Name: ___________________________ Email address (write legibly): ________________________________

AP Physics 1 Summer Assignment

Packet 3

The assignments included here are to be brought to the first day of class to be submitted. They are:

Problems from Conceptual Physics

Find the Mistake

Straightening Curves

Other materials due on the first day of class:

Topics 3, 4, and 5 from The Physics Classroom

Your list of Big Ideas

Read the instructions packet for all the details about these assignments.

Name: ___________________________

AP Physics 1 Summer Assignment – Problems From Conceptual Physics

All problems are taken directly from Conceptual Physics, 3rd

edition by Paul G. Hewitt

Use a separate page to show solutions. For all problems that need it, use g = 9.8 m/s2 for the acceleration

due to gravity.

1) Why is it that an object can accelerate while traveling at constant speed, but not at constant

velocity?

2) A ball is thrown straight up. What will be the instantaneous velocity at the top of its path? What

will be its acceleration at the top? Why are your answers different?

3) a) Find the speed required to throw a ball straight up and have it return 6.00 seconds later.

b) How high does the ball go?

4) If a salmon swims straight upward in the water fast enough to break through the surface at a speed

of 5.00 meters per second, how high can it jump above water?

5) In the absence of air resistance, why does the horizontal component of velocity for a projectile

remain constant while the vertical component changes?

6) At the instant a ball is thrown horizontally over a level range, a ball held at the side of the first is

released and drops to the ground. If air resistance is neglected, which ball strikes the ground first?

7) A projectile is fired straight up at 141 m/s. How fast is it moving at the top of its trajectory?

Suppose it is fired upward at 45.0o above the horizontal plane. How fast is it moving at the top of

its curved trajectory?

8) The force of gravity is twice as great on a 2 kg rock as on a 1 kg rock. Why does the 2 kg rock not

fall with twice the acceleration?

9) a) Calculate the acceleration if you push with a 20.0 N horizontal force on a 2.0 kg block on a

horizontal friction-free air table.

b) What acceleration occurs if the friction force is 4.0 N?

10) A horizontal force of 100 N is required to push a crate across a factory floor at a constant speed.

What is the net force acting on the crate? What is the force of friction acting on the crate?

11) A 10.0 kg mass on a horizontal friction-free air track is accelerated by a string attached to another

10.0 kg mass hanging vertically from a pulley as shown. What is the force due to gravity of the

hanging mass? What is the acceleration of the system of both masses?

12) Suppose the masses described in the preceding problem are 1.00 kg and 100.0 kg, respectively.

Compare the accelerations when they are interchanged, that is, for the case where the 1.00 kg mass

dangles over the pulley, and then for the case where the 100.0 kg mass dangles over the pulley.

What does this indicated about the maximum acceleration of such a system of masses?

13) When you walk along the floor, what pushes you along?

14) When you jump up, the world really does recoil downward. Why can’t this motion of the world

be noticed?

15) When a rifle is fired, how does the size of the force of the rifle on the bullet compare with the

force of the bullet on the rifle? How does the acceleration of the rifle compare with that of the

bullet? Defend your answer.

16) If a bicycle and a massive truck have a head-on collision, upon which vehicle is the impact force

greater? Which vehicle undergoes the greater change in its motion? Defend your answers.

17) Since the force that acts on a bullet when a gun is fired is equal and opposite to the force that acts

on the gun, does this imply a zero net force and therefore the impossibility of an accelerating

bullet? Explain.

18) A bug and the windshield of a fast-moving car collide. Indicate whether each of the following

statements is true or false.

a) The forces of impact on the bug and on the car are the same size.

b) The impulses on the bug and on the car are the same size.

c) The change in speed of the bug and the car is the same.

d) The changes in momentum of the bug and of the car are the same size.

19) If a mouse and an elephant both run with the same kinetic energy, can you say which is running

faster? Explain in terms of the equation for KE.

20) A hammer falls off a rooftop and strikes the ground with a certain KE. If it fell from a roof that

was four times higher, how would its KE of impact compare? Its speed of impact? (Neglect air

resistance.)

10.0

kg

10.0

kg

Name: ___________________________

AP Physics 1 – Find the Mistake

Each problem has the correct answer in italics, and an incorrect approach to solving is shown. For each, you need

to:

describe, in words, what is wrong about the given solution (e.g., simply saying the equation is wrong is not

enough – if that is what is wrong, you must say why that equation is wrong);

and then solve it correctly. But, do NOT take a different approach. (e.g., if the original solution uses

kinematics, do not solve it using energy)

For some, more than one incorrect solution is shown. Address the mistakes in both. Throughout this worksheet,

the mistakes are never mathematical, nor are the significant figures incorrect. There is something wrong with the

physics in each situation. (For any problems involving the acceleration due to gravity, use g = 9.8 m/s2.)

1) You drop a rock from rest off a 24.0 m tall cliff. It hits the ground 2.21 s later. What is its speed as it hits the ground?

(21.7 m/s)

You solve by doing: v = d / t = 24.0 m / 2.21 s = 10.9 m/s

2) Gina is driving behind Paul, and notices a police car. To make sure she is under the speed limit, she slows down from 22.4

m/s to 19.8 m/s in 5.65 seconds. How much distance does she cover in this time? (119 m)

You solve for the acceleration by doing: a = v / t = (vf – vo) / t = (22.4 m/s – 19.8 m/s) / 5.65 s = 0.460 m/s2

Then, you solve for the distance by doing: d = vot + ½ at2 = (22.4 m/s)(5.65 s) + ½(0.460 m/s

2)(5.65 s)

2 = 134 m

3) Robin Hood shoots an arrow horizontally with an initial speed of 38 m/s from a height of 1.8 m. Assuming it does not hit

anything along the way, how much time does it take to land, and how far away does it hit the ground from where Robin is

standing? (t = 0.61 s; d = 23 m)

You correctly solve for the time as 0.61 s, but then to solve for the distance it travels, you do:

d = vot + ½ at2 = (38.0 m/s)(0.61 s) + ½ (9.8 m/s

2)(0.61 s)

2 = 25 m

Find the Mistake – Page 2

4) Your friend stands on the roof of her house, which is 12.2 m off the ground, and kicks a soccer ball to you so that it is

initially moving horizontally at 13.9 m/s. What is its speed as it hits the ground? (20.8 m/s)

You correctly solve for the time as 1.58 s, but then to solve for the speed, you do:

vf = vo + at = 0 m/s + (9.8 m/s2)(1.58 s) = 15.5 m/s

5) As shown to the right, a 6.00 kg mass is pulled

upwards by a string. It is being accelerated upwards at a

rate of 0.500 m/s2. Find the tension in the string.

(T = 61.8 N)

v a = 0.500 m/s

2

You solve by assuming the tension must be equal to the weight, so T = 58.8 N.

6.00 kg

Name: ___________________________

AP Physics 1 Summer Assignment – Straightening Curves

Graphical analysis is a huge part of your studies in physics. When data is plotted, relationships are seen

in ways that numbers alone cannot express. The shape of the graph gives information about the situation

being studied, and that shape is often manipulated in order to calculate physical quantities that describe

the event being studied.

The easiest curve to analyze is a straight line. By definition, a straight line is a curve with constant slope.

From algebra, we know that the general equation of a straight line is y = mx + b, where m is the slope

and b is the y-intercept. If the data we plot gives a straight line, it is easy for us to calculate the slope of

the line or to read the value of the y- or x-intercept – all of which may be pertinent to analyzing the

problem. We’ll start with some easy examples.

1) vf = vo + at This gives the final speed (or velocity) after an object has undergone constant

acceleration for some time. You can see that the equation maps to the equation for a line very

clearly:

vf = vo + a t

y = b + mx

It is obvious that, when vf is plotted on the vertical axis and t is plotted on the horizontal axis,

the slope will represent the acceleration and the y-intercept will represent vo. While this is an easy

example, there are some things to note. First, when a straight line is plotted, the slope must be constant

and so must the y-intercept. (If the slope were not constant, the line would not be straight, and if the y-

intercept were not constant – well, what would that even look like?) So, when looking at the equation and

choosing what to plot, you must arrange it so that whatever lines up with m and b are constant values.

For this particular situation – where an object starts at some initial speed, vo, and undergoes acceleration

– everything is fine because vo cannot change and because we assume the acceleration is uniform.

2) F = qvB It is fine if you do not know what all these terms mean – what is important is if you

can map it to the straight line equation. Let’s say that the goal is to find q, (which happens to

represent the charge on a small particle). A good idea is to pull the desired quantity out in front

so it physically appears where m, the slope, appears. Whatever is left is then plotted on the

horizontal. Even then, you have some choices:

F = q(vB) F = (qv)B

y = m x + b y = m x + b

plot F vs. vB plot F vs. B

slope = q slope = qv

Either choice is perfectly fine. In the first, you have to calculate the product of v and B to plot it

on the horizontal, but the slope is what you were seeking (q), which is nice. In the second, the

slope does not directly get you your answer. You would have to manipulate the slope to get q ,

the charge, by dividing the slope by v , the speed at which the particle travels.

Here is sample data for this experiment. The first graph is done for you to start you off.

Calculate vB and fill in the missing column on the data chart. On the second set of axes, plot F vs. vB,

draw the trend line, and calculate the slope, which should be very close to my value of q.

Data: v = 1.6 x 10

4 m/s

B = Magnetic Field (T) F = Force (N) vB (T.m/s)

0.040 2.4 x 10-3

0.055 3.7 x 10-3

0.062 4.5 x 10-3

0.080 5.4 x 10-3

0.097 6.3 x 10-3

0.11 7.4 x 10-3

Before plotting, now is a very good time to review the handout you received called,

“Expectations For Graphing,” since all those details come into play here.

F vs. B F vs. vB

slope = 6.4 x 10-3

N – 2.0 x 10-3

N

0.095 T – 0.030 T

= 0.068 N/T

slope = qv

q = slope = 0.068 N/T

v 1.6 x 104 m/s

q = 4.3 x 10-6

C

0 0.02 0.04 0.06 0.08 0.10 B (T)

0

F ( x 10-3 N)

0.8

1.6

2.4

3.2

4.0

4.8

5.6

6.4

7.2

8.0

0 vB (T.m/s)

0

F ( x 10-3 N)

0.8

1.6

2.4

3.2

4.0

4.8

5.6

6.4

7.2

8.0

Straightening Curves – Page 2

The examples so far already looked like linear functions – ones that produce a straight line, no

matter what. But not all equations are so simple. You have already been exposed to this idea

when you did the Falling Masses Activity. When you plotted d vs. t , you got a parabola. Then,

you were told to plot 2d vs. t 2, and doing so “straightened” the curve. So, you can manipulate

equations into looking like straight lines through careful selection of what you plot.

Certain types of equations pop up with enough regularity that we can look at the general form and

then match future equations to the standard examples. (For all of the equations that follow, the

variable k is simply a constant.)

Parabola: y = k x2 plot y vs. x

2

If you simply plot y vs. x, you will get a parabolic curve. But, if you want to get a value for k,

you straighten the curve by plotting y vs. x2. Simply square the x values. By doing so, you are

forcing the parabolic equation to look like a linear one:

y = k (x2)

y = m x + b (in other words, this is what you did in the Falling Masses Activity)

Root Curve: y = k x plot y vs. x or plot y2 vs. x (do you see how that works,

too?)

Hyperbolic Curve: y = k plot y vs. 1 or plot 1 vs. x

x x y

Inverse-Square Curve: y = k plot y vs. 1

x2 x

2

The shapes and their generic equations:

Parabola: y = k x2

Root Curve: y = k x

Hyperbola: y = k

x

Inverse-Square Curve: y = k

x2

(keeps increasing) (asymptotically

approaches both axes)

(asymptotically approaches both

axes; drops off more quickly than

the hyperbolic curve)

Below are sets of data. Plot each normally (y vs. x). From the shape of the curve, determine what

category it belongs to (see previous page), and then determine what you could plot to straighten the curve.

Use the blank column to calculate the manipulated value of y or x. (For example, if you get a parabola,

you calculate x2 , and then plot y vs. x

2 .) Plot a new graph, calculate the slope of the line, and use the

slope to find the constant k in the relationship. (The slope itself may be k, or you may have to

manipulate the slope, depending on what you chose to plot.) Acceptable answers for all the graphs that

follow appear on the back of “Answers to Numerical Problems from Conceptual Physics.”

Plot 1 Plot 2 Plot 3

y x y x y x

3.50 2 12.6 2 0.354 2

1.75 4 50.2 4 0.500 4

0.875 8 201 8 0.707 8

0.778 9 254 9 0.750 9

0.467 15 707 15 0.968 15

0.389 18 1020 18 1.06 18

Plot 1: y vs. x Plot 1: Straightened

0 x

0

y

0

0

Straightening Curves – Page 3

Plot 2: y vs. x

Plot 2: Straightened

Plot 3: y vs. x

Plot 3: Straightened

0 x

0

y

0

0

0 x

0

y

0

0

3) vf = 2ad This experiment would be allowing an object to slide from rest down a frictionless

ramp of constant inclination, from different heights, so that the object moves different distances d

along the ramp and then finding its speed vf at the end of the ramp. Again, what is important is if

you can map the relationship to the straight-line equation, y = mx + b.

So, what would be plotted? There are a lot of choices. Let’s say the goal of the experiment was

to measure a, the acceleration along the ramp. A good idea is to pull the desired quantity out in

front so it physically appears where the slope appears in the straight-line equation. Whatever is

left is then plotted on the horizontal.

vf = a 2d

y = m x + b

so, plot vf vs. 2d

and now your slope = a

However, other options exist by squaring both sides of the equation:

(vf 2 ) = 2a (d) (vf

2 ) = a (2d) and there are more . . .

so, plot (vf 2) vs. (d) plot (vf

2) vs. (2d)

slope = 2a slope = a

Although it should be obvious, you cannot plot (vf 2 ) vs. (a) because the point of the experiment

was to find acceleration – you do not know what a is yet. (And how would you calculate it,

anyway? The point is that you cannot just play any algebra tricks you want.)

The point is: if d was inside a square root, to straighten the curve, you need to take the square

root of it and use those values. Or, if you square the equation and you have vf 2 , then you square

the speeds and use those values.

Basically, whatever is done to the variable in the equation – DO THAT and then PLOT

THAT as one of your variables.

Sample data for this experiment follows. Blank columns are left for you to manipulate the data.

Choose two of the three methods shown above, fill in the top of the data column with what you are

plotting and its units, plot the data, calculate the slope, and find the acceleration from each. (Ideally, the

two values of acceleration that you find should match or be very close.)

For each graph, label the axes with quantity and units (e.g.: d (m) or 2d ( m ) ), and put a title

on the top of the graph that states what is being plotted. Titles for graphs are always listed as, “(Vertical)

vs. (Horizontal).” (Notice the titles of the graphs of F vs. B and F vs. vB are both listed that way.)

Straightening Curves – Page 4

Data:

vf = final speed (m/s) d = distance (m)

2.5 1.1

2.3 0.90

1.9 0.70

1.7 0.50

1.5 0.30

0.75 0.10

Method 1 Method 2

0

0

0

0

Want some more examples? Sure . . . (don’t worry about recognizing the symbols)

B = onI

E =

2o

v = 3RT

M

v = 3RT

M

I = c Bmax2

2o

fo = 1 .

2 LC

plot B vs. I

plot E vs. plot v vs. T plot v

2 vs. 3T plot I vs. Bmax

2

2o

plot fo 2 vs. 1

L

slope = on slope = 1

2o

slope = 3R

M

slope = R

M

slope = c

slope = 1 .

4C

One more for you to try:

4) W = I 2Rt In this experiment, an electrical device with resistance R is turned on for a given

amount of time. For different amounts of current I, the energy given off, W, is measured.

Sample data for this experiment follows. Your goal is to find the constant resistance in the circuit. Blank

columns are left for you to manipulate the data. Fill in the top of the column with what you are plotting

and its units, plot the data, calculate the slope, and use it to find an experimental value of the resistance in

ohms (). Make sure to label the axes with quantity and units (e.g.: I (A)), and put a title on the top of

the graph that states what is being plotted.

Data: t = 30.0 s

I = current (A) W = energy (J)

0.50 19

0.85 52

1.1 98

1.4 150

1.7 200

2.1 330

0

0

AP Physics 1 Summer Assignment

Answers to Numerical Problems from Conceptual Physics

Below, you are given the numerical answers so that you know if you are solving the problems correctly.

This in no way means that you can skip writing out the solution. For each of these, you must show work

in the proper problem-solving format if you expect to get credit.

2) v = 0 m/s; a = 9.8 m/s2 down (you must explain)

3) a) 29.4 m/s; b) 44.1 m

4) 1.28 m

7) 0 m/s; 99.7 m/s (show work)

9) a) 10 m/s2

b) 8.0 m/s2

11) F = 98.0 N; a = 4.90 m/s2

12) 0.0970 m/s2; 9.70 m/s

2 (you must explain)

20) KEnew = 4 KEold;

vnew = 2 vold (you must explain both of these results)

Answers to “Straightening Curves”

The answers that follow are the ones I got. If your answer varies a little, you probably still did it

correctly. If your answer is much different from mine, you should look for mistakes in your work.

2) Your value of q should be close to the one I calculated from the first graph.

Plot 1, straightened: k = 7

Plot 2, straightened: k = 3.1

Plot 3, straightened: k = 0.25

3) vf = 2ad (it’s a root curve if you plot vf vs. d), straightened: a = 2.9 m/s

2

4) W = I 2Rt (it’s a parabola if you plot W vs. I ), straightened: R = 2.5