Version 064 – Test 1 – swinney – (58385)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
a
t
a
2.
a
t
5.
correct
Michael stands motionless holding a baseball in his hand. After a while he tosses it
upwards, and it travels up for a while before turning about and heading toward the
ground. Define upwards to be positive.
Which diagram can describe the vertical
acceleration of the ball, while it is in Michael’s
hand and after he lets it go, assuming it has
not yet hit the ground?
1.
1
a
t
6.
Explanation:
The ball first experiences a period of zero
acceleration when Michael is just holding the
ball. Then, as the ball is thrown upward, it
feels an upward acceleration. Quickly the ball
is released. Once the ball leaves Michael’s
hand it is in free-fall. In free-fall all objects feel a downward (here negative) constant
gravitational acceleration.
a
t
t
a
3.
t
002 10.0 points
A proton of mass mp is moving along the -ẑ
direction and undergoes a change in its speed
from 0.993c to 0.997c. What is the magnitude
and direction of the impulse acting on the
proton?
1. 4.47mp c along +ẑ
a
4.
t
2. 4.47mp c along -ẑ correct
3. 0.004mp c along -ẑ
4. 5.2mp c along -ŷ
5. 6.7mp c along +ẑ
6. 8.1mp c along +x̂
7. 0.004mp c along +ẑ
Version 064 – Test 1 – swinney – (58385)
8. 5.2mp c along +ẑ
003
9. 8.1mp c along -ẑ
Let : mp = 1.7 × 10−27 kg ,
v1 = 0.993 c ,
v2 = 0.997 c , and
c = 3 × 108 m/s .
1
v2
1− 2
c
,
so
1
(0.993 c)2
1−
c2
= 8.46637 and
γ2 = s
1
(0.997 c)2
c2
= 12.9196 .
1−
We use the formulae for the impulse and
the momentum
Impulse = p~f − p~i
= pf − pi (−ẑ)
M omentum = γ m v
Thus
(γ2 v2 − γ1 v1 )
× mp c
c
(12.9196) (0.997 c)
=
× mp c
c
(8.46637) (0.993 c)
× mp c
−
c
= 4.47 mp · c .
mV
Explanation:
γ1 = s
10.0 points
A stationary object explodes, breaking into
three pieces of masses m, m, and 2 m. The
two pieces of mass m move off at right angles
to each other with different momenta, m V
1
and m V , as shown below.
3
10. 6.7mp c along -ẑ
γ=s
2
1
mV
3
What is the magnitude of the velocity of
the piece having mass 2 m?
√
3V
1. k~v k =
3
V
2. k~v k =
2
√
10 V
3. k~v k =
3
V
4. k~v k =
3
√
10 V
5. k~v k =
2
√
10 V
correct
6. k~v k =
6
4V
7. k~v k =
3
√
5 3V
8. k~v k =
3
√
2 3V
9. k~v k =
3
√
5 3V
10. k~v k =
2
I=
Explanation:
The vector sum of all three momenta must
be zero. Since momenta of the two masses m
are a right angle, the momentum of the mass
2m forms the hypotenuse of a right triangle.
Version 064 – Test 1 – swinney – (58385)
Thus the magnitude of the momentum of the
mass 2m is just given by
r
√
10
1
2
2
p2m = (m V ) + ( m V ) =
m V.
3
3
Hence,
√
10
mV
3
√
10 V
v=
6
(2 m) v =
004 10.0 points
Consider the following diagram
~
D
~
A
~
C
~
B
Which of the following pairs of statements
is true?
~ −B
~ =C
~
Ia. A
~
~
~
Ib. A − D = C
~ −C
~ =B
~
Ic. A
~
~
~
Id. B − A = C
3
~ −A
~ = C
~ is a basic property of vector
B
subtraction. The law of vector addition yields
~ +B
~ =D
~ which is the same as A
~ =D
~ −B
~
A
005 10.0 points
What is the unit vector that points from the
point (1, 2, 3) toward the point (3, 2, 1)?
1
1. √ h1, 1, −1i
3
1
2. √ h0, 1, −1i
2
1
3. √ h−1, 0, 1i
2
1
4. √ h1, 1, 0i
2
5. h1, 0, 1i
6. h1, 0, −1i
1
7. √ h1, 0, −1i correct
2
8. h−1, 0, −1i
9. h−1, 0, 1i
1
10. √ h1, 1, 1i
3
~ =D
~ −C
~
IIa. A
~ =D
~ −B
~
IIb. A
1. Ib, IIa
2. Ib, IIb
3. Ic, IIb
4. Ic, IIa
5. Id, IIa
6. Ia, IIb
7. Id, IIb correct
8. Ia, IIa
Explanation:
Explanation:
If ~r = h(3 − 1),q(2 − 2), (1 − 3)i
=
√
22 + (−2)2 = 2 2
h2, 0, −2i, then |~r | =
~r
1
and r̂ ≡
= √ h1, 0, −1i.
|~r |
2
006
10.0 points
A book is at rest on an incline as shown
above. A constant force vertically downward
is in contact with the book.
Version 064 – Test 1 – swinney – (58385)
F
B
oo
k
5.
force
6.
The following figures show several attempts
at drawing free-body diagrams for the book.
Which figure has the correct directions for
each force? The magnitudes of the forces are
not necessarily drawn to scale.
friction
force
normal
7.
normal
force
1.
4
normal
weight
friction
weight
friction
correct
weight
friction
force
normal
weight
8.
weight
friction
force
normal
2.
force
weight
friction
normal
3.
force
normal
friction
weight
4.
normal
force
friction
weight
Explanation:
The normal force points perpendicular to
the surface of the inclined plane. The weight
force points down. The Fhand also points
down. The friction force keeps the block from
sliding and consequently points up the incline.
007 10.0 points
If the mass of a planet is 0.17 times that of
the Earth and its radius is 0.757 times that of
the Earth, estimate the gravitational field g
at the surface of the planet. The gravitational
acceleration on Earth is 9.8 m/s2 .
1. 2.90726
2. 6.19144
3. 3.31019
4. 11.363
5. 11.1535
6. 13.8527
7. 17.1725
8. 7.15583
Version 064 – Test 1 – swinney – (58385)
9. 7.95411
10. 1.93065
5
2. The ball is in free-fall and there is no
acceleration at any point on its path.
Correct answer: 2.90726 m/s2 .
Explanation:
3.
Let :
Mp = 0.17 ME
Rp = 0.757 RE ,
correct
and
gE = 9.8 m/s2 .
4.
Since
M
M
∝ 2,
2
R
R
the ratio of the gravitational fields is
g=G
2
Mp RE
gp
=
gE
ME Rp2
5.
6.
2
RE
0.17 ME
gE
ME (0.757 RE )2
0.17
=
(9.8 m/s2 )
0.7572
gp =
7.
8.
2
= 2.90726 m/s .
9.
008
10.0 points
A ball is thrown and follows the parabolic
path shown. Air friction is negligible. Point
Q is the highest point on the path. Points P
and R are the same height above the ground.
Explanation:
Since air friction is negligible, the only acceleration on the ball after being thrown is
that due to gravity, which acts straight down.
Q
P
R
009 10.0 points
A block of mass m is on a frictionless plane
inclined at θ with the horizontal and is pushed
by a horizontal force F at a constant velocity
up the plane.
The acceleration of gravity is g .
F
m
θ
Which of the following diagrams best indicates the direction of the acceleration, if any,
on the ball at point P ?
1.
What is the magnitude of the normal force
N the plane exerts on the block?
Version 064 – Test 1 – swinney – (58385)
1. N = F tan θ
N=
mg
cos θ
The first choice appears in the answer list
while the second one does not. And since
none of the other answers are equivalent to
the above identities,
N=
F
cos θ
4. N = mg tan θ
5. N = F
N=
6. N = F sin θ
F
sin θ
is the correct choice.
7. N = mg cos θ
F
correct
sin θ
mg
9. N =
sin θ
8. N =
10. N = mg sin θ
Explanation:
Consider the free body diagram for the
block
N
F
W = mg
Basic Concepts: We apply the Momentum Principle
~ net = m~a
F
010 10.0 points
Margie (of mass 42 kg) and Bill (of mass
63 kg), both with brand new roller blades,
are at rest facing each other in the parking
lot. They push off each other and move in
opposite directions, with Margie moving at a
constant speed of 12 ft/s .
At what speed is Bill moving?
1. 7.74194
2. 10.8088
3. 8.0
4. 6.47059
5. 10.25
6. 11.9429
7. 11.6129
8. 8.91429
9. 12.25
10. 10.3676
Correct answer: 8 ft/s.
Explanation:
.
For convenience, we choose the x-direction
to be to the right and the y-direction to be
up. The net force in each direction of this
coordinate system is
and
F
sin θ
and
2. N = mg
3. N =
6
Fx,net = F − N sin θ
Fy,net = N cos θ − m g
We apply the Momentum Principle and
since the velocity is constant, the acceleration is ~a = ~0. Thus, each equation gives
separately
Let : mM = 42 kg ,
mB = 63 kg , and
∆vM = 12 ft/s .
By Newton’s third law, Margie (M) and Bill
(B) experience the same force. We can write
∆vM
tM
∆vB
FB = mB aB = mB
.
tB
FM = mM aM = mM
and
Version 064 – Test 1 – swinney – (58385)
Since FM = FB ,
mM
7
L1 = L2 = L, so for the aluminum wire,
∆vM
∆vB
= mB
tM
tB
∆L1 =
M1 g L
M1 g L
=
A1 YAl
π r12 Y1
The times are the same, so
and for the steel wire
mM ∆vM = ∆vB mB
mM
∆vM
∆vB =
mB
42 kg
(12 ft/s)
=
63 kg
= 8 ft/s .
011 10.0 points
Masses M1 and M2 are supported by wires
that have equal lengths when unstretched.
The wire supporting M1 is an aluminum wire
0.75 mm in diameter, and the one supporting
M2 is a steel wire 0.5 mm in diameter.
M1
if the two wires
What is the ratio
M2
stretch by the same amount? Young’s modulus for aluminum is 7 × 1010 N/m2 and for
steel 2 × 1011 N/m2 .
1. 1.27562
2. 0.896
3. 3.98672
4. 1.0115
5. 4.46429
6. 1.4
7. 3.77857
8. 0.7875
9. 1.95537
10. 0.83595
∆L2 =
M2 g L
M2 g L
.
=
A2 Ysteel
π r22 Y2
The wires stretch by the same amount (and
L1 = L2 ), so
M2 g L
M1 g L
=
2
π r1 Y 1
π r22 Y2
M1
r 2 Y1
= 12
M2
r2 Y 2
(0.375 mm)2 (7 × 1010 N/m2 )
=
(0.25 mm)2 (2 × 1011 N/m2 )
= 0.7875 .
012 10.0 points
A target lies flat on the ground 11 m from the
side of a building that is 10 m tall, as shown
below.
The acceleration of gravity is 10 m/s2 . Air
resistance is negligible.
A student rolls a 11 kg ball off the horizontal
roof of the building in the direction of the
target.
v
Let :
r1
r2
Y1
Y2
10 m
Explanation:
10 m
Correct answer: 0.7875.
= 0.375 mm ,
= 0.25 mm ,
= 7 × 1010 N/m2 ,
= 2 × 1011 N/m2 .
Young’s Modulus is
Y =
FL
.
A ∆ℓ
11 m
and
The horizontal speed v with which the ball
must leave the roof if it is to strike the target
is most nearly
√
3
1. v =
m/s.
11
Version 064 – Test 1 – swinney – (58385)
√
2. v = 3 11 m/s.
√
3. v = 2 11 m/s.
√
11 2
m/s. correct
4. v =
2
√
5
5. v =
m/s.
11
√
6. v = 11 5 m/s.
√
11 3
m/s.
7. v =
3
√
2
8. v =
m/s.
11
√
9. v = 11 2 m/s.
√
10. v = 11 3 m/s.
013
Observe the motion in the vertical direction
only and it is a purely 1-dimension movement
with a constant acceleration. So the time
need for the ball to hit the ground is
s
2h
t=
g
and the horizontal speed should be
x
v=
t
for the ball to hit the target. Therefore
r
g
v=x
2h
s
10 m/s2
= (11 m)
2 (10 m)
11
= √ m/s
2
√
11 2
m/s .
=
2
10.0 points
Consider a man standing on a scale which
is placed in an elevator. When the elevator is
stationary, the scale reading is his weight W .
Scale
Explanation:
m = 11 kg , not required
h = 10 m ,
x = 11 m , and
g = 10 m/s2 .
8
Find S, the scale reading when the elevator
is moving downward with acceleration a =
1
g.
3
2
1. S = W correct
3
1
2. S = W
2
3. S = 0 m/s2
3
W
4
4
5. S = W
3
4. S =
6. S = 2 W
5
W
4
1
8. S = W
3
5
9. S = W
3
3
10. S = W
2
7. S =
Version 064 – Test 1 – swinney – (58385)
Explanation:
We consider the forces acting on the man.
Taking up (̂) as positive, we know that m g
acts on the man in the downward (−̂) direction. The only other force acting on the
~ s from the scale. By
man is the normal force S
the law of action and reaction, the force on
the scale exerted by the man (i.e., the scale
reading) is equal in magnitude but opposite
~ s vector. Initially, the
in direction to the S
elevator is stationary (no acceleration) so
W − mg = 0,
or W = m g as we would expect.
Call the scale reading for this part S (in
units of Newtons). Consider the free body
diagram for each the case where the elevator
is accelerating down (left) and up (right). The
man is represented as a sphere and the scale
reading is represented as S.
S
a
Sdown
a
mg
mg
Equation 1 in the upward direction reads
S − mg = ma,
1
and since a = − g at this particular instant,
3
1
S − mg = − mg
3
2
2
1
S = mg − mg = mg = W .
3
3
3
Thus the scale gives a smaller reading when
the elevator is accelerating downward.
014
10.0 points
Given: The battleship and enemy ships 1
and 2 lie along a straight line. Neglect air
friction.
battleship
9
1
2
Consider the motion of the two projectiles
fired at t = 0. Their initial speeds are different
and they reach different maximum heights h1
and h2 .
What is the ratio of the time of flight, t1
and t2 respectively, that the shells reach?
r
h1
1
t1
=√
1.
t2
2 h2
t1
h1
2.
=
t2
h2
r
h2
t1
=2
3.
t2
h
r1
h1
t1 √
= 2
4.
t2
h2
t1
h2
5.
=
t2
h1
r
1 h1
t1
=
6.
t2
2 h2
r
h1
t1
=2
7.
t2
h
r 2
t1
h2
=
8.
t2
h
r1
t1
1 h2
9.
=
t2
2 h
r 1
h1
t1
=
correct
10.
t2
h2
Explanation:
At a shell’s maximum height h, the y component of a shell’s velocity vector is 0 and
when a shell hits a ship its height is 0, so
1
0 = h − g t2
s 2
2h
t=
g
Version 064 – Test 1 – swinney – (58385)
10
Therefore, the ratio of the flight times which
is the same as the ratio of half-flight times is
q2
F4x
F4y
F4
F1
r
2 t1
t1
=
=
2 t2
t2
h1
h2
015 10.0 points
Four point charges are placed at the four corners of a square. Each side of the square has
a length L.
q4 =
q
q3 =
q
b
b
Pb
b
F3
From the figure,
F1x = −
k q2
L2
F3y = −
k q2
L2
L
F4x
b
q1 = −q L q2 = q
Find the magnitude of the electric force on
q2 due to all three charges q1 , q3 and q4 in
terms of L, q, and the Coulomb constant k.
1.
k q2 1
√
=
2 L2 2
F4y = −
k q2 1
√
2 L2 2
For the net force,
5 k q2
4 L2
3 k q2
2.
correct
2 L2
r
5 k q2
3.
2 L2
1 k q2
4.
2 L2
5 k q2
5.
2 L2
r
1 k q2
6.
2 L2
√
3 k q2
7.
2 L2
r
3 k q2
8.
2 L2
√
5 k q2
9.
2 L2
3 k q2
10.
4 L2
Explanation:
Fx = F1x + F4x
k q2 1
k q2
√
=− 2 +
L
2 L2 2
Fy = F3y + F4y
k q2 1
k q2
√
=− 2 −
L
2 L2 2
~k=
kF
=
q
Fx2 + Fy2
3 k q2
2 L2
016 10.0 points
Two 100 kg boxes are dragged along a frictionless surface with a constant acceleration
of 1.10848 m/s2 , as shown in the figure.
Each rope has a mass of 1 kg.
1 kg
100 kg
100 kg
1 kg F
Version 064 – Test 1 – swinney – (58385)
Find the tension in the leftmost rope at the
point where it is attached to the rightmost
box.
1. 110.768
2. 106.334
3. 90.516
4. 100.872
5. 117.339
6. 78.1929
7. 141.376
8. 124.909
9. 83.1361
10. 111.957
Explanation:
F43
m3
F34 − F32 = m3 a
(3)
F − F43 = m4 a
(4)
Adding these equations, we obtain
F = (m1 + m2 + m3 + m4 ) a
= (100 kg + 1 kg + 100 kg + 1 kg)
× 1.10848 m/s2
= 223.913 N .
= 111.957 N .
The free-body diagrams for the boxes and
the ropes are below.
m1
F12
F32
(2)
F23 = F21 + m2 a
= 110.848 N + (1 kg) 1.10848 m/s2
Let : m1 = 100 kg ,
m2 = 1 kg ,
m3 = 100 kg ,
m4 = 1 kg , and
a = 1.10848 m/s2 .
m2
F23 − F21 = m2 a
Using Eq. 2 at point B,
Correct answer: 111.957 N.
F21
11
017 10.0 points
An ice skater whose mass is 50 kg moves with a
constant momentum of (400, 0, 300) kg · m/s.
During this period of constant momentum,
she passes the location (0, 0, 3) m. What was
her location at a time 3 s earlier?
1. (10, 3, 7) m
F23
F34
2. (−10, 3, −7) m
F
3. (−20, 0, −14) m
m4
Because the vertical forces have no bearing
on the problem they have not been included.
~ 34 = tension force exerted by m4 on m3 ,
F
~ 43 = tension force exerted by m3 on m4 ,
F
~ 23 = tension force exerted by m3 on m2 ,
F
~ 32 = tension force exerted by m2 on m3 ,
F
~ 12 = tension force exerted by m2 on m1 ,
F
~ 21 = tension force exerted by m1 on m2 ,
F
The equal and opposite pairs of forces are
~
~ 12 , F
~
~
~
~
F21 = −F
X32 = −F23 , F43 = −F34 .
~ = m~a to the boxes and the
F
Applying
ropes, going from left to right (the accelerations are equal),
F12 = m1 a
(1)
4. (24, 0, 15) m
5. (−24, 0, −15) m correct
6. (−8, 0, −5) m
7. (8, 0, 5) m
Explanation:
Given that
Let : ~p = (400, 0, 300) kg · m/s ,
r~f = (0, 0, 3) m ,
m = 50 kg , and
t = 3 s.
Version 064 – Test 1 – swinney – (58385)
We use the formula for the momentum to
obtain the velocity
~p
m
= (8, 0, 6) m/s
We make use of the relation between velocity
and displacement
r~f = r~i + ~v ∆t
which yields
r~i = r~f − ~v ∆t
= (−24, 0, −15) m
018 10.0 points
Consider the six equally-spaced objects shown
in the diagram below. The points A, B, C and
D represent locations in empty space and the
large circles represent planets with masses m
and 2m.
m
B
C
9. A; B and C
10. C; A
~v =
A
2m
D
1. At which location(s) would the magnitude of the net gravitational force on an
object be a maximum?
2. At which location(s) would the magnitude of the net gravitational force on an
object be a minimum?
Separate your answers by a semicolon.
Explanation:
At A and D, the planets are on the same
side of you, so they pull you in the same direction, causing you to weigh more. At B and C,
you would weigh less because the planets pull
on you in opposite directions. Since the gravitational force is proportional to mass and
inversely proportional to distance squared,
standing closer to the more massive planet
maximizes the force while standing further
from the more massive one planet minimizes
the force. Hence the magnitude of net force is
largest at D and smallest at B.
019 10.0 points
A 280 kg load is hung on a wire of length
4.8 m and cross sectional area 2.5 × 10−5 m2 ,
What is its increase in length? The acceleration of gravity is 9.8 m/s2 and Young’s
modulus is 9.4 × 1010 N/m2 .
1. 9.94737
2. 8.06071
3. 8.38393
4. 6.86763
5. 10.5195
6. 13.9167
7. 5.60477
8. 11.3474
9. 6.26613
10. 10.2083
1. A; C
Correct answer: 5.60477 mm.
2. D; B correct
Explanation:
3. B and C; A
4. B and C; A and D
5. A and D; B and C
6. D; B and C
Let : m = 280 kg ,
g = 9.8 m/s2 ,
A = 2.5 × 10−5 m2 ,
ℓ = 4.8 m , and
Y = 9.4 × 1010 N/m2 .
7. B; D
8. B; A and D
12
∆ℓ
F
=Y
A
ℓ
Version 064 – Test 1 – swinney – (58385)
mg ℓ
A Y
4.8 m
(280 kg)(9.8 m/s2 )
=
2.5 × 10−5 m2 9.4 × 1010 N/m2
1000 mm
×
1m
= 5.60477 mm .
∆ℓ =
020 10.0 points
Seven forces of equal magnitude (24 N) act
on a 8 kg object as shown.
What is the magnitude of the acceleration
of the object?
1. 6.0
2. 3.75
3. 2.5
4. 1.0
5. 1.57143
6. 1.71429
7. 0.9
8. 1.4
9. 1.5
10. 3.0
Correct answer: 3 m/s2 .
Explanation:
Let :
F = 24 N and
m = 8 kg .
The up and down forces cancel, as do the
right and left forces, and as do the up-right
and down-left forces, so the net force is 24 N
downward and to the right.
a=
24 N
F
=
= 3 m/s2 .
m
8 kg
13