Version 062 – Test 1 – swinney – (58535)
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
An alpha particle of mass m = 6.6 × 10−27 kg
has a speed of 0.94 c, where the speed of light
is c = 3 × 108 m/s. The alpha particle collides
with a gold nucleus and rebounds with the
same speed in the opposite direction.
If the collision lasted for t = 5 × 10−7 s,
what is the magnitude of the force that the
gold nucleus exerted on the alpha particle?
1. 1.54592e-11
2. 1.13466e-11
3. 1.46737e-11
4. 2.18211e-11
5. 3.19487e-11
6. 6.02402e-11
7. 2.73485e-11
8. 2.40961e-11
9. 4.34579e-11
10. 1.74688e-11
Correct answer: 2.18211 × 10
−11
N.
Explanation:
The magnitude of the relativistic momentum is
pr = γmv
mv
=r
v2
1− 2
c
.
We can use the momentum principle to
determine the force as
~ net | = |∆P
~ |/∆t
|F
= 2 γm v/∆t
2mv
r
=
v2
∆t 1 − 2
c
= 2.18211 × 10−11 N
002
10.0 points
1
When you are moving up at constant speed
in an elevator, there are two forces acting on
you: the floor pushing up on you (F1 ) and
gravity pulling down (F2 ).
What is the relationship between the magnitude of F1 and F2 and the physical principle
that explains this relationship?
1. F1 > F2 from the principle of reciprocity.
2. It depends on which direction the elevator
is moving.
3. F1 = F2 from the principle of reciprocity.
4. F1 > F2 from the momentum principle.
5. F1 < F2 from the momentum principle.
6. F1 < F2 from the principle of reciprocity.
7. F1 = F2 from the momentum principle.
correct
Explanation:
Since the speed is constant, there is no
change in momentum. The momentum principle states that the net force must therefore
be zero, which requires F1 = F2 .
003 10.0 points
A box is knocked off a 1.1 m high table and
then strikes the floor 2.6 m from the edge of
the table. Assume that wind resistance is
negligible.
The acceleration of gravity is 9.81 m/s2 .
What was the speed of the box when it slid
off the table?
1. 5.49031
2. 3.37865
3. 3.10466
4. 4.01215
5. 2.2082
6. 6.66553
7. 4.22568
8. 6.04046
9. 4.15912
Version 062 – Test 1 – swinney – (58535)
2
10. 2.52517
Correct answer: 5.49031 m/s.
Earth2
Explanation:
Basic Concept:
D
∆x = vx ∆t
RE
1
∆y = − g (∆t)2
2
Earth
Fscale
Given:
∆y = −1.1 m
∆x = 2.6 m
g = 9.81 m/s2 .
1. D = 10RE
4. D
∆t =
s
2 ∆y
∆x
=
−g
vx
5. D
6. D
vx =
=
r
s
−g
∆x
2∆y
−(9.81 m/s2 )
(2.6 m)
2 (−1.1 m)
= 5.49031 m/s .
7. D
8. D
9. D
10. D
004 10.0 points
How close would another Earth have to be to
our Earth (center-to-center distance) so that
you would weigh 10% more your current
weight? Note that this other Earth is on the
opposite side of our Earth from you. Neglect
rotation. (In the figure, Fscale is the normal
force which measures your weight. The con11
dition of the problem is that Fscale =
FE1 ,
10
where FE1 is the force of gravity from the
Earth on you.)
!
11
RE
10
√
=
11 − 1 RE
√ = 1 + 11 RE
r !
4
RE
= 1+
5
√ = 1 + 10 RE
√
= 5RE
√
10 − 1 RE correct
=
r !
9
RE
= 1+
10
s !
2
= 1+
RE
3
2. D =
3. D
Solution:
r
1+
Explanation:
For you to stand stationary on the surface
of the Earth,
FE2 + Fscale + FE1 = 0.
We are solving for your scale weight to be
11/10 times force you usually feel due to the
Earth.
FE2 +
11
FE1 + FE1 = 0
10
Version 062 – Test 1 – swinney – (58535)
⇒ FE2 =
FE1
10
k
x
We know that
GME2 m
GME1 m
=
2
(D + RE )2
10RE
2
⇒ (D + RE )2 = 10RE
⇒D=
√
3
10RE − RE > 2RE
We see that our result is greater than 2RE ,
so it is sensible.
005 10.0 points
If you drop an object in the absence of air resistance, it accelerates downward at 9.8 m/s2 .
If instead you throw it downward, its downward acceleration after it leaves your hand is
1. more than 9.8 m/s2 , but only if thrown
very hard.
2. zero if thrown at the right velocity.
3. always equal to 9.8 m/s2 . correct
4. always less than 9.8 m/s2 .
5. cannot be determined without more information.
6. always more than 9.8 m/s2 .
Explanation:
The acceleration of gravity is a constant,
independent of initial velocity.
006 10.0 points
A block is attached by a taught cord to an unstretched spring. The block is slowly lowered
down the incline until it comes to rest and is
released. It stays at rest.
m
µ=
θ
0
How far x is the block in its final position
from its initial position when it was first attached to the cord? Suppose the incline is
frictionless. The angle of inclination is θ, the
spring constant is k the mass of the block is
m and the acceleration of gravity is g .
k
m g cos θ
k
2. x =
m g sin θ
m g tan θ
3. x =
k
mg
4. x =
k
r
m g cos θ
5. x =
k
1. x =
6. None of these
m g sin θ
correct
k
m g cos θ
8. x =
k
k
9. x =
m g tan θ
Explanation:
The block is not moving so the net force is
zero, i.e. the spring force is equal and opposite
to the component of the gravitational force
along the plane:
7. x =
Fs = m g sin θ
m g sin θ
kx =
k
m g sin θ
x=
k
Version 062 – Test 1 – swinney – (58535)
007
10.0 points
Given: The battleship and enemy ships A
and B lie along a straight line. Neglect air
friction.
battleship
A
B
Consider the motion of the two projectiles
fired at t = 0. Their initial speeds are both v0
but they are fired with different initial angles
θ1 and θ2 .
What is the ratio of the maximum heights,
h1 and h2 respectively, that the shells reach?
sin2 θ2
h1
=
h2
sin2 θ1
h1
=1
2.
h2
h1
sin2 θ1
correct
3.
=
h2
sin2 θ2
h1
tan θ2
4.
=
h2
tan θ1
h1
cos θ1
5.
=
h2
cos θ2
tan2 θ1
h1
=
6.
h2
tan2 θ2
cos2 θ1
h1
=
7.
h2
cos2 θ2
h1
sin θ2
8.
=
h2
sin θ1
h1
sin θ1
9.
=
h2
sin θ2
cos2 θ2
h1
=
10.
h2
cos2 θ1
Explanation:
At its maximum height, the y component
of its velocity vector is 0, so
1.
vy2
v 2 sin2 θ
h=
=
2g
2g
Therefore, the ratio of the heights is
h1
=
h2
v02 sin2 θ1
2g
v02 sin2 θ2
2g
=
sin2 θ1
sin2 θ2
008 10.0 points
A spherical mass rests upon two wedges, as
seen in the figure below. The sphere and the
wedges are at rest and stay at rest. There is no
friction between the sphere and the wedges.
M
The following figures show several attempts
at drawing free-body diagrams for the sphere.
Which figure has the correct directions for
each force?
Note: The magnitude of the forces are not
necessarily drawn to scale.
1.
normal
normal
correct
weight
2. Since the sphere is not moving, no forces
act on it.
3.
normal
weight
4.
normal
weight
0 = vy2 = vy2 − 2 g h
4
normal
weight
Version 062 – Test 1 – swinney – (58535)
5
+q
̂
normal
weight
normal
weight
a
5.
ı̂
60◦
6.
normal
normal
friction
friction
weight
7.
normal
weight
8.
weight
weight
normal
normal
9.
weight
Explanation:
Weight – the force of gravity – pulls the
sphere down. The normal force of the left
wedge upon the sphere acts perpendicular to
(normal to) their surfaces at the point of contact; i.e., diagonally upward and rightward.
Likewise, the normal force of the right wedge
upon the sphere acts diagonally upward and
leftward.
009
10.0 points
Three point charges are placed at the vertices of an equilateral triangle as shown.
−q
−q
What is the magnitude of the electric force
on the charge at the bottom left-hand vertex
of the triangle due to the other two charges?
√
2
~k = 3 kq
1. kF
2 a2
√
2
~k = 2 kq
2. kF
3 a2
2
~ k = k q correct
3. kF
a2
2
√
~k = 3 kq
4. kF
a2
2
~ k = √3 k q
5. kF
2 a2
2
~ k = √2 k q
6. kF
3 a2
2
√
~k = 2 kq
7. kF
a2
1 k q2
~
√
8. kF k =
2 3 a2
2
~k = 1 kq
9. kF
2 a2
2
~ k = √1 k q
10. kF
2 a2
Explanation:
Each force has a magnitude
k q2
.
a2
The x-component of the net force is
F21 = F31 =
Fx = (+F31 cos 60◦ − F21 ) ı̂
k q2
k q2
◦
= + 2 cos 60 − 2
ı̂
a
a
1 k q2 k q2
− 2
ı̂
= +
2 a2
a
1 q2
= − k 2 ı̂
2 a
Version 062 – Test 1 – swinney – (58535)
6
and the y-component is
√
k q2
3 q2
Fy = − 2 sin 60◦ ̂ = −
k ̂ ,
a
2 a2
a=
F
22 N
=
= 3.66667 m/s2 .
m
6 kg
so the magnitude of the net force is
~k=
kF
q
Fx2 + Fy2 = k
=k
q 2
a
r
1 3
+
4 4
q2
a2
011 10.0 points
A particle moves in the x-direction as indicated in the position vs. time plot below.
.
010 10.0 points
Five forces of equal magnitude (22 N) act
on a 6 kg object as shown.
At point P, the mass has
1. negative velocity and negative acceleration.
2. zero velocity and zero acceleration.
3. negative velocity and zero acceleration.
What is the magnitude of the acceleration
of the object?
1. 1.33333
2. 5.0
3. 3.0
4. 3.66667
5. 7.5
6. 2.8
7. 3.25
8. 2.0
9. 2.4
10. 1.2
Correct answer: 3.66667 m/s2 .
Explanation:
Let :
F = 22 N and
m = 6 kg .
The top and bottom forces cancel, as do the
right and left, so the net force is 22 N upward
and to the right.
4. positive velocity and positive acceleration.
5. negative velocity and positive acceleration.
6. zero velocity but is accelerating (positively or negatively).
7. positive velocity and zero acceleration.
8. positive velocity and negative acceleration. correct
Explanation:
The velocity is positive because the slope of
the curve at P is positive. The acceleration
is negative because the curve is concave down
at P.
012 10.0 points
Consider the four locations A, B, C and D
shown in the diagram below. The circles are
planets that have masses m and 2m.
Version 062 – Test 1 – swinney – (58535)
A
m
B
C
2m
D
1. At which location(s) would the gravitational force from both planets’ pull you in the
same direction?
2. At which location(s) would the total
gravitational force on you be a maximum?
4. h1.4, 2.8, 1.4i
5. h0.3333, 0.3333, 0.3333i
6. h
1.4 2.8 1.4
,
,
i
3 3 3
7. h0.4472, 0.7746, 0.4472i
1. C; A
8. h1, 2, 1i
2. D; D
9. h0.5773, 0.5773, 0.5773i
3. A and B; D
4. None of these
5. B; D
6. B and C; D
7. A and D; A
8. B and C; C
9. A and D; D correct
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.
The closer you stand to the more massive
one, the more you weigh, so you would feel a
maximum weight at D.
013 10.0 points
A rectangular block is 1.4 feet on two sides
and 2.8 feet on the other side. It has one
corner at the origin. What is the unit vector
pointing from the origin to the diagonally
opposite corner located at h1.4, 2.8, 1.4i?
1. h0.4083, 0.8165, 0.4083i correct
2. h
1.4 2.8 1.4
,
,
i
2 2 2
3. h0, 2, 0i
7
10. h0, 0, 1i
Explanation:
If ~r = hd, e, di = hd, 2, di, then |~r | =
q
√
~r
=
d2 + (2d)2 + d2 = 6d and r̂ ≡
|~r |
1 2 1
h √ , √ , √ i.
6 6 6
014 10.0 points
Two ice skaters approach each other at right
angles. Skater A has a mass of 73.8 kg and
travels in the +x direction at 2.17 m/s. Skater
B has a mass of 33.5 kg and is moving in the
+y direction at 2.31 m/s. They collide and
cling together.
Find the final speed of the couple.
1. 0.691107
2. 1.50787
3. 1.21219
4. 1.69531
5. 1.1207
6. 1.2633
7. 1.43353
8. 1.65762
9. 1.83852
10. 1.01713
Correct answer: 1.65762 m/s.
Explanation:
From conservation of momentum ∆p = 0
mA vA ı̂ + mB vB ̂ = (mA + mB ) vf
Therefore
p
(mA vA )2 + (mB vB )2
vf =
mA + mB
Version 062 – Test 1 – swinney – (58535)
p
(160.146 kg m/s)2 + (77.385 kg m/s)2
=
73.8 kg + 33.5 kg
= 1.65762 m/s
8
016 10.0 points
~ and B.
~
Consider the vectors A
~
A
015 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. (−20, 0, −14) m
2. (8, 0, 5) m
~
B
~ −B
~ to the
Which sketch shows the vector A
same scale?
1.
3. (24, 0, 15) m
4. (−8, 0, −5) m
5. (−24, 0, −15) m correct
2.
6. (−10, 3, −7) m
7. (10, 3, 7) 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.
We use the formula for the momentum to
obtain the velocity
3.
4.
~p
m
= (8, 0, 6) m/s
~v =
We make use of the relation between velocity
and displacement
5.
r~f = r~i + ~v ∆t
6.
which yields
r~i = r~f − ~v ∆t
= (−24, 0, −15) m
7.
correct
Version 062 – Test 1 – swinney – (58535)
9
F
8.
θ
M
m
Explanation:
~ −B
~
A
~
A
~
−B
017 10.0 points
The earth attracts a large flying bug toward
its center with a force of 0.10 Newtons.
With what force does the bug attract the
earth itself?
1. We need the mass of the bug.
2. Trillions of Newtons
3. 0.10 N correct
4. 10 N
5. 1.0 N
6. Zero; the earth attracts the bug, not the
other way around.
7. A very, very, very small fraction of a
Newton, since the bug’s mass is negligible
compared to the earth’s mass
What is the magnitude of the force that
block 2 exerts on block 1?
1. None of these
F sin θ
correct
3
F cos θ
3. F1,2 =
3
2 F cos θ
4. F1,2 =
3
F sin θ
5. F1,2 =
2
2 F sin θ
6. F1,2 =
3
2F
7. F1,2 =
3
F cos θ
8. F1,2 =
2
Explanation:
Let F1,2 be the magnitude of the force of
the small box pushing on the large block, but
by reciprocity that is equal in magnitude to
the force of the large block on the small block,
i.e. F1,2 = F2,1 . Let right be the positive
x-direction.
For the combined system of both blocks
2. F1,2 =
Explanation:
The force with which the earth attracts the
bug is the same in magnitude as the force
with which the bug attracts the earth; the
magnitude is 0.10 N in both cases. The forces
differ only in direction, according to Newton’s
Third Law and Newton’s Law of Gravity.
The contact force supplies the acceleration
on the smaller box:
018 10.0 points
Two blocks lay on a surface that has negligible
friction. A force F is applied to block 1 at
angle θ as shown in the picture and block 1
pushes against block 2. The mass of block 1
is twice the mass of block 2, i.e. M = 2 m.
F2,1 = m a
F sin θ
,
=
3
F sin θ
⇒ F1,2 =
3
F sin θ = (M + m) a
F sin θ
a=
3m
Version 062 – Test 1 – swinney – (58535)
10
4. a) and c)
019
10.0 points
5. a), d), and f)
Identify all graphs that represent motion
with constant nonzero acceleration (note the
axes carefully).
7. f) only
x
t
a)
6. b), d), and e)
8. d) only
9. b), d) and f) correct
10. c) only
v
t
b)
a
t
c)
x
t
d)
v
t
e)
a
f)
1. a), b), and f)
2. b) only
3. None of these
t
Explanation:
For constant speed, a = 0 .
a) x = k t, k > 0, increases at a constant
rate, so it has a constant velocity and zero
acceleration.
b) v = k t, k > 0, increases at a constant
rate, so it has a constant acceleration.
c) a = k t, k > 0, increases at a constant
rate.
d) x = −k t2 , k > 0, is parabolic with
constant negative acceleration.
e) v = k, k > 0, represents a constant
velocity.
f) a = k, k > 0, represents a constant,
nonzero acceleration.
020
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 .
Version 062 – Test 1 – swinney – (58535)
11
~ s vector. Initially, the elin direction to the S
evator is moving upward with constant speed
(no acceleration) so
W − mg = 0,
Scale
Find S, the scale reading when the elevator
1
is moving upward with acceleration a = g.
6
2
1. S = W
3
7
2. S = W correct
6
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
Equation 1 in the upward direction reads
S − mg = ma,
3. S = 0 m/s2
5
W
6
6
5. S = W
7
6
6. S = W
5
8
7. S = W
7
4
8. S = W
3
7
9. S = W
5
5
10. S = W
7
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
4. S =
mg
and since a =
S − mg =
1
g at this particular instant,
6
1
mg
6
S = mg +
7
7
1
mg = mg = W .
6
6
6
Thus the scale gives a higher reading when
the elevator is accelerating upwards.