Download Version 073 – midterm 1 v1 – shih – (58505) 1

Version 073 – midterm 1 v1 – shih – (58505)
This print-out should have 16 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
may the force be with you
001 10.0 points
A rope has a cross section A = 8.18 m2 and
density ρ = 2270 kg/m3 . The linear density
of the rope µ, defined to be the mass per unit
length, can be written in the form µ = ρx Ay .
Based on dimensional analysis, determine
the powers x and y by choosing an expression
below.
A2
ρ2
ρ
2. µ =
A
1
3. µ =
ρA
1. µ =
6. µ =
7. µ =
8. µ =
9. µ =
2y − 3x = −1
2y − 3 = −1
y = 1.
Thus µ = ρ1 A1 = ρ A .
002 10.0 points
Despite a very strong wind, a tennis player
manages to hit a tennis ball with her racquet
so that the ball passes over the net and lands
in her opponent’s court.
Consider the following forces:
1. A downward force of gravity,
2. A force by the hit, and
3. A force exerted by the air.
Which of the above forces is (are) acting on
the tennis ball after it has left contact with
the racquet and before it touches the ground?
1. 1 and 3. correct
2. 1, 2, and 3.
4. µ = ρ A2
5. µ =
1
A2
ρ
ρ
A2
A
ρ2
A
ρ
1
ρ A2
3. 1 only.
4. 2 and 3.
5. 1 and 2.
Explanation:
The forces acting on the tennis ball after it
has left contact with the racquet are gravity
and air resistance. A force by the hit is no
longer acting on the ball.
10. µ = ρ A correct
Explanation:
[x] means “the units of x”.
[µ] = [ρx Ay ]
x
y
[M]
M
=
L2
3
[L]
L
+1
−1
[M] [L] = Mx L−3x L2y
= Mx L2y−3x
Equating the exponents,
x=1
003 10.0 points
The diagram describes the acceleration vs
time behavior for a car moving in the xdirection.
a
Pb
Q
b
0
At the point Q, the car is moving
t
Version 073 – midterm 1 v1 – shih – (58505)
dv
>0 .
which means positive derivative
dt
In other words, the car has positive acceleradv
tion a =
> 0 . Since v > 0, this means the
dt
car is accelerating.
1. with a decreasing speed.
2. with an increasing speed. correct
3. with a constant speed.
Explanation:
dv
a =
. As long as the acceleration is
dt
positive the velocity is always increasing.
004 10.0 points
A car moves along a straight line. The graph
below shows its velocity in the x-direction vx
as a function of time.
6
vx (m/s)
5
P
4
3
2
1
0
0
1
2
3
4 5
time (s)
2
6
7
8
9
005 10.0 points
A jet airliner moving initially at 786 mph to
the east where there is no wind moves into a
region where the wind is blowing at 414 mph
in a direction 79◦ north of east.
What is the new speed of the aircraft with
respect to the ground?
1. 635.595
2. 865.317
3. 605.287
4. 1432.06
5. 1030.55
6. 737.237
7. 1538.03
8. 1004.33
9. 955.705
10. 1157.37
Correct answer: 955.705 mph.
Explanation:
Which statement correctly describes the
car’s motion at point P on the graph?
Vwind
φ
1. The car is stationary.
2. The car is accelerating forward. correct
3. The car has zero acceleration.
6. The car’s direction is about 30◦ off the x
axis.
Explanation:
The graph depicts the velocity v(t) as a
function of time and NOT the position x(t).
At point P on the graph, the velocity is positive (indeed, v ≥ 0 at all times depicted on
the graph) and the curve v(t) slopes upward,
θ
Vjet
The new velocity of the aircraft is the sum
of the initial velocity and the velocity of the
wind:
4. The car is decelerating.
5. The car is climbing a hill.
V
~vf = ~vi + ~vwind = vx ı̂ + vy ̂
where
vx = vi + vwind cos φ
vy = vwind sin φ
with φ being the angle between vwind and vjet .
So the speed is
q
|~v | = vx2 + vy2 .
006
10.0 points
Version 073 – midterm 1 v1 – shih – (58505)
10 m
y
A student rolls a 5 kg ball off the horizontal
roof of the building in the direction of the
target.
v
10 m
Denote the initial speed of a cannon ball fired
from a battleship as v0 . When the initial
projectile angle is 45◦ with respect to the
horizontal, it gives a maximum range R.
3
θ′
45◦
x
5m
R/2
R
The time of flight tR of the cannonball for
this maximum range R is given by
2 v0
3 g
1 v0
2. tR = √
3 g
√ v0
3. tR = 3
g
1 v0
4. tR =
2 g
1 v0
5. tR =
4 g
1 v0
6. tR = √
2 g
v0
7. tR = 4
g
v0
8. tR =
g
v0
9. tR = 2
g
√ v0
correct
10. tR = 2
g
Explanation:
The cannonball’s time of flight is
2 v0y
2 v0 sin 45◦ √ v0
tR =
=
= 2
.
g
g
g
1. tR =
007 10.0 points
A target lies flat on the ground 5 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.
The horizontal speed v with which the ball
must leave the roof if it is to strike the target
is most nearly
√
1. v = 2 5 m/s.
√
2. v = 5 2 m/s.
√
5 3
3. v =
m/s.
3
√
5
m/s.
4. v =
5
√
2
m/s.
5. v =
5
√
6. v = 5 3 m/s.
√
7. v = 3 5 m/s.
√
5 2
8. v =
m/s. correct
2
√
3
9. v =
m/s.
5
√
10. v = 5 5 m/s.
Explanation:
m = 5 kg , not required
h = 10 m ,
x = 5 m , and
g = 10 m/s2 .
Observe the motion in the vertical direction
only and it is a purely 1-dimension movement
Version 073 – midterm 1 v1 – shih – (58505)
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
v=
x
t
for the ball to hit the target. Therefore
r
g
v=x
2h
s
10 m/s2
= (5 m)
2 (10 m)
5
= √ m/s
2
√
5 2
=
m/s .
2
008 10.0 points
A block of mass 9 m can move without friction
on a horizontal table. This block is attached
to another block of mass 6 m by a cord that
passes over a frictionless pulley, as shown.
4
2
g correct
5
13
g
6. k~ak =
20
2
7. k~ak = g
9
18
8. k~ak =
g
23
6
9. k~ak =
g
11
13
10. k~ak =
g
30
Explanation:
5. k~ak =
Let :
m1 = 9 m ,
m2 = 6 m , and
m1 + m2 = 15 m .
a
m1
g
T
m2
9m
6m
Let the tension of the cord be T .
Apply Newton’s second law to the two
masses:
T = m1 a .
(1)
m2 g − T = m2 a ,
If the masses of the cord and the pulley
are negligible, what is the magnitude of the
acceleration of the descending block?
14
g
17
8
2. k~ak =
g
11
17
3. k~ak =
g
19
4
g
4. k~ak =
11
so
(2)
m2 g = m1 a + m2 a so
m2
a=
g
m1 + m2
2
6m
g = g.
=
15 m
5
1. k~ak =
keywords:
009 10.0 points
A toy cannon fires a 0.0714 kg shell with initial
Version 073 – midterm 1 v1 – shih – (58505)
velocity vi = 9.3 m/s in the direction θ = 44 ◦
above the horizontal. The shell’s trajectory
curves downward because of gravity, so at the
time t = 0.535 s the shell is below the straight
line by some vertical distance ∆h.
5
1
Thus, x = x̂ but y = ŷ − gt2 , or in other
2
words, the shell deviates from the straight-line
path by the vertical distance
∆h = ŷ − y =
9.
3
m
∆h
/s
∆y
◦
44
y
g t2
.
2
Note: This result is completely independent on the initial velocity vi or angle θ of
the shell. It is a simple function of the flight
time t and nothing else (besides the constant
g = 9.8 m/s2 ).
g t2
2
(9.8 m/s2 ) (0.535 s)2
=
2
= 1.4025 m .
∆h =
∆x
Find this distance ∆h in the absence of
air resistance. The acceleration of gravity is
9.8 m/s2 .
1. 0.586608
2. 3.01181
3. 1.4025
4. 1.63701
5. 0.696432
6. 0.34151
7. 0.252492
8. 0.935748
9. 0.889232
10. 2.88262
Correct answer: 1.4025 m.
Explanation:
In the absence of gravity, the shell would fly
along the straight line at constant velocity:
x̂ = t vi cos θ ,
ŷ = t vi sin θ .
The gravity does not affect the x coordinate
of the shell, but it does pull its y coordinate
at constant downward acceleration ay = −g,
hence
x = t vi cos θ,
g t2
.
y = t vi sin θ −
2
010 10.0 points
A car makes a 310 km trip at an average
speed of 35.7 km/h. A second car starting
1 h later arrives at their mutual destination
at the same time.
What was the average speed of the second
car for the period that it was in motion?
1. 37.7808
2. 55.5063
3. 59.6118
4. 58.5872
5. 47.9567
6. 42.5061
7. 40.3463
8. 35.481
9. 41.9242
10. 34.226
Correct answer: 40.3463 km/h.
Explanation:
Let :
d = 310 km ,
v = 35.7 km/h ,
∆t = 1 h .
Average velocity is v =
d
.
t
and
Version 073 – midterm 1 v1 – shih – (58505)
∆t = t1 − t2 =
v = 2 m/s ,
r = 8 m.
d
d
−
v1 v2
d
d
=
− ∆t
v2
v1
v1
v2
1
·
=
d
d
v
− ∆t 1
v1
d v1
v2 =
d − v1 ∆t
(310 km)(35.7 km/h)
=
310 km − (35.7 km/h)(1 h)
v2
r
(2 m/s)2
=
(8 m)
= 0.5 m/s2 .
012 10.0 points
The mass of the worker m1 = 50 kg. The
mass of the block at the end of the rope,
m2 = 100 kg.
10.0 points
b
A student is swinging a 6 kg ball in a circular path in the vertical plane. Consider the
instant when the string is horizontal as the
ball is on its way up as shown in the figure.
T
T
m2
a
8m
2 m/s
a
y
m1
x
M
and
ac =
= 40.3463 km/h .
011
6
Determine the acceleration.
At this instant, the instantaneous tangential speed is 2 m/s , and the radius is 8 m.
What is the centripetal acceleration?
1. 10.125
2. 5.33333
3. 40.5
4. 1.33333
5. 6.0
6. 1.0
7. 2.77778
8. 0.5
9. 9.14286
10. 4.0
Correct answer: 0.5.
m2
g = 2g
m1
m2 − m1
2. a =
g=g
m1
m2 − m1
1
3. a =
g = g correct
m1 + m2
3
Explanation:
Apply “F = m a”on the m1 + m2 mass
system. The net force is
1. a =
m2 g − m1 g = (m1 + m2 ) a
This leads to
a=
1
m2 − m1
g= g
m1 + m2
3
Explanation:
013
Let :
M = 6 kg ,
10.0 points
Version 073 – midterm 1 v1 – shih – (58505)
Consider a man standing on a scale which
is placed in an elevator. When the elevator is
stationary, the scale reading is Ss .
7
~ 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 elin direction to the S
evator is moving upward with constant speed
(no acceleration) so
Ss − m g = 0 ,
Scale
Find Sup , the scale reading when the elevator is moving upward with acceleration
1
~a = g ̂, in terms of Ss .
3
5
1. Sup = Ss
4
2. Sup = 2 Ss
3
Ss
2
1
= Ss
2
or Ss = m g as we would expect.
Call the scale reading for this part Sup (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.
Sup
a
Sdown
a
mg
mg
Equation 1 in the upward direction reads
Sup − m g = m a ,
3. Sup =
4. Sup
5. Sup = 0 m/s2
6. Sup =
7. Sup =
8. Sup =
9. Sup =
10. Sup =
1
Ss
3
5
Ss
3
3
Ss
4
4
Ss correct
3
2
Ss
3
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
and since a =
Sup − m g =
1
g at this particular instant,
3
1
mg
3
Sup = m g +
4
4
1
m g = m g = Ss .
3
3
3
Thus the scale gives a higher reading when
the elevator is accelerating upwards.
Comments for Matter Interaction readers:
Apply the derivative form of the momentum
dp
principle
= Fnet . The left hand side is
dt
1
m a = m g, and the right hand side is Sup −
6
1
m g. This leads to Sup − m g = m g.
6
014
10.0 points
An object was suspended in a fixed place
(y = 0) and then allowed to drop in a free fall.
Version 073 – midterm 1 v1 – shih – (58505)
Taking up as the positive vertical direction,
which of the following graphs correctly represents its vertical motion as displacement vs
time?
8
Explanation:
The object is undergoing a constant downward gravitational acceleration g. The slope
of a position vs time curve represents the
velocity.
y
t
1.
Let :
y
t
2.
a = −g ,
v0 = 0 m/s ,
y0 = 0 m .
y = y0 + v0 t +
y
t
3.
t
015 10.0 points
Two 30 N forces and a 60 N force act on a
hanging box as shown.
y
30 N
t
5.
1 2
1
a t = − g t2 .
2
2
This is a parabolic shaped curve starting at
y = 0 with a continuously decreasing slope as
time increases.
y
4.
and
30 N
y
t
6.
correct
y
t
7.
60 N
Will the box experience acceleration?
y
t
8.
1. Yes; upward.
2. Yes; downward. correct
y
t
9.
3. Unable to determine without the angle.
4. No; it is balanced.
y
10.
t
Explanation:
The horizontal components of the two 30 N
forces cancel, leaving an upward force that is
less than 60 N. Thus, the net force on the box
is down, causing it to accelerate downward.
Version 073 – midterm 1 v1 – shih – (58505)
016 10.0 points
A particle undergoes two displacements. The
first has a magnitude of 161 cm and makes an
angle of 138.1◦ with the positive x axis. The
resultant of the two displacements is 105 cm
directed at an angle of 33.6◦ to the positive x
axis.
Find the magnitude of the second displacement.
1. 213.099
2. 178.491
3. 226.149
4. 198.204
5. 161.148
6. 245.214
7. 201.354
8. 189.272
9. 174.295
10. 176.24
Correct answer: 213.099 cm.
Explanation:
Let :
|A| = 161 cm ,
θa = 138.1◦ ,
|R| = 105 cm ,
θR = 33.6◦ .
and
A
138.1◦
R
33.6◦
x
B
R=A+B
B = R−A.
The components of the second displacement B are
Bx = Rx − Ax = R cos θR − A cos θA
= (105 cm) cos 33.6◦
− (161 cm) cos 138.1◦
= 207.291 cm and
By = Ry − Ax = R sin θR − A sin θA
= (105 cm) sin 33.6◦
− (161 cm) sin 138.1◦
= −49.4149 cm ,
so the magnitude of B is
q
|B| = Bx2 + By2
q
= (207.291 cm)2 + (−49.4149 cm)2
= 213.099 cm .
9