Columbia University Medical Center
SMDEP/HCOP Physics
Summer 2009
Instructor:
Luke Ward, ldw2105@columbia.edu
Course Objective:
This course is intended to prepare pre-medical and pre-dental
students for college-level physics and the MCAT. Problem solving
is emphasized.
Prerequisites:
No previous experience with physics is assumed, but algebra and
trigonometry will be used extensively.
Grading:
0%
25%
25%
50%
Reading materials:
There is no printed textbook for this course. Two free online
resources encompass the material covered and provide extra
problems:
Pre-test
Homework (weekly)
Midterm
Final
The Physics Classroom by Tom Henderson
http://www.physicsclassroom.com/
SparkNotes: SAT Physics
http://www.sparknotes.com/testprep/books/sat2/physics/
UNIT 1
Objective:
Kinematics
To understand the relationship between position, velocity,
and acceleration
Review:
Vector math (adding, decomposing, trigonometry)
Skills:
Drawing and interpreting motion vs. time graphs
representing 1D motion
Solving 1D motion problems using kinematics equations
Drawing and interpreting vectors representing 2D motion
Solving 2D motion problems using vector math
Applications: Acceleration and the human body (“g-tolerance”)
Reading:
Physics Classroom: 1D Kinematics (all)
Vectors – Motion and Forces in 2D (all)
UNIT 2
Objectives:
Forces
To identify the forces acting on an object and describe them
qualitatively
To understand the relationship between force and motion
Review:
Checking answers: reasonableness, units
Skills:
Drawing and interpreting vectors representing 2D forces
(free-body diagrams)
Connecting force to motion using Newton's Laws
Applications: Drag and electrophoresis
Reading:
Physics Classroom: Newton’s Laws (all)
Circular/Planetary Motion (Lessons 1-2)
UNIT 3
Objectives:
Work-Energy
To identify various types of potential energy
To understand motion as kinetic energy
To appreciate the basic subject matter of thermodynamics:
energy and its conversion to work and heat
Applications: Thermodynamics of the human body (metabolism,
homeostasis, heat/cold injury)
Reading:
Physics Classroom: Work, Energy, and Power (all)
SparkNotes:
Dot Product: Vector Multiplication
UNIT 4
Objective:
Reading:
UNIT 5
Objective:
Reading:
UNIT 6
Objective:
Skills:
Collisions
To use the concepts of conservation of momentum and energy to predict
how objects will behave after a collision
Physics Classroom: Momentum and its Conservation (all)
Fields
To extend the concept of gravitational potential to gravitational fields, and
by analogy electric and magnetic fields
Physics Classroom: G field: Circular/Planetary Motion (Lesson 3)
E field: Static Electricity (Lessons 3-4)
G and E potential: Current Electricity (Lesson 1)
SparkNotes:
B field: Magnetism (15.1, 15.2, 15.4 only)
Cross Product: Vector Multiplication
Waves
To describe and understand mechanical and electromagnetic waves
Predicting the type and energy of radiation associated with
changes to nuclei and electrons
Converting between wavelength and frequency
Applications: Ionizing vs. non-ionizing radiation in health and medicine
Reading:
Physics Classroom: The Wave Equation and EM Spectrum
SparkNotes: Quantum Physics (only material before “Wave-Particle
Duality” and no equations besides E=hf!)
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
Homework Problem Set #1
DUE 7/1/09
Directions: Homework problems are solved on your own and are due at the beginning of lecture,
one week after they are assigned. Show all of your work, and use proper units and significant
figures at every step. Attach extra paper if you need it. During study group, you may get help
solving the problems from the TAs.
1. Sketch simple position vs. time, velocity vs. time, and acceleration vs. time graphs for
the following scenarios (just draw the shape of the curve, don’t worry about units):
a. You jog down Riverside Drive, stop for a while to catch your breath, and then
walk back in the opposite direction to where you started.
b. You are driving on an interstate, then slow down as you pass an accident, then
quickly speed up again and cruise at your previous speed.
c. You jump straight up in the air and land on your feet.
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
2. The graph of x vs. t shown below is for a particle in straight-line motion.
a. At what times is the car at rest?
b. At what times is the car moving forward (i.e. in the positive x direction)?
c. At what times is the car moving backwards (i.e. in the negative x
direction)?
d. At what position did the car start? At what position did it end?
e. Invent a story of what this car might be doing.
3.
While barbecuing on the roof of Bard Hall, you toss a burger off of the roof while
flipping it straight up. The burger rises 1 m, before turning to fall straight down
onto Riverside Drive, which is 30 m below where the burger started. Find the initial
vertical speed of the burger. Then find how long the burger spends in the air. You
will actually find two possible answers for this time value. Only one is physically
correct. What could the other correspond to?
4.
A rock is dropped from a 100 m high cliff. How long does it take to fall (a) the first
50 m, and (b) the second 50 m?
5.
You throw a water balloon vertically downward with an initial speed of 12 m/s
from the roof of a building, 30 m above the ground.
a. How long does it take the balloon to reach the ground?
b. What is the speed of the balloon at impact?
6.
The Indy 500 is a 500-mile stock car race on a loop. The record average winning
speed was 186 mile/hr. (a) In what time did the winner finish? (b) What was his
average velocity (if the start and finish lines are the same place?)
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
Homework Problem Set #1
DUE 7/1/09
Directions: Homework problems are solved on your own and are due at the beginning of lecture,
one week after they are assigned. Show all of your work, and use proper units and significant
figures at every step. Attach extra paper if you need it. During study group, you may get help
solving the problems from the TAs.
1. Sketch simple position vs. time, velocity vs. time, and acceleration vs. time graphs for
the following scenarios (just draw the shape of the curve, don’t worry about units):
a. You jog down Riverside Drive, stop for a while to catch your breath, and then
walk back in the opposite direction to where you started.
b. You are driving on an interstate, then slow down as you pass an accident, then
quickly speed up again and cruise at your previous speed.
c. You jump straight up in the air and land on your feet.
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
(a)
(b)
x
(c)
x
t
v
x
t
v
t
a
v
t
a
t
t
t
a
t
t
SMDEP Physics B 2008
June 17, 2008
2. The graph of x vs. t shown below is for a particle in straight-line motion.
a. At what times is the car at rest?
All of the places where the slope is zero:
[15 m, 20 m] ∪ [30 m, 35 m] ∪ [40 m, 60 m]
b. At what times is the car moving forward (i.e. in the positive x direction)?
All of the places where the slope is positive:
[0 m, 15 m] ∪ [35 m, 40 m]
c. At what times is the car moving backwards (i.e. in the negative x
direction)?
All of the places where the slope is negative:
[20 m, 30 m]
d. At what position did the car start? At what position did it end?
x = -6 m, x = 1 m
e. Invent a story of what this car might be doing.
Answers will vary. Involves going forward then
braking, going backwards and then braking, then
forwards again. Perhaps parallel parking?
3.
While barbecuing on the roof of Bard Hall, you toss a burger off of the roof while
flipping it straight up. The burger rises 1 m, before turning to fall straight down
onto Riverside Drive, which is 30 m below where the burger started. Find the initial
vertical speed of the burger. Then find how long the burger spends in the air. You
will actually find two possible answers for this time value. Only one is physically
correct. What could the other correspond to?
For the first part, we use the fact that it takes 1 second to reach its
maximum height, at which v = 0.
We can use vf2 = vo2 + 2aΔx:
SMDEP Physics B 2008
June 17, 2008
(0 m/s)2 = vo2 + 2(-9.8 m/s2)(1 m)
vo ≈ 4.43 m/s
For the second part, there are several ways to solve it. The simplest:
xf = xo + vo Δt + ½aΔt2
xf = (0 m) + (4.43 m/s)Δt + (0.5)(-9.8 m/s2)Δt2
Quadratic with a = -4.9, b = 4.43, c = -xf
Δt = (-4.43 ± √(4.432 – 4(-4.9)(-xf))/2(-4.9)
(Only the positive solution makes sense!)
For xf = -30 m, Δt ≈ 2.97 s
4.
A rock is dropped from a 100 m high cliff. How long does it take to fall (a) the first
50 m, and (b) the second 50 m?
xf = xo + vo Δt + ½aΔt2
-50 m = 0 + (0 m/s)Δt + ½(-9.8 m/s2)Δt2
Δt ≈ 3.19 s for the first 50 m
xf = xo + vo Δt + ½aΔt2
-100 m = 0 + (0 m/s)Δt + ½(-9.8 m/s2)Δt2
Δt ≈ 4.52 s for 100 m,
So 4.52 s – 3.19 s ≈ 1.33 s for the second 50 m
5.
You throw a water balloon vertically downward with an initial speed of 12 m/s
from the roof of a building, 30 m above the ground.
a. How long does it take the balloon to reach the ground?
xf = xo + vo Δt + ½aΔt2
-30 m = 0 m + (-12 m/s)Δt + ½(-9.8 m/s2)Δt2
Quadratic with a = -4.9, b = -12, c = 30
Δt = (-4.43 ± √(4.432 – 4(-4.9)(-xf))/2(-4.9)
(Only the positive solution makes sense!)
Δt ≈ 1.54 s
b. What is the speed of the balloon at impact?
vf = vo + aΔt
vf = (-12 m/s) + (-9.8 m/s2)(1.54 s)
vf ≈ -27.09 m/s
6.
The Indy 500 is a 500-mile stock car race on a loop. The record average winning
speed was 186 mile/hr. (a) In what time did the winner finish? (b) What was his
average velocity (if the start and finish lines are the same place?)
t = 500 mi * (1 hr / 186 mi) ≈ 2.69 hr
Average velocity = 0 m/s
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
In-Class Problem Set #1
Directions: In-class problems are solved during normal lecture times with the help of the
instructor and TAs. If your answers turn out wrong when we discuss them, be sure to
write the correct answers and explanation for future reference. HOLD ON TO THESE,
because you should study for tests with them, and we may refer back to them during later
lectures. Use lots of extra paper if you need it!
1.
A person walks forward 2 m, turns right, walks forward 2 m, turns right, walks
forward 2 m, and finally turns right and walks forward 1 m.
a. Draw the path that the person took.
b. What was the person’s total distance traveled?
c. What was the her total displacement?
d. If it took her 7 s, what was her average speed?
e. What was her average velocity?
2.
Sketch simple position vs. time, velocity vs. time, and acceleration vs. time
graphs for the following scenarios (just draw the shape of the curve, don’t worry
about units):
a. A dog walks for a while,
and then runs for a while.
b. You are driving down
I-95, screech to a halt at
stopped traffic, and then
gently push on the
accelerator, slowly coming
back to your original
speed.
c. You ride a local 1 train
from City College at 137th
street to 168th street,
stopping at 145th and 157th
streets.
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
3. The graph of x vs. t shown below is for a particle in straight-line motion.
a. State, for each of the intervals AB, BC, CD, and DE, whether the velocity v is
positive, negative, or 0 and whether the acceleration a is positive, negative, or 0.
b. From the curve, is there any interval over which the acceleration is obviously not
constant?
c. If the axes are shifted upward together such that the time axis ends up running
along the dashed line, do any of your answers change?
4.
During a hard sneeze, your eyes might shut for 0.50 s. If you are driving a car at 90
km/h, how far does it move in that time?
5.
An object has an acceleration of 3.2 m/s2. At a certain instant its velocity is 9.6
m/s. What is its velocity (a) 2.5 s earlier and (b) 2.5 s later?
6.
A world’s land speed record was set by Colonel John P. Stapp when, on March 19,
1954, he rode a rocket-propelled sled that moved down a track at 1020 km/h. He
and the sled were brought to a stop in 1.4 s. What acceleration did he experience?
How many “g’s” is this?
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
In-Class Problem Set #1
Directions: In-class problems are solved during normal lecture times with the help of the
instructor and TAs. If your answers turn out wrong when we discuss them, be sure to
write the correct answers and explanation for future reference. HOLD ON TO THESE,
because you should study for tests with them, and we may refer back to them during later
lectures. Use lots of extra paper if you need it!
1.
A person walks forward 2 m, turns right, walks forward 2 m, turns right, walks
forward 2 m, and finally turns right and walks forward 1 m.
a. Draw the path that the person took.
b. What was the person’s total distance traveled? 7 m
c. What was the her total displacement? 1 m to the right
d. If it took her 7 s, what was her average speed? 1 m/s
e. What was her average velocity? 1/7 m/s to the right
2.
Sketch simple position vs. time, velocity vs. time, and acceleration vs. time
graphs for the following scenarios (just draw the shape of the curve, don’t worry
about units):
a. A dog walks for a while,
and then runs for a while.
b. You are driving down
I-95, screech to a halt at
stopped traffic, and then
gently push on the
accelerator, slowly coming
back to your original
speed.
c. You ride a local 1 train
from City College at 137th
street to 168th street,
stopping at 145th and 157th
streets.
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
Answers will vary – 1,2 were covered in class, check with TAs or instructor for 3
3. The graph of x vs. t shown below is for a particle in straight-line motion.
a. State, for each of the intervals AB, BC, CD, and DE, whether the velocity v is
positive, negative, or 0 and whether the acceleration a is positive, negative, or 0.
velocity
acceleration
AB
+
0
BC
+
CD
0
0
DE
+
b. From the curve, is there any interval over which the acceleration is obviously not
constant? Bad question – maybe CD or DE, but hard to tell.
c. If the axes are shifted upward together such that the time axis ends up running
along the dashed line, do any of your answers change? No.
4.
During a hard sneeze, your eyes might shut for 0.50 s. If you are driving a car at 90
km/h, how far does it move in that time?
v = Δx/Δt
Δx = vΔt
Δx = 0.5 s * 90 km/h = 45 s-km/h
45 s-km/h * 1/60 h/min * 1/60 min/s * 1000 m/km = 12.5 m
5.
An object has an acceleration of 3.2 m/s2. At a certain instant its velocity is 9.6
m/s. What is its velocity (a) 2.5 s earlier and (b) 2.5 s later?
Columbia University SMDEP/HCOP Physics 2009
June 24, 2009
a = Δv/Δt
Δv = aΔt
(vf – v0) = a(tf – t0)
vf = v0 + a(tf – t0)
For (a), v0 = 9.6 m/s, a = 3.2 m/s2, tf = -2.5 s, t0 = 0 s
vf = 9.6 m/s + (3.2 m/s2)(-2.5 s) = 1.6 m/s
For (b), v0 = 9.6 m/s, a = 3.2 m/s2, tf = 2.5 s, t0 = 0 s
vf = 9.6 m/s + (3.2 m/s2)(2.5 s) = 17.6 m/s
6.
A world’s land speed record was set by Colonel John P. Stapp when, on March 19,
1954, he rode a rocket-propelled sled that moved down a track at 1020 km/h. He
and the sled were brought to a stop in 1.4 s. What acceleration did he experience?
How many “g’s” is this?
a = Δv/Δt
a = (-1020 km/h)/(1.4 s)
a = 728.6 km/h-s
728.6 km/h-s * 1/3600 h/s * 1000 m/km = 202.4 m/s2
202.4 m/s2 ÷ 9.8 m/s2 = 20.7 g
Columbia University SMDEP/HCOP Physics 2009
June 29, 2009
In-Class Problem Set #2
Directions: In-class problems are solved during normal lecture times with the help of the
instructor and TAs. If your answers turn out wrong when we discuss them, be sure to
write the correct answers and explanation for future reference.
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
1.
g = 9.8 m/s2
(WARMUP FROM CLASS #1) A graph of x vs. t is shown for a ball thrown
straight up (a) and a yo-yo (b). Describe the acceleration that each experiences.
(a) x
(b)
x
t
t
2.
You throw a ball straight up at 20 m/s, releasing the ball 1.5 m above the ground.
What is the maximum height of the ball? What is its impact speed as it hits the
ground?
3.
A sprinter accelerates at 2.5 m/s/s until he reaches his top speed of 15 m/s, and
continues to run at this speed. How long does it take him to run the 100-m dash?
4.
A heavy object is raised by sliding it 12.5 m along a plank oriented at 20 degrees to
the horizontal.
a. How high above the ground is it raised?
b. How far is it moved horizontally?
Columbia University SMDEP/HCOP Physics 2009
June 29, 2009
5.
A wheel with radius 45 cm rolls without slipping along a horizontal floor as
shown. At the beginning, a dot painted on the rim of the wheel is at the point of
contact between the wheel and the floor. Later, the wheel has rolled through onehalf of a revolution. What is the displacement of the dot?
6. Two vectors a and b have equal magnitudes of 10 units and are oriented as shown.
If we call their vector sum r, find:
a. the x and y components of r,
b. the magnitude of r, and
c. the angle r makes with the x-axis.
7. A golfer takes three putts to get the ball into the hole. The first putt displaces the
ball 12 ft north, the second 6 ft southeast, and the third 3 ft southwest. What
displacement was needed to get the ball into the hole on the first putt?
Columbia University SMDEP/HCOP Physics 2009
June 29, 2009
In-Class Problem Set #2
Directions: In-class problems are solved during normal lecture times with the help of the
instructor and TAs. If your answers turn out wrong when we discuss them, be sure to
write the correct answers and explanation for future reference.
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
1.
g = 9.8 m/s2
(WARMUP FROM CLASS #1) A graph of x vs. t is shown for a ball thrown
straight up (a) and a yo-yo (b). Describe the acceleration that each experiences.
(a) x
(b)
x
t
t
(a) the acceleration is constant and negative
(b) the acceleration is constant and positive
2.
You throw a ball straight up at 20 m/s, releasing the ball 1.5 m above the ground.
What is the maximum height of the ball? What is its impact speed as it hits the
ground?
You can use vf2 = v02 + 2aΔx for both parts of the problem.
For the first part, v0 = 20 m/s, a = -9.8 m/s2, vf = 0 m/s (at max height), solve for Δx:
vf2 = v02 + 2aΔx
Δx = (vf2 - v02)/2a
Δx = (- 400 m2/s2) / (-19.6 m/s2)
Δx ≈ 20.41 m
The final height at the top is x0 + Δx = 1.5 m + 20.41 m = 21.91 m
For the second part, you could do two things: only consider the second half of the time
(as if the ball were dropped from 21.91 m) or consider the entire time.
To consider only the second half of the time, v0 = 0 m/s, a = -9.8 m/s2, Δx = -21.91 m,
solve for vf:
Columbia University SMDEP/HCOP Physics 2009
June 29, 2009
vf2 = v02 + 2aΔx
vf = sqrt(v02 + 2aΔx)
vf = sqrt(2(-9.8 m/s2)(-21.91 m))
vf ≈ -20.7 m/s
Only the negative square root makes sense in this case!
To consider the entire time, v0 = 20 m/s, a = -9.8 m/s2, Δx = -1.5 m, solve for vf:
vf2 = v02 + 2aΔx
vf = sqrt(v02 + 2aΔx)
vf = sqrt(400 m2 + 2(-9.8 m/s2)(-1.5 m))
vf ≈ -20.7 m/s
3.
A sprinter accelerates at 2.5 m/s/s until he reaches his top speed of 15 m/s, and
continues to run at this speed. How long does it take him to run the 100-m dash?
We have to break this up into the part during which he’s accelerating, and the part
afterwards (since we can only use our equations on pieces of problems with constant
acceleration.)
How long is the first part?
a = ∆v/∆t
∆t = ∆v/a
∆t = (15 m/s) / (2.5 m/s2)
∆t = 6 s
How much of the 100 m did he finish? (i.e. how much is left to cover in the second part
of the problem?)
xf = xo + vo Δt + ½aΔt2
xf = 0 m + 0 m-s/s + ½(2.5 m/s2)(36 s2 )
xf = 45 m
How long does it take to finish the other 55 m?
v = Δx/Δt
Δt = Δx/v
Δt = (55 m) / (15 m/s)
Δt = 3.67 s
Since a is 0..
Add up the time from the two halves of the problem:
∆t = 6 s + 3.67 s = 9.67 s
Columbia University SMDEP/HCOP Physics 2009
June 29, 2009
4.
A heavy object is raised by sliding it 12.5 m along a plank oriented at 20 degrees to
the horizontal.
a. How high above the ground is it raised?
b. How far is it moved horizontally?
a) 2.5 sin 20˚ = 0.86 m
b) 2.5 cos 20˚ = 2.35 m
5.
A wheel with radius 45 cm rolls without slipping along a horizontal floor as
shown. At the beginning, a dot painted on the rim of the wheel is at the point of
contact between the wheel and the floor. Later, the wheel has rolled through onehalf of a revolution. What is the displacement of the dot?
Right triangle with base = πr and height = 45 cm; solve for hypotenuse
a2 = b2 + c2
a = sqrt(b2 + c2)
a = sqrt((45 cm)2 + ((3.14)(45 cm))2)
a = 148.3 cm
6. Two vectors a and b have equal magnitudes of 10 units and are oriented as shown.
If we call their vector sum r, find:
a. the x and y components of r,
b. the magnitude of r, and
Columbia University SMDEP/HCOP Physics 2009
June 29, 2009
c. the angle r makes with the x-axis.
r = <10 cos 30˚ + 10 cos 135˚ , 10 sin 30˚ + 10 sin 135˚>
r = <1.59, 12.07>
|r| = sqrt(1.592 + 12.072) = 12.18
angle = tan-1 (12.07/1.59) = 82.50˚
7. A golfer takes three putts to get the ball into the hole. The first putt displaces the
ball 12 ft north, the second 6 ft southeast, and the third 3 ft southwest. What
displacement was needed to get the ball into the hole on the first putt?
Three putts are
<0, 12> + <6 cos 315 , 6 sin 315> + <3 cos 225, 3 sin 225>
<0, 12> + <6/√2 , -6/√2> + <-3/√2, -3/√2>
<2.12, 5.64>
Or in other words, using magnitude and direction:
Magnitude = sqrt(2.122 + 5.642) = 6.02 ft
Direction = tan-1 (5.64/2.12) = 69.4 degrees
Columbia University SMDEP/HCOP Physics 2009
July 1, 2009
Homework Problem Set #2
DUE 7/8/09
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
g = 9.8 m/s2
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
1.
Draw a vector from the origin (0,0) to (3,7). What is the length of this vector?
What is the angle of the vector relative to the x-axis? The y-axis?
2.
Consider two vectors: a = (5,3) and b = (3,5). Draw a and b. What are the
magnitudes and directions of a and b? Redraw b so that it starts from the tip of a.
If c = a + b, draw c. Find c in component form. Find the magnitude and direction
of c.
3.
Let c = (4,-7) and d = (5,2). Draw these vectors first, with the tail of each at the
origin. Then draw the vectors c + d, c - d, and d - c, and give the value of each
written as (x,y).
4.
Let e = (-1,-3) and f = (3,6). Draw these vectors, starting at the origin. Calculate
and draw the following vectors: 2e – ⅓f and 2(e – f). What is the angle between e
and f?
Columbia University SMDEP/HCOP Physics 2009
July 1, 2009
5.
Three forces act on a block as shown below:
y
3N
10 N
30º
8N
x
a. Make a guess as to how the net force will look. Draw it as a vector acting on
the block.
b. Add the forces in component-vector notation. Then find the magnitude and
direction of the net force. Was your drawing close?
6.
Vector h has magnitude 3 and vector k has magnitude 2.
a.
b.
7.
What is the range of possible magnitudes of h + k?
What is the range of possible magnitudes of h – k?
A plane is trying to fly 500 mi/h due east, but there is a wind blowing at 70 mi/h
southwest. What is the plane’s actual speed?
Columbia University SMDEP/HCOP Physics 2009
July 1, 2009
Homework Problem Set #2
DUE 7/8/08
1.
Draw a vector from the origin (0,0) to (3,7).
What is the length of this vector?
r = √(x² + y²) = √(3² + 7²) = √(58) ≈ 7.62
What is the angle of the vector relative to the x-axis? The y-axis?
θ = tan-1 (y/x) = tan-1 (7/3) ≈ 66.80º with the x-axis
90º - tan-1 (7/3) ≈ 23.20º with the y-axis
2.
Consider two vectors: a = (5,3) and b = (3,5). Draw a and b.
What are the magnitudes and directions of a and b?
r = √(x² + y²) = √(5² + 3²) = √(34) ≈ 5.83 for both a and b
θ = tan-1 (y/x) = tan-1 (3/5) ≈ 30.96º for a
θ = tan-1 (y/x) = tan-1 (5/3) ≈ 59.04º for b
Redraw b so that it starts from the tip of a. If c = a + b, draw c. Find c in
component form. Find the magnitude and direction of c.
c = (5+3, 3+5) = (8,8)
r = √(x² + y²) = √(8² + 8²) = √(128) ≈ 11.31
θ = tan-1 (y/x) = tan-1 (1) = 45º
3.
Let c = (4,-7) and d = (5,2). Draw these vectors first, with the tail of each at the
origin. Then draw the vectors c + d, c - d, and d - c, and give the value of each
written as (x,y).
c+d = (9,-5)
c-d = (-1,-9)
d-c = (1, 9)
4.
Let e = (-1,-3) and f = (3,6). Draw these vectors, starting at the origin. Calculate
and draw the following vectors: 2e – ⅓f and 2(e – f). What is the angle between e
and f?
2e – ⅓f = (-3,8)
2(e – f) = (-8, -18)
θ for e = tan-1 (y/x) = tan-1 (-3/-1) ≈ calculator gives 71.57º, but we
know we are in Quadrant III, so this is really 180º + 71.57º
(measured counterclockwise from the x-axis.)
θ for f = tan-1 (y/x) = tan-1 (6/3) ≈ 63.43º. This answer from your
calculator is in the right quadrant (I).
Columbia University SMDEP/HCOP Physics 2009
July 1, 2009
(180º + 71.57º) - 63.43º = 188.14º measured the long way, 360º –
188.14º ≈ 171.86º measured the short way.
5.
Three forces act on a block as shown below:
y
3N
10 N
30º
8N
x
a. Make a guess as to how the net force will look. Draw it as a vector acting on
the block.
b. Add the forces in component-vector notation. Then find the magnitude and
direction of the net force. Was your drawing close?
x = -8 + 10 cos 30º ≈ 0.66
y = 3 + 10 sin 30º = 8
r = √(x² + y²) = √((-8+10cos30)² + 8²) ≈ 8.03
θ = tan-1 (y/x) = tan-1 (8 / (-8+10cos30)) ≈ 85.28º
6.
Vector h has magnitude 3 and vector k has magnitude 2.
a.
b.
7.
What is the range of possible magnitudes of h + k?
What is the range of possible magnitudes of h – k?
For both, [1, 5]
A plane is trying to fly 500 mi/h due east, but there is a wind blowing at 70 mi/h
southwest. What is the plane’s actual speed?
If east is the positive x direction, the resultant trajectory is
(500 - 70sin45, -70sin45) = (500 – 70/√2, -70/√2).
The magnitude is 453.21 mi/h.
Columbia University SMDEP/HCOP Physics 2009
July 1, 2009
In-Class Problem Set #3
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
g = 9.8 m/s2
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
F = ma
1. A 1000 kg block hangs on a rope. Find the tension on the rope if the block is
stationary, then if it’s moving upward at a steady speed of 5 m/s, finally if it’s
accelerating upward at 5 m/s². Draw a motion diagram and a free-body diagram
for each situation.
2. Compute the initial upward acceleration of a rocket of mass 1.3 x 104 kg if the
initial upward force produced by its engine (the thrust) is 2.6 x 105 N. Do not
forget the weight of the rocket.
3. A skier (75 kg mass) starts down a 50-m high, 10º slope on frictionless skis.
What is his speed at the bottom? Draw a motion diagram and a free-body
diagram as part of your problem solving.
6. Burglars are trying to pull a 1000 kg safe up a frictionless ramp to their getaway
truck. The ramp is tilted at angle θ. How hard are they pulling if the safe is at
rest? If it moves up the ramp at a steady 1 m/s? If it accelerates up the ramp at 1
m/s2 ? Do these answers have the expected behavior at the limits of θ  0º and θ
 90º ?
Columbia University SMDEP/HCOP Physics 2009
July 1, 2009
In-Class Problem Set #3
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
g = 9.8 m/s2
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
F = ma
1. A 1000 kg block hangs on a rope. Find the tension on the rope if the block is
stationary, then if it’s moving upward at a steady speed of 5 m/s, finally if it’s
accelerating upward at 5 m/s². Draw a motion diagram and a free-body diagram
for each situation.
If it’s stationary, F = 0 and T = mg = (1000 kg x 9.8 m/s/s) = 9800 N
If it’s moving upward at steady speed, same. 9800 N
If it’s accelerating upward at 5 m/s, F = ma = 1000 kg x 5 = 5000 N net
5000 N more than mg. 9800 + 5000 = 14800 N
2. Compute the initial upward acceleration of a rocket of mass 1.3 x 104 kg if the
initial upward force produced by its engine (the thrust) is 2.6 x 105 N. Do not
forget the weight of the rocket.
260000 N (up) - mg (down)
260000 N – (13000 x 9.8) = 132600 N up
F =132600 = ma
a = 132600/m = 132600/13000 = 10.2 m/s/s
3. A skier (75 kg mass) starts down a 50-m high, 10º slope on frictionless skis.
What is his speed at the bottom? Draw a motion diagram and a free-body
diagram as part of your problem solving.
Component of gravity downhill is mg sin 10º.
Acceleration due to gravity downhill is a=F/m = mg sin 10º / m = g sin 10º.
Distance to go = 50/sin 10º.
v2 = 2ax = 2(g sin 10)(50/sin 10) = 100g.
v = 10 sqrt (g) = 31.30 m/s
Columbia University SMDEP/HCOP Physics 2009
July 1, 2009
6. Burglars are trying to pull a 1000 kg safe up a frictionless ramp to their getaway
truck. The ramp is tilted at angle θ. How hard are they pulling if the safe is at
rest? If it moves up the ramp at a steady 1 m/s? If it accelerates up the ramp at 1
m/s2 ? Do these answers have the expected behavior at the limits of θ  0º and θ
 90º ?
Component of gravity downhill is mg sin θ.
To keep it at rest, have to pull 9800 sin θ.
To keep it moving steady 1 m/s, have to pull 9800 sin θ.
To accelerate up 1 m/s/s, need an additional F = ma = (1000)(1) = 1000 N.
Total needed in this case is 9800 + 1000 = 10800 N uphill.
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
Homework Problem Set #3
DUE 7/15/09
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
g = 9.8 m/s2
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
Newton’s Second Law: Fnet = ma
Problem-solving strategy:
1.
Draw picture of the whole situation.
2.
For each object:
a. Draw a free-body diagram with the forces acting on the object. (These
should be the only vectors in your drawing that touch the object.)
b. Define x-y axes. (Easiest if the acceleration is along one of these!)
c. Off to the side, draw the acceleration direction. If it’s not obvious
from the problem, you can figure this out with the “oil-drip” vector
method we discussed (make dots for positions at every second, connect
the dots with velocity vectors, then subtract each velocity vector from
the following one to get the acceleration between pairs of velocities.)
d. Use Newton’s Second Law, Fnet = ma, first in the x-direction.
1. On the left hand, add up the x-components of the
various forces. Call forces in the same direction as the
acceleration positive, and forces against negative.
2. On the right hand, use the correct variables for the
object’s mass and the x-component of its acceleration.
e. Repeat (d) for the y-direction.
3.
For each equation: identify the unknowns. What are you given? What are
you solving for?
4.
Solve for the variable you want.
5.
Plug in numbers (do this at the very end!)
1.
In a modified tug-of-war game, two people pull in opposite directions, not on a
rope, but on a 25 kg sled resting on (frictionless) ice. If the participants exert
forces of 90 N and 92 N, what will be the magnitude of the sled’s acceleration?
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
2.
An elevator weighing 6240 kg is pulled upward by a cable with acceleration 4
m/s².
a.
Calculate the tension in the cable.
b.
What is the tension when the elevator is decelerating at the rate 4
m/s² but still moving upward? (Draw the acceleration vector for
each case. Remember that the direction of the acceleration vector
is the same as the direction of the net force!)
3.
You push on a block with mass m with a force F, causing it to accelerate on a
frictionless surface.
a.
Draw a complete free-body diagram for this situation. What will
the acceleration of the block be while you are pushing it?
b.
Now, rather than pushing on the block horizontally, you push on it
at an angle θ:
θ
It still accelerates in the horizontal direction. What will the
magnitude of the acceleration be? Draw a free-body diagram.
(Hint: the normal force will be different. How do you know that
must be the case?)
4.
A block of mass m is on a ramp which makes an angle of θ with the ground. It
accelerates down the surface of the ramp.
a.
Define the ramp (the axis of acceleration) as the x-axis. What is
the x-component of the total weight mg? (This is the part of the
weight that lets the block move down the ramp.) At what rate will
it accelerate?
b.
What is the y-component of the total weight mg?
c.
What is the block’s acceleration in the y-dimension? Use this to
find the normal force.
5.
A sliding block on a frictionless surface with mass M = 3.3 kg is attached by a
cord strung around a pulley to a hanging block with mass m = 2.1 kg:
a. Draw a free-body diagram for each,
including the direction of acceleration for
each. (The magnitude of the tension force T
and the acceleration vector a will be the same
on both blocks!)
b. Write an equation Fnet = ma for each block.
c. Combine the two equations. What is the
acceleration of each block? What is the tension
T?
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
6.
What is the acceleration of a 3 kg block down a 30º plane if the coefficient of
kinetic friction is µ = 0.1? First find a formula, then plug in the numbers.
7.
You order a refrigerator weighing 1000 N and it arrives at your door in a
cardboard box. To get it to your kitchen, you tie a rope to it and pull it
horizontally. You find that to get it to move at all, you have to pull it horizontally
with a force of magnitude 460 N. Once it breaks loose and starts to move, you
can keep it moving at a constant velocity with only 200 N. (a) What is the mass
of the refrigerator? (b) What are the coefficients of static and kinetic friction
between the cardboard and your floor? (c) If you pull with a force of 50 N, what is
the friction force? (d – EXTRA CREDIT) It starts to get easier to pull if you pull
up at an angle. What is the optimal angle? (You may solve this with a graphing
calculator, or calculus.)
8.
If a ball with mass 0.09 kg is thrown straight up in a vacuum, it reaches a height
of 10 m. If you throw it upward with the same initial velocity but in air instead of
a vacuum, the maximum height is 7.8 m because of air drag. What is the average
air drag force on the ball during its upward motion?
SMDEP Physics 2009
1.
2.
Homework Problem Set #3
KEY
In a modified tug-of-war game, two people pull in opposite directions, not on a
rope, but on a 25 kg sled resting on (frictionless) ice. If the participants exert
forces of 90 N and 92 N, what will be the magnitude of the sled’s acceleration?
Fnet = ma
a = Fnet /m
|a| = (92 N – 90 N) / (25 kg) = 0.08 m/s²
An elevator weighing 6240 kg is pulled upward by a cable with acceleration 4
m/s².
a.
Calculate the tension in the cable.
Your free-body diagram should show two vectors touching the
box: the weight down (which we know is mg) and the tension up
(unknown; call it T.)
Off to the side, draw an acceleration vector: up at 4 m/s/s.
From Newton’s second law, we know that an upward acceleration
vector means a net force up.
Add your forces on the left and put mass & acceleration on the
right. We’ll call the direction it’s accelerating the positive
direction.
Fnet = ma
T – mg = ma
T = ma + mg = (6240 kg)(4 m/s²) + (6240 kg)(9.8 m/s²) = 86112 N
b.
What is the tension when the elevator is decelerating at the rate 4
m/s² but still moving upward? (Draw the acceleration vector for
each case. Remember that the direction of the acceleration vector
is the same as the direction of the net force!)
We can keep the same coordinate system since we already worked
everything out:
T = ma + mg = (6240 kg)(-4 m/s²)+(6240 kg)(9.8 m/s²) = 36192 N
Keep in mind that if the elevator is just moving between floors at a
constant velocity, then the net forces on it are zero, so T = w = mg.
This is the same as if it is sitting still!
3.
You push on a block with mass m with a force F, causing it to accelerate on a
frictionless surface.
a.
Draw a complete free-body diagram for this situation. What will
the acceleration of the block be while you are pushing it?
SMDEP Physics 2009
You should have w down and N up (you can write that both are
equal to mg.) In addition you should have F in some horizontal
direction. Since w and N cancel, Fnet = F. Also draw an
acceleration vector off to the side of your diagram.
Fnet = ma
F = ma
a = F/m
b.
Now, rather than pushing on the block horizontally, you push on it
at an angle θ:
θ
It still accelerates in the horizontal direction. What will the
magnitude of the acceleration be? Draw a free-body diagram.
(Hint: the normal force will be different. How do you know that
must be the case?)
You should set up a coordinate system where the x-axis is
horizontal, since this is the axis that the block accelerates in. Draw
this acceleration vector again off to the right, horizontal, to remind
yourself that Fnet will need to be horizontal.
Decompose the force of your push into an x- and y-component and
use those as vectors on your free diagram The x-component will
be F cos θ to the right, and the y-component will be F sin θ down.
Draw the weight w and the normal force N as before, but think a
minute before setting both to mg. We know that w is always mg,
but now the normal force must be more; otherwise the overall
acceleration wouldn’t be horizontal (it would be into the ground!)
We can now do the math in each dimension. In the x dimension:
Fnet(x) = max
F cos θ = ma
|a| = (F cos θ) / m
How about the y dimension? The problem doesn’t ask for the
normal force, but this is how you would find it:
Fnet(y) = may
N – mg – F sin θ = may
N – mg – F sin θ = m(0)
N = mg + F sin θ
SMDEP Physics 2009
4.
A block of mass m is on a ramp which makes an angle of θ with the ground. It
accelerates down the surface of the ramp.
a.
Define the ramp (the axis of acceleration) as the x-axis. What is
the x-component of the total weight? (This is the part of the
weight that lets the block move down the ramp.) At what rate will
it accelerate?
The x-component of the weight is mg sin θ
Using F=ma, the acceleration is g sin θ down the ramp
b.
What is the y-component of the total weight?
The y-component of the weight is mg cos θ
c.
What is the block’s acceleration in the y-dimension? Use this to
find the normal force.
The block’s acceleration in the y-dimension is zero. This means
that Fnet(y) = 0.
Fnet(y) = ma
N – mg cos θ = ma
N = mg cos θ
5.
(No pulley problems on final) A sliding block on a frictionless surface with mass
M = 3.3 kg is attached by a cord strung around a pulley to a hanging block with
mass m = 2.1 kg:
a. Draw a free-body diagram for each,
including the direction of acceleration for
each. (The magnitude of the tension force T
and the acceleration vector a will be the same
on both blocks!)
b. Write an equation Fnet = ma for each block.
c. Combine the two equations. What is the
acceleration of each block? What is the tension T?
The free-body diagram for M should have w down, N up, and T to the right. Off
to the side, note that the acceleration is to the right.
The diagram for m should have w down and T up. Off to the side, note that the
acceleration is down.
These can each be in just one dimension (the dimension that each is moving in:)
Block M (sliding)
Block m (falling)
Fnet = ma
Fnet = ma
T = Ma
mg – T = ma
Combining the two, T cancels out: mg = Ma + ma
mg = a(M + m)
a = mg/(M + m)
a = (2.1 kg)(9.8 m/s/s)/(3.3 kg + 2.1 kg) = 3.81 m/s/s
SMDEP Physics 2009
T = Ma = (3.3 kg)(9.6 m/s/s) = 31.68 N
6.
What is the acceleration of a 3 kg block down a 30º plane if the coefficient of
kinetic friction is µ = 0.1? First find a formula, then plug in the numbers.
(a) The x forces should be: mg sin θ down the ramp, and a frictional force of µ(mg cos θ)
up the ramp. The net force down the ramp is mg sin θ - µ(mg cos θ).
F=ma, so a=F/m. a = (mg sin θ - µ(mg cos θ))/m = g sin θ - µg cos θ.
a = (9.8 m/s/s)(1/2) – (0.1)(9.8 m/s/s)(√3/2) = 4.05 m/s/s down the ramp
7.
You order a refrigerator weighing 1000 N and it arrives at your door in a
cardboard box. To get it to your kitchen, you tie a rope to it and pull it horizontally. You
find that to get it to move at all, you have to pull it horizontally with a force of magnitude
460 N. Once it breaks loose and starts to move, you can keep it moving at a constant
velocity with only 200 N. (a) What is the mass of the refrigerator? (b) What are the
coefficients of static and kinetic friction between the cardboard and your floor? (c) If you
pull with a force of 50 N, what is the friction force? (d – EXTRA CREDIT) It starts to
get easier to pull if you pull up at an angle. What is the optimal angle? (You may solve
this with a graphing calculator, or calculus.)
(a) w = mg
m = w/g = 1000 N / (9.8 m/s/s) = 102.04 kg
(b) Matching static friction: F = µstaticN
µstatic = F/N = 460 N / 1000 N = 0.46
(c) Matching kinetic friction: F = µkineticN
µkinetic = F/N = 200 N / 1000 N = 0.20
(d) You want to maximize the net force, which is F cos θ - µ(mg – F sin θ) to the right.
As long as you can get this equation, you are fine on the material!
You can graph this function, substituting values for everything but θ (it shouldn’t matter
which values you pick for anything besides µ.) Then graph this between θ = 0˚ and 90˚,
and find the value of θ that maximizes the net force.
Using calculus: find the derivative with respect to θ, and solve for dFnet/dθ = 0:
d/dθ = -F sin θ + µF cos θ = 0
-µ = tan θ
θ = arctan(-µ) = 11.3 degrees
8.
If a ball with mass 0.09 kg is thrown straight up in a vacuum, it reaches a height
of 10 m. If you throw it upward with the same initial velocity but in air instead of a
SMDEP Physics 2009
vacuum, the maximum height is 7.8 m because of air drag. What is the average air drag
force on the ball during its upward motion?
Draw a free-body diagram first! The ball has two forces down on it while it heads
towards its maximum height: its weight (=mg) and the air drag.
Solve for initial velocity given that the acceleration in a vacuum is just g:
vf2 = vo2 + 2aΔx
(0 m/s) 2 = vo2 + 2(9.8 m/s/s)(10 m)
vo = 14 m/s up
Now, solve for the acceleration with drag, given the smaller distance traveled in air and
the same initial velocity as in vacuum:
vf2 = vo2 + 2a’Δx
0 = (14 m/s) 2+ 2a(7.8 m) = -12.56 m/s/s
An additional 2.76 m/s/s acceleration. F = ma = (0.09 kg)(2.76 m/s/s) = 0.25 N down
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
In-Class Problem Set #4
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
g = 9.8 m/s2
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN
1. You shoot a bullet horizontally due uptown off of the deck of the Empire State
Building (a height of 381 m) at 1000 m/s. How long does it take to land? How
long would a dropped penny take to land? Where does it land? A NYC block is
about 80 m north/south. Neglect air resistance.
2. A passenger on a Ferris wheel moves in a vertical circle of radius R with a
constant speed v. Assuming that the seat remains upright during the motion, what
force does the seat exert on the passenger (i.e. how much does the passenger feel
like she weighs) at the top of the circle and at the bottom?
3. You are pushing a 2 kg book to the right across a tabletop. The coefficient of
kinetic friction between the book and the table is 0.25. How much force do you
need to apply to keep the book moving at a constant velocity?
4. Say you want to press a 2 kg block of wood against a wall to keep it from falling.
How much force do you need to apply if the coefficient of static friction between
the block and the wall is 0.3?
5. Two blocks are connected as shown. As M falls it drags m up the ramp. The
coefficient of kinetic friction between the block and the ramp is µ.
a. Draw a free body diagram for each
block (don’t do any math yet.)
b. Use Newton’s Second Law to relate
the net force on each block to its
acceleration.
c. Combine the two equations from (b)
and solve for the acceleration.
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
In-Class Problem Set #4
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
g = 9.8 m/s2
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN
1. You shoot a bullet horizontally due uptown off of the deck of the Empire State
Building (a height of 381 m) at 1000 m/s. How long does it take to land? How
long would a dropped penny take to land? Where does it land? A NYC block is
about 80 m north/south. Neglect air resistance.
It’s important to treat the x- and y- coordinates independently. There is no
acceleration in the x-direction – just a constant velocity uptown. In the y
direction, it is as if you dropped a penny.
In the y direction, find how long it takes to fall: yf = yo + vo Δt + ½aΔt2
-381 m = ½ (-9.8 m/s/s)Δt2
t = 8.82 s
In the x direction, find out how far it gets in that time:
xf = xo + xo Δt + ½aΔt2
xf = (1000 m/s)(8.82 s) = 8820 m
About 110 blocks, so 144th St (and 5th Av.. actually right in the Harlem River)
NOTE: NO CIRCULAR MOTION PROBLEMS ON THE MIDTERM:
2. A passenger on a Ferris wheel moves in a vertical circle of radius R with a
constant speed v. Assuming that the seat remains upright during the motion, what
force does the seat exert on the passenger (i.e. how much does the passenger feel
like she weighs) at the top of the circle and at the bottom?
The acceleration that she experiences both at the top and bottom is v2/R towards the
center, so the net force on her must be F = ma = mv2/R down while at the top, and
mv2/R up while at the bottom. If you draw the free body diagrams, you’ll see that
this means that she gets mv2/R less normal force than usual at the top, and mv2/R
more normal force than usual at the bottom. So (a) N = m(g - v2/R) and (b) N =
m(g + v2/R).
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
3. You are pushing a 2 kg book to the right across a tabletop. The coefficient of
kinetic friction between the book and the table is 0.25. How much force do you
need to apply to keep the book moving at a constant velocity?
Free-body diagram should have N (=mg) up, w (=mg) down, friction
(=0.25mg) left, and push to the right. What is the acceleration? 0. So the
net force must be zero. The push to the right must exactly cancel the
friction to the left, so it is 0.25mg = (0.25)(2 kg)(9.8 m/s/s) = 4.9 N.
4. Say you want to press a 2 kg block of wood against a wall to keep it from falling.
How much force do you need to apply if the coefficient of static friction between
the block and the wall is 0.3?
From a free-body diagram you will see that in order for the block to stay pinned
against the wall, the force of static friction (up!) must cancel the weight down.
And in the horizontal dimension, the normal force must equal the force with
which you push.
Fnet = ma
Fnet = 0
Fgravity - Ffriction = 0
mg = µN
mg = µ Fpush
Fpush = mg/µ = (2 kg) (9.8 m/s/s)/0.3 = 65.33 N
NO PULLEY PROBLEMS IN THE MIDTERM OR FINAL
5. Two blocks are connected as shown. As M falls it drags m up the ramp. The
coefficient of kinetic friction between the block and the ramp is µ.
a. Draw a free body diagram for each block (don’t do any math yet.)
M should have weight (=mg) down, T up.
m should have weight (=mg) decomposed into coordinates parallel and
perpendicular to the ramp: mg sin θ down parallel to the ramp, mg cos θ down
perpendicular to the ramp, N (= mg cos θ) up perpendicular to the ramp, and T up
parallel to the ramp. Also we need to add friction (=µN = µmg cos θ) down
parallel to the ramp
b. Use Newton’s Second Law to relate the net force on each block to its
acceleration.
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
In each block’s coordinate system, use the direction of acceleration as the
positive direction.
For M: Fnet = Ma
Mg – T = Ma
For m: Fnet = ma
T – µmg cos θ - mg sin θ = ma
c. Combine the two equations from (b) and solve for the acceleration.
We can eliminate T by “adding” the two equations:
Mg – µmg cos θ - mg sin θ = ma + Ma
a = g(M – µm cos θ - m sin θ) / (m + M)
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
Review for Midterm on 7/16/09
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
v = Δx/Δt
a = Δv/Δt
2
g = 9.8 m/s
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN
arad = v2/R
Solve all of these problems using symbols, and don’t plug in numbers until the very end!
This is important for several reasons. Often, variables cancel out and expressions
become much simpler. Also, since some problems rely on values plugged in from
previous problems, you might get one numeric answer wrong simply because you got an
earlier one wrong, and not because you didn’t understand. Finally, it makes it easier to
check your answers against the key (and for graders, who will award partial credit),
since you then diagnose whether you made an error in reasoning, in algebraic
manipulation, or in arithmetic.
I have a GPS device that measures its height above sea level and records it every second.
The height y vs time t is shown for a period of time that I was playing with it:
y
(5 s, 10 m)
(15 s, 5 m)
25 s
27 s
34 s
46 s
t
v
t
a
t
1. (Day 1) Draw basic v vs. t and a vs. t graphs showing the y-component of the velocity
and acceleration. Don’t worry about units, just the shapes.
2. (Day 1) What was the sign of the y-acceleration at t = 16 s?
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
3. What was the y-velocity at t = 2 s?
4. From t = 25 s until t = 27 s, the device was in free fall. How high was it at t = 26 s?
5. I was driving at 60 mi/hr up an inclined road from t = 0 s until t = 5 s. What was the
slope of the incline in degrees? (A mile is about 1.61 km.)
6. At t = 34 s, I threw the device at the same initial vertical velocity as I did at 26 s, but
this time it took longer to land because I had attached a small fan to it that blew air
straight down, providing it with a constant upward thrust force the whole time. How
strong was this force as a percentage of the device’s weight?
7. From t = 10 s until t = 15 s, I rolled the device back down the road that I drove up, on a
skateboard. (Rolling objects experience a force of rolling friction opposing their motion.)
What was the coefficient of rolling friction?
8. At t = 25 s, I had the device inside a football of mass m, which I threw up at a 45
degree angle. How far did it get horizontally (in the x-direction) before landing?
9. After collecting these data, I went to a 100 m cliff overlooking the ocean. I attached
the device, which has a mass of 1 kg, to the end of 1 m of fishing line. I used “35-pound”
fishing line, which means that the line breaks if it gets more than 35 pounds of tension on
it. (A mass of 1 kg weighs about 2.20 lbs.) I swung the mass/device in a circle above my
head, faster and faster, until it finally broke, sending it flying out into the ocean. At what
velocity does it shoot out over the ocean? Neglect gravity for right now – assume that all
of the tension goes into accelerating the device.
10. The device felt a force of gravity totaling 9.8 N while it was falling; the Earth
experienced an equal force in the opposite direction (9.8 N towards the device) as it fell.
Ignoring any other forces on the Earth, how far did the Earth move up as a result of the
mass falling? The Earth’s mass is about 5.97 x 1024 kg.
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
Review for Midterm on 7/16/09 KEY
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
v = Δx/Δt
a = Δv/Δt
2
g = 9.8 m/s
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN
arad = v2/R
Solve all of these problems using symbols, and don’t plug in numbers until the very end!
This is important for several reasons. Often, variables cancel out and expressions
become much simpler. Also, since some problems rely on values plugged in from
previous problems, you might get one numeric answer wrong simply because you got an
earlier one wrong, and not because you didn’t understand. Finally, it makes it easier to
check your answers against the key (and for graders, who will award partial credit),
since you then diagnose whether you made an error in reasoning, in algebraic
manipulation, or in arithmetic.
I have a GPS device that measures its height above sea level and records it every second.
The height y vs time t is shown for a period of time that I was playing with it:
y
(5 s, 10 m)
(15 s, 5 m)
25 s
27 s
34 s
46 s
t
v
t
a
t
1. (Day 1) Draw basic v vs. t and a vs. t graphs showing the y-component of the velocity
and acceleration. Don’t worry about units, just the shapes.
2. (Day 1) What was the sign of the y-acceleration at t = 16 s?
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
Positive. (Look at the slope of y vs. t during the time – this is the velocity. Then ask
yourself how the velocity changes.)
3. What was the y-velocity at t = 2 s?
v = dx/dt = 10 m/5 s = 2 m/s
4. From t = 25 s until t = 27 s, the device was in free fall. How high was it at t = 26 s?
a = Δv/Δt
-g = vf – v0/Δt
v0 = gΔt + vf
vf2 = vo2 + 2aΔy
vf2 = (gΔt + vf )2 - 2gΔy
Δy = (vf2 - (gΔt + vf )2)/-2g
Δy = g/2
Δy = 4.9 m
(If you replace t with 1 and vf with 0, you notice at this
point that v0 = g)
(If you replace t with 1 and vf with 0…)
5. I was driving at 60 mi/hr up an inclined road from t = 0 s until t = 5 s. What was the
slope of the incline in degrees? (A mile is about 1.61 km.)
Draw a right triangle. The vertical leg is 10 m, since that’s how far you go vertically.
You drive 60 mi/hr up the hypotenuse, and it takes you 5 s. From this you can figure out
the length of the hypotenuse, and the angle.
H = vΔt
sin θ = O/H
60 mi/hr * 1.61 km/mi * 1000 m/km 1 hr/3600 sec = 26.83 m/s
sin θ = y/vΔt
θ = arcsin(y/vΔt)
θ = arcsin(10 m/(26.83 m/s * 5 s)) = 4.27º
6. At t = 34 s, I threw the device at the same initial vertical velocity as I did at 26 s, but
this time it took longer to land because I had attached a small fan to it that blew air
straight down, providing it with a constant upward thrust force the whole time. How
strong was this force as a percentage of the device’s weight?
You can draw a free-body diagram to convince yourself that the net force on the object
will be the weight minus the unknown force.
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
Fnet = Fthrust - mg
We have to find Fnet (actually, Fthrust as a percentage of the device’s weight, so we
solve for Fthrust/mg.) using the acceleration that we observe. It loses all of its upwards
velocity in 6 s.
a = Δv/Δt = -v0/Δt
Fnet = ma
Fthrust - mg = ma
Fthrust - mg = m(-v0/Δt)
Fthrust/mg - 1 = (1/g)(-v0/Δt)
Fthrust/mg = 1 - (v0/gΔt)
Fthrust/mg = 1 - (9.8)/(9.8 * 6) = 0.83 = 83%
7. From t = 10 s until t = 15 s, I rolled the device back down the road that I drove up, on a
skateboard. (Rolling objects experience a force of rolling friction opposing their motion.)
What was the coefficient of rolling friction?
Draw a free-body diagram to convince yourself that the force downhill is mg sin θ and
the force uphill is µ mg cos θ. Draw that acceleration vector off to the side so that you
can invoke Newton’s Second Law. The acceleration is 0 (look at the graph!)
So Fnet = 0.
mg sin θ = µ mg cos θ
µ = tan θ = tan (4.27º) = 0.075
8. At t = 25 s, I had the device inside a football of mass m, which I threw up at a 45
degree angle. How far did it get horizontally (in the x-direction) before landing?
Draw a picture. You figured out the v0y in Problem 4: it’s g. Since you’re at a 45 degree
angle, v0y = v0x. And the total magnitude of the velocity is √(2) * v0y. This all follows
from the geometry.
I got from 0 to √(2) * v0y in 1 s. That means a = √(2) * v0y while I was throwing it.
F = ma = m√(2) * v0y = 13.86m N
How far did it get horizontally? We know it was in the air for 2 s, during which its
horizontal velocity was constant. So it traveled 2 s * v0x = 19.6 m.
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
NOTE: THIS IS A CIRCULAR MOTION PROBLEM, SO IT WILL NOT BE ON
THE MIDTERM, BUT MAY BE ON THE FINAL.
9. After collecting these data, I went to a 100 m cliff overlooking the ocean. I attached
the device, which has a mass of 1 kg, to the end of 1 m of fishing line. I used “35-pound”
fishing line, which means that the line breaks if it gets more than 35 pounds of tension on
it. (A mass of 1 kg weighs about 2.20 lbs.) I swung the mass/device in a circle above my
head, faster and faster, until it finally broke, sending it flying out into the ocean. At what
velocity does it shoot out over the ocean? Neglect gravity for right now – assume that all
of the tension goes into accelerating the device.
35 pounds means the tension can’t exceed 35 lbs. A kg weighs 9.8 N.
35 lbs* 9.8N/kg * 1 kg/2.20 lbs = 155.91 N
That’s how much tension I was applying when it broke. That tension was accelerating
the stuff at the end in a circle.
We know that the acceleration is v2/R. That means we were applying a force of mv2/R.
T = mv2/R
v = √(TR/m) = √(TR/m) = √(155.91 N * 1 m/1 kg) = 12.49 m/s
10. The device felt a force of gravity totaling 9.8 N while it was falling; the Earth
experienced an equal force in the opposite direction (9.8 N towards the device) as it fell.
Ignoring any other forces on the Earth, how far did the Earth move up as a result of the
mass falling? The Earth’s mass is about 5.97 x 1024 kg.
This horizontal velocity has nothing to do with the vertical falling. You drop the 1 kg
device from 100 m, which means that it exerts a force of Fg (= mg = 9.8 N) on the Earth
while it is falling. How long does it take to fall?
xf = ½aΔt2
Δt = √(2xf /a) (= 4.52 s)
What acceleration does the earth experience during that time?
Fg = mearthaearth
aearth = Fg / mearth
And the distance the Earth travels as a result of this acceleration is
Δxearth = ½aearthΔt2
Δxearth = ½ (Fg / mearth)Δt2
Δxearth = ½ (mdeviceg / mearth) (2Δxdevice /g)
Columbia University SMDEP/HCOP Physics 2009
July 8, 2009
Δxearth = (mdevice Δxdevice / mearth) = (1 kg)(100 m)/( 5.97 x 1024 kg) = 1.67 x 10-23 m
(8 orders of magnitude smaller than the diameter of a nucleus.)
Midterm KEY
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
v = Δx/Δt
a = Δv/Δt
2
g = 9.8 m/s
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN
arad = v2/R
This test should not take longer than 30 minutes, but you may use up to 90 minutes.
Attach your work on extra sheets of paper.
The velocity of a car vs. time is shown below over seven time intervals, labeled A-G:
v
(m/s)
5
0
A
B
C
D
E
F
G
t (s)
-10
1.(3 x 1 pt each)
a.
b.
c.
During which interval(s) is the car not moving? D
During which interval(s) is the car’s speed the greatest? F
During which interval(s) is the car experiencing the greatest
net force in the positive direction? G
Some people did not understand that an interval of time is BETWEEN two points
of time (i.e., between tick marks) rather than at an instant in time. If
people thought that the letters were labels for single times at ticks to their
left, then the correct answers would be:
(a) A, D, E, and unnamed time at right (b) between F and G inclusive (c)
between G and unnamed time at right, not inclusive
If people thought that the letters were labels for single times at ticks to
their right, then the correct answers would be:
(a) 0 (?), between C and D inclusive, G (b) between E and F inclusive (c)
between F and G, not inclusive
If someone was consistent in making this error, I marked it correct.
If someone mistook the graph for an x vs. t plot and the answers were right in the
context of that mistake, I only took off 2 pts instead of 3. [If you got (a) B,D,F
(b) E,G (c) between F and G]
2.
You push a book weighing 9.8 N to the right across a tabletop. The coefficient of
kinetic friction between the book and the table is 0.25. How much force do you
need to apply to keep the book accelerating to the right at a constant rate of 1
m/s²? (2 pts for correct diagram/math, 1 pt for correct answer.) (- ½ pt for wrong
units)
A complete free body diagram should show
w=mg=9.8 N down,
N=mg=9.8 N up,
Ffriction = µN = µ mg = (0.25 * 9.8) = 2.45 N to the left,
unknown push force F to the right (=3.45 N.)
Net acceleration should be drawn off to the side to show 1 m/s/s to the right.
Net force may be shown off to the side. It should not be shown as a separate
force on the free-body diagram.
Invoke Fnet = ma: the net force to the right must be 1 N. So Fpush = 3.45 N.
3.
You throw a ball horizontally off of a 30 m high building and it lands 40 m from
the foot of the building. What is the displacement between the bullet’s starting
point and the place where it hits the ground? Give a magnitude and direction. (1
pt for correct math/diagram, 1 pt for correct magnitude, 1 pt for correct angle) (½ pt for wrong units)
Displacement magnitude from Pythagoras: 50 m.
Angle to vertical is arctan(4/3) = 53.13º.
Angle to the horizontal is 36.87º. Either is OK if diagram is clear.
Many people said “Southeast” in addition to the bearing.
This is incorrect but I did not count off for it.
EXTRA CREDIT: What was the ball’s initial velocity? (1 pt) (- ½ pt for wrong units)
First, how long did everything take? Something dropped from 30 m:
xf = xo + vo Δt + ½aΔt2
-30 = -½gΔt2
Δt = sqrt(60/g) = 2.47 s
If it went 40 m horizontally in that time with a constant velocity:
v = 40 m/2.47 s = (16.17 m/s horizontal)
Columbia University SMDEP/HCOP Physics B 2009
July 13, 2009
In-Class Problem Set #5
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
p = mv
v = Δx/Δt
a = Δv/Δt
2
g = 9.8 m/s
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN
arad = v2/R
p conserved in all collisions
1.
(Review) A block sits on a steep incline of 60º to the horizon. The coefficient of
kinetic friction between the block and the incline is 0.20. What is the acceleration
of the block if (a) it is sliding down the slope, and (b) it has been given an upward
shove and is still sliding up the slope?
2.
(Review) In emergency situations with major blood loss the doctor will order the
patient placed in the Trendelberg position, in which the foot of the bed is raised to
get maximum blood flow to the brain. If the coefficient of static friction between
the typical patient and the bedsheets is 0.80, what is the maximum angle the bed
can be tilted with respect to the floor before the patient begins to slide?
3.
Two crates, one with mass 4 kg and the other with mass 6 kg, sit on the
frictionless surface of a frozen pond, connected by a light rope. A woman
wearing golf shoes (so that she can get traction on the ice) pulls horizontally on
the 6-kg crate with a force F = 50 N. (a) Draw free-body diagrams for both carts
and the woman. (b) Find the tension on the rope and the acceleration of the 6-kg
block.
4.
Drosophila melanogaster is an important model organism whose genetics allowed
Thomas Hunt Morgan, while working at Columbia, to win a Nobel Prize for
creating the first genetic maps (based on recombination rates.) An average
Drosophila fly weighs 10 mg. You are rollerblading in Central Park at 16 km/h
when a hovering fly hits your face, sticking to you and dying immediately. (a)
The fly applies the same force to you as you apply to the fly; why does it kill the
fly but not you? (b) You weigh 60 kg. What are the momenta of you and the fly
both before and after the collision? (c) How much do you slow down as a result of
the collision?
Columbia University SMDEP/HCOP Physics B 2009
July 13, 2009
In-Class Problem Set #5
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
p = mv
1.
v = Δx/Δt
a = Δv/Δt
2
g = 9.8 m/s
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN
arad = v2/R
p conserved in all collisions
(Review) A block sits on a steep incline of 60º to the horizon. The coefficient of
kinetic friction between the block and the incline is 0.20. What is the acceleration
of the block if (a) it is sliding down the slope, and (b) it has been given an upward
shove and is still sliding up the slope?
In both cases friction opposes the motion; that means that in (a) it goes against
gravity, and in (b) it goes with gravity!
In (a), the forces parallel to the ramp are: mg sin θ down the ramp, and µ mg cos
θ up the ramp.
Fnet = mg sin θ - µ mg cos θ
a = Fnet / m
a = g(sin θ - µ cos θ) = (9.8 m/s/s)(√3/2 – (0.20)(1/2))
a = 7.65 m/s2
In (b), the forces parallel to the ramp are: mg sin θ down the ramp, and µ mg cos
θ down the ramp.
Fnet = mg sin θ + µ mg cos θ
a = Fnet / m
a = g(sin θ + µ cos θ) = (9.8 m/s/s)(√3/2 + (0.20)(1/2))
a = 9.47 m/s2
2.
(Review) In emergency situations with major blood loss the doctor will order the
patient placed in the Trendelberg position, in which the foot of the bed is raised to
get maximum blood flow to the brain. If the coefficient of static friction between
the typical patient and the bedsheets is 0.80, what is the maximum angle the bed
can be tilted with respect to the floor before the patient begins to slide?
In this case the force downhill parallel to the bed, mg sin θ, just matches the static
friction uphill parallel to the bed, µ mg cos θ.
Columbia University SMDEP/HCOP Physics B 2009
July 13, 2009
mg sin θ = µ mg cos θ
µ = sin θ / cos θ = tan θ
θ = arctan µ ≈ 38.66°
3.
Two crates, one with mass 4 kg and the other with mass 6 kg, sit on the
frictionless surface of a frozen pond, connected by a light rope. A woman
wearing golf shoes (so that she can get traction on the ice) pulls horizontally on
the 6-kg crate with a force F = 50 N. (a) Draw free-body diagrams for both carts
and the woman. (b) Find the tension on the rope and the acceleration of the 6-kg
block.
There are actually two ropes: the first is between the woman and the first crate (we
already know this – it’s 50N), and the second is between the two crates. Let’s call
the tension on the first T1 and on the second T2.
The woman feels T1 towards the blocks. Since she doesn’t accelerate, we know
her net force is 0, so the friction of her shoes is in equal magnitude in the other
direction and cancels it out.
The 6-kg block feels T1 towards the woman but also T2 back towards the other
block, and the 4-kg block just feels T2 towards the other block. The acceleration
of the two blocks is the same (this does NOT mean that the net force on both is
the same, though! Why?)
The net force on the 6kg block is T1-T2, so the acceleration is (T1-T2)/6. The net
force on the 4 kg block is T2, so its acceleration is (T2)/4.
a1 = a2
(T1-T2)/6 = (T2)/4
(50 N – T2)/(6 kg) = (T2)/(4 kg)
(25/3) – (1/6)T2 = (1/4)T2
T2 = 20 N
a1 = a2 = 5 m/s/s
4.
Drosophila melanogaster is an important model organism whose genetics allowed
Thomas Hunt Morgan, while working at Columbia, to win a Nobel Prize for
creating the first genetic maps (based on recombination rates.) An average
Drosophila fly weighs 10 mg. You are rollerblading in Central Park at 16 km/h
when a hovering fly hits your face, sticking to you and dying immediately. (a)
The fly applies the same force to you as you apply to the fly; why does it kill the
fly but not you? (b) You weigh 60 kg. What are the momenta of you and the fly
both before and after the collision? (c) How much do you slow down as a result of
the collision?
Columbia University SMDEP/HCOP Physics B 2009
July 13, 2009
a) The same force will accelerate something less massive, like the fly, much more
than something more massive, like you. The fly experiences an overwhelming
acceleration; you are likely to barely feel it (although you will decelerate a tiny
bit.)
b)
c)
pfly = mflyvfly = (1 x 10-5 kg)(0 m/s) = 0 kg-m/s
phuman = mhumanvhuman = (60 kg)(16 m/s) = 960 kg-m/s
pfly + phuman = pafter
960 kg-m/s = pafter
960 kg-m/s = maftervafter
vafter = 960 kg-m/s / mafter = 15.999997333333778 m/s
vhuman - vafter = 2.67 x 10-6 m/s slower
Columbia University SMDEP/HCOP Physics 2009
July 15, 2009
In-Class Problems #6
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
v = Δx/Δt
a = Δv/Δt
2
g = 9.8 m/s
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN Fspring = -kx
arad = v2/R
p = mv
Emech = K + U
K = ½mv2
Ugrav = mgh
W = ΔEmech
a · b = |a| |b| cos θ
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
W = F · Δx (= FΔx cos θ)
1.
Conservation of momentum. A car of mass 1000 kg traveling at 20 m/s rear-ends
a stopped truck with mass 3000 kg. What is the momentum of the car and truck
right before the collision? If the car and truck stick together (an inelastic
collision), what is the momentum of the wreckage right after the collision? If the
car and truck bounce off each others’ bumpers (an elastic collision), what is the
total momentum of the car and truck right after the collision?
2.
Energy lost from the system during collisions. In the case of a totally elastic
collision, the total kinetic energy of the system is conserved. Otherwise, it is lost
to the environment (either to do work or to generate heat/noise/radiation.) In the
case of the car and truck sticking together afterwards, for example, energy is lost.
What is the kinetic energy of the system before and after the collision?
3.
Conservation of mechanical energy. A ball is kicked horizontally off of the roof
with a speed of 20 m/s. What is its speed when it hits the ground if the building is
10 m tall?
4.
Dot products. Evaluate: (a) <5, 1> · <0, 4> (b) (10, 330º) · (10, 0º) (c) [1 N due
SW] · [1 m due S]
5.
Conservation of mechanical energy. How much work does the force of gravity do
in pulling a 10-kg block down a 30º inclined plane of length 8 m? Use both the
force method and conservation of mechanical energy method to calculate the
velocity of the block at the bottom of the plane if it is released from rest at the top.
6.
Energy lost from the system through work. Energy can also leave the system to
do work. Assume that after the collision in question #2, all of the kinetic energy
of the wreckage is lost through friction. What is the distance that the wreckage
will skid in terms of the coefficient of kinetic friction µ?
Columbia University SMDEP/HCOP Physics 2009
July 15, 2009
In-Class Problems #6
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
x = r cos θ
y = r sin θ
r = √(x² + y²) θ = tan-1 (y/x)
v = Δx/Δt
a = Δv/Δt
2
g = 9.8 m/s
Newton’s Second Law: Fnet = ma
Fgravity = mg Ffriction = -µN Fspring = -kx
arad = v2/R
p = mv
Emech = K + U
K = ½mv2
Ugrav = mgh
W = ΔEmech
a · b = |a| |b| cos θ
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
W = F · Δx (= FΔx cos θ)
1.
Conservation of momentum. A car of mass 1000 kg traveling at 20 m/s rear-ends
a stopped truck with mass 3000 kg. What is the momentum of the car and truck
right before the collision? mv=(1000 kg * 20 m/s) = 20000 kg-m/s If the car and
truck stick together (an inelastic collision), what is the momentum of the
wreckage right after the collision? 20000 kg-m/s If the car and truck bounce off
each others’ bumpers (an elastic collision), what is the total momentum of the car
and truck right after the collision? 20000 kg-m/s
2.
Energy lost from the system during collisions. In the case of a totally elastic
collision, the total kinetic energy of the system is conserved. Otherwise, it is lost
to the environment (either to do work or to generate heat/noise/radiation.) In the
case of the car and truck sticking together afterwards, for example, energy is lost.
What is the kinetic energy of the system before and after the collision?
First use conservation of momentum to find the velocity of the wreckage.
mv = 20000 kg-m/s
(4000 kg)v = 20000 kg-m/s
v = 5 m/s
K0 = ½ mv2 = (.5)(1000 kg)(20 m/s)2 = 200000 J.
Kf =½ (4000)(5 m/s)2 = 50000 J.
150000 J lost to the environment during the collision.
3.
Conservation of mechanical energy. A ball is kicked horizontally off of the roof
with a speed of 20 m/s. What is its speed when it hits the ground if the building is
10 m tall?
E0 = Ef
Columbia University SMDEP/HCOP Physics 2009
July 15, 2009
mgh0 + ½mv02 = mghf + ½mvf2
gh0 + ½v02 = ghf + ½vf2
vf = √(2gh0 + v02- ghf)
v = √ ((2 * 9.8 m/s2 * 10 m) + (20 m/s)2) = 24.41 m/s
4.
Dot products. Evaluate: (a) <5, 1> · <0, 4> (b) (10, 330º) · (10, 0º) (c) [1 N due
SW] · [1 m due S]
(a) Draw them; the angle between the x-axis and <5, 1> is arctan(1/5) = 11.3º , so the
angle between the two vectors is the complement: 90º - 11.3º. |a| = sqrt(25 + 1) =
5.1. |a| |b| cos θ = (5.1)(4)(cos (90º - 11.3º)) = 4.
Alternately, you can just look at the part of <5, 1> that is parallel to <0, 4> - its ycomponent, 1 – and multiply that by the magnitude of <0, 4>, which is 4. 4 * 1 =
4.
(b) Draw them: the angle between them is 30º. |a| |b| cos θ = (10)(10)(cos 30º) = 86.6
(c) Draw them: the angle between them is 45º. |a| |b| cos θ = (1)(1)(cos 45º) = 0.71 J
5.
Conservation of mechanical energy. How much work does the force of gravity do
in pulling a 10-kg block down a 30º inclined plane of length 8 m? Use both the
force method and conservation of mechanical energy method to calculate the
velocity of the block at the bottom of the plane if it is released from rest at the top.
W = F · d = (mg)(d)(sin θ) = (10 kg)(9.8 m/s/s)(8 m)(1/2) = 392 N-m
W = ΔEmech = ½mv²
v = √(2W/m) = 78.4 m/s
mgh = ½mv²
v = √(2gh) = 78.4 m/s
6.
Energy lost from the system through work. Energy can also leave the system to
do work. Assume that after the collision in question #2, all of the kinetic energy
of the wreckage is lost through friction. What is the distance that the wreckage
will skid in terms of the coefficient of kinetic friction µ?
50000 J left in the system after collision
50000 J = FΔx = µmgΔx
Δx = 50000 J / µmg
Δx = 1.28/µ m
Columbia University SMDEP/HCOP Physics
July 20, 2008
Homework #4
Potentially useful info:
v = Δx/Δt
g = 9.8 m/s2
x = r cos θ
Fnet = ma
p = mv
Emech = K + U
K = ½mv2
W = ΔEmech
W = F · Δx (= FΔx cos θ)
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
r = √(x² + y²)
θ = tan-1 (y/x)
Fgrav(two objects) = GmM/r2
FE (uniform E field) = qE
FE(two objects) = kq1q2/r2
Ffriction = -µN
Ugrav(two objects) = -GmM/r
UE(uniform E field) = qEd
UE(two objects) = kq1q2/r
a = Δv/Δt
arad = v2/R
y = r sin θ
G=6.7×10-11Nm2kg-2
k =9×109 Nm2C-2
1.
Momentum. A 5-kg and 3-kg block slide towards each other on a frictionless
surface, each with a speed of 2 m/s. After they collide, the 3 kg block is going in
the opposite direction (away from the collision) at a speed of 2 m/s. What is the
velocity of the 5-kg block after the collision? Does it move towards or away from
the collision? How much energy is lost from the two-block system during the
collision?
2.
Work-energy theorem. The human heart is a powerful pump that moves about
285 L of blood daily. Assume that the work done is the work required to lift this
amount of blood to the height of an average American female (1.63 m.) A liter of
blood weighs about 0.65 kg. How much work does the heart do in a day? What
is the rate of work done in N/s? (This is the power in watts.)
3.
Work-energy theorem. You push with a force of 10 N, to the right and down at an
angle 10 degrees below horizontal, on a 5-kg box as it moves 10 m to the right
along a frictionless surface. How much work do you do on it? How fast is it
moving by the end of the push if it starts at rest?
4.
How much work would you have to do on Venus to put it into Earth’s orbit?
Mvenus = 2.87 x 1024 kg; Msun = 1.99 x 1030 kg; Dsun-venus = 1.08 x 108 km; Dsun-earth
= 1.50 x 108 km.
5.
Electrostatics. A balloon with a charge of -4.0 mC is held a distance of 1 m from
a second balloon having the same charge. Calculate the magnitude of the
repulsive force. Draw the electric force field around the balloons.
6.
Electrical potential. In the Bohr model of the atom, electrons orbit the nucleus.
Draw the proton of a hydrogen atom. Draw electric field lines and equipotential
lines. If you give a electron energy, will it move to a higher or a lower orbit?
Columbia University SMDEP/HCOP Physics
July 20, 2008
Homework #4
Potentially useful info:
v = Δx/Δt
g = 9.8 m/s2
x = r cos θ
Fnet = ma
p = mv
Emech = K + U
K = ½mv2
W = ΔEmech
W = F · Δx (= FΔx cos θ)
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
r = √(x² + y²)
θ = tan-1 (y/x)
Fgrav(two objects) = GmM/r2
FE (uniform E field) = qE
FE(two objects) = kq1q2/r2
Ffriction = -µN
1.
Ugrav(two objects) = -GmM/r
UE(uniform E field) = qEd
UE(two objects) = kq1q2/r
a = Δv/Δt
arad = v2/R
y = r sin θ
G=6.7×10-11Nm2kg-2
k =9×109 Nm2C-2
Momentum. A 5-kg and 3-kg block slide towards each other on a frictionless
surface, each with a speed of 2 m/s. After they collide, the 3 kg block is going in
the opposite direction (away from the collision) at a speed of 2 m/s. What is the
velocity of the 5-kg block after the collision? Does it move towards or away from
the collision? How much energy is lost from the two-block system during the
collision?
Let's draw our axis so that the 5kg is on the left and starts out moving to the right
(positive) direction, and the 3kg is on the right and moving to the left (negative.)
mvbefore = mvafter
(5 kg)(2 m/s) + (3 kg)(-2 m/s) = (5 kg)v + (3 kg)(2 m/s)
10 kg-m/s - 6 kg-m/s = 5v kg-m/s + 6 kg-m/s
v = -2.5 m/s or 2.5 m/s to the left
2.
Work-energy theorem. The human heart is a powerful pump that moves about
285 L of blood daily. Assume that the work done is the work required to lift this
amount of blood to the height of an average American female (1.63 m.) A liter of
blood weighs about 0.65 kg. How much work does the heart do in a day? What
is the rate of work done in N/s? (This is the power in watts.)
W = Fd
W = (285 L)(0.65 kg/L)(1.63 m)
It's ok to just multiply since the vectors are parallel.
W = 301.96 J
P = (301.96 J/day)(1 day / 60 * 60 * 24 sec) = 0.003 W
(NOTE: this was taken from a textbook - but it turns out that the power of
the heart is on the order of several W. This is a bad example and that's
Columbia University SMDEP/HCOP Physics
July 20, 2008
why we skipped it in class.)
3.
Work-energy theorem. You push with a force of 10 N, to the right and down at an
angle 10 degrees below horizontal, on a 5-kg box as it moves 10 m to the right
along a frictionless surface. How much work do you do on it? How fast is it
moving by the end of the push if it starts at rest?
W=F⋅d
W = (10 N)(10 m)(cos 10°) = 98.5 J
K = (1/2)mv2
v = sqrt(2K/m)
v = sqrt(2 * 98.5 J / 5 kg)
v = 6.3 m/s
4.
How much work would you have to do on Venus to put it into Earth’s orbit?
Mvenus = 2.87 x 1024 kg; Msun = 1.99 x 1030 kg; Dsun-venus = 1.08 x 108 km; Dsun-earth
= 1.50 x 108 km.
U of Venus in Earth’s orbit minus U of Venus in its own orbit
-GMvenusMsun/Dsun-earth – (-GMvenusMsun/Dsun-venus)
= 9.92 x 1035 J
5.
Electrostatics. A balloon with a charge of -4.0 mC is held a distance of 1 m from
a second balloon having the same charge. Calculate the magnitude of the
repulsive force. Draw the electric force field around the balloons.
F = kq1q2/r2
F = (9 x 109 Nm2C-2)(-.004)(-.004)/1 m = 144000 N
Force field should consist of vector which point to whichever balloon is closer to
the vector; along the perpendicular bisector of the line between the two balloons,
vectors should be pointing straight to the midpoint
6.
Electrical potential. In the Bohr model of the atom, electrons orbit the nucleus.
Draw the proton of a hydrogen atom. Draw electric field lines and equipotential
lines. If you give a electron energy, will it move to a higher or a lower orbit?
Field lines should point away from the proton and get weaker farther away.
Equipotential lines should be concentric circles that get farther apart farther away.
Giving an electron energy moves it to a higher orbit; a finite amount of energy
(the ionization energy) is required to move it to an infinite distance. The
ionization energy is analogous to the escape velocity.
Columbia University SMDEP/HCOP Physics 2009
July 20, 2009
In-Class Problems #7
1.
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
r = √(x² + y²)
θ = tan-1 (y/x)
v = Δx/Δt
g = 9.8 m/s2
x = r cos θ
Fnet = ma
p = mv
Emech = K + U
K = ½mv2
Ugrav(surf) = mgh
Ugrav(Univ) = -GmM/r
Uspring = ½kx2
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
W = ΔEmech
W = F · Δx (= FΔx cos θ)
Fgrav (uniform G field at surface of earth) = mg
FE (uniform E field) = qE
Ffriction = -µN
Fgrav(two objects) = GmM/r2 G=6.7×1011m3kg-1s-2
FE(two objects) = kq1q2/r2
k =9×109 Nm2C-2
a = Δv/Δt
arad = v2/R
y = r sin θ
New material: equipotential lines. Below is a drawing of a 3-story office
building. The first-story windows are 10 m high, the second-story windows are
20 m, the third-story windows are 30, and the roof is 40 m high.
a. Draw vectors representing the gravitational field
in this picture. These should be vectors pointing
down, anywhere on the diagram, with the same
300 J/kg
magnitude.
b. If I drop a 1-kg book from the third to the first
200 J/kg
floor, how much kinetic energy does it gain due to
gravity? ΔK = ΔU = Δmgh = mg(Δh) = (1kg)(9.8
m/s/s)(20 m) = 196 J
100 J/kg
c. What is the potential energy of the book on the
roof (relative to the ground?) mgh = (1kg)(9.8
m/s/s)(40 m) = 392 J Indicate on the drawing all of
the places where a 1-kg mass would have a
potential energy of 400 J. Also indicate on the
drawing places where the gravitational potential is
300 J/kg, 200 J/kg, and 100 J/kg. See diagram
These are called equipotential lines.
d. Imagine you are playing air hockey on the second floor. You hit the 1-gram puck and
1 sec passes before your partner hits it back. While it is gliding, what is the net force on
the puck? 0 N Does its kinetic energy change? No. What is the force of gravity on the
puck? mg = (0.001 kg)(9.8 m/s/s) = 9.8x10-3 N How much work does gravity do on the
puck? None; F (dot) d = 0, since they are perpendicular
400 J/kg
Columbia University SMDEP/HCOP Physics 2009
July 20, 2009
2.
New material: equipotential lines. Now we zoom out of the previous picture,
until we can see the whole Earth (and the moon!)
a. Draw the gravitational field. In contrast to the last
picture, show that the gravitational force gets weaker
as you get farther away. Draw a force line on the
moon and indicate its strength (G=6.7×1011m3kg-1s-2,
m
MEarth=6.0×1024m3kg-1s-2, mmoon=7.3×1022kg, and the distance
between their centers is 3.8×108m.) These should all point
towards the earth, but farther away from the earth, they should
be smaller. The force on the moon should be GmM/r2 =
2.03x1042 N
b. Astrophysicists use a strange convention to refer to
gravitational potential energy: the potential energy of
an object is zero at a distance of infinity from the
gravitational source, so all objects have a negative
gravitational potential. It is given by our formula
Ugrav(two obj) = -GmM/r. What is the moon’s gravitational
potential energy? -GmM/r = -7.7x1050 J
c. How much more potential energy does a 1-kg book
have at the moon’s height than at the surface of the
earth? (Use -GmM/r and the fact that the Earth’s
radius is 6.4×106m.) Label an equipotential line at the
moon’s height and give its value in J/kg. 6.18x1029
J/kg, should be a circle around the Earth passing
through the moon.
d. If you kept drawing equipotential lines at equal
J/kg intervals like you did in 1c, would they be
equally spaced on this diagram? No; they would get
farther apart farther away from the Earth.
E
4.
Electrostatics. A balloon with a charge of 4.0 µC is held a distance of 0.70 m
from a second balloon having the same charge. Calculate the magnitude of the
repulsive force. Draw electric field lines showing how a positive charge would
move in the space between the balloons.
FE(two objects) = kq1q2/r2
-6
k =9×109 Nm2C-2
F = k(4.0 x 10 C)(4.0 x 10-6 C) / (0.70 m)2 = 0.29 N
Field lines should be away from both balloons. If the balloons are drawn on the
positive and negative x-axis, there should be charges going straight up on the
positive y axis and straight down on the negative y axis.
Columbia University SMDEP/HCOP Physics
July 22, 2009
In-Class Problems #8
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
g = 9.8 m/s2
arad = v2/R
x = r cos θ
y = r sin θ
θ = tan-1 (y/x)
Fnet = ma
|a × b| = |a| |b| sin θ
p = mv
Emech = K + U
K = ½mv2
W = ΔEmech
W = F · Δx (= FΔx cos θ)
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
r (aka |a|)= √(x² + y²)
a · b = |a| |b| cos θ
Fgrav(two objects) = GmM/r2
FE (uniform E field) = qE
FE(two objects) = kq1q2/r2
Ffriction = -µN
FB(uniform B field) = qv × B
Ugrav(two objects) = -GmM/r
UE(uniform E field) = qEd
UE(two objects) = kq1q2/r
G=6.7×10-11Nm2kg-2
k =9×109 Nm2C-2
1. Electrostatics. A positively charged object with a charge of +85 nC is being used
to balance the downward force of gravity on a 1.8-gram balloon which has a
charge of -63 nC. How high above the balloon must the object be held in order to
balance the balloon?
2. Electrical force. Two electrons are fixed on a two-dimensional plane at (-3 Å, -5
Å) and at (4 Å, 4 Å). An electron is then placed at the origin. What force does it
feel (magnitude and direction?) qelectron = 1.6 x 10-19 C; one ångström = 1.0 x 10-10
m.
3. Evaluate the following cross products and report the answer in <x,y,z> notation:
(a) <0, 1, 0> × <1, 0, 0> (b) (r = 10, θ = 20°) × (r = 7, θ = 265°)
4. Force felt in an electric and magnetic field. A particle with positive charge of 3 C
moves upward at a speed of 10 m/s. It passes simultaneously through a magnetic
field of 0.2 T directed into the page and an electric field of 2 N/C directed to the
right. How is the motion of the particle affected? Draw a free-body diagram.
Columbia University SMDEP/HCOP Physics
July 22, 2009
In-Class Problems #8
Potentially useful info:
v = Δx/Δt
a = Δv/Δt
g = 9.8 m/s2
arad = v2/R
x = r cos θ
y = r sin θ
θ = tan-1 (y/x)
Fnet = ma
|a × b| = |a| |b| sin θ
p = mv
Emech = K + U
K = ½mv2
W = ΔEmech
W = F · Δx (= FΔx cos θ)
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
r (aka |a|)= √(x² + y²)
a · b = |a| |b| cos θ
Fgrav(two objects) = GmM/r2
FE (uniform E field) = qE
FE(two objects) = kq1q2/r2
Ffriction = -µN
FB(uniform B field) = qv × B
Ugrav(two objects) = -GmM/r
UE(uniform E field) = qEd
UE(two objects) = kq1q2/r
G=6.7×10-11Nm2kg-2
k =9×109 Nm2C-2
1. Electrostatics. A positively charged object with a charge of +85 nC is being used
to balance the downward force of gravity on a 1.8-gram balloon which has a
charge of -63 nC. How high above the balloon must the object be held in order to
balance the balloon?
The downward force of gravity is
Fweight = mg = (0.0018 kg)(9.8 m/s2) = 0.017 N
The upward force from the positively charged object must equal this. Solve for r.
Fweight = FE
Fweight = kq1q2/r2
r = √(kq1q2/Fweight) = √(kq1q2/mg)
r = 0.052 m = 5.2 cm
2. Electrical force. Two electrons are fixed on a two-dimensional plane at (-3 Å, -5
Å) and at (4 Å, 4 Å). An electron is then placed at the origin. What force does it
feel (magnitude and direction?) qelectron = 1.6 x 10-19 C; one ångström = 1.0 x 10-10
m.
The force from the first electron comes from the third quadrant and pushes the
electron towards the first quadrant. The angle of the first electron is
arctan(-5/-3), whose solution in the first quadrant is 59.1° - this is the
direction of the force. The angle of the second electron is clearly 45° - if
you need convincing, arctan(4/4) = 45°. That means that the force is
pushing it towards the third quadrant at an angle of 225°.
Columbia University SMDEP/HCOP Physics
July 22, 2009
The magnitudes of the two forces are given by kq1q2/r2 for each.
For the first, F = (9 x 109 Nm2C-2)(1.6 x 10-19 C)(1.6 x 10-19 C)/((-3 x 10-10 m)2 +
(-5 x 10-10 m)2) = 6.8 x 10-10 N at an angle of 59.1°
For the second, F = (9 x 109 Nm2C-2)(1.6 x 10-19 C)(1.6 x 10-19 C)/((4 x 10-10 m)2
+ (4 x 10-10 m)2) = 7.2 x 10-10 N at an angle of 225°.
Decompose each force into x- and y-components in order to add them:
<6.8 x 10-10 cos 59.1°, 6.8 x 10-10 N sin 59.1°>
<7.2 x 10-10 cos 225°, 7.2 x 10-10 N sin 225°>
+_____________________________________
<-1.6 x 10-10, 7.4 x 10-11>
Magnitude = sqrt(x2 + y2) = 1.8 x 10-10 N = 180 pN
Direction = arctan(y/x) = 155.2°
3. Evaluate the following cross products and report the answer in <x,y,z> notation:
(a) <0, 1, 0> × <1, 0, 0> (b) (r = 10, θ = 20°) × (r = 7, θ = 265°)
(a) <0, 0, -1>.
(b) Magnitude is |a||b| sin θ = (7)(10) sin 65° = 63.4. Direction is in the positive z
direction. So <0, 0, 63.4>
4. Force felt in an electric and magnetic field. A particle with positive charge of 3 C
moves upward at a speed of 10 m/s. It passes simultaneously through a magnetic
field of 0.2 T directed into the page and an electric field of 2 N/C directed to the
right. How is the motion of the particle affected? Draw a free-body diagram.
FB = qv x B = (3 C)(10 m/s up) x (0.2 T into page) = 6 N to the left
FE = qE = (3 C)(2 N/C to the right) = 6 N to the right
No net force; particle keeps moving up at 10 m/s
Columbia University SMDEP/HCOP Physics 2009
July 27, 2009
Review for Final Exam
Potentially useful info:
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
r = √(x² + y²)
θ = tan-1 (y/x)
a · b = |a| |b| cos θ
v = Δx/Δt
a = Δv/Δt
g = 9.8 m/s2
arad = v2/R
x = r cos θ
y = r sin θ
Fnet = ma
|a × b| = |a| |b| sin θ
p = mv
Emech = K + U
K = ½mv2
W = ΔEmech
W = F · Δx (= FΔx cos θ)
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
Power = W/t
Fspring = -kx
Fgrav (uniform G field at surface of earth) = mg
Fgrav(two objects) = GmM/r2
FE (uniform E field) = qE
FE(two objects) = kq1q2/r2
FB (uniform B field) = qv × B
E = hf
h=6.6x10-34 J-s
Uspring = ½kx2
k = spring constant
Ugrav(uniform G field at surface of earth) = mgh
Ugrav(two objects) = -GmM/r
G=6.7×10-11Nm2kg-2
UE(uniform E field) = qEd
UE(two objects) = kq1q2/r
k =9×109 Nm2C-2
Ffriction = -µN
v = λf
c = 3x108 m/s
Readings for each section (these links are clickable if you view this in Word):
UNIT 1: Kinematics
http://www.physicsclassroom.com/Class/1DKin/
http://www.physicsclassroom.com/Class/vectors/
UNIT 2: Force
http://www.physicsclassroom.com/Class/newtlaws/
UNIT 3: Work-Energy
http://www.physicsclassroom.com/Class/energy/
http://www.sparknotes.com/testprep/books/sat2/physics/chapter4section6.rhtml
UNIT 4: Collisions
http://www.physicsclassroom.com/Class/momentum/
UNIT 5: Fields
http://www.physicsclassroom.com/Class/circles/ Lesson 3
http://www.physicsclassroom.com/Class/estatics/ Lesson 3 and 4
http://www.physicsclassroom.com/Class/circuits/ Lesson 1
http://www.sparknotes.com/testprep/books/sat2/physics/chapter15.rhtml Lessons 1, 2, 4
UNIT 6: Waves
http://www.physicsclassroom.com/Class/waves/u10l2e.cfm
http://www.sparknotes.com/testprep/books/sat2/physics/chapter19section3.rhtml
http://www.physicsclassroom.com/Class/light/u12l2a.cfm
http://www.epa.gov/rpdweb00/understand/ionize_nonionize.html
Columbia University SMDEP/HCOP Physics 2009
July 27, 2009
1. (Unit 1) A snowball rolls off a barn roof that slopes downward at an angle of 40º to
the horizontal. The edge of the roof is 14.0 m above the ground, and the snowball has a
speed of 7.00 m/s as it rolls off the roof. Air resistance may be ignored. How far from
the base of the barn does the snowball strike the ground if it doesn’t strike anything else
while falling?
2. (Unit 2) You take an elevator from the first floor of Hammer to the seventh floor. The
elevator starts from rest and speeds up until about the second floor, then travels at a
constant speed, then decelerates once it gets to about the fifth floor. Draw a free-body
diagram of yourself as the elevator decelerates. Show the relative magnitudes of the
vectors and your net acceleration and force.
3. (Unit 2) A waitress shoves a ketchup bottle with mass 0.45 kg toward the right along a
smooth, level lunch counter. As the bottle leaves her hand, it has an initial velocity of 2.8
m/s. As it slides, it slows down because of the constant horizontal friction force exerted
on it by the counter top. It slides a distance of 1.0 m before coming to rest. What is the
magnitude of the friction force?
4. (Unit 2) Two blocks, one on the left with mass x kg and one on the right with y kg, are
contacting each other side-to-side on a frictionless surface. You push on the left block
with a force of F Newtons, and both blocks accelerate as a result. Draw a free body
diagram of the left block.
5. (Unit 3) Calculate the dot product <3, 5> · <10, 0>.
6. (Unit 3) Two tugboats pull a disabled supertanker. Each tug exerts a constant force of
1.50 x 106 N, one 16º north of west and the other 16º south of west, as they pull the
tanker 0.65 km toward the west. What is the total work that they do on the supertanker?
7. (Unit 4) Suppose you catch a baseball, and then someone invites you to catch a
bowling ball with either the same momentum or the same kinetic energy as the baseball.
Which would you choose? Explain.
8. (Unit 4) A small compact car with a mass of 1000 kg travels north into an intersection
at 15 m/s. It collides with an SUV with a mass of 2000 kg, which was traveling east
through the intersection at 10 m/s. Nobody is hurt. The wreckage moves as one mass
afterwards. What is its velocity after the wreck? What fraction of the kinetic energy is
lost during the collision? (Hint: the north-south and east-west momentum of the system
both stay the same.)
9. (Unit 5) Your mass is 70 kg. What is your gravitational potential energy due to the
Earth, (a) relative to the Earth’s surface? (b) Relative to infinitely far from the Earth? (c)
How much energy would you have to gain right now to leave the Earth and never fall
back? (d) What velocity would you have to have? MEarth=6.0×1024m3kg-1s-2 and rEarth =
6.4×106m.
Columbia University SMDEP/HCOP Physics 2009
July 27, 2009
10. (Unit 5) An electric current runs up the left side of this page. To the left of this
question, draw the magnetic field that would be induced in this part of the paper. If you
dropped a proton straight down onto the page and it started feeling the magnetic field as it
hit the page, what direction would the magnetism push it?
11. (Unit 5) Calculate the cross product <1, 1> x <0, 1>.
12. (Unit 5-6) The following is a diagram of the electron transport chain in chloroplasts.
P680 and P700 are so-named because they are best at absorbing 680-nm and 700-nm
light, respectively. (a) Which type of light has more energy per photon? (b) Which
molecule would be more likely to reduce another molecule: excited P680* or unexcited
P700? (c) If an electron in this chain were removed from the cell and put into outer space,
would it have more or less energy? (d) These wavelengths both appear red. Which is
closer to being infrared?
13. (Unit 6) Medical ultrasound makes use of sound with a frequency on the order of 2
MHz. The speed of sound in body tissue is 1500 m/s. What is the distance between
compressions induced by ultrasound in the body?
14 (Unit 6) Describe each of the following as either ionizing or non-ionizing: (a) radio
waves (b) gamma rays (c) yellow light (d) microwaves
Columbia University SMDEP/HCOP Physics 2009
July 27, 2009
KEY
1. The initial x-velocity is v cos θ and the y-velocity is –v sin θ. How long does it take to
drop 14 m? yf = y0 + vy0t + ½ at2 ; 0 m = 14 m – (7 m/s sin (40º))t + ½ (-9.8 m/s/s)t².
Positive quadratic solution for t = 1.29 s. How far does it move in the x-direction in that
time? (7 m/s cos (40º))*1.29 s = 6.92 m.
2. The forces on you are the gravity down and normal force up. Your net acceleration is
down so your net force must be down. So the normal force is less than gravity. This is
consistent with the fact that you feel less heavy.
3. vf2 = vo2 + 2aΔx; Solve for a first. a = -3.9 m/s/s. Then from F=ma, F = -1.8 N. The
magnitude is 1.8 N.
4. What is the net force on the left block? The blocks together accelerate as if they were
one object of (x + y) kg. So the acceleration of both blocks is a = F/m = F/(x + y). Fnet on
left block = mleft blocka = xa = xF/(x+y). So Fright block on left block = xF/(x+y) – F. Your diagram
should have the weight xg, the normal force xg, a push to to the right of F, and a push to
the left of xF/(x+y) – F.
5. You can do |a||b| cos θ, where |a| = sqrt(34), |b| = 10, and cos θ = 3/sqrt(34). This
comes out to 30. You can also say, “which component of a is parallel to b?” and then
simply multiply the magnitude of the component of a and the magnitude of b. The
component of a parallel to b is the x-component: 3. 3 * 10 = 30.
6. The westerly components of the two forces are each 1.50 x 106 N * (cos 16º), so
altogether it is a force of 2.88 x 106 N west. Luckily, the force is parallel to the
displacement already, so the work done is 2.88 x 106 N * 650 m = 1.87 x 109 J.
7. Same momentum (try with some numbers.)
8. Px = 20 x 104 kg-m/s before and after. Py = 1.5 x 104 kg-m/s before and after. Total P
vector is 2.5 x 104 kg-m/s at 37º N of W. When the mass becomes 3000 m/s moving all
together, V = P/M = 8.3 m/s, 37º N of W.
9. (a) 0 (you’re on the surface.) (b) Ugrav(two objects) = -GmM/r = (-6.7×10-11 )(10 kg)(
6.0×1024m3kg-1s-2 )/( 6.4×106m) = -6.28 × 108 J. (c) 6.28 × 108 J (d) ½ mv2 = 6.28 × 108
J; v = sqrt(2 * 6.28 × 108 J / 10 kg) = 1.12 x 104 m/s. This is the escape velocity for any
mass.
10. B is into the paper. A proton entering this field with a velocity into the paper will not
feel a force from B because it is parallel to its velocity.
11. The result is in the positive z direction only. The magnitude is |a||b| sin θ =
(sqrt(2))(1)(sin 45º) = 1. So the answer is <0, 0, 1>.
12. 680-nm has more energy. (b) excited P680* because the electrons are higher energy
(easier for them to leave.) (c) More energy (=0.) (d) Infrared is lower frequency, longer
wavelength, so 700-nm.
13. v = λf. (1500 m/s) = λ(2 x 106 s-1 ). λ = 7.5 x 10-4 m.
14. (a) non-ionizing (b) ionizing (c) non-ionizing (d) non-ionizing
Columbia University SMDEP/HCOP Physics 2009
July 29, 2009
Name:________________
Final Exam
Potentially useful info:
v = Δx/Δt
g = 9.8 m/s2
x = r cos θ
Fnet = ma
xf = xo + vo Δt + ½aΔt2
vf2 = vo2 + 2aΔx
r = √(x² + y²)
θ = tan-1 (y/x)
a ⋅ b = |a| |b| cos θ
p = mv
Emech = K + U
K = ½mv2
W = ΔEmech
W = F · Δx (= FΔx cos θ)
Fgrav (uniform G field at surface of earth) = mg
Fgrav(two objects) = GmM/r2
FE (uniform E field) = qE
FE(two objects) = kq1q2/r2
FB (uniform B field) = qv × B
a = Δv/Δt
y = r sin θ
p conserved when no external forces
p conserved in all collisions
E conserved unless there is flow of work/heat
E only conserved in elastic collisions
Ugrav(uniform G field at surface of earth) = mgh
Ugrav(two objects) = -GmM/r
G=6.7×10-11Nm2kg-2
UE(uniform E field) = qEd
UE(two objects) = kq1q2/r
k =9×109 Nm2C-2
Ffriction = -µN
E = hf
h=6.6x10-34 J-s
v = λf
c = 3x108 m/s
Directions: Please answer all questions here, on the front page. There are a total of 13
numeric answers and three drawing answers. Please use units in your answer where
appropriate.
1. (a)_____________
(b) Draw in the space below.
4. (a) _____________
(b) _____________
(c) _____________
(d) _____________
5. (a) Draw lines on the diagram below.
(c) _____________
2. (a) _____________
(b) _____________
(c) _____________
3. (a) _____________
(b) _____________
5. (b) _______________
6. Draw a vector here: ____________
7. _______________
Columbia University SMDEP/HCOP Physics 2009
July 29, 2009
Name:________________
1. A pool ball leaves a 1.2-meter high table with an initial horizontal velocity of 2.4 m/s.
(a) Predict the time required for the pool ball to fall to the ground.
(b) Draw a free-body diagram showing the forces acting on the ball 0.10 s after
leaving the table. Indicate the strength of the forces on your diagram.
(c) What is the total displacement between the table’s edge and the ball’s landing
location?
2. You push on a 10-kg block on a frictionless surface with a force of 10 N, directed
downwards at an angle of 60º to the horizontal, as it moves 1 m. You then let go of it
and stop applying force to it as it starts sliding over a rough surface, and it quickly
decelerates due to the friction, sliding an additional 1 m over the rough surface before it
comes to a stop.
(a) How much work do you do on the block?
(b) How fast is the block moving at the instant you stop pushing it?
(c) What is the coefficient of kinetic friction of the rough surface?
3.
Two ice hockey players from opposing teams skate directly towards each other,
each with a speed of 3 m/s. The player on the left weighs twice what the player on the
right weighs. They grab each other when they collide, so they move together after the
collision (to the right, obviously.)
(a) What is their speed after colliding?
(b) If the player on the left exerts x N on the player on the right, what force does
the player on the right exert on the player on the left? Express in terms of x.
4. Perform the following vector operations on the vectors a = <1, 2> and b = <0, 2>.
(a) Express a in terms of its magnitude (r) and direction (θ).
(b) Calculate a + b.
(c) Calculate a ⋅ b.
(d) Calculate the magnitude of a × b.
Columbia University SMDEP/HCOP Physics 2009
July 29, 2009
Name:________________
5.
Shown below is a uniform electric field, with electric field lines pointing to the
right, and three points in space, labeled A, B, and C.
A
B
C
(a) Draw equipotential lines on this diagram so that each point has an equipotential line
passing through it. Copy your lines onto the smaller version of this picture on the front
page.
(b) In which case does the electric field do more work: (1) as a proton moves from point
A to point C, or (2) as an electron moves from point B to point A? Mark your answer as
either “1” or “2” on the front page.
Columbia University SMDEP/HCOP Physics 2009
July 29, 2009
Name:________________
6. Electromagnetic waves are made up of electric fields and magnetic fields that oscillate
in intensity and are perpendicular to each other. One property of an electromagnetic
wave is its Poynting vector, named after the British physicist John Poynting (1852-1914),
and it points in the direction that the wave is traveling. He discovered that this direction
can be given by the cross product E × B, where E is the direction of the electric field and
B is the direction of the magnetic field.
If an electromagnetic wave is passing through this page and its electric (E) field is
pointing out of the page towards you and its magnetic (B) field is pointing straight down,
what direction is the wave traveling? Draw your answer as a vector on the first page.
Electric field (E)
Magnetic field (B)
7.
The following image is from a recent Science publication by Schreier et al. (2007)
in which they report the mechanism of thymine dimerization. Thymine dimers are the
most common damage caused to DNA by UV light; therefore, this is likely the chemical
mechanism that causes skin cancer. UV light at 266 nm provides the activation energy to
cause C=C double bonds of neighboring bases to react to form a cyclobutane ring (made
of single bonds.)
Assuming that each electron needs to
absorb one photon of UV light during
this reaction, what is the difference in
energy between the electrons in their
excited state (shown as a ππ*
molecular orbital in the figure) and
electrons in their ground state before
the reaction (shown as S0 in the
figure)?