Physics 130
General Physics - McColgan
Fall 2012
Midterm Exam 1
October 2, 2012
Name:
Instructions
1. This examination is closed book and closed notes. All your belongings
except a pen or pencil and a calculator should be put away and your
bookbag should be placed on the floor.
2. You will find one page of useful formulae on the last page of the exam.
3. Please show all your work in the space provided on each page. If you
need more space, feel free to use the back side of each page.
4. Academic dishonesty (i.e., copying or cheating in any way) will
result in a zero for the exam, and may cause you to fail the
class.
In order to receive maximum credit,
each solution should have:
1.
2.
3.
4.
5.
6.
7.
8.
9.
A labeled picture or diagram, if appropriate.
A list of given variables.
A list of the unknown quantities (i.e., what you are solving for).
Graphical representations of the motion (e.g., position, velocity, acceleration vs. time), if appropriate.
A labeled 1D or 2D coordinate axis system, if appropriate.
The full equation or equations needed to solve the problem
(e.g., appropriate equations of motion).
An algebraic solution of the unknown variables in terms of the
known variables.
A final numerical solution, including units, with a box around it.
An answer to additional questions posed in the problem, if any.
1
Physics 130
General Physics - McColgan
Fall 2012
1. One game at the amusement park has you push a puck up a long, frictionless ramp. You
win a stuffed animal if the puck, at its highest point, comes to within 10 cm of the end
of the ramp without going off. You give the puck a push, releasing it with a speed of
6.0 m{s when it is 9 m from the end of the ramp. The puck’s speed after traveling 2.5 m
is 5.0 m{s. Are you a winner?
Solution:
This is a 1D kinematics question. We are given the puck’s initial velocity vDi , and
it’s velocity after it has traveled 2.5 m. We are also given the length of the ramp,
xf , and we are asked to find whether the puck stops within 10 cm of the end of the
ramp. To solve this problem we need the 1D kinematic equation relating velocities,
distance, and acceleration. We need to find the acceleration from the two velocities
and the two speeds at the start of the puck’s trip up the slide:
vi2 2apxf xiq
2apxf xi q vi2 vf2
vi2 vf2
a 2pxf xi q
p6 m{sq2 p5 m{sq2
a 2p2.5q
a 2.2 m{s2
vf2
(1)
(2)
(3)
(4)
(5)
(6)
We can use the same equation again, using the starting and ending velocity and the
acceleration that we just found, where vf 0:
vf2
2
vi2 2apxf
xi q
(7)
Physics 130
General Physics - McColgan
2apxf xi q p xf xi q 0
vi2 2apxf xi q
vi2
vi2
2a
p6 m{sq2
2p2.2 m{s2 q
8.2 m
Fall 2012
(8)
(9)
(10)
(11)
(12)
(13)
Unfortunately, you do not win! The puck stops well before the winning distance. To
win, the puck has to stop at a distance between 8.9 and 9 m.
3
Physics 130
General Physics - McColgan
Fall 2012
2. As a science project, you drop a watermelon off the top of the Empire State Building,
350 m above the sidewalk. It so happens that Superman flies by at the instant you
release the watermelon. Superman is headed straight down with a constant speed of
40 m{s. How fast is the watermelon going when it passes Superman?
Solution:
This is a 1D kinematics question. We are given the watermelon and Superman’s
starting position at 350 m and Superman’s constant velocity of 40 m{s. We also
know that the acceleration of gravity is 9.8m{s2 .
To solve this problem we need use the 1D kinematic equations for the watermelon
and Superman:
xSf
xSi
xW f
xW i
1
aS p∆tq2 xSi vSi ∆t,
2
1
1
vW i ∆t
aW p∆tq2 xW i
aW p∆tq2 .
2
2
vSi ∆t
(14)
(15)
We’re trying to find the speed of the watermelon just as it passes Superman, or when
the xSf xW f . Setting these two equations equal to each other, we have:
xSi
vSi ∆t xW i
1
aW p∆tq2
2
(16)
(17)
Since the initial positions are the same, xSi
xW i, and this equation becomes:
1
vSi ∆t aW p∆tq2
2
(18)
(19)
4
Physics 130
General Physics - McColgan
We can factor ∆t, and substitute aW
vSi
Fall 2012
g, and the equation simplifies to:
12 g∆t
(20)
(21)
Solving for the time, ∆t:
∆t 2vSi
g
40 mq
2p
8.2 sec
9.8 m{s2
(22)
(23)
To find the velocity of the watermelon:
vi a∆t
vf g∆t
vf 9.8 m{s2 p8.2 secq 80.4 m{s
vf
5
(24)
(25)
(26)
(27)
Physics 130
General Physics - McColgan
Fall 2012
~ 9ı̂ 3̂, B
~ 7ı̂ 5̂, and D
~ 3A
~ 2B.
~
3. Let A
~ in component form.
(a) Write vector D
~ B,
~ and D.
~
(b) Draw a coordinate system and on it show vectors A,
~
(c) What are the magnitude and direction of vector D?
Solution:
~ in component form we first multiplyA
~ by 3 and B
~ by 2.
(a) To write D
~ 3 9ı̂ 3 3̂ 27ı̂ 9̂
3A
~
2B
2 p7qı̂
2 5̂ 14ı̂
10̂
(28)
(29)
(30)
Next we add the x- and y-components:
3A~ 2B~
p27ı̂ 9̂q p14ı̂ 10̂q
p27 14qı̂ p9 10q̂
41ı̂ 19̂.
~
D
(31)
(32)
(33)
(34)
(b)
~ is given by
(c) The magnitude of the vector D
|D~ | a
412
6
p19q2
45.2,
(35)
(36)
Physics 130
General Physics - McColgan
Fall 2012
and the direction θ (measured positive counter-clockwise from the x-axis) is equal to
θ tan1
Dy
Dx
tan1
7
19
41
24.9or 335.1.
(37)
Physics 130
General Physics - McColgan
Fall 2012
4. The figure shows three ropes tied together in a knot. One of your friends pulls on a
rope with 3.0 units of force and another pulls on a second rope with 5.0 units of force.
How hard and in what direction must you pull on the third rope to keep the knot from
moving?
Solution:
This is a vector problem that uses x- and y- components of vectors. The three vectors
must be balanced. This means that the x- and y - components of the three vectors
must be balanced.
¸
F 0
¸ x
Fy 0
(38)
(39)
(40)
The 3 unit force vector is only in the x-direction. The 5 unit force vector has xand y- components. The x-component of the 5 unit force vector is found from SOHCAH-TOA:
cos θ
F5x
5 cos θ 5 cosp60q 2.5 units
Fy
sin θ 5
Fy 5 sin θ 5 sinp60q 4.33 units
Fx
8
(41)
(42)
(43)
(44)
(45)
Physics 130
General Physics - McColgan
Fall 2012
Now we add the x-components:
¸
Fx
2.5
3 0.5
(46)
(47)
For the unknown vector to balance the other forces on the knot,
Fx 0.5
Fy 4.33
(48)
(49)
(50)
To find the magnitude of the force,
a
|D~ | p.5q2 p4.33q2
4.36 units
(51)
(52)
and the direction θ (measured positive counter-clockwise from the x-axis) is equal to
θ tan1
Dy
Dx
tan1
4.33
.5
9
83.4 S of W or θ 263.4.
(53)
Physics 130
General Physics - McColgan
Fall 2012
5. A stunt-man drives a car at a speed of 35 m{s off a 25 m high cliff. The road leading to
the cliff is inclined upward at an angle of 15 .
(a) How far from the base of the cliff does the car land?
(b) What is the car’s impact speed?
Solution:
This is a 2D projectile motion problem. We are given the initial height of the car
above the ground, yi 25 m, its initial velocity, vi 35.0 m{s, and the angle above
the horizontal with which the car leaves the cliff, θ 15 . We are asked about the
final distance from the base of the cliff and the car’s impact speed.
a) To solve the problem we first determine the initial velocity of the ball in the
horizontal (x) and vertical (y) dimensions:
vxi
vyi
vi cos θ p35 m{sq cos p15 q 33.8 m{s,
vi sin θ p35 m{sq sin p15 q 9.1 m{s.
(54)
(55)
Next, we use the vertical kinematic equation of projectile motion to find the time it
takes the car to hit the ground, resulting in yf 0:
yf
0 yi
1
vyi ∆t g p∆tq2 ,
2
1
g p∆tq2 vyi ∆t yi 0,
2
4.9∆t2 9.1∆t 25 0
10
(56)
(57)
(58)
(59)
Physics 130
General Physics - McColgan
Fall 2012
Using the quadratic equation and substituting in values,
b ∆t ∆t ?2
b 4ac
2a
a
2
9.1 p9.1q 4p4.9qp25q
2p4.9q
∆t 3.4 sec
(60)
(61)
(62)
(63)
Next, we use the horizontal kinematic equation of projectile motion to find the
distance the car lands from the bottom of the cliff:
xf
xi vxi∆t
vxi∆t
p33.8 m{sqp3.4 secq
114 m
(64)
(65)
(66)
(67)
(68)
b)To find the impact speed, we need to find the final x- and y- components of velocity.
The x-component of the final velocity is:
vf x
vix 33.8 m{s
(69)
(70)
The y-component of the final velocity is:
vf y
viy g∆t 9.1 m{s p9.8 m{s2qp3.4 secq
24.2 m{s
(71)
(72)
(73)
To find the magnitude of the velocity or the impact speed,
a
|~v| p33.8q2 p24.2q2
41.6 m{s
11
(74)
(75)
(76)
Physics 130
General Physics - McColgan
Fall 2012
6. You are watching an archery tournament when you start wondering how fast an arrow
is shot from the bow. Remembering your physics, you ask one of the archers to shoot
an arrow parallel to the ground. You find the arrow stuck in the ground 50 m away,
making a 2 angle with the ground. How fast was the arrow shot?
Solution:
This is a 2D projectile motion problem. We are told that the arrow is pointed in the
horizontal direction initially, that the range, or horizontal distance the arrow travels
is xf 50 m, and that the angle at which the arrow hits the ground with respect to
the horizontal is θ 15 . We are asked to find the initial speed of the arrow.
a) To start the problem, let’s use the kinematic equation for x where the initial
position, xi 0:
xf
xi
vix ∆t vix ∆t
(77)
(78)
Then, let’s use the kinematic equation for y where the initial velocity viy
vyf
0:
vyi g∆t g∆t
Since the velocity in the x-direction is a constant, let’s set vix
(79)
(80)
vf x vx.
Now, use the angle at which the arrow hits the ground to relate vx and vy .
tan θ
vvy
(81)
x
vx
vy
tan
θ
(82)
(83)
12
Physics 130
General Physics - McColgan
From the equation above, we know that vy
vx
Fall 2012
g∆t, which gives
g∆t
vy
tan
θ
tan θ
(84)
(85)
Plugging this into the equation for xf , we get
xf
g∆t
∆t
vx∆t tan
θ
2
g∆t
(86)
(87)
tan θ
(88)
Find the time by rearranging the equation to get
∆t d
xf tan θ
∆t g
d
θ
xf tan
g
(89)
p50 mq tanp2q ∆t 0.4 sec
9.8 m{s2
(90)
∆t2
To find the initial velocity,
vy
g∆t
tan
θ
tan θ
vy
p
9.8 m{s2 qp0.4 secq
vx tan θ
tanp2q
vx 112 m{s
vx
(91)
(92)
(93)
(94)
Note: We drop the negative sign on g because we’re keeping track of the positive
and negative values, and we know that vx is positive. The negative in the equation
would result from the fact that the velocity is in the negative y-direction and the
angle is in the 4th quadrant.
13
Physics 130
General Physics - McColgan
Fall 2012
7. As the earth rotates, what is the speed of (a) a physics student in Miami, Florida, at
latitude 25 , and (b) a physics student in Fairbanks, Alaska, at latitude 66 ? Ignore the
revolution of the earth around the sun. The radius of the earth is 6390 km.
Solution:
This is a uniform angular acceleration problem in which we are given the radius of
the earth, re 6390 km, and the latitude (angle θ with respect to the equator) of
each physics student.
(a) From the geometry of the problem, the linear distance of the student from the
rotational axis of the earth is
r
re sin p90 θq
p6390 kmq sinp65q
5791 km.
(95)
(96)
(97)
The speed of the student in Miami is just equal to the circumference divided by the
period, T , which is just one day:
v
2πr
T
2π 5791 km
1 day
24 3600 sec
1 day
0.421 km s1
421 m{s.
(b) This solution proceeds just like above but with θ
is 2600 km, and v 0.189 m{s 189 km s1 .
14
66.
(98)
(99)
(100)
(101)
In this case the radius
Physics 130
General Physics - McColgan
Fall 2012
8. A rock stuck in the tread of a 56.0 cm diameter bicycle wheel has a tangential speed of
2.80 m{s. When the brakes are applied, the rock’s tangential deceleration is 1.30 m{s2 .
(a) What are the magnitudes of the rock’s angular velocity and angular acceleration at
t 1.70 s?
(b) At what time is the magnitude of the rock’s acceleration equal to g?
Solution:
This is a non-uniform angular acceleration problem. We are given the size of the
wheel, r 28 cm 0.28 m, the initial tangential speed of the rock, vi 2.80 m{s, and
the tangential deceleration of the rock when the brakes are applied, aT 1.30 m{s2 .
(a) In order to determine the final angular velocity, ωf of the rock after t 1.70 sec,
we first have to determine the initial angular velocity:
wi
m{s
10 rad{sec.
vri 2.8
0.28 m
(102)
The angular acceleration is a constant and given by:
α
aT
r
m{s
1.30
4.64 rad{sec2.
0.28 m
2
(103)
Finally, using the appropriate rotational kinematic equation and substituting, we
obtain
ωf
ωi α∆t
10 rad{sec p4.64 rad{sec2 q p1.7 secq
2.11 rad{sec.
15
(104)
(105)
(106)
Physics 130
General Physics - McColgan
Fall 2012
(b) In order to find the time at which the magnitude of the angular acceleration of
the rock equals g, wea
first need to find the linear velocity of the rock at that time.
We want |~a| g a2T a2r , where aT is given in the problem and ar v 2 {r.
Substituting and solving for v we get
g
g
2
ñ v4
v
d
2 2
v
r
a2T
(107)
v4
r2
r2pg2 a2Tq
r1{2pg2 a2Tq1{4
p0.28 mq1{2 rp9.8 m{s2q2 p1.30 m{s2q2s1{4
1.65 m{s.
a2T
(108)
(109)
(110)
(111)
(112)
To find the time at which the rock reaches this velocity, we use the 1D kinematic
equation for velocity and solve for the time:
vf
ñ ∆t
vi aT ∆t
vi vf
aT
2.80 m{s 1.65 m{s
1.30 m{s2
0.88 sec.
16
(113)
(114)
(115)
(116)
Physics 130
General Physics - McColgan
Kinematics and Mechanics
xi vxi∆t 12 axp∆tq2
vxf vxi ax ∆t
2
vxf
vxi2 2axpxf xiq
1 2
yf yi vyi t
ay t
2
vyf vyi ay t
2
vyf
vyi2 2ay pyf yiq
1
θf θi ωi ∆t
α∆t2
2
ωf ωi α∆t
ωf 2 ωi 2 2α∆θ
s rθ
c 2πr
2πr
v
T
vt ωr
v2
ar ω2r
r
at rα
xf
17
Fall 2012