Physics 221 2006 S Exam 2 Solutions
Answer key
Solutions
Questions
1
2
9
31
C
41
A
51
A
32
C
42
B
52
B
33
D
43
C
53
D
34
C
44
A
54
E
35
B
45
C
55
E
36
B
46
A
56
E
37
E
47
D
57
C
38
A
48
C
58
E
39
E
49
B
40
B
50
D
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Physics 221 2006 S Exam 2 Solutions
[31] (C)
The total mechanical energy is the sum of potential and kinetic energy so
E = U + K = 12 kx 2 + 12 mv 2
= 12 (100 Nm )(0.3m) 2 + 12 (2kg )(3 ms ) 2
= 4.5 J + 9 J = 13.5 J
[32](C)
Since p = mv and K = 12 mv 2 so that v =
vP = 2
vR =
K0
p0
K0
p0
vQ = 4
2K
p
. Going through each of the four particles:
K0
p0
vS = 2 Kp00
Thus vQ > vP = vS > vR
[33](D)
The total kinetic energy is the same before and after an elastic collision so (D) is false.
[34](C)
The interval between t=2s and t=8s goes from a maximum to a minimum and so the 6s
interval is half a period. The period is therefore T=12s.
[35](B)
Take the peak at t=2s as the “first peak”. It is 2s to the right of the origin which is 1/6 of a
period so φ is negative and φ = −(2π )(1/ 6) = − π3 .
Another way to do it is to note that if we plug t=0 into the general expression where
x=1m we get
1m = A cos(φ ) = (2m) cos(φ )
∴φ = ± arccos ( 12 ) = ± π3
to resolve the trig ambiguity note that the velocity is positive and since
v(t = 0) = − A sin(φ )
only the − π3 solution gives the correct velocity sign. (of course these solutions are
modulo 2π)
[36](B)
When the mass is at a maximum displacement, then the kinetic energy is 0 so the
mechanical energy is equal to the potential energy. Such a case is t=2s where x=2m so
-2-
Physics 221 2006 S Exam 2 Solutions
E = K + U = 0 + 12 kx 2 = 12 (10 Nm )(2m) 2 = 20J
This mechanical energy does not change with time because in a simple harmonic
oscillator the force is conservative.
[37](E)
This is an inelastic collision so momentum is conserved. The final momentum of the
system is p = (3000kg )(10 ms ) = 30000 Ns . All of this momentum must be carried by the
30000 Ns
initial car so its velocity is v =
= 30 ms
1000kg
-3-
Physics 221 2006 S Exam 2 Solutions
[38](A)
Taking the zero of potential at equilibrium, at that position the mechanical energy is
E = U + K = 0 + 10J = 10J . At the maximum of the oscillation the kinetic energy is zero.
Thus
E = 12 kx 2
∴x =
2E
2(10 J )
=
= 45cm
k
100 Nm
[39](E)
Let F1 be the force of scale 1 and F2 be the force of scale 2 with W the weight of the
plank. The force condition in the vertical direction is F1 + F2 = W . Taking the torque
condition about the left end of the plank,
0 = τ net = 23 LF2 − 12 LW
∴ F2 = 34 W = 45 N
By the force condition F1 = 15 N .
[40](B)
The velocity of the center of mass does not change during the collision. We can find this
velocity from the initial state:
vcm =
(200kg )(10 ms iˆ) + (400kg )(0 ms )
= (3.3 ms )iˆ
400kg + 200kg
[41](A)
The moment of inertia of the horizontal rod is I h = 13 mL2 = 13 (5kg )(2m) 2 = 6.6 kg m 2 . All
of the mass of the vertical rod is at a distance of 2m from the axis so
I v = mL2 = (5kg )(2m)2 = 20 kg m 2
thus the total moment of inertia is I = I v + I h = 26.7 kg m 2 .
The rotational kinetic energy is therefore:
K = 12 Iω 2 = 12 (26.7 kg m 2 )(3s −1 ) 2 = 120 J
[42](B)
The angular momentum is
G G G
L = r × p = (3iˆ + 2 ˆj + kˆ) × (−iˆ − ˆj + kˆ) Js
= [(2 ⋅1 − (−1⋅1))iˆ + (1⋅ (−1) − 3 ⋅1) ˆj + (3 ⋅ (−1) − 2 ⋅ (−1))kˆ]Js
= (3iˆ − 4 ˆj − kˆ)Js
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Physics 221 2006 S Exam 2 Solutions
[43](C)
The kinetic energy when the book hits the ground is the same as the difference in
potential energy of the book between the top of the campanile and the bottom.
Thus K = mgh . The momentum of the book on impact is thus
p = mv = m
2K
m
= 2mK = 2m 2 gh = m 2 gh
This is also the magnitude of the impulse delivered by the ground on the book so the
magnitude of the average force is
m
m 2 gh (2kg ) 2(9.8 s 2 )(94m)
=
= 860 N
∆t
(0.1s )
F = J / ∆t =
[44](A) The mechanical energy of the snowball does not change as the snowball travels.
Thus
1
2
mv12 + mgh1 = 12 mv22 + mgh2
∴ 12 v12 + gh1 = 12 v22 + gh2
From the question we know all the inputs except for v2. Solving for this we get
v2 = v12 − 2 g (h2 − h1 )
= (20 ms ) 2 − 2(9.8 sm2 )(8.0m − 1.5m) = 16.5 ms
[45](C)
The impulse which the rocket delivers to the exhaust per unit time is
∆J / ∆t = vexhaust (∆mexhaust / ∆t ) . This will be the same as the average force the engine
exerts on the exhaust and by Newton’s third law, the force exerted on the rocket. Thus:
∆J / ∆t = vexhaust (∆mexhaust / ∆t ) = (300 ms )(3 kgs ) = 900 N .
[46](A) To find the center of mass, just plug into the formula for center of mass:
xcm =
∑xm
i
M
i
=
(0m)(1kg + 1kg ) + (2m)(1kg + 2kg + 1kg ) + (4m)(1kg )
= 1.7m
7 kg
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Physics 221 2006 S Exam 2 Solutions
[47](D)
The masses along the axis of rotation do not contribute to the moment of inertia.
Calculating the moment of inertia of the remaining masses:
I = ∑ mi ri 2 = (1kg )(2 2) 2 + (1kg )( 2) 2 + (1kg )( 2) 2 + (1kg )(2 2) 2 = 20 kg m 2
[48](C)
The angular momentum is given by
dθ
L = Iω = I
= I dtd  Ae − Bt  = − IABe − Bt
dt
Plugging in t=0:
−1
| L(t = 0) |= (2kg m 2 )(4)(3s −1 )e − (3s )(0 s ) = 24Js
[49](B) The initial kinetic energy is Kinit=16J when it is at the bottom of the ramp. As it
slides up the ramp, if we take the x-axis up the ramp then the x-component of the force
directed opposite to its motion is:
Fx = Wx + f k = − mg sin θ − mg µ k cos θ
= − mg (sin θ + µ k cos θ )
= − mg ( 12 + 12 )
= − mg
Using the work energy theorem we can therefore find the distance up the ramp that the
block slides
d =−
K init K init
16 J
=
=
= 82cm
Fx
mg (2kg )(9.8 sm2 )
[50](D)
Take the zero of gravitational potential energy at the initial position of the ball and the
zero of the spring potential energy at the point where the spring is at its relaxed length.
Mechanical energy is conserved so
1
2
kx 2 = mgh
∴k =
2mgh 2(0.5kg )(9.8 sm2 )(1.1m)
=
= 2200 Nm
(0.07 m) 2
x2
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Physics 221 2006 S Exam 2 Solutions
[51](A)
Taking the torque condition about the hinge, take L the length of the rod. Wrod the weight
of the rod, and Wbox the weight of the box:
τ net = 0 = − 12 LWrod − LWbox + T0 L sin θ
∴T0 =
1
2
Wrod + Wbox 12 (40 N ) + (20 N )
=
= 57 N
sin θ
sin 45o
[52](B)
The torque on the rod about the hinge due to the rod’s weight is:
τ = 12 Lmg
The moment of inertial of the rod about the hinge is I = 13 mL2
Using Newton’s second law for rotation of a rigid body, τ = Iα ,
τ
Lmg 3g 3(9.8 sm2 )
=
=
= 7.35 rad/s 2
α= =
2
2L
2(2m)
I
mL
1
2
1
3
[53](D)
The angular momentum about the axle is the same before and after. Thus
1=
Lf
Li
=
I fω f
I iωi
=
I f Ti
I iT f
so the final period is given by
I
T f = f Ti
Ii
The initial moment of inertia of the system is the moment of inertia of the disk is
I i = 12 mdisk r 2
the final moment of inertial includes the ball on the rim
I f = 12 mdisk r 2 + mball r 2
thus
Tf =
1
2
1
mdisk + mball
2 (2 kg ) + 1kg
T
=
(4s ) = 8s
i
1
1
m
(2
kg
)
disk
2
2
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Physics 221 2006 S Exam 2 Solutions
[54](E)
Since K = p 2 /(2m) , the magnitude of momentum of each of the fragments is
p1 = 2m1 K1 = 2(0.5kg )(25 J ) = 5 Ns ; p2 = 2m2 K 2 = 2(0.5kg )(16 J ) = 4 Ns and
p3 = 2m3 K 3 = 2(0.5kg )(100 J ) = 10 Ns . The initial momentum is 0 so the three
fragments must add to 0 by conservation of momentum and the three momentum vectors
must form a triangle. Since p3 > p1 + p2 this cannot happen so this scenario is not
physically possible (see written assignment 1 question 1 and written assignment 8)
[55](E) The total mechanical energy at release is just the potential at x=6m which is 1J.
the maximum KE will be at the point where the potential is minimum (if it can get there).
In this case this is at x=3m. Thus K = E − U (3m) = 1J − (−3J) = 4J .
[56](E)
Using the result of example 10.7 in the text, v2 = v3 =
4
3
gh . In case 1, using
conservation of energy the kinetic energy at the bottom of the ramp is K = mgh so
v1 = 2 gh . Thus v1 > v2 = v3 . In particular v2 = v3 because they have the same shape and
mass distribution so they have the same value of c in the notation of ex. 10.7 while v1
must be the fastest velocity because in that case none of the initial potential energy is
converted to rotational kinetic energy, it is all converted to translational kinetic energy.
[57](C)
First let us seem if the system is over or under damped.
2 km = 2 (150 Nm )(0.25kg ) = 12.2 kgs
this is bigger than b=1.5kg/s. The system oscillates with exponential damping. The
kinetic energy is always positive and, since, the system starts at rest, the kinetic energy is
initially 0. The only graph like that is graph C.
[58](E)
To the right of the dashed line the force is constant and therefore conservative. Thus the
force does 0 work in loops Q and S. Path R has 0 work because the positive work done
in the lower loop is exactly canceled by the work done in the upper loop. In case P the
force does a positive amount of work. On the top and bottom of the square, no work is
done since the path is perpendicular to the force but on the right side a there is a large
force parallel to the path which is not canceled by the small force opposite to the path on
the left side.
-8-
Physics 221 2006 S Exam 2 Solutions
Exam Questions
[31] A 2kg block is attached to a spring with force constant k=100N/m. The block is
initially moving at a speed of 3m/s and the spring is compressed 0.3m from its relaxed
length. What is the total mechanical energy of this system taking the potential energy of
the spring to be 0 when the spring is relaxed?
(A) 4.5J
(B) 9.0J
(C) 13.5J
(D) 18.0J
(E) 22.5J
[32] Particles P, Q, R and S are moving in the +x direction. The momenta and kinetic
energies of these particles are given in the table below. Which is the correct ranking of
the speeds of these particles?
Particle
P
Q
R
S
Kinetic Energy
K0
2K0
K0
2K0
Momentum
p0
p0
2p0
2p0
(A) vR > vP > vS > vQ
(B) vR > vS > vP > vQ
(C) vQ > vP = vS > vR
(D) vQ > vS > vP > vR
(E) None of the above
[33] If a collision between two particles is elastic, which of the following statements is
false?
(A) The total kinetic energy is conserved.
(B) The total momentum is conserved.
(C) The velocity of the center of mass of the system does not change.
(D) The total kinetic energy is reduced by the collision.
(E) The total angular momentum of the system is conserved.
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Physics 221 2006 S Exam 2 Solutions
This graph applies to questions 34, 35 and 36
t=−4s
t=−2s
t=0s
t=2s
t=4s
t=6s
t=8s
t=10s
t=12s
x=2m
x=0m
x=−2m
[34] The graph above is a sketch of the position versus time graph for a mass attached to
a spring undergoing simple harmonic motion. What is the period of this motion?
(A) 6s
(B) 11s
(C) 12s
(D) 24s
(E) 0s
[35] In the graph above, what is the phase angle φ if we write the solution in the form
x(t ) = A cos(ω t + φ ) ?
(A) −
π
2
(B) −
π
3
(C) 0
(D) +
π
3
(E) +
π
2
[36] In the graph above, if the spring constant is k=10N/m, what is the mechanical
energy of the system?
(A) 10J
(B) 20J
(C) 40J
(D) 80J
(E) The mechanical energy changes with time.
- 10 -
Physics 221 2006 S Exam 2 Solutions
[37] A car with mass 1000kg is driving along a road. It collides with a stationary truck
with mass 2000kg. The two vehicles fuse together and the moment after the collision, the
fused wreckage of the two vehicles is moving at a speed of 10m/s. What was the initial
speed of the car?
(A) 10m/s
(B) 15 m/s
(E) None of the above.
(C) 17 m/s
(D) 25 m/s
[38] A particle with mass m=5kg resting on a frictionless horizontal surface is connected
to an ideal spring with force constant k=100N/m. At t=0s, the mass is at the equilibrium
position and has a kinetic energy of 10J. What is the amplitude of the subsequent
oscillation?
k=100 N/m
5kg
(A) 45cm
(B) 89cm
Kinitial=10J
(C) 63cm
- 11 -
(D) 32cm
(E) 98cm
Physics 221 2006 S Exam 2 Solutions
[39] A 3m long plank of weight 60N and uniform density rests on two scales which
measure force in Newtons. Scale #1 supports the left hand end of the plank while scale #2
supports a point 1m from the right end of the plank. What are the readings of the two
scales?
3m
1m
2m
Plank weight=60N
Scale 1
Scale 2
(A) Scale 1 reads 45N; Scale 2 reads 15N
(B) Scale 1 reads 40N; Scale 2 reads 20N
(C) Scale 1 reads 30N; Scale 2 reads 30N
(D) Scale 1 reads 20N; Scale 2 reads 40N
(E) Scale 1 reads 15N; Scale 2 reads 45N
[40] A rock of mass 200kg is traveling through space at a speed of 10m/s in the +x
direction. It collides with a stationary rock of mass 400kg. The 200kg rocks bounces off
at an angle of 10º with respect to the x-axis. What is the velocity of the center of mass of
the two rock system after the collision?
(B) (3.3m/s) iˆ
(C) (20m/s) iˆ
(A) (10m/s) iˆ
(E) Cannot be determined from the information given.
- 12 -
(D) (30m/s) iˆ
Physics 221 2006 S Exam 2 Solutions
[41] Consider the configuration of two thin rods shown
below. Each rod has a mass of 5kg uniformly distributed
over a length of 2m. If it rotates at an angular velocity of
ω=3rad/s about the axis shown, what is the rotational
kinetic energy of the system?
(A) 120J
(D) 40J
(B) 240J
(E) 135J
Axis
2m; 5kg
(C) 27J
G
2m; 5kg
[42] If a particle has momentum p = (−i − ˆj + kˆ)Ns and radius
G
G
vector R = (3i + 2 ˆj + kˆ)m , what is the angular momentum vector L of the particle about
the origin?
G
(A) L = (−3,+4,+1) Js
G
(D) L = (+3,+4,−1) Js
G
(B) L = (+3,−4,−1) Js
G
(E) L = (−3,−6,+1) Js
- 13 -
G
(C) L = (−3,−4,+1) Js
Physics 221 2006 S Exam 2 Solutions
[43] A physics book (2 kg) is dropped from the top of ISU Campanile (about 94 m high).
What is the magnitude of the average force it exerts on the ground if it is in contact with
the ground for 0.1s before coming to a complete stop? Neglect air resistance.
(A) 86N
(B) 196N
(C) 860N
(D) 980N
(E) 1960N
[44] Sarah throws a 50g snowball at John who is in a tree. Sarah releases the snowball at
an angle of 60º to the horizontal at a speed of 20m/s from a point 1.5m above the ground.
The snowball strikes John who is 8m above the ground. How fast is the snowball
traveling when it strikes John? Neglect air resistance.
(A) 16.5m/s
(B) 15.6m/s
(C) 18.3m/s
(D) 13.1m/s
(E) The snowball cannot reach John given the information in the problem.
[45] A rocket engine operates by ejecting exhaust gasses at a speed of 300m/s. If it
produces exhaust at a rate of 3kg/s, what is the magnitude of force that the rocket engine
generates?
(A) 300N
(B) 600N
(C) 900N
- 14 -
(D) 1350N
(E) 45000N
Physics 221 2006 S Exam 2 Solutions
This figure applies to questions 46 and 47
y-axis
Axis
4m
1kg
2m
1kg
2m
1kg
2kg
1kg
1kg
x-axis
4m
2m
[46] Consider the system of point masses shown above. What is the x-coordinate of the
center of mass of this system?
(A) 1.7m
(B) 2.0m
(C) 0.0m
(D) 4.0m
(E) 2.3m
[47] Consider the system of point masses shown above. What is the moment of inertia of
this system of masses about the indicated axis that passes through the origin at an angle
of 45° to the x-axis?
(A) 5 kg m²
(B) 10 kg m²
(C) 12 kg m²
- 15 -
(D) 20 kg m²
(E) 24 kg m²
Physics 221 2006 S Exam 2 Solutions
[48] The angular position of a wheel with moment of inertia I=2kg m² is given as a
function of time by θ = Ae − Bt where A=4 radians and B=3s−1. At t=0s, what is the
magnitude of the angular momentum of the wheel?
(A) 72 Js
(B) 18 Js
(C) 24 Js
(D) 144 Js
(E) 54Js
[49] A 2kg block is sliding to the right along a frictionless horizontal surface. Initially the
kinetic energy of the block is 16J. It encounters an incline that slopes upwards at an angle
of 30º. The coefficient of kinetic friction between the block and the incline is µ k = 0.58 .
What is the maximum distance, d, which the block slides up the ramp before coming to a
stop?
2kg
µ k = 0.58
d=?
K.E.=16J
2kg
30º
Frictionless
(A) 41cm
(B) 82cm
(C) 163cm
- 16 -
(D) 71cm
(E) 52cm
Physics 221 2006 S Exam 2 Solutions
[50] A small 0.5-kg ball is pressed against a vertical spring of negligible mass as shown
below. The spring is compressed x= 7.0 cm from its relaxed position. When the system is
released from rest, the ball is shot up in the air and reaches a maximum height of h = 1.1
m above its initial position. Determine the force constant k of the spring. Neglect air
resistance.
h
x
(A) 500 N/m
(B) 1100 N/m
(E) The ball can never reach that height.
(C) 1500 N/m
- 17 -
(D) 2200 N/m
Physics 221 2006 S Exam 2 Solutions
[51] In the figure at right, a horizontal rod of
uniform density with weight 40N and length
2m is attached to a wall with a frictionless
ideal hinge. An ideal massless string connects
the end of the rod to the wall at a point 2m
above the hinge. A box with weight 20N is
hung from the end of the rod with an ideal
massless string. If the system is in
equilibrium, what is the tension, T0, in the
string connecting the end of the rod to the
wall?
2m
T0
40N
2m
Hinge
20N
Wall
(A) 57N
(B) 85N
(C) 40N
(D) 60N
(E) 42N
[52] In the figure at right, a horizontal rod of uniform
density with weight 40N and length 2m is attached to a
wall with a frictionless ideal hinge. If the rod is released at
rest, what is the angular acceleration of the rod about the
hinge just after it is released?
(A) 4.90 rad/s²
(B) 7.35 rad/s²
(C) 8.40 rad/s²
(D) 9.80 rad/s²
(E) 14.70 rad/s²
40N
2m
Hinge
Wall
- 18 -
Physics 221 2006 S Exam 2 Solutions
[53] A disk of uniform density with mass 2kg and radius 1m can rotate without friction
about a vertical axle. It is initially rotating with a period of 4s. A sticky ball of mass 1kg
(which can be treated as a point mass) is dropped vertically downwards. The ball sticks to
the rim of the disk. After the ball sticks, what is the period of rotation of the disk?
Disk
radius = 1m
mass=2kg
Vertical axle
Ball
Mass=1kg
(edge view)
(A) 2s
(B) 4s
(C) 6s
(D) 8s
(E) 12s
[54] A 1.5 kg rock initially at rest undergoes an explosion into three fragments of mass
0.5kg each. Fragment #1 has a kinetic energy of 25J; Fragment #2 has a kinetic energy of
16J and fragment #3 has a kinetic energy of 100J. What is the angle between the velocity
vectors of fragment #1 and fragment #2?
(A) 49°
(B) 42°
(D) There is not enough information given to determine this angle
(E) The scenario described is not physically possible.
- 19 -
(C) 90°
Physics 221 2006 S Exam 2 Solutions
[55] Consider the graphs of potential energy of a 4kg particle versus position shown
below. If the particle is released at rest from the position x=6m, what is the maximum
kinetic energy that the particle assumes thereafter?
Potential
(J)
6
4
2
0
2
4
6
8
−2
−4
(A) 0J
(B) 1J
(C) 2J
(D) 3J
- 20 -
(E) 4J
Position
(m)
Physics 221 2006 S Exam 2 Solutions
[56] In the three systems below, the inclines have the same angle with the horizontal and
the objects are released from the same height h.
1. Block of mass M on a
frictionless incline.
h
θ
2. Solid cylinder of mass M
and radius R rolling down
without slipping.
h
θ
3. Solid cylinder of mass M
and radius 2R rolling down
without slipping.
h
θ
How do their speeds at the bottom of the incline compare?
A.
B.
C.
D.
E.
v1 < v2 = v3
v1 < v2 < v3
v1 > v2 > v3
v1 > v3 > v2
v1 > v2 = v3
- 21 -
Physics 221 2006 S Exam 2 Solutions
[57] A 250-g particle attached to a spring with k = 150 N/m is also subject to a damping
force F = −bv, where v is the velocity of the particle and b = 1.5 kg/s. Which of the
graphs shown below best represents the kinetic energy of the particle as a function of
time?
- 22 -
Physics 221 2006 S Exam 2 Solutions
[58] (Note: Extra Credit Problem)
The figure below depicts a position dependent non-conservative force acting on a
particle. To the right of the dashed line the force is constant while to the left it varies as
shown. P, Q, R and S are closed path direction indicated. Note that R is a “figure eight”
path that crosses itself. In which case is the work done by the field on a particle moving
around the closed path not zero.
(A) All of the paths
(B) Paths P and R
(C) Paths Q and S
(D) Paths P and Q
(E) Path P only
- 23 -