Physics 221.
Final Exam
Spring 2003
The situation below refers to the next two questions:
A block of mass m = 0.2 kg starts from rest at point A and slides down a frictionless
curved surface to point B, where its speed is v = 8 m/s. Then, it slides along a rough
horizontal surface for d =10 m and comes to rest at point C.
A
h
B
C
d
1. What is the initial height h of the block?
a. h = 1.2 m
b. h = 2.5 m
*c. h = 3.3 m
d. h = 4.6 m
e. h = 5.1 m
Mechanical energy is conserved while no friction is present:
1
mgh + 0 = 0 + mv 2
2
2
v
h=
= 3.3m
2g
2. Find the coefficient of kinetic friction between the block and the horizontal surface
between B and C.
a. µk = 0.1
b. µk = 0.2
*c. µk = 0.3
d. µk = 0.4
e. µk = 0.5
The missing mechanical energy must be equal to the work done by friction:
∆E = Wnon −conservative
1 2
mv − 0 = f k d
2
where f k = µ k N = µ k mg
⇒
µk =
v2
= 0.3
2 gd
Page 1 of 28
Physics 221.
Final Exam
Spring 2003
The situation below refers to the next two questions:
The graph below shows the potential energy U and the total mechanical energy E of a
particle as a function of position x. The only force acting on this particle is the force
associated to this potential energy.
U (J)
E = 5.0 J
x (m)
1
2
3
4
5
6
3. At which position does the particle have the greatest kinetic energy?
a. At x = 1 m
b. At x = 2 m
*c. At x = 4 m
d. At x = 5 m
e. At x = 6 m
Max KE ↔ Min U → At x = 4 m
4. Which of the following statements about the force on the particle at x = 3 m is true?
a. It is zero
b. It points in the −x direction
*c. It points in the +x direction
d. The particle can never be at x = 3 m, so there is no force to be discussed.
e. The direction of the force depends on the direction of the velocity of the particle.
dU
dx . At x = 3 m, the curve has a
The force is minus the slope of the U(x) curve,
negative slope, so the force must point in the +x direction.
Fx = −
Page 2 of 28
Physics 221.
Final Exam
Spring 2003
The situation below refers to the next two questions:
A rope is attached to the handle of a bucket with a stone of mass m inside. A person
swirls the whole system in vertical circles as shown in the figure below. The distance
between the stone and the center of the circle is L = 1.5 m.
L
5. What is the minimum speed that the system must have at the highest point of the
trajectory if the stone is to stay in contact with the bottom of the bucket?
*a. vmin = 3.8 m/s
b. vmin = 4.2 m/s
c. vmin = 5.7 m/s
d. vmin = 6.1 m/s
e. vmin = 7.8 m/s
The condition for minimum speed is that the normal force exerted by the bottom of the
bucket on the stone is equal to zero when the system is at the highest point of the
trajectory. This means that the only force providing the required centripetal acceleration
to the stone is its weight mg.
v2
mg = m min
L
vmin = gL = 3.8m / s
6. As the bucket moves from the highest to the lowest point in its trajectory, the work
done by the tension on the rope is ________ and the work done by gravity is ________.
a. negative, negative
b. zero, negative
c. positive, negative
d. negative, positive
*e. zero, positive
The tension on the rope does zero work along any portion of the circle because it is
always perpendicular to the displacement.
Gravity does a positive work along this displacement, because the force points down and
the y-component of all the differential displacement vectors dl that this displacement is
Page 3 of 28
Physics 221.
Final Exam
Spring 2003
made of also point down. In other words, the angle between the force and the
G G
dW
=
F
g ⋅ dl = Fg dy > 0
displacement is always acute so
Or: Wg = -∆Ug, where U=mgh. As the stone moves from top to bottom of the trajectory,
its gravitational potential energy decreases. Therefore ∆U<0, so W>0.
The situation below refers to the next two questions:
A firecracker rocket is flying vertically up when it explodes in two fragments of masses
m and M (> m). The bigger mass is ejected to the left and the small mass to the right.:
7. During this process, it is reasonable to assume that:
a. total kinetic energy is conserved, but total linear momentum is not conserved
*b. total linear momentum is conserved, but total kinetic energy is not conserved
c. both total kinetic energy and total linear momentum are conserved
d. total linear momentum is conserved and the kinetic energy of each part is conserved.
e. both kinetic energy and linear momentum of each part is conserved
The external forces acting on this system (gravity, air resistance) are small compared to
the forces involved in the explosion (or the change in momentum of each part due to the
external forces ∆pext =Fext ∆t is negligible compared to the change in momentum due to
the internal forces ∆pint =Fint ∆t), so Fext = 0 is a reasonable approximation. The, total
linear momentum is conserved.
Kinetic energy is not conserved because some energy is used to restructure the system
(one object goes to two). An explosion is an inverted perfectly inelastic collision.
8. Immediately after the explosion, the center of mass of the system moves:
*a. Up
b. Down
c. Up and to the left
d. To the left
e. It does not move.
Fext = macm, so acm = 0. Therefore, vcm is constant. (Or: ptotal is constant, but ptotal =
mtotalvcm, so vcm = constant). Before the explosion, the center of mass was moving up.
Thus after the explosion it is also moving up.
Page 4 of 28
Physics 221.
Final Exam
Spring 2003
9. A marble of mass m = 8.0 g is fired into a block of soft wood hung from two wires as
shown below.The block has a mass M = 2.0 kg and the marble imbeds itself in the
block causing it to rise to a maximum height h = 0.12 m. Find the speed v of the
marble just before impact. (Ignore the mass of the wires)
m
v
M
h
a. v = 290 m/s
*b. v = 385 m/s
c. v = 400 m/s
d. v = 425 m/s
e. v = 440 m/s
When the marble hits the block, we have a perfectly inelastic collision:
m+M
v=
v'
m
mv= (m+M)v’, so
After the collision, the system moves as a pendulum and the total mechanical energy is
conserved:
1
( M + m)v '2 = ( M + m) gh
2
v ' = 2 gh
Thus,
m+M
v=
2 gh = 385m / s
m
Page 5 of 28
Physics 221.
Final Exam
Spring 2003
10. A car must achieve an average speed of 250 km/h on a track of total length of 1500
m. The car travels the first half of the track at an average speed of 200 km/h. Therefore,
the average speed needed in the second half of the track has to be:
*a. 333 km/h
b. 300 km/h
c. 275 km/h
d. 265 km/h
e. 260 km/h.
If the average speed must be 250 km/h, the time to cover the complete track must be:
∆xtotal
1.5km
∆ttotal =
=
= 6 × 10−3 h
vave,total 250km / h
The time used in the first half is
∆x
0.75km
∆t1 = 1 =
= 3.75 × 10−3 h
vave,1 200km / h
So the second half must be covered in
∆t2 = ∆ttotal − ∆t1 = 2.25 × 10 −3 h
Thus,
∆x
0.75km
vave ,1 = 1 =
= 333km / h
∆t1 2.25 × 10−3 h
11. Two balls are thrown from the roof of a building with an initial speed of 10 m/s. Ball
number 1 is thrown vertically downward and ball number 2 vertically upward. Ball 1 hits
the street below the building in 4 seconds. How much later does the second ball hit the
ground? (Take g=10 m/s2).
a. ∆t = 1s
*b. ∆t = 2s
c. ∆t = 3s
d. ∆t = 4s
e. ∆t = 5s
Using the first ball, we can figure out the initial height for both:
1
0 = h − (10m / s )(4 s ) − (10m / s 2 )(4 s ) 2
2
h = 120m
For the second ball,
1
0 = 120m + (10m / s )t − (10m / s 2 )t 2
2
The (positive) solution to this quadratic equation is t = 6 s, so ball 2 hits the ground 2
seconds after ball 1.
Page 6 of 28
Physics 221.
Final Exam
Spring 2003
12. An object is thrown at ground level with an initial speed of 20 m/s. It hits the ground
2 seconds later. At what angle above the horizontal was the object thrown?
a. 10o
b. 20o
*c. 30o
d. 40o
e. 45o
Consider the y(t) equation for this motion:
1
0 = 0 + v0 y t − gt 2
2
1
0 = v0 sin θ t − gt 2
2
gt
= 0.5 ⇒ θ = 30D
sin θ =
2v0
13. A point on the rim of a 25-cm-radius wheel has a constant centripetal acceleration of
4.0 m/s2. The tangential acceleration of that point is
a. 6.0 m/s2
b. 4.0 m/s2
c. 2.0 m/s2
d. 1.0 m/s2
*e. None of the above.
If the centripetal acceleration in a circular motion is constant, the speed must be constant
as well (ac = v2/R), so the tangential acceleration is zero.
Page 7 of 28
Physics 221.
Final Exam
Spring 2003
14. When a man holding weights and spinning on a frictionless rotating stool extends his
arms horizontally and thereby doubles its moment of inertia, the rotational kinetic energy
is:
a. twice as much as before
*b. half as much as before
c. the same as before
d. four times as much as before
e. one fourth as much as before.
There is no sizable external torque on this system (friction on the axis of the stool and air
resistance are quite small), so angular momentum is conserved.
I iωi = I f ω f
Ii
I
ω
ωi = i ωi = i
If
2Ii
2
Compare the kinetic energies:
1
KEi = I iωi2
2
ωf =
2
ω  1
1
1
1
KE f = I f ω 2f = (2 I i )  i  = I iωi2 = KEi
2
2
4
2
 2 
Page 8 of 28
Physics 221.
Final Exam
Spring 2003
The situation below refers to the next two questions:
A uniform solid sphere rolls down an incline. The magnitude of the acceleration of the
g
center of mass is observed to be 10 .
15. What is the angle θ between the incline plane and the horizontal?
a. θ = 4°
*b. θ = 8°
c. θ = 15°
d. θ = 20°
e. θ = 33°
We need a free-body diagram:
N
fs
mg
translation of the cm along the incline: mg sin θ − f s = macm
rotation about the cm: f s R = I cmα
rolling without slipping: acm = Rα
3 equations with 3 unknowns (a,α,fS)
I cmα I cm acm
=
R
R2
I a
mg sin θ − cm 2cm = macm
R
mg sin θ
g sin θ
acm =
=
I
I
m + cm2 1 + cm2
R
mR
fs =
For a uniform solid sphere,
g sin θ 5
acm =
= g sin θ
2
7
1+
5
If this is to be equal to g/10,
I cm =
2
mR 2
5
, so
Page 9 of 28
Physics 221.
Final Exam
Spring 2003
g
5
g sin θ =
7
10
7
sin θ =
50
o
θ =8
16. If a frictionless block were to slide down the same incline, its acceleration a would
be:
g
10
*a.
g
a=
10
b.
g
a<
10
c.
d. The answer depends on the ratio of the block’s and sphere’s masses
e. The answer depends on the radius of the sphere and the height of the block
a>
Mass never matters when the only force doing work (i.e., speeding up or slowing down
the object) is gravity.
The size of an object does not matter either for sliding objects.
For rolling objects, only the shape matters, not the actual dimensions: We always
encounter the ratio I/mR2, where I ∝ mR2, so the R always cancels out.
Therefore, d and e are false.
When an object slides down a frictionless incline, all the gravitational potential energy
becomes translational kinetic energy.
When an object rolls down an incline, the gravitational potential energy becomes both
translational kinetic energy and rotational kinetic energy. So the translational kinetic
energy is less, and therefore the final speed is less than for the sliding object.
In terms of acceleration, this means that the acceleration is greater for the sliding object
(it attains a larger speed).
Page 10 of 28
Physics 221.
Final Exam
Spring 2003
17. A uniform rod is free to rotate about its center of mass, that we shall take as the
origin, as shown in the figure. Its initial position is such that the end labeled P is at rP =
(2i − 3j) m. A force F = (2i + 5j) N is applied at that point.
y
●
x
P
What is the torque due to this force on the rod about the origin?
a. τ = (−27 k) Nm
b. τ = (−11 k) Nm
c. τ = (−3 k) Nm
*d. τ = (16 k) Nm
e. τ = (21 k) Nm
iˆ ˆj kˆ
G G
τ = r × F = 2 −3 0 = 16kˆ
2 5 0
G
Or
G G G
τ = r × F = (2iˆ − 3 ˆj ) × (2iˆ + 5 ˆj ) = 4iˆ × iˆ − 6 ˆj × iˆ + 10iˆ × ˆj − 15 ˆj × ˆj = 0 + 6kˆ + 10kˆ + 0 = 16kˆ
Page 11 of 28
Physics 221.
Final Exam
Spring 2003
18. The mobile shown below is made of four identical objects of mass m. The rods and
the strings are massless. The upper rod has length L and is supposed to be horizontal.
Find the distance d between the left end of the rod and the point from which the mobile
should be hung from the ceiling.
T
L
d=
4
*a.
L
d=
3
b.
2L
d=
3
c.
3L
d=
4
d.
L
d=
2
e.
L
3mg
mg
d
Net force = 0
T − mg − 3mg = 0
⇒
T = 4mg
Net torque (about the left end) = 0
Td − mgL = 0
⇒
d=
mgL mgL L
=
=
T
4mg 4
Or: Take torque about the point where T is applied. Then net torque equal zero means:
3mgd − mg ( L − d ) = 0
⇒
d=
mgL L
=
4mg 4
Page 12 of 28
Physics 221.
Final Exam
Spring 2003
19. A small dust particle with mass m = 10−9g and electrostatic charge q = 3×10–17 C has
been suspended between the plates of a horizontal parallel plate capacitor so that the
electrostatic force compensates the gravitational force. The magnitude of the electric
field inside the capacitor is:
a. E = 105 V/m
*b. E = 3×105 V/m
c. E = 3×10–5 V/m
d. E = 5×105 V/m
e. E = 5×10–3 V/m
qE
E
mg
The net force must be zero: qE – mg = 0
E=
mg
= 3 ×105 N / C
q
Page 13 of 28
Physics 221.
Final Exam
Spring 2003
20. Two point charges Q1 and Q2 are placed at (x = a, y = 0) and (x = 0, y = 2a),
respectively, as shown in the figure below. Find the work done by the electric field when
a third charge q is brought from infinity to the origin.
y
(x = 0, y = 2a)
Q2
Q1 = 3.0 µC
Q2 = 2 Q1 = 6 µC
q = –2.0 µC
a = 5.0 cm
Q1
q
x
(x = a, y = 0)
*a. W = 2.2 J
b. W = 1.1 J
c. W = 0.31 J
d. W = −1.3 J
e. W = −5.4 J
W = − (U f − U i ) = − (U f − 0 ) = −U f
(where the initial point is infinity, so Ui = 0 )
Q
Q
Q
2Q
qQ
U f = qVorigin = q ( ke 1 + ke 2 ) = q (ke 1 + ke 1 ) = 2ke 1 = −2.2J
a
2a
a
2a
a
Page 14 of 28
Physics 221.
Final Exam
Spring 2003
21. A pear-shaped conductor (gray shape in the figures) has a certain net charge Q ≠ 0.
Which of the figures is a reasonable representation of the electric field lines in the region
around the conductor?
(b)
(a)
(c)
*(d)
(e)
In a isolated conductor in equilibrium, there cannot be positive charges (E-lines out) on
one side and negative charges (E-lines in)on the other side (they are free to move and
they attract each other), (a) and (e) are bad options.
The electric field (and therefore the electric field lines) must be perpendicular to the
surface at all points, so (c) is a bad option.
The electric field is stronger near the regions with higher curvature (or smaller curvature
radius), which means that there must be a higher density of lines in the region with
higher curvature (the right-hand side). So (b) is a bad option.
Option (d), instead, verifies all these criteria.
Page 15 of 28
Physics 221.
Final Exam
Spring 2003
22. Charge is distributed uniformly with the volume charge density ρ throughout the
volume of an infinitely long cylinder of radius R. Find the magnitude of the electric field
at a distance r>R from the cylinder axis.
E=
ρR
2ε 0 r
E=
ρR
2πε 0 r 2
E=
ρ
2ε 0 r 2
a.
b.
c.
E=
*d.
ρ R2
2ε 0 r
ρ
4πε 0 r 2
E=
e.
r
L
Use a Gaussian cylinder of length L and radius r:
Φ=
∫
G G
E ⋅ ds =
cylinder
∫
G G
E ⋅ ds +
side
∫
caps
G G
E ⋅ ds =
∫
G G
E ⋅ ds
side
(no flux through the caps because E and ds are perpendicular there). So:
G G
Φ = ∫ E ⋅ ds = ∫ Eds =E ∫ ds =E 2π rL
side
side
side
Using Gauss’s law,
qenclosed ρπ R 2 L
Φ=
=
ε0
ε0
Therefore,
E 2π rL =
E=
ρπ R 2 L
ε0
ρ R2
2ε 0 r
Page 16 of 28
Physics 221.
Final Exam
Spring 2003
23. Three capacitors with the same shape and dimensions are made of three different
conductors: gold, copper and aluminum. The conductivity of gold is greater than the
conductivity of copper, and the conductivity of copper is greater than the conductivity of
aluminum. Rank the capacitances of these capacitors.
a. Cgold > Ccopper > Caluminum
b. Cgold < Ccopper < Caluminum
c. Cgold = Ccopper = Caluminum (if they are parallel plate capacitors only)
*d. Cgold = Ccopper = Caluminum (always)
e. Not enough information to answer the question.
Capacitance only depends on the geometry of the capacitor (same for all of them) and on
the dielectric between the plates, but not on the material of the plates (it just needs to be
a conductor).
24. If you stretch a cylindrical wire and it remains cylindrical, how does this affect the
resistance of the wire (measured end to end along its length)?
Before
After
*a. It increases.
b. It decreases.
c. It remains the same.
d. The result could be different for different materials.
e. The result could be different for different cylinders and different stretches.
Resistivity is the same for both, since that depends only on the material used. But
l
R=ρ
A , so increasing l (length) and decreasing A (cross section area)
resistance is
both increase the resistance.
Page 17 of 28
Physics 221.
Final Exam
Spring 2003
The situation below refers to the next two questions:
A copper wire has cross-sectional area 2.0×10–6 m2 and length 4.0 m. The resistivity of
copper is ρ = 1.7×10−8 Ωm. A current of 2.0 A is uniformly distributed across the crosssection.
25. What is the magnitude of the electric field along the wire?
*a. E = 1.7×10–2 V/m
b. E = 2.7×10–2 V/m
c. E = 3.7×10–2 V/m
d. E = 4.7×10–2 V/m
e. E = 5.7×10–2 V/m
Electric field and current density J are related through resistivity:
I
E = ρ J = ρ = 1.7 × 10 −2 V / m
A
26. How much electric energy is transferred to thermal energy in 30 min?
a. 185 J
*b. 245 J
c. 375 J
d. 435 J
e. 525 J
Power is energy per unit time, so the total energy for an interval ∆t is ∆U=P∆t.
P = I 2R = I 2ρ
∆U = I 2 ρ
l
A
l
∆t = 245 J
A
(Note: Minutes must be converted to seconds)
Page 18 of 28
Physics 221.
Final Exam
Spring 2003
27. The figure shows a circuit containing one ideal battery of emf ε =12 V, and four
resistances with the following values: R1= 20 Ω, R2 = 20 Ω, R3 = 30 Ω, R4 = 8.0 Ω. What
is the magnitude of the current through the battery?
ε
R3
R1
R2
R4
a. I = 0.10 A
b. I = 0.20 A
*c. I = 0.30 A
d. I = 0.40 A
e. I = 0.50 A
This circuit can be reshaped as follows:
R2
R4
R1
R3
ε
The equivalent resistance for the two resistors in parallel is
1
1
1
=
+
⇒
R2−3 = 12Ω
R2−3 R2 R3
So the equivalent resistance of the whole circuit is: Req = R1 + R2-3 + R4 = 40 Ω
I=
Current through the battery:
ε
Req
= 0.30 A
Page 19 of 28
Physics 221.
Final Exam
Spring 2003
28. In the circuit shown below, ε =12.0 V, R = 1.40 MΩ, and C = 2.00 µF. The capacitor
is initially uncharged and the switch is open. The switch is closed at t = 0. How long does
it take for the charge to build up to 16.0 µC?
a. t = 1.1 s
b. t = 1.4 s
c. t = 2.1 s
d. t = 2.8 s
*e. t = 3.1 s
R
ε
C
The charge as a function of time for this circuit is:
Q = Q∞ (1 − e − t /τ )

 Q∞ 
Q 
t = −τ ln 1 −
 = τ ln 

 Q∞ 
 Q∞ − Q 
so
The time constant is τ = RC = 2.8 s
The final charge is determined by the maximum voltage across the capacitor. When the
current is zero, the plates are connected to the battery, so VC = ε. Therefore,
Q∞= εC=24 µC.
 Q∞ 
t = τ ln 
 = 3.1s
Q
Q
−
 ∞

If Q = 16 µC,
Page 20 of 28
Physics 221.
Final Exam
clamp
force
probe
spring
X
lab
table
135 cm
plate
motion
detector
mesh
Spring 2003
29. In lab, you studied the vertical oscillation of a
plate. Position data collected by the computer is
referenced to the X axis shown in the figure at the
right.
Assume that the plate hangs at its equilibrium
position of X = 50 cm. Consider two trials of its
motion which differ only in the way in which they are
begun, namely,
1. In trial #1, the plate is displaced
downward to X = 60 cm and released at
rest;
2. In trial #2, the plate is displaced
downward to X = 70 cm and released at
rest.
For these two trials, compare:
1. the time required for the plate to first return to its release point;
2. the maximum acceleration experienced by the plate.
For trial #2 (as compared to trial #1),
a.
the time to return is twice as long, while the maximum acceleration is the
same;
b. the time to return is twice as long, while the maximum acceleration is twice
as large;
c. the time to return is twice as long, while the maximum acceleration is four
times as large;
d. the time to return is the same, and the maximum acceleration is the same;
*e. the time to return is the same, while the maximum acceleration is twice as
large.
The period of the oscillations is independent of the amplitude (10 cm and 20 cm,
respectively), so the time to return must be the same.
The maximum acceleration is proportional to the maximum displacement from
equilibrium or amplitude. Since the amplitude is twice as large in trial 2, the maximum
acceleration will also be twice as large.
Page 21 of 28
Physics 221.
Final Exam
X
0
Spring 2003
As you did in lab, consider the motion of a cart which
is given an initial velocity up an inclined track toward a
"motion detector". Assume the cart goes part of the
way up, and then returns back down the track. (As was
done in lab, assume the sensor gives positions relative
to the X axis illustrated above).
30. Which of the following best graphs illustrates the velocity, VX , of the cart (after
leaving your hand) versus time?
Vx
Vx
t
0
Vx
Vx
t
A
Vx
t
B
t
C
t
D
E
The cart moves in the negative-x direction, so we must start with a negative
velocity. This already leaves only d!
Also, the cart turns around. This means that the speed must be 0 at some point.
This happens in d and e only.
31. Which of the following best illustrates the acceleration, aX, of the cart versus time
(during the same period)?
ax
ax
ax
t
0
A
ax
t
B
ax
t
C
t
D
t
E
The acceleration points down the ramp all the time and is constant (a=g sin θ). The
positive x is taken down along the ramp, so the acceleration must be positive.
Page 22 of 28
Physics 221.
Final Exam
Spring 2003
32. Consider the collision of two identical hockey pucks on a smooth ice rink. Assume
that by analyzing the videotape of the collision, various but not all components of the
momenta of the two pucks before and after collision have been determined and are
recorded in the table below.
Assume that there is negligible friction with the ice, but that the pucks are soft and that
considerable kinetic energy is “lost” in the collision. Under these assumptions, what is
the value (if it can be determined) of the entry marked with the “X”?
Px
Py
Before Collision
Puck #1 Puck #2
0.42
-0.92
0.70
0.30
After Collision
Puck #1 Puck #2
-0.26
X
0.60
a) -0.02
*b) -0.24
c) +0.26
d) +0.76
e) Cannot be determined from the information given.
Conservation of linear momentum in the x direction:
0.42 − 0.92 = −0.26 + X
X = 0.42 − 0.92 + 0.26 = −0.24
33. The trajectory of a puck sliding
freely on a level air table is recorded
(as done in 221 lab) on a piece of
(metric) graph paper, shown at the
right. Assume that the spark
generator was operating at a rate of
60 Hertz, and that the figure at the
right is approximately full scale.
Y
X
Estimate the magnitude of the
X component of the velocity
of the puck.
a) vx = 103 cm/s
*b) vx =84 cm/s
c) vx =60 cm/s
d) vx = 0.02 cm/s
e) It is not possible to calculate an estimate without a protractor to measure
angles.
The horizontal distance between the first and the fourth points is 4.2 cm.
Page 23 of 28
Physics 221.
Final Exam
Spring 2003
The time elapsed between the first and the fourth points is 3 time 1/60 = 0.05 s.
4.2cm
vx =
= 84cm / s
0.05s
So the x-component of velocity is
.
Page 24 of 28
Physics 221.
Final Exam
Spring 2003
34. In the rotational motion experiment, the following graph was obtained when the
wheel of the apparatus which you used was rotated by hand. Provide the t likely
explanation for this particular graph.
a.
The wheel was rotated rapidly then stopped for a second or two, then rotated
rapidly, then stopped again, and so on.
b. The wheel was rotated at different velocities, increasing in steps as time went
on.
c. The wheel was rotated slowly, and the angular encoder on the wheel axis has
a resolution of 1/20 of a revolution.
*d. The wheel was rotated slowly, and the angular encoder on the wheel axis has
a resolution of 1/200 of a revolution.
e. The apparatus likely was malfunctioning.
Any rapid rotation would produce fluctuations in the graph, so a and b not good options.
You saw this graph when the wheel was rotated slowly. The angle “jumps” correspond to
the angular precision of the apparatus. Between 0.1 rad and 0.3 rad, there are 6 jumps. So
each jump corresponds to 0.2/6=0.03 rad.
1turn
1
0.033rad
turns
= 5.3 × 10−3 turns =
2π rad
188
This is closer to answer (d) than to answer (c).
Page 25 of 28
Physics 221.
#1
Final Exam
Spring 2003
35. Consider two conducting spheres mounted on insulating
rods and located (with the relative positions shown) in a region
of large electric field, near a negatively charged rod as shown.
#2
Assume that both of the spheres are initially uncharged. Then
one of the two spheres is displaced slightly so that the two
spheres touch each other briefly. Assume that neither touches
the rod, and that their relative positions always are as shown.
Which of following are possible values for the resulting charge
on each sphere, #1 and #2, respectively, after the sequence of events described?
A
*B
C
D
E
#1
+3.0 nC
−3.0 nC
−3.0 nC
+3.0 nC
0 nC
#2
−3.0 nC
+3.0 nC
−3.0 nC
+3.0 nC
0 nC
When the spheres touch, the negative charges in sphere #2 flow to sphere #1 (repelled by
the negatively charged rod). When the spheres are separated (in the presence of the rod),
sphere #1 remains therefore negatively charged and sphere #2 remains positively charged.
The total charge must be zero, because both spheres were initially uncharged.
Page 26 of 28
Physics 221.
Final Exam
Spring 2003
36 As done in lab, a student, seat, chain, and two load cells are at rest and supported by
cables attached to the upper (sensitive) fittings of the two load cells. The cell readings
and those of the large protractors are
given in the table below.
protractors
load cells
0
0
chain
Y
seat
X
LEFT
RIGHT
Cell reading
531 N
600 N
Protractor
reading
60°
50°
Use this information to estimate the X component of the force exerted by the left cell
upon the end of the left chain.
a. +266 N
b. −266 N
*c. −460 N
d. +460 N
e. 0 (The force must be zero, since nothing is accelerating.)
L
θ1
R
θ2
W
The x-component of force L is negative according to the x-y axis given in the figure.
Lx = −L sin θ1 = −(531 N) (sin 60°) = −460N
Page 27 of 28
Physics 221.
Final Exam
Spring 2003
Did you bubble in your exam version?
Page 28 of 28