Physics 221.
Exam 2 (Midterm)
Spring 2003
51. This booklet is version A. Enter A on question 51.
The situation below refers to the next two questions:
Two particles A and B move along the x-axis according to the graph shown below.
x (m)
A
0
1
B
2
3
4
5
t (s)
52. Which of the following statements about the x-component of their accelerations
between t = 0 and t = 5 s is true?
a.
b.
c.
d.
e.
ax,A > 0 , ax,B > 0
ax,A > 0 , ax,B = 0
ax,A = 0 , ax,B > 0
ax,A < 0 , ax,B > 0
ax,A < 0 , ax,B = 0
Particle B moves at constant speed (x(t) has a constant slope), so ax,B = 0.
Particle A starts off with a certain positive velocity (positive slope) and slows down
(slope decreases) between 0 and 2 s (approx.). This corresponds to a negative
acceleration. After t =2 s, it starts moving in the negative x-direction (i.e., negative
velocity) and speeds up, so the acceleration is still negative. Mathematically, the second
derivative of this curve is always negative.
3. When is their relative velocity equal to zero?
a.
b.
c.
d.
e.
At some point between t = 0 and t = 1 s.
At some point between t = 1 and t = 2 s.
At some point between t = 2 and t = 3 s.
At some point between t = 3 and t = 4 s.
At some point between t = 4 and t = 5 s.
Page 1 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
Their relative velocity will be zero whenever their velocities are equal. This is, whenever
the slopes of the curves are equal. This happens sometime between 1 and 2 s (see dotted
red line in figure).
54. Two vectors A and B have the same magnitude. Vector A is in the positive xdirection. Vector B is in the negative y-direction. Which of these figures correctly depicts
the angle between vectors A and (A-B)? (Ignore length of vectors in the figure).
A−B
a.
A−B
b.
90°
c.
45°
A
A
A
45°
A−B
A−B
d.
60°
A
e.
60°
A
A−B
y
A−B
−B
A
x
B
55. Force vector FA has the magnitude of 5.0 N and an angle of 300° with respect to the
positive x-axis (counter-clockwise). Force vector FB has a magnitude of 5.0 N and an
angle of 60° with respect to the positive x-axis (counter-clockwise). What is the
magnitude of the third force vector, which must be added to FA and FB in order to make
the net force equal to zero? (All forces act on the same object).
a.
b.
c.
d.
e.
2.5 N
5.0 N
7.5 N
10.0 N
None of the above
Page 2 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
y
FB
60°
FC
x
60°
FA
You can see the result right away from the symmetry of the problem (all three vectors are
separated by angles of 120°) or you can do the calculation.
Fc,y = 0 (due to the symmetry)
Fc,y = −(5 N) cos 60° − (5 N) cos 60° = − 5 N
56. Let A = i + 2 j + α k and B = 3 i − j + β k. Find the values of α and β that will make
A×B = 0.
a.
b.
c.
d.
e.
α = −2, β = 0
α = 0, β = 2
α = −2, β = −2
α = 3, β = −2
There are no values of α and β that will make A×B = 0.
A×B = 0 means that the vectors are parallel to each other. In this case, there should be
some scalar k such that A = k B. This is clearly impossible considering the first two
components.
You can also perform the cross product. A×B = (2β+α)i +(3α−β)j −7k. The z-component
is never zero.
Page 3 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
57. If you go from the origin to x = 10.0 m in 5.0 s, stay there for 2.0 s, and then go to
x = −2.0 m in 3.0 s, what will be the x-component of your average velocity during this
trip?
a.
b
c.
d.
e.
vx,average = –2.0 m/s
vx,average = –1.2 m/s
vx,average = –0.2 m/s
vx,average = 0.2 m/s
vx,average = 1.2 m/s
vaverage =
∆x (−2 − 0)m
=
= −0.2 m/s
∆t (5 + 2 + 3)s
58. Three blocks A, B and C rest on a frictionless, horizontal surface. A force F = 18 N
is applied on block A. What is the force exerted on block C by block B?
F
A
B
C
mA = 2 kg
mB = 3 kg
mC = 4 kg
a.
b.
c.
d.
e.
FC,B = 5 N
FC,B = 8 N
FC,B = 12 N
FC,B = 14 N
FC,B = 18 N
The only horizontal force acting on C is the normal by B, so FC,B = FC,net.
F
a=
mA + mB + mC . So the net force on C is
The acceleration of the whole system is
FC,net = mC a =
mC
4 kg
F=
18 N = 8 N
mA + mB + mC
9 kg
.
The situation below refers to the next two questions:
A particle of mass m = 1.0 kg travels in a circle of radius R = 3.0 m at a constant speed of
2.0 m/s.
59. Find the work done by the net force acting on it during half a revolution.
a.
b.
c.
d.
e.
W=0
W = 4π J
W = 8π J
W = 16π J
None of these.
Page 4 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
Since the particle moves in a circle at constant speed, the net force must point towards
the center of the circle and is therefore perpendicular to the trajectory at all points.
Hence the work done by this force is zero.
Also: The net work (or work done by the net force) must be zero because the kinetic
energy is constant and Wnet = ∆KE..
60. Find the angular displacement of the particle in 5.0 s.
a.
b.
c.
d.
e.
∆θ =
∆θ =
∆θ =
∆θ =
∆θ =
35°
78°
120°
150°
190°
v
2.0 m/s
360D
5.0 s = 3.33 rad
∆ϕ = ω∆t = ∆t =
= 191D
R
3.0 m
2π rad
61. For a particle in circular motion at constant speed v, which of the five plots best
shows the magnitude of the acceleration a as a function of radius r?
a
a
a
r
a.
a
r
b.
a
r
c.
r
r
d.
e.
ac =
v2
r or
A particle in uniform circular motion has centripetal acceleration only, so
ac = rω 2 . For a given linear speed v, the acceleration changes with radius as in the first
expression,
ac =
constant 

v2
y=

x
.
r . So the curve is a hyperbola 
Page 5 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
The situation below refers to the next two questions:
A block of mass m = 9.0 kg is released from rest at the top of a frictionless ramp of length
d = 1.5 m. The speed of the block at the bottom is 2.5 m/s.
N
d
mg sinθ
θ
mg cosθ
mg
62. What is the angle θ between the ramp and the horizontal?
a.
b.
c.
d.
e.
θ = 5°
θ = 8°
θ = 12°
θ = 30°
θ = 40°
This problem can be solved using Newton’s laws or conservation of energy.
Newton’s laws:
The only two forces acting on the box are its weight and the normal by the incline.
Along the incline: mg sinθ =ma → a = g sinθ
2
2
For constant acceleration, v − v0 = 2a∆x , so
v − 0 = 2dg sin θ
2
⇒
v2
sin θ =
2dg
⇒
 v2 
D
θ = sin 
 = 12
 2dg 
−1
Conservation of energy
Etop = mgh = mgd sin θ
1 2
mv
2
1
mgd sin θ = mv 2
2
Ebottom =
⇒
 v2 
D
 = 12
 2dg 
θ = sin −1 
Page 6 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
63. A force F is applied on the box, perpendicularly to the box top surface, as shown in
the figure below. How does this affect N, the magnitude of the normal force exerted by
the ramp on the box, and a, the magnitude of the acceleration of the box? The incline is
still frictionless.
F
θ
a.
b.
c.
d.
e.
N
mg
N increases, a stays the same.
N increases, a decreases.
N stays the same, a stays the same.
N stays the same, a decreases.
N stays the same, a increases.
Newton’s 2nd law in the direction perpendicular to the incline: N − mg cosθ – F = 0, so N
= mg cosθ + F. So N is larger when F is there.
This does not affect the acceleration, because the equation in the direction along the
incline is independent. If there was friction, there would be some interdependence
because the magnitude of the kinetic friction is proportional to the magnitude of the
normal. But in this case there is no friction.
The situation below refers to the next two questions:
The figure below shows the overhead view of two identical stones that travel in circles
over a frictionless horizontal surface. Each stone is tied to a massless string whose
opposite end is anchored at the center of the circle. Both stones have the same period.
64. Compare TL and TS, the tensions on the long and the short string, respectively.
a. TL < TS
b. TL = TS
c. TL > TS
d. You cannot tell without knowing the magnitude of the centripetal acceleration in each
case.
e. You cannot tell without knowing the mass of the stone.
Page 7 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
The tension in this case is the only force in the radial direction, so it must be responsible
for the centripetal acceleration: T = mac . In this case, we want to write the acceleration
2
in terms of the angular speed, ac = rω , because both stones have the same period, i.e.,

2π 
ω =

period  . So T = mrω 2 , and therefore the larger the
the same angular speed 
radius, the larger the tension.
65. Let ωL and ωS be the angular velocity vectors of the stone with the long and the short
string, respectively.. Which of the following is true?
a.
b.
c.
d.
e.
ωL points into the page, ωS points out of the page, and ωL= ωS
ωL points out of the page, ωS points into the page, and ωL= ωS
They both point out of the page, and ωL< ωS
They both point into the page, and ωL> ωS
They both point into the page, and ωL= ωS
Direction is obtained using the right-hand rule.
Magnitudes must be equal since the periods are equal and ω=2π/period.
Page 8 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
66. A frictionless roller coaster cart of mass m tops the hill at point A with speed v0. Find
the work done on the cart by the gravitational force as it moves from point A to point B.
v0
B
A
h
h
h/3
a. Wg = mgh
1
mgh
b. Wg = 3
c. Wg = 2mgh
2
mgh
d. Wg = 3
e. Wg = 0
Gravity is a conservative force, so Wg = −∆U g where Ug is the gravitational potential,
Ug =mhy + constant. Since the height y is the same in points A and B (y=h for both), the
work by gravity for this displacement is zero.
67. A small 0.20-kg rock is released from rest at point A, which is at the top edge of a
large hemispherical bowl with radius R = 0.50 m. Assume that the size of the rock is
small compared to R, so the rock can be treated as a particle, and assume that the rock
slides rather than rolls. The work done by friction on the rock when it moves from point
A to point B at the bottom of the bowl is –0.08 J. What is the speed of the rock when it
reaches point B?
A
a.
b.
c.
d.
e.
vB =
vB =
vB =
vB =
vB =
2.00 m/s
2.26 m/s
3.00 m/s
3.13 m/s
3.26 m/s
B
Page 9 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
When there are non-conservative forces (like friction) in a system, ∆E = Wnon −conservative .
1
Einitial = mgR
E final = mv 2
2
1 2
mv − mgR = Wn −c
2
v=
2 (Wn −c + mgR )
= 3.00 m/s
m
The situation below refers to the next four questions:
Three velocities from a collision of objects A and B are drawn to scale in the figure
below. The initial velocity of object B, vB,i, is not shown, but it is known to have no
component in the y-direction. Note that | vA,i | = | vA,f |. There are no external forces acting
on the system.
vA,f
y
vA,i
vB,f
x
68. The ratio between the masses of the objects is:
mA 2
=
a. mB 5
mA 2
=
b. mB 3
mA
=1
c. mB
mA 4
=
d. mB 3
mA
=2
e. mB
From the figure and the text, it follows that pinitial,y = 0. Therefore, pfinal,y = 0.
v
mA
p A, y + pB , y = 0 ⇒
mAv A, y + mB vB , y = 0 ⇒
= − B, y
mB
v A, y
vB , y
From the figure we see that v A, y
=−
2
mA 2
=
3 , so mB 3 .
Page 10 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
69. Which of the following is true about vB,i, the initial velocity of object B?
a. It points in the +i direction.
b. It points in the −i direction.
c. It is zero.
d. Its direction cannot be determined without knowing whether the collision is elastic or
inelastic.
e. Its direction cannot be determined without the actual value of each mass.
Momentum is conserved in the x-direction: p A,i , x + pB ,i , x = p A, f , x + pB , f , x
pB , f , x = 0 and p A,i , x > p A, f , x (both positive), because v
A,i, x is 5-squares long and vA,f, x is 4squares long.
So pB ,i , x = p A, f , x − p A,i , x < 0
70. Which of these arrows best represents
the direction of the impulse on ball A?
a. A
b. B
c. C
B
d. D
e. E
C
D
E
A
G
G
G
G
I
The impulse is A = ∆p A = p A, f − p A,i , which corresponds to:
pf - pi
pf
pi
pf
- pi
71. Which of the following correctly describes the motion of the center of mass of this
system during this process?
a.
b.
c.
d.
e.
Always in the +i direction, at constant speed.
Always in the −i direction, at constant speed.
Always in the +i direction, but the speed is different before and after the collision.
At constant speed but in different directions before and after the collision.
It is at rest all the time.
There are no external forces acting on the system, so the velocity of the CM is constant.
By looking at the final velocities, we can conclude that the CM has a net positive x
component. And we know from the initial situation that the y-component of the velocity of
the CM is zero.
Page 11 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
The situation below refers to the next two questions:
Two blocks are connected through an ideal, massless string that goes over an ideal,
massless pulley, as shown in the figure below. The mass of block A is mA = 10.0 kg and
the coefficient of kinetic friction between block A and the incline is µk = 0.2. The angle
between the incline and the horizontal is θ = 30°.
T
N
T
f
A
mA g
B
θ
mB g
72. Block A slides down the incline at constant speed. What is the mass of block B?
a.
b.
c.
d.
e.
2.0 kg
3.3 kg
5.0 kg
6.7 kg
10.0 kg
From the free-body diagram we can write Newton’s second law for mass A (both along
the incline and perpendicularly to it) and for mass B:
m A g sin θ − f − T = 0
(constant speed ⇒ a = 0)
m A g cos θ − N = 0
T −mBg = 0
(constant speed ⇒ a = 0)
where f = µ k N = µ k m A g cos θ . So
T = mBg
T = m A g sin θ − f = m A g ( sin θ − µ k cosθ )
mB = m A (sin θ − µ k cosθ ) = 3.3 kg
Page 12 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
73. If friction was not there but the masses and the angle of the incline remained the
same, the tension on the string would be:
a.
b.
c.
d.
e.
Larger.
Larger on the side of block A only.
The same.
Smaller
Smaller on the side of block A only.
If friction disappears, the system acquires an acceleration (down for A, up for B). If we
look at B, that means that T > mBg, so T must be larger that before. And it’s larger on
both sides, because in a massless string the tension is the same at all points.
The situation below refers to the next three questions:
A box stands in an elevator that travels a distance of 20m. During the displacement,
work done on the box by the normal force exerted by the elevator floor is WN = 12 kJ,
and the work done on the box by gravity is Wg = –10kJ. Use g = 10 m/s2 for simplicity.
74. What is the mass of the box?
a.
b.
c.
d.
e.
mbox = 20 kg
mbox = 40 kg
mbox = 50 kg
mbox = 80 kg
mbox = 100 kg
The work done by gravity is
m=
later). So
Wg
g ∆h
Wg = mg ∆h
(in magnitude, we’ll worry about the sign
= 50 kg
.
75. What is the magnitude of the normal force that the elevator exerts on the box?
a.
b.
c.
d.
e.
N = 100N
N = 200N
N = 300N
N = 400N
N = 600N
The magnitude of the work done by normal force is WN = N ∆h . So
N=
WN
∆h
= 600 N
.
Page 13 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
76. The elevator is:
a.
b.
c.
d.
e.
Moving up and slowing down.
Moving up at constant speed.
Moving up and speeding up.
Moving down and slowing down.
Moving down and speeding up.
The work done by gravity is negative. This means that the weight vector and the
displacement vector point in opposite directions (in this case it can only be 0° or 180°).
Thus the elevator is going up.
The only two forces acting on the box are the normal and the weight. The net work on the
box is Wnet = Wg + WN = +2kJ > 0 . Since Wnet = ∆KE , the box −and the elevator− are
speeding up.
77. A motorist drives south at 20.0 m/s for 3.00 minutes, then turns west and travels at
25.0 m/s for 2.00 minutes, and finally travels northwest at 30.0 m/s for 1.00 minute.
Determine the magnitude of total displacement.
N
W
E
a.
b.
c.
d.
e.
6.25 km
5.53 km
2.50 km
4.87 km
3.75 km
S
If we take the +x axis along the E and the +y axis along the N, the displacements for
each time interval are the following:
G
∆r1 = ( 20.0 m/s ) (3.00 × 60 s)(− ˆj ) = − ( 3600 m ) ˆj
G
∆r2 = ( 25.0 m/s ) (2.00 × 60 s)(−iˆ) = − ( 3000 m ) iˆ
−iˆ + ˆj
G
∆r3 = ( 30.0 m/s ) (1.00 × 60 s)(
) = (1273 m ) −iˆ + ˆj
2
G
G
G
G
∆r = ∆r1 + ∆r2 + ∆r3 = −(4273 m)iˆ − (2327 m) ˆj
G
∆r = (4273 m) 2 + (2327 m) 2 = 4866 m = 4.8 km
(
)
Page 14 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
78. A car, initially at rest, accelerates at a constant rate a1 for ∆t1 = 5.0 s and then
accelerates at a2 = −5.0 m/s2 for ∆t2 = 10.0 s after which the car is going with speed
vf = 5.0 m/s. Find a1, the acceleration of the car for the first 5 seconds.
a.
b.
c.
d.
e.
a1 = 5.0 m/s2
a1 = 11 m/s2
a1 = 14 m/s2
a1 = 18 m/s2
a1 = 20 m/s2
Speed after the first 5 seconds: vt =5 s = 0 + a1 (5 s)
2
Speed after the other 10 seconds: 5.0 = vt =5 s − (5 m/s )(10 s)
So
a1 =
->
vt =5 s = 55 m/s
vt =5 s
= 11 m/s 2
5s
79. A car is initially moving at a speed of 25 km/h when the brakes are applied, resulting
in a constant deceleration that brings the car to rest in 15 m. Assuming the same
deceleration, what would the stopping distance ∆x be if the initial speed were 50 km/h?
a.
b.
c.
d.
e.
∆x = 30 m
∆x = 45 m
∆x = 60 m
∆x = 75 m
∆x = 90 m
2
2
2
When the acceleration is constant, v − v0 = 2a∆x , or −v0 = 2a∆x when the final speed is
zero. Even without substituting the numbers, it is clear that for twice the initial speed, a
distance four (22) times larger will be required. This the answer is 4 times 15 m = 60 m.
Page 15 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
80. You can keep a book immobile against a vertical wall by applying a force
perpendicular to the book. If you press twice as hard, the magnitude of the static friction
between the book and the wall changes by a factor of:
fS
Nby
Nby hand
mg
a.
b.
c.
d.
e.
1 (It does not change).
2
1/2
4
1/4
As seen in the free-body diagram, fs= mg for the book to stay where it is. This is totally
independent of the two normal forces! If we push harder (i.e., increase Nby hand), Nby wall
also increases (they must cancel each other). This will increase the maximum value of the
static friction µsN, but not the actual value at that moment. (So if we push harder we’ll be
able to keep a heavier book immobile against the wall).
The situation below refers to the next two questions:
Two billiard balls (same mass) collide head-on on a billiard table. The collision can be
considered elastic. The initial velocity of ball A is vA,i = 1.3 m/s i and the initial velocity
of ball B is vB,i = −0.85 m/s i.
81. What is vB,f, the final speed of ball B?
a.
b.
c.
d.
e.
vB,f = 0.58 m/s
vB,f = 0.69 m/s
vB,f = 0.85 m/s
vB,f = 1.3 m/s
vB,f = 1.9 m/s
In an elastic collision, both linear momentum and kinetic energy are conserved.
In this case (head-on collision), the whole process is 1D.
mv A,i + mvB ,i = mv A, f + mvB , f
v A ,i + vB ,i = v A , f + vB , f
0.45 m/s = v A, f + vB , f
1 2 1 2
1
1
mv A,i + mvB ,i = mv A2 , f + mvB2 , f
2
2
2
2
v A2 ,i + vB2 ,i = v A2 , f + vB2 , f
2.4 ( m/s ) = v A2 , f + vB2 , f
2
If you solve the system, you find vA,f = −0.85 m/s and vB,f = 1.3 m/s. Because of the
symmetry (they are identical balls), the only option that allows them to keep both
momentum and kinetic energy constant is switching their velocities.
Page 16 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
82. If someone put gum (of negligible mass) on one of the balls and they stick together
after they collide, what is their final momentum pf? Each ball has a mass m = 0.34 kg.
a.
b.
c.
d.
e.
pf = −(0.31 kg m/s) i
pf = −(0.25 kg m/s) i
pf = 0.15 kg m/s i
pf = 0.45 kg m/s i
pf = 0.60 kg m/s i
Now kinetic energy is not conserved (inelastic collision), but linear momentum is still
conserved:
p f = m ( v A,i + vB ,i ) = 0.15 kg m/s
mv A,i + mvB ,i = p f
The situation below refers to the next two questions:
The graph below shows the potential energy U as a function of position x for a particle of
mass m = 0.5 kg. The only force acting on this particle is the force associated to this
potential energy.
U (J)
50
10
−8
−6
−4
−2
2
4
6
8
x (m)
83. Determine the force on the particle at x = 0 m.
a.
b.
c.
d.
e.
F (x = 0 m) = − 40 N i
F (x = 0 m) = − 6 N i
F (x = 0 m) = 0
F (x = 0 m) = 6 N i
F (x = 0 m) = 40 N i
The force at x = 0 m is minus the slope of the U(x) curve at x = 0. The slope is – 6, so the
force is + 6 N.
Page 17 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
84. If the mechanical energy of the particle is E = 30 J when its position is x = 4 m,
determine at which position will the particle achieve its maximum speed.
a.
b.
c.
d.
e.
x = −6 m
x = −3 m
x = 1.5 m
x=5m
x=6m
For E = 30 J and being initially at x = 4 m, the particle is confined to the region between
x = 1.5 m (approx.) and x = 6 m. The maximum kinetic energy and therefore the
maximum speed is achieved when the potential energy is minimum (in the allowed
region), which happens at x = 5 m.
85. Two balls are thrown simultaneously from point P and follow the trajectories shown
in the figure below. Which ball hits the ground first?
Ball A
Ball B
P
a.
b.
c.
d.
e.
Ball A
Ball B
They hit the ground simultaneously.
You cannot tell without knowing the initial velocities.
You cannot tell without knowing the angles at which they are thrown.
The time to reach the bottom is twice the time to reach the top or to come down from the
top. Ball A goes higher, so it takes longer.
86. A pilot must fly due north, to reach the destination. The speed of the airplane is
300 km/h (relative to the air) and wind blows from the northeast at a speed of
90 km/h relative to the ground. Find the speed of the airplane relative to the ground.
a.
b.
c.
d.
e.
vplane, ground = 230 km/h
vplane, ground = 260 km/h
vplane, ground = 290 km/h
vplane, ground = 300 km/h
vplane, ground = 310 km/h
Page 18 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
N
vplane,ground
vplane,air
φ
E
W
45°
vair,ground
S
G
G
G
v
=
v
+
v
plane,ground
plane,air
air,ground . This is
From the Galilean transformations, we know that
G
v
= 300 km/h
graphically expressed in the diagram above. We know that plane,air
and
G
vair,ground = 90 km/h
.
Write the relation for the x- and y-components separately:
x : 0 = 300sin ϕ − 90sin 45D
y : vplane,ground = 300 cos ϕ − 90 cos 45D
−1
D
From the first equation we get ϕ = sin (0.21) = 12.2 . So
vplane,ground = 300 cos12.2D − 90 cos 45D = 230 km/h
Page 19 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
87. Determine the center of mass of the following surface, made with 5 identical, uniform
squares of side a.
y
a
a
x
a.
b.
c.
d.
e.
( xCM , yCM ) = 
3 9 
a, a 
 2 10 
( xCM , yCM ) = 
3

a, a 
2

( xCM , yCM ) = 
2

a, a 
3

( xCM , yCM ) =  a,
9 
a
 10 
( xCM , yCM ) =  a,

2 
a
3 
You can find the CM with positions of the CM of each square (slow), or with the positions
of the CM of 2 rectangles and one square (a little faster) or use the “hole” trick (fast).
CM of a 6-square object: (3/2a,a)
CM of the missing square: (3/2a, 3/2a)
3

3 3 
6 M  a, a  − M  a, a   18 − 3 6 − 3 
2 2  = 2 2,
2 a =  3, 9 a
( xCM , yCM ) =  2 




5M
5 
 2 10 
 5


(Also, if you think about it, the xcm was clearly 3a/2 (symmetry), so it was either a or b.
And b couldn’t be, because ycm = a is the result for the 6 square rectangle! )
Page 20 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
88. A bullet of mass m = 20 g is shot against a wooden block of mass M = 4 kg that is
attached on the opposite side to a strong spring of constant k = 4 ×104 N/m, which is itself
fixed to a rigid, immovable support (see figure below). The bullet strikes the block and
embeds itself in it. The spring is compressed a maximum distance x = 0.1 m. Before the
shot, the spring is perfectly relaxed. Friction is negligible.
k
vi
M
m
Find vi, the initial speed of the bullet.
a.
b.
c.
d.
e.
vi = 120 m/s
vi = 750 m/s
vi = 900 m/s
vi = 1500 m/s
vi = 2000 m/s
We need to use conservation of momentum for the inelastic collision (step 1) and
conservation of energy for the compression of the spring (step 2).
Step 1: mvi = ( M + m)V
1
1
( M + m)V 2 = kx 2
2
Step 2: 2
M +m
vi =
V=
m
Therefore,
⇒
V=
( M + m ) kx 2
m2
kx 2
M +m
= 2000 m/s
Page 21 of 22
Physics 221.
Exam 2 (Midterm)
Spring 2003
89. A 320 N bag hangs from three ropes as shown. The ropes make angles θ1 = 60° and
θ2 = 30° with the ceiling. The system is in equilibrium. Find T2, the tension on the righthand rope.
θ1
θ2
T1
T2
T3
mg
a.
b.
c.
d.
e.
T2 = 130 N
T2 = 150 N
T2 = 160 N
T2 = 210 N
T2 = 280 N
T3 = mg, so I’ll use mg instead.
Horizontal forces: T1 cos θ1 − T2 cos θ 2 = 0
Vertical forces : T1 sin θ1 + T2 sin θ 2 − mg = 0
Here’s an elegant way to solve this system:
T1 = T2
cos θ 2
cos θ1
T2
cos θ 2
sin θ1 + T2 sin θ 2 − mg = 0
cos θ1
T2
sin θ1 cosθ 2 + sin θ 2 cosθ1
= mg
cos θ1
T2
sin(θ1 + θ 2 )
= mg
cos θ1
mg
cos θ1
cos 60D
= 160 N
T2 =
mg =
mg =
D
sin(θ1 + θ 2 )
sin 90
2
Did you bubble in your exam version?
Page 22 of 22