Physics Tutor In Chicago-AP Physics (C) 2022 MechanicsPaper Solution

2022 AP Physics C: Mechanics Free-Response ANSWERS a ne-I

T=

mg

- >

aW =

F.dz

One-2

Impulse

=

F. A t

OnesMgh-ho" ==Fx1

v

ADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATION CONSTANTS AND CONVERSION FACTORS

Neutron mass, mn

27

kg

10

27

kg

9.11 10

31

1.67

Electron mass, me Avogadro’s number, N 0

1

8.31 J (mol K)

Boltzmann’s constant, k B

1.38

10

23

e

1.60

10

19

C

1 electron volt, 1 eV

1.60

10

19

J

Electron charge magnitude,

kg

10 23 mol

6.02

R

Universal gas constant,

10

1.67

Proton mass, m p

Speed of light, Universal gravitational constant, Acceleration due to gravity at Earth’s surface,

c

3.00

10 m s

G

6.67

10

11

1u

1.66

10

27

kg

931 MeV c 2

Planck’s constant,

h

6.63

10

34

Js

4.14

10

25

8.85 10

12

e0

Vacuum permittivity, Coulomb’s law constant, k Vacuum permeability, 1 atmosphere pressure,

PREFIXES Factor Prefix Symbol

mole, hertz, newton, pascal, joule,

1 4 pe0 m0

Magnetic constant, k

m kg s A K

kg 2

9.8 m s2

g

1 unified atomic mass unit,

meter, kilogram, second, ampere, kelvin,

N m2

J K

hc

UNIT SYMBOLS

8

m0 4 p

1 atm mol Hz N Pa J

1.99

Jm C

2

1.24 Nm

109 N m 2

9.0

7

(T m) A

1 10

7

(T m) A

1.0

105 N m 2

watt, coulomb, volt, ohm, henry,

W C V W H

15

eV s

3

10 eV nm

2

C2

10

4p

10

1.0

105 Pa

farad, tesla, degree Celsius, electron volt,

F T C eV

VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES q 0 30 37 45 53 60 90

109

giga

G

sin q

0

12

35

2 2

45

3 2

1

106

mega

M

cos q

1

3 2

45

2 2

35

12

0

103

kilo

k

tan q

0

3 3

34

1

43

3

10 2

centi

c

10 3

milli

m

6

micro

m

10 9

nano

n

12

pico

p

10

10

The following assumptions are used in this exam.

I. The frame of reference of any problem is inertial unless otherwise stated.

II. The direction of current is the direction in which positive charges would drift.

III. The electric potential is zero at an infinite distance from an isolated point charge.

IV. All batteries and meters are ideal unless otherwise stated.

V. Edge effects for the electric field of a parallel plate capacitor are negligible unless otherwise stated.

ADVANCED PLACEMENT PHYSICS C EQUATIONS MECHANICS

Ãx

x

ax t

Ãx 0

x0

1 2 at 2 x

Ãx 0 t

Ãx2

Ãx 0 2

2ax x

a

F m

Fnet m

F

dp dt

J

p

F dt

m FN

DE

W

K

1 2 mà 2

P

dE dt

P

F v

DUg

Ã2 r

r

F t net I

I 2

I

r dm

mr

kDx

Us

1 k Dx 2

x

xmax cos( wt

T

2p w

p

Iw

2p

Tp

2p

1 2 Iw 2

w

w0

at

q

q0

w0 t

FG UG

1 2 at 2

2

1 f

Ts

rw r

acceleration energy force frequency height rotational inertia impulse kinetic energy spring constant length angular momentum mass power momentum radius or distance period time potential energy velocity or speed work done on a system position coefficient of friction angle torque angular speed angular acceleration phase angle

Fs

2

mi xi mi

x cm

K

w2r

t

a

L

F dr

mg Dh

ac

Ã

Dp

mv

Ff

t

x0

a = E= F = f = h = I = J = K = k = = L = m= P = p = r = T = t = U= v = W= x = m = q = t = w = a = f =

ELECTRICITY AND MAGNETISM

m k

FE q

E

DV

DV

Q C

UC

r2

Gm1m2 r

i

i

= = = = = = = = = = = =

N =

qi ri

P Q q R r t U

= = = = = = = V= v = r = F = k =

1 q1q2 4pe0 r

k e0 A d

1 Cs

Ci

1 C i i

area magnetic field capacitance distance electric field emf force current current density inductance length number of loops of wire per unit length number of charge carriers per unit volume power charge point charge resistance radius or distance time potential or stored energy electric potential velocity or speed resistivity flux dielectric constant

FM

qv

B d

dQ dt

1 QDV 2

1 C DV 2

2

B m0 I

dB

m0 I d r 4p r 2

R

r A

F

I d

E

rJ

Bs

m0 nI

I

Nevd A

FB

I

DV R

e

E d

e

L

UL

1 2 LI 2

g

Gm1m2

1 4pe0

qV

Cp

n

E dr

UE

C

F I J L

dV dx

Ex

V

e

Q e0

E dA

I f

A B C d E

1 q1q2 4pe0 r 2

FE

Rs 1 Rp

P

Ri

i

i

I DV

1 Ri

B

B dA

dI dt

d FB dt

ADVANCED PLACEMENT PHYSICS C EQUATIONS GEOMETRY AND TRIGONOMETRY Rectangle A bh Triangle

A

1 bh 2

Circle

A

pr2

C

2p r

s rq Rectangular Solid V wh Cylinder 2

V

pr 

S

2p r 

CALCULUS

A = area C = circumference V = volume S = surface area b = base h = height  = length w = width r = radius s = arc length q = angle

df dx

d f du du dx

d n x dx

nx n

d ax e dx

aeax 1 x

d ln ax dx d sin ax dx

a cos ax

d cos ax dx

s

x n dx

r

q 2p r 2

eax dx

Sphere

dx

a sin ax 1

n

ln x

V

x

S

4p r 2

cos ax dx sin ax dx

Right Triangle

b2

c2

sin q

a c

cos q

b c

tan q

a b

a

c q

a 90°

b

xn 1, n

1

1 ax e a

4 3 pr 3

a2

1

a 1 sin ax a 1 cos ax a

VECTOR PRODUCTS   A B AB cos q   A B AB sin q

1

Begin your response to QUESTION 1 on this page.

PHYSICS C: MECHANICS SECTION II Time—45 minutes

3 Questions Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight.

Show all your work in this booklet in the spaces provided after each part.

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1. A block of mass m is pulled across a rough horizontal table by a string connected to a motor that is attached to the floor. The string passes over a pulley with negligible friction that is vertically aligned with the left edge of the table as shown. The string and pulley both have negligible mass. The pulley is at height H above the table. The motor exerts a constant force of tension FT on the string, and the block remains in contact with the table at all times as the block slides across the table from x = L to x = 0. The coefficient of kinetic friction bet ween the table and the block is mk. Express all algebraic answers in terms of m, H, FT , x, mk, L, and physical constants as appropriate.

MAN

FT

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(a) On the dot below that represents the block, draw and label the forces (not components) that act on the block when the block is at x = L. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

Tension FT =

FN

=

Normal reaction

&

Ff

Eg

=

=

Friction

frice

mg: weight

(b) Derive an expression for the angle M q that the string makes with the horizontal as a function of x.

tano-t 0=

fan"

(#)

= 1

-

(c)

Continue your response to QUESTION 1 on this page. i. Derive an expression for the normal force AN FN exerted on the block by the table as a function of the block’s position x.

tano:

FN+ FFN

=

=

31n0-8g

Fg-F> Fgotr

=

0

t

SiO-H

318

H2+32

(Huxas)

nFT

TV A

If

Eg

ii. Derive an expression for the magnitude of the net horizontal force NC FnetTexerted on the block as a function of the position x.

FNET

=

FTCD8

=F+ -

-Ff

(sGries)

FT)Prers)

Jos @ =

-

-

&

H2+22

MkFN

Mc.

(Tg-F3 ) Finter

(d) Write, but do not solve, an integral expression that could be used to solve for the work W done by the string on the block as the block moves from x = L to x = 0. an

= Fixedze

=

&= O

=

-Fdxcoso

Stti (ertHn)

are

(e) Does the string do more, less, or the same amount of work on the block as the block moves from x = L to x = L1 compared to when the block moves from x = L to x = 0 ?

W 22 = v More work when the block moves from x = L to x = L 2 ______ ______ Less work when the block moves from x = L to x = L 2 ______ The same amount of work when the block moves from x = L to x = L 2 Justify your answer.

⑪

2=

W A

To

Los

6 to

c

1/2,

=

Op 02

I

>

I

x

=

Wz

=

1

/

Fed

to

Cos

0

O2.

On

Wp > W2. more

work

when

block moves

from <=Ltox-Ly

Begin your response to QUESTION 2 on this page.

<

2. Cart 1 of massI m1 is held at rest above the bottom of an incline. Cart 2 has mass M m2, where m2 > m1, and is at rest at the bottom of the incline. At time t = 0, Cart 1 is released and then travels down the incline and smoothly transitions to the horizontal section. The center of mass of Cart 1 moves a vertical distance of H, as shown. At time k tC, Cart 1 reaches the bottom of the incline and immediately collides with and sticks to Cart 2.

After the collision, the t wo-cart system moves with constant speed v. Frictional and rotational effects are negligible. (a) During the collision, is the impulse on Cart 1 from Cart 2 greater than, less than, or equal to the magnitude of the impulse on Cart 2 from Cart 1 ? ___ Greater than -> Justify your answer.

W Equal to ___ Less than ___ F. DT- same, exert same

justify your and wer

force

on

same

Ame

Continue your response to QUESTION 2 on this page. (b) On the following axes, draw graphs of the magnitude of the momentum of each cart as a function of time t, before and after Alk tC. The collision occurs in a negligible amount of time. The grid lines on each graph are drawn to the same scale.

2 ART-2

CART-I ·

b

St

<

+c

increasing

at

t= f@ maximum

momentum

1

x

to

smaller momentum

0

<t <+c

P=O at t>

to

momentum increases

(c) Show that the velocity v of the t wo-cart system after the collision is given by the equation & ⎛m⎞ v=2g⎜ 1 ⎜H.

=> oRg ⎜ ⎜⎝ m 1 + m 2 ⎜⎠

*

pply

MI

M.+ Mz

conservation

#.gH Apply

↓f

law

=

↓

M, UF

=

=

U

of

conservation

of

M.G

Of

=

first

of every

=

2g

linear

(mitmn) Of mid,

sm.tmen) proved

=>

1eH (mitmn) my

H

momentum.

(d) A group of students use the setup to perform an experiment. They measure the mass of Cart 1 to be m1 = 0.250 kg. The mass of Cart 2 is unknown. The students perform several trials and in each trial, # Cart 1 is released from a different height H and the final velocity of the t wo-cart system is measured.

The students graph v as a function of H , as shown below. &

slope

the of

graph

=

It

=

I

=

(

120

-

(m/seb) JTn

Tot

06 0

0 T0-0.3 5

S

1.72

i Draw a line that represents the best fit to the data points shown. ii. Use the best-fit line to calculate the mass of Cart 2.

Scope=(Mitm) Rg= -

(0.25)

Mz=

2(9.8)

M.IE -

scope -

1072

0.25

=

MI

0.39

Kg.

(e) After the experiment, the students use a balance to measure the mass of Cart 2 and find it to be less than what was determined in part (d). To explain this discrepancy, one of the students proposes that the mass of Cart 1 was incorrectly measured at the beginning of the experiment. The students measure the mass of Cart 1 again and record a new value, m1′.

Should the students expect that m1′ will be greater than 0.250 kg, less than 0.250 kg, or equal to 0.250 kg ? _____m1′ > 0.250 kg Justify your answer. _____m1′ < 0.250 kg _____mai m1′ = 0.250 kg

a

-gm:

A

smaller

Me

and momentum

slope of mass

my

indicates was

height #, must

that

initial

-

be

smaller,

energy

with Identical

this means

smaller

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that

the

(a) Derive an expression for the mass ⑭MB mB of the block.

MB8

MBg

=

=

kz

el2= 8

4R(0).

=

aUk -

Radi us aNC

R

arc

=>

=

=

RCO)

R.

Mr=LDO (b) At time t = 0, the string on the right side of the disk is cut and the block falls to the ground. On the circle below, which represents the disk, draw and label the forces (not components) that act on the disk immediately after the string is cut and the block is falling to the ground. Each force should be represented by an arrow that starts on and is directed away from the point of application.

Facele

/

c

string force

Fo

Fg-gravity

(c) Derive an expression for the angular acceleration a of the disk immediately after the string is cut.

Take

about

Torque

Fs (R)

=

1

Point P

Is

(a) 2=

=

RCO)

&

3R P

-M9R"(a)

2K(5(0) (M =

Nd

a

2

FAXLE

=> se

Fs=1(k) K(R) (0)(R)

A

KO

=

it d

4

v

mg

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(e) The disk is adjusted on the support so that the axle does not pass through the center of mass of the disk. The block is again hung on the right side of the disk and the spring-disk-block system comes to equilibrium, as shown in Figure 3. The axle does not exert a torque on the disk. For each force on the disk, indicate whether the magnitude of the torque about the axle caused by that force increases, decreases, or stays the same relative to part (b).

IN

Torque Hence

due to

equilibrium to

Toveve

due

magnitude of toreve we

spring force -Forewe

to make

-

mains same

due to

gravity force Toveve

+

&

sbuing therecreases

due to

tension

due to block

-

↓

Increases

Fg -

magnitude of foreue

due to

gravity

Increases

AP® Physics C: Mechanics 2022 Free-Response Questions

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