Physics 221 2007S Exam 2 SOLUTIONS
[28] Assume that the moon can be modeled as a solid sphere of uniform density. The
mass of the moon is 7.4 ×1022 kg and the radius of the moon is 1800 km. Which is the
best estimate of the moment of inertia of the moon about an axis through the moon’s
center of mass?
(A) 2.4 × 1035 kg m 2
(D) 9.6 ×1034 kg m 2
(B) 1.6 × 1035 kg m 2
(E) 4.8 × 1034 kg m 2
(C) 1.2 × 1035 kg m 2
Answer [D]: For a solid sphere of uniform density, I = 52 MR 2 , so plugging in the
numbers from the problem:
I = 52 MR 2 = 0.4(7.4 ×1022 kg)(1800000m) 2 = 9.6 ×1034 kg m 2
[29] Three objects of mass m are dropped from a height h. One falls straight down, one
slides down a frictionless incline and one swings at the end of a pendulum. What is the
relationship between their speeds when they reach the ground? Neglect air resistance.
(A) vF > vI > vP
(D) vF < vI < vP
Fall
(B) vF > vP > vI
(E) vF < vP < vI
(C) vF = vI = vP
Incline
Pendulum
h
vF
vI
vP
Answer: [C] vF = vI = vP because the only force which does work is the conservative
force of gravity. Thus the net work done on the object, hence the increase in kinetic
energy, is the same in all cases. Since all of the objects have the same mass, it follows
that the final velocity is the same.
Page 1 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[30] A truck with mass 2000kg is driving along a road. It collides with a stationary car
with mass 1000kg. 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 truck?
(A) 6.7m/s
(B) 15 m/s
(E) None of the above.
(C) 13 m/s
(D) 25 m/s
Answer [B]: Using conservation of momentum, if vi is the initial velocity and v f is the
final velocity, then
pi = mtruck vi
= p f = (mtruck + mcar )v f
∴ vi =
mtruck + mcar
vf
mtruck
2000 kg+1000 kg m
10
s
2000 kg
m
= 15
s
=
[31] ] Particle B has twice the mass of particle A and particle B is moving with linear
momentum twice as big as particle A. What is the ratio:
Kinetic energy of B : Kinetic energy of A?
(A) 4:1
(B) 2:1
(C) 1:1
(D) 1:2
(E) 1:4
Answer [B] : The kinetic energy is given in terms of momentum and mass by
K=
p2
2m
Thus
KB : K A =
pB2
p2
: A = pB2 mA : p A2 mB
2mB 2mA
= (2 p A ) 2 : p A2 (2mA )
= 4:2
= 2 :1
Page 2 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[32] Consider the assembly depicted below of
nine 1m×1m tiles. Six of the tiles have mass
1kg while the other three have mass 2kg as
shown. Using the axes in the figure, what is
the location of the center of mass for the
system?
Piece A: mass=4kg
CM=(1.5m,2.5m)
Piece B: mass=4kg
CM=(1.0m,1.0m)
(A) (2.2m , 1.5m)
(B) (1,5m , 2.2m)
(C) (1.5m , 1.5m)
(D) (1.5m , 1.7m)
(E) (1.7m , 1.5m)
Piece C: mass=4kg
CM=(2.5m,1.0m)
Answer [E]: The total mass of the system is 12 kg. The simplest way to approach the
problem is do divide it up into three pieces where the CM is obvious and then combine
those results into a grand center of mass. The division is shown in red. The grand center
of mass for the whole system is thus
G
G
G
mA rA + mB rB + mC rC
G
rcm =
mA + mB + mC
(4kg)(1.5m iˆ + 2.5m ˆj ) + (4kg)(1.0m iˆ + 1.0m ˆj ) + (4kg)(2.5m iˆ + 1.0m ˆj )
=
(4kg) + (4kg) + (4kg)
= (1.7m iˆ + 1.5m ˆj )
Another way of simplifying the problem is by noting that the y component of the CM has
to be in the middle (yCM =1.5 m) because the system is made of three horizontal slices of
4 kg each.
Then we only need to compute the x component of the CM. We can then look at the
horizontal slices, which have 3 kg, 4 kg and 5 kg respectively.
xcm =
mA x A + mB xB + mC xC
mA + mB + mC
(3kg)(0.5m) + (4kg)(1.5m) + (5kg)(2.5m)
12kg
= 1.7m
=
[33] A horizontal disk 1 of radius R and mass M rotates with an angular speed ω. Disk 2
with the same axis, same radius, mass M/3 and no initial angular speed is dropped on top
of the first one. What is the final angular velocity, ωf ,of the two disk system when the
Page 3 of 24
Physics 221 2007S Exam 2 SOLUTIONS
disks stop rubbing and rotate as a single unit? Assume that there is no friction on the axle
or other external torques.
(A) ωf =4ω
(B) ωf =4ω/3
(C) ωf =ω
(D) ωf =3ω/4
(E) ωf =ω/4
Disk 2
Answer [D]: Angular momentum about the axle is
conserved since there is no net torque about the
axle. Thus
ω
Disk 1
Linitial = ω I1 = L final = ω f ( I1 + I 2 )
I1
ω
I1 + I 2
If R is the common radius of the two disks, then:
I1 = MR 2
∴ω f =
I2 =
M 2
R
3
thus
ωf =
MR 2
MR + MR
2
1
3
2
ω = 34 ω
[34] A particle with mass m=5.0kg resting on a frictionless horizontal surface is
connected to an ideal spring with force constant k=100N/m. If the amplitude of the
oscillations is 10cm, what is the mechanical energy of the system? Take the zero for the
spring potential energy at the point where the spring is in equilibrium.
k=100 N/m
5kg
(A) 50 J
(B) 10 J
(C) 5.0 J
(D) 1.0 J
(E) 0.50 J
Answer [E]:
The mechanical energy is given by
E = 12 kA2 = 12 (100 mN )(0.10m) 2 = 0.50J
[35] The graph below represents the position of a block attached to a spring. If the force
constant of the spring is k=10N/m, what is the kinetic energy of the system at time t=0s?
Page 4 of 24
Physics 221 2007S Exam 2 SOLUTIONS
(A) 10J
(B) 15J
(C) 20J
(D) 30J
(E) 40J
Answer [B]: The amplitude from the graph is A=2m. The mechanical energy is given by
E = 12 kA2 = 12 (10 mN )(2.0m) 2 = 20J . At t=0s, the displacement is x=1m, so the potential
energy is U (t = 0s) = 12 kx 2 = 12 (10 mN )(1m) 2 = 5J . The kinetic energy is thus
K (t = 0s) = E − U (t = 0s)
= 20J − 5J
= 15J
[36] If a collision is elastic then which of the following must be true?
(A) The total kinetic energy is conserved
(B) The total linear momentum is not conserved
(C) The total kinetic energy is not conserved
(D) The velocity of the center of mass of the system changes.
(E) The relative speed of the two particles is reduced after the collision
Answer [A]: An elastic collision is defined to be a collision where the total kinetic energy
is the same before and after, so A is true. Items B and D are false for all collisions. C and
E are only true for inelastic collisions.
Page 5 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[37] A bullet of mass 1g with velocity of 200m/s collides with a iron block of mass 1kg
resting on a frictionless ice surface. After the collision, the bullet bounces off and is
moving at 150m/s in the +y direction. What is the final speed of the block after the
collision with the bullet?
(A) 0.05 m/s
(B) 0.10 m/s
(C) 0.15 m/s
(D) 0.20 m/s
(E) 0.25 m/s
Top View
150 m/s
y
x
200 m/s
1kg
block
Answer [E]: There is no net external force acting on this system, so total linear
momentum is conserved. In this case:
G
pinit = mbullet vbullet ,initial
G
G
p final = mbullet vbullet , final + mblock vbullet , final
Equating the initial and final momentum:
G
G
G
mbullet vbullet ,initial = mbullet vbullet , final + mblock vbullet , final
m
G
G
G
∴ vblock , final = bullet ( vbullet ,initial − vbullet , final )
mblock
Using the coordinate system indicated, from the question we have:
G
vbullet ,initial = 200 ms iˆ
G
vbullet , final = 150 ms ˆj
so
Page 6 of 24
Physics 221 2007S Exam 2 SOLUTIONS
m
G
G
G
vblock , final = bullet ( vbullet ,initial − vbullet , final )
mblock
0.001g
200 ms iˆ − 150 ms ˆj
1kg
= (0.20iˆ − 0.15 ˆj ) m
=
(
)
s
the speed is thus
vblock , final = (0.20) 2 + (0.15) 2
m
s
= 0.25 ms
[38] A uniform pole is 4.00m long with negligible diameter. It has a mass of 200kg and it
is stuck in the ground tilted at an angle 30º from the vertical. What is the magnitude of
the torque due to the weight of the pole about the point at the bottom of the pole?
(A) 1960 Nm
(B) 3395 Nm
(C) 3920 Nm
(D) 6790 Nm
30°
Pole
Mass=200kg
P
(E) 7840 Nm
θ=150°
Length=4m=L
mg
Torque by
weight=?
Answer [A]: The weight of the pole can be considered to be acting on the center of mass.
L
The magnitude of the torque due to the weight about the point P is τ = mg sin θ where
2
L is the length of the pole and θ is the angle between the weight and the line of the pole
as shown. Plugging in the numbers:
L
sin θ
2
= (200 kg)(9.81 sm2 ) 12 (4m) sin(150o )
τ = mg
= 1960 Nm
Page 7 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[39]
In the figure at right, a horizontal massless rod of 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
y
equilibrium, what is the magnitude of the net force
exerted on the hinge by the rod?
2m
G
T
G
Fh
(A) 0N
(D) 40N
(B) 20N
(E) 57N
(C) 28N
Hinge
45°
●PG
mg
2m
20N
Wall
Answer [B]: By Newton’s third law, the force
exerted by the rod on the hinge is the same as the force exerted by the hinge on the rod,
G
which I will denote Fh .
From the geometry of the problem, since the rope is at a 45° angle from the rod, in the
coordinate system shown,
Tx = −Ty
The rod is in equilibrium, so the net torque on it is 0 about any point. I will take P as the
center of torque for this calculation.
Both the weight of the block and the tension in the rope exert no torque about P because
G
they act on that point. The only force which could exert a torque is Fh , but since the net
torque is 0, this means that this force of the hinge must be along the axis of the lever arm
as shown in the figure.
Since the rod is in equilibrium, the net force in the y direction is 0. Thus,
0 = Ty − mg ⇒ Ty = mg
while in the x-direction:
0 = Fh + Tx
⇒ Fh = −Tx = Ty = mg because of the angle of the rope.
The magnitude of the force which the rod exerts on the hinge is therefore also equal to
mg=20 N.
Page 8 of 24
x
Physics 221 2007S Exam 2 SOLUTIONS
[40] 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 mechanical energy of the particle as a function of
time? (Take the zero of potential energy to be when the spring is in equilibrium)
Answer [B] A, C and D cannot be correct because using the convention given above
for potential energy, once E=0 the system is at rest in equilibrium and there is no
further motion.
Furthermore, D has E<0, which is not allowed in the convention given (although this
is not a problem if we choose a different convention) since with this convention U is
always positive and K by definition is always positive.
E cannot be right because drag forces should be doing negative work on the system
and the mechanical energy should be degrading as a function of time.
B is as one would expect since the drag forces are dissipating the initial mechanical
energy and therefore it exponentially decays to 0. There is a slight undulation on the
decaying exponential curve (which cannot easily be seen with the numbers given and
the resolution of the plot) due to the fact that more energy is dissipated when the mass
in moving more quickly.
Page 9 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[41] Several forces F are applied to the 2D objects depicted below. All forces have equal
magnitude. All the objects have the same mass, uniformly distributed. Which of the
following statements about the magnitude of the angular acceleration about the center of
mass of each system is true?
(A) α P < α Q < α R
(B) α P < α R < α Q
(D) α Q < α R < α P
(E) α R < α Q < α P
(C) α Q < α P < α R
P
Q
τ= −FR
F
τ=+FRsin60°
= 23 FR
τ=+FR
F
Rod
F
F
τ=+FRsin60°
= 23 FR 60°
τ=0
Rod
F
60°
F
τ=0
2R
2R
Solid Disk
F
τ=+FR
2R
τ=+FR
F
R
Answer [B]: To determine the angular acceleration, use Newton’s second law for rigid
body rotation: α =
τ
. So we need to know the torque and moment of inertia for each of
I
the systems. The torque about the CM for each of the forces is shown above. Thus the net
torques are:
τP = 0
τ Q = 3FR
τ R = 2 FR
While the moments of inertia are
I P = 13 MR 2
I Q = 13 MR 2
I R = 12 MR 2
Thus:
F
F
αQ = 4
MR
MR
so the ordering is α P < α R < α Q
αP = 0
α Q = 5.2
Page 10 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[42] The figure below is the energy diagram for a particle that can move along the x axis.
Compare the magnitude of the force acting on the particle and the speed of the particle at
x1 and x2.
U
E
x
x1
A.
B.
C.
D.
E.
x2
|F1| > |F2|
| v1 |<| v2 |
|F1| > |F2|
| v1 |>| v2 |
|F1| < |F2|
| v1 |<| v2 |
|F1| < |F2|
| v1 |>| v2 |
It is not possible to answer the question without knowing the initial position
and velocity of the particle.
dU
, so the magnitude of the force is the
dx
magnitude of the slope of the U vs. x curve. From the graph above we see that therefore
|F1| < |F2| . The relation between the potential energy at the two points is U 2 < U1 and
Answer [C]: The force is given by F = −
since K = E − U it follows that K 2 > K1 . Kinetic energy is tied to speed by K = 12 mv 2 ,
hence | v1 |<| v2 | .
Page 11 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[43] The figure below shows blocks of mass m attached to pulleys of mass m with an
ideal string which pulls without slipping. Which is the correct ranking for the
acceleration of the blocks?
(A) aP < aQ < aR
(B) aQ < aP < aR
(D) aP = aQ > aR
(E) aP = aQ = aR
Solid uniform disk of mass m,
radius R/2.
(C) aP = aQ < aR
Solid uniform disk of mass m,
radius R
Hoop of mass m; radius R: all
mass concentrated on rim.
T
T
m
m
m
mg
P
Q
Page 12 of 24
R
Physics 221 2007S Exam 2 SOLUTIONS
Answer [D]: Define I to be the moment of inertia of the pulley and R the radius of the
pulley. As show in the diagram P above, Newton’s second law applied to the block is:
ma = mg − T
The second law for rotation applied to the pulley is:
Iα = τ = TR
but α =
a
R
(no slipping of the rope on the pulley) and therefore
I
a
R2
Plugging this back into the equation for the block:
T=
ma = mg −
∴a =
I
a
R2
1
I
1+
mR 2
g
For both case P and Q, I = 12 mR 2 , so a = 23 g in both cases. In case R, I = mR 2 , so
a = 12 g , hence aP = aQ > aR .
Page 13 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[44] Consider the two situations depicted. In each a block of mass M=1kg is suspended
from the ceiling by a spring with force constant k=10 N/m. In case 1 the block is pulled
down a distance of 10cm from equilibrium and then released. In case 2 the block is pulled
down a distance of 20cm from equilibrium and then released. If T1 is the period of
oscillation in case 1 and T2 the period in case 2, what is the ratio T1:T2?
Case 1
Case 2
k
k
M
M
20 cm
10 cm
(A) 4:1
(B) 2:1
(C) 1:1
(D) 1:2
(E) 1:4
Answer [C]: The period of a harmonic oscillator only depends on the mass and the force
constant, which is the same in both cases. Thus the period is the same in case 1 and case
2.
[45] Which of the following is true about a conservative force?
(A) The force does a positive amount of work when a particle moves around a closed
loop.
(B) The work done by the force as a particle moves from point A to point B depends
on the path of the motion.
(C) Kinetic friction is a conservative force.
(D) When a conservative force does work on an object, its temperature will change.
(E) None of the above.
Answer [E]: (A) is false because a conservative force does no work on a particle moving
around a closed path.
(B) is false because the work done by a conservative force is path independent.
(C) is false because kinetic friction dissipates mechanical energy into heat and therefore
is not conservative.
(D) is false because a conservative force does not convert mechanical to or from nonmechanical heat.
The answer is therefore (E), none of the above.
Page 14 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[46] A 1.00m long rod of uniform mass distribution weighing 200N is supported at its
ends by wires A and B. A 100N weight is attached to the rod 20cm from wire A. What is
the tension in wire A if the system is in equilibrium?
TA
1.00m=L
A
200N
100N
(A) 300N
(B) 180N
(C) 150N
(D) 130N
0.80m=a
(E) 75N
Answer [B]: Looking at the free-body diagram for the rod above, if we take the torque
condition about point P, we find:
τ net = 0 = (100 N )a + (200 N )
∴TA = (100 N )
L
− LTA
2
a
1
+ (200 N )
L
2
where a=80cm is the distance from P to the attachment point of the weight and L=1m is
the length of the rod. Plugging in the numbers:
TA = (100 N )
a
1
1
+ (200 N ) = (100 N )(0.8) + (200 N )
L
2
2
= 180N
Page 15 of 24
TB
B
●P
Physics 221 2007S Exam 2 SOLUTIONS
[47] Romeo (100 kg) sits in the left end of a 4.0 m-long symmetric boat (100 kg) that
floats still at some distance from the shore and plays a romantic song with his guitar to
Juliet (50 kg), who sits in the right end. After the serenade, Juliet carefully moves to
Romeo’s side to plant a kiss on his cheek. How far to the right does Romeo move during
this process? Assume negligible friction between the boat and the water.
(A) 0.0m
(B) 0.8m
(C) 1.0m
(D) 2m
(E) 3.2m
x
4.0 m=L
Answer [B]: The center of gravity should remain fixed since the water exerts little
external force on this system. Let us use a coordinate system show above where the
origin is Romeo’s initial position. The initial center of mass is
xcm,i =
xRomeo ,i mRomeo + xBoat ,i mBoat + xJuliet ,i mJuliet
M tot
(0m)(100kg) + (2m)(100kg) + (4m)(50kg)
250kg
= 1.6m
=
Note that I take the position of the boat to be the center of mass of the boat which is
always 12 L = 2m to the right of Romeo. Juliet’s final position is the same as Romeo’s.
Therefore:
Page 16 of 24
Physics 221 2007S Exam 2 SOLUTIONS
xcm, f = xcm ,i =
∴ xcm ,i =
xRomeo, f mRomeo + xBoat , f mBoat + xJuliet , f mJuliet
M tot
xRomeo , f (mRomeo + mJuliet ) + mBoat ( xRomeo , f + 12 L)
M tot
Where we have used M tot = mRomeo + mJuliet + mBoat .
Solving this for xRomeo , f :
xRomeo , f = xcm ,i − 12 L
mBoat
M tot
= (1.6m) − 12 (4m)
100kg
250kg
= 0.8m
Page 17 of 24
= xRomeo , f + 12 L
mBoat
M tot
Physics 221 2007S Exam 2 SOLUTIONS
[48] Cylinder 1 is released on an incline and rolls down (without slipping).
Cylinder 2 has an initial angular speed at the bottom of the ramp and thus starts rolling up
(again without slipping).
What is the direction of the static friction force in each case as they roll along the ramp?
Answer[B]:
Page 18 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[49] A system consists of two identical solid disks of mass M and radius R attached
together as shown below. What is the moment of inertia of the system about an axis
perpendicular to the page at point P?
(A) 7MR²
(B) 6MR²
(C) 5MR²
(D) 4MR²
(E) 3MR²
P
●
R
90°
R
Solid disk
Mass M
Radius R
Disk1
R
Solid disk
Mass M
Radius R
Disk2
Answer [A]: Each of the solid disks has a moment of inertia about its center of mass of
I cm = 12 MR 2 . Using the parallel axis theorem on disk 1, where the center is a distance R
away from P:
I1 = I cm + Md 2 here d = R so
= 12 MR 2 + MR 2
= 32 MR 2
Likewise, using the parallel axis theorem on disk 2, where the center is
origin:
I 2 = I cm + Md 2 here d = 5R so
= 12 MR 2 + 5MR 2
= 112 MR 2
Thus
I tot = I1 + I 2 = 7 MR 2
Page 19 of 24
5R from the
Physics 221 2007S Exam 2 SOLUTIONS
[50] Consider the three forces of magnitude 1.00N exerted on an equilateral triangle of
uniform density depicted below with side length 2.00m. What is the magnitude of the
torque about the center of mass of the system?
(A) 1.30 Nm
(B) 2.25 Nm
(C) 2.60 Nm
(D) 3.00 Nm
(E) 5.20 Nm
60°
1N
90°
2m
R⊥ = 1m
☻
2m
2m
60°
90°
90°
1N
60°
1N
Answer[D]: The center of mass is at the center of the triangle ☻. Focusing initially at the
force on the lower left corner, we see by extending the line of the arrow as shown that
R⊥ = 1m (i.e. the distance of closest approach of the line to the center of mass. The torque
of that force is τ = FR = (1 N m )kˆ (i.e. out of the page). By symmetry, the torques at
1
⊥
the other corners are identical so the net torque is τ net = 3τ 1 = 3Nm , again in the direction
out of the page.
Page 20 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[51] A particle of mass 5kg is located at the point (1m, 2m, 3m) and has a velocity of
m
G
v = (3iˆ + 2 ˆj + 1kˆ) . What is the y-component of the angular momentum vector of the
s
particle about the origin?
(A) Ly = −50 Js
(B) Ly = +50 Js
(D) Ly = −40 Js
(E) Ly = +40 Js
(C) Ly = +130 Js
G
G G
Answer [E]: Angular momentum is given by L = mr × v , so the y-component of angular
momentum is given by:
Ly = m( zvx − xvz )
= (5kg)((3m)(3 ms ) − (1m)(1 ms ))
= +40Js
Page 21 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[52] A 20-g rubber ball bounces on a table as shown in the figure below. The ball and the
table are in contact for 15 ms. What is the magnitude of the average force on the ball by
the table?
Ball
60 m/s
60 m/s
60°
(A) 0 N
(B) 30 N
(C) 80 N
60°
(D) 140 N
(E) 160 N
Answer [D]:
G
G G
G
Δp p f − pi
=
The average force is given in terms of the impulse by Fav =
Δt
Δt
In this case:
G
pi = (.020kg)(60 ms )(iˆ cos 60o − ˆj sin 60o )
G
p f = (.020kg)(60 ms )(iˆ cos 60o + ˆj sin 60o )
G G
G
∴Δp = p f − pi = (1.2Ns)(2 ˆj sin 60o )
= 2.1Ns
(i.e. the impulse) Since this happens over Δt = 15ms , Fav =
Page 22 of 24
2.1Ns
= 138N
.015s
Physics 221 2007S Exam 2 SOLUTIONS
[53] A particle is released from rest at x = x0 in the presence of a conservative force with
the potential energy shown below. No other forces are present. Which of the following
best describes the subsequent motion of the particle?
U
Mechanical Energy
x
x0
(A) The particle oscillates about the first minimum.
(B) The particle oscillates about the second minimum.
(C) The particle has a complex oscillation that includes both minima.
(D) The particle moves in the positive x direction all the time.
(E) None of the above.
Answer[D]: The mechanical energy shown with the line in the above diagram is equal to
the potential energy at x = x0 . Since there is no turning point to the right, the particle
moves in the positive x direction for all time after being released.
[54] A 1.0-kg block is pressed against a horizontal ideal spring with spring constant k =
400 N/m. The spring is compressed 10 cm from its relaxed position. When released, the
block slides on a horizontal surface for 50 cm and then up an incline, where it reaches a
maximum height of 15 cm vertically above the initial level. What is the magnitude of the
work done by the friction forces?
(A) 0.5 J
(B) 1.0 J
(C) 1.5 J
(D) 2.0 J
(E) 2.5 J
Answer[A]: The only non-conservative force present is kinetic friction so the work done
by friction is equal to the change in the mechanical energy.
Both at the start and at the finish the block is at rest so the mechanical energy is all
potential energy. The initial mechanical energy is Ei = 12 kx 2 = 12 (400 mN )(0.1m) 2 = 2J . The
final mechanical energy is E f = mgh = (1.0kg)(9.81 sm2 )(0.15m) = 1.47J . The work done
by friction is thus W fric = 1.47J − 2.00J = −0.53J ∴ | W fric |= 0.53J .
Page 23 of 24
Physics 221 2007S Exam 2 SOLUTIONS
[55 Extra Credit] An 18g bullet is shot vertically upwards into a 10kg block and quickly
becomes imbedded in the block. If the block lifts upwards 9.0mm, what is the initial
kinetic energy of the bullet?
(A) 0.0016 J
(B) 0.88 J
(C) 49 J
(D) 330 J
(E) 490 J
Answer[E]:
Let us denote the mass of the bullet=m and the mass of the block=M. Taking the initial
position of the block as the reference point for gravitational potential, the final
mechanical energy of the block is:
E f = (m + M ) gh
where h is the height lifted (=9mm). This will also be the kinetic energy of the blockbullet system the instant after the bullet is embedded. Neglecting the impulse due to
gravity over the short period of time the bullet is burrowing into the block, momentum is
conserved during the inelastic collision between the bullet and the block. On the one
hand, this momentum is given in terms of the post-collision KE of the block (i.e. E f ):
pafter = 2(m + M ) E f
and, on the other hand, in terms of the initial kinetic energy ( K i ) of the bullet:
pbefore = 2mK i
Equating these two:
2mK i = 2(m + M ) E f = 2(m + M ) 2 gh
(m + M )2
(10kg) 2
gh =
(9.81 sm2 )(0.0090m) = 490J
m
(0.018kg)
Note that it is reasonable to take the approximation M + m ≈ M throughout. Doing so
does not change the answer significantly.
∴ Ki =
Page 24 of 24