60N. The force on the right block is
+150N. In both cases we applied the expression for a charged object in an electric field
F=qE.
At this point the problem is similar to question 20 on exam 1: The total force on the
system is 90N. This results in an acceleration of 9m/s². The net force on the two blocks
on the left is (3kg)( 9m/s²)=27N. Since the charged block exerts a force of –60N the
tension of the string must be 87N.
Correct response=48%
I wonder how many people will get the question similar to this correct on the final.
[68] A uniform electric field E=100 N/C is directed in the plane of a circular plate of
radius R=1 m. What is the magnitude of electric flux
through the plate?
(A) Φ=1000 Nm²/C
(B) Φ=628 Nm²/C
(C) Φ=314 Nm²/C
(D) Φ=100 Nm²/C
(E) Φ= 0
Since the electric field is parallel to the surface, the
flux through the surface is 0.
Correct Response=62%
1m
E=100N/C
[69] A conducting sphere of radius 10 cm has an unknown charge. If the electric field
15 cm from the center of the sphere has the magnitude 3x103 N/C is directed radially
inward, what is the net charge on the sphere?
(A) +5.0x10-9 C
(B) -5.0x10-9 C
(C) +7.5x10-9 C
(D) -7.5x10-9 C
(E) -2. 5x10-9 C
From the Gauss’s law argument discussed in class, the field outside a charged sphere is
the same as a Coulombic field for the same charge at the origin. The point in question is
outside the sphere so
q
r2
q = Er 2 / k
E=k
Plugging in the numbers, q=-7.5x10-9 C. We know the sign is negative since the field is
directed inwards.
Correct response=59%
[70] Three charges of size 1nC, 2nC and 1nC are placed on the x-axis spaced 1m apart
with the 2nC in the middle. Point P is half way between the left 1nC and 2nC while
point Q is half way between the 2nC and the right 1nC. What is the ratio of the xcomponent of the electric field at point P ( EPx )to the x-component of the electric field
at point Q ( EQx ).
(A) EPx : EQx = +1:1
(B) EPx : EQx = −1:1
(C) EPx : EQx = +1:2
(D) EPx : EQx = −2:1
+1nC
+2nC
(E) EPx : EQx = +4:3
The figure has mirror symmetry about a vertical line through the 2nC sphere. The x
component of the electric field at P will therefore be the opposite of the field at Q. The
correct answer is therefore –1:1.
+1nC
[71] Consider two parallel charged sheets of charge with charge density + 3nC / m 2
and − 2nC / m 2 respectively. The two sheets are perpendicular to the x-axis where the
positively charged sheet is to the left of the negatively
y
charged sheet as shown below. What is x-component of
the electric field at a point half way between the two
E=?
sheets?
nC
(A) E x = +1
2ε 0 m 2
nC
(B) E x = −1
+3nC/m²
−2nC/m²
2ε 0 m 2
nC
(C) E x = +5
2ε 0 m 2
nC
(D) E x = −5
2ε 0 m 2
nC
(E) E x = +3
2ε 0 m 2
Using the expression for the electric field from a charged sheet, the electric field due to
the left sheet within the gap is:
ELx = +3
nC
2ε 0 m 2
while the field due to the right sheet is
ERx = +2
nC
2ε 0 m 2
1) The total is thus E x = ELx + ERx = +5
nC
2ε 0 m 2
x
[72]
is
The magnitude of the electric field at a distance 3.0m from a 1.0 µC point charge
(A) 200 N/C
(B) 400 N/C
(C) 600 N/C
(D) 800 N/C
(E) 1000N/C
Using coulombs law, E=kQ/r²=1000N/C
Correct Response=91%
[73] Four point charges of charge Q are arranged in a square with side length L. What
is the magnitude of the net electrostatic force on one of the charges.
F=?
2
k Q
1
4
(A) E 2 1 + 12 2
L
Q
Q
L
kEQ2
(B) 2 2 − 2
L
L
L
L
k Q2
(C) E 2 1 + 2
Q
Q
L
2
3
kEQ 2
3 2
(D)
L2
kEQ2 1
+ 2
(E)
L2 2
[
]
[ ]
[ ]
[ ]
[ ]
Using coulombs law, the contributions to the force on charge 4 caused by charges 1-3 are
kEQ2 ˆ
i
L2
kEQ 2 ˆ
F3 = 2 j
L
F1 =
F2 =
k E Q 2  1  iˆ + ˆj  k E Q 2  iˆ + ˆj 

= 2 

L2  2  2 
L  2 2 
The total force is therefore
G G G
G k Q2
F = F1 + F2 + F3 = E 2
L
(
1 ˆ ˆ

1 + 2 2  i + j


)
The magnitude of F is therefore
F=
kEQ 2
L2
1
1  ˆ ˆ kEQ2 

+
1
 2 2  | i + j |= L2  2 + 2 




Correct response=32%
Most popular wrong answer is (C ), I suspect calculation error.
[74] A two identical cylindrical surfaces of length 1m and diameter 2m are in a
uniform electric field directed in the +x direction. Cylinder #1 has its axis parallel to the
x-axis while cylinder #2 has its axis
Cylinder #1
Cylinder #2
parallel to the y-axis. In which case is
there the greatest total flux though the
surface.
(A) Cylinder #1
(B) Cylinder #2
(C) Both the same
(D) Depends on the magnitude of E
y
x
The same number of field lines go into each cylinder as come out so the net flux out of
each is 0.
Correct Response=42%
G
E
[75] A conducting spherical shell has an
inner radius R and an outer radius 2R. The
shell has a total charge of +Q. If a point
charge of +q placed at the center of the shell,
What is the total charge on the outer surface
of the shell?
(A) −q
(B) +Q
(C) Q+q
(D) –Q-q
(E) Q-q
Charge on outer
surface =?
2R
Total Charge Q
Charge q
There is no electric field within a conductor so a gaussian sphere of radius (3/2)R will
have no flux and therefore must enclose no net charge. The charge on the inner surface is
thus –q. By conservation of charge then, the total charge on the outer surface is Q+q.
Correct response=57%
R