AP® Physics C: Mechanics Free-Response Questions Set 1 ANSWERS 2019 & we-I a = wo, F= ma mVzQu-2c=p M. U1 + Mc z = M, Vit McVz mgh=I mot Oue-3 WE Not a t = oCFW1ETECC I W
CONSTANTS AND CONVERSION FACTORS
Proton mass, 271.6710 kgmp
Neutron mass, 271.6710 kgmn Electron mass, 319.1110 kgme
Avogadro’s number, 231 0 6.0210 molN
Universal gas constant, 8.31 J(molK)R
Boltzmann’s constant, 23 1.3810JKkB
Electron charge magnitude, 191.6010 Ce
1 electron volt, 191 eV1.6010 J
Speed of light, 3.0010 msc 8 Universal gravitational constant, 11226.6710 NmkgG Acceleration due to gravity at Earth’s surface, 29.8 msg
1 unified atomic mass unit, 27 21 u1.6610 kg931 MeV c Planck’s constant, 34 156.6310 Js4.1410 eVsh 25 31.9910 Jm1.2410 eVnmhc
Vacuum permittivity, 1222 0 8.8510CNme Coulomb’s law constant, 922 0 149.010 NmCk pe
Vacuum permeability, 7 0 410 (Tm)Amp Magnetic constant, 7 0 4110 (Tm)Ak mp
1 atmosphere pressure, 5251 atm1.010 Nm1.010 Pa
UNIT SYMBOLS
meter, m kilogram, kg second, s ampere, A kelvin, K
mole, mol hertz, Hz newton, N pascal, Pa joule, J
watt, W coulomb, C volt, V ohm, W henry, H
farad, F tesla, T degree Celsius, C electron volt, eV PREFIXES Factor Prefix Symbol 109 giga G
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.
-2-
ADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATION
106 mega M 103 kilo k 210 centi c 310 milli m 610 micro m 910 nano n 1210 pico p
VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES q 0 30 37 45 53 60 90 sin q 0 12 35 22 45 32 1 cos q 1 32 45 22 35 12 0 tan q 0 33 34 1 43 3
MECHANICS
a = acceleration E = energy F = force f = frequency h = height I = rotational inertia J = impulse K = kinetic energy k = spring constant = length L = angular momentum m = mass P = power p = momentum r = radius or distance T = period t = time U = potential energy v = velocity or speed W = work done on a system x = position m = coefficient of friction q = angle t = torque w = angular speed a = angular acceleration f = phase angle
AND MAGNETISM
A = area B = magnetic field C = capacitance d = distance E = electric field e = emf F = force I = current J = current density L = inductance = length n = number of loops of wire per unit length N = number of charge carriers per unit volume P = power Q = charge q = point charge R = resistance r = radius or distance t = time U = potential or stored energy V = electric potential v = velocity or speed r = resistivity F = flux k = dielectric constant
-ADVANCED PLACEMENT PHYSICS C EQUATIONS
0ÃÃ x x at x 2 00 1 2 Ã xxx xta t 22 0 2ÃÃxxxaxx
aFFnet mm Fdp dt DJ Fdtp p mv m FFfN DE WFdr 21 2 ÃKm PdE dt PFv DD g Umgh 2 2Ã wac r r t rF tt a net II 2 Irdmm 2 r ii cm i mx x m Ãw r wLrpI 21 2 KI w 0 twwa 2 00 1 2 tt qqwa DFs kx 21 2 Ds Ukx max cos( xxtwf 21T f p w 2s Tm kp 2Tp g p 12 2G Gmm F r 12 G Gmm U r ELECTRICITY
0
2 12 0 1 4 peE Fqq r EFE q 0e EdAQ x EdV dx DVEdr 0 1 4 pe i ii Vq r 12 0 1 4 peE UqVqq r DVQ C 0ke A C d pi i CC 11 s CCii IdQ dt 211 22C UQVCV D D R A r r EJ d INevA D IV R i si RR 11 i pi RR D PIV FM qvB 0mB dI 0 24 m p dBIdr r F IdB 0Bs nIm F B B dA e F EddB dt LdI dt e 21 2L ULI
ADVANCED PLACEMENT PHYSICS C EQUATIONS
GEOMETRY AND TRIGONOMETRY
Rectangle A bh Triangle 1 2 A bh Circle 2A p r 2 Cr p sr q Rectangular Solid Vw h Cylinder 2Vr p 222 Srr pp Sphere 34 3 Vr p 24 Sr p
Right Triangle 222 abc sin a c q cos b c q tan a b q
A = area C = circumference V = volume S = surface area b = base h = height = length w = width r = radius s = arc length q = angle
s r q c a b 90°q
CALCULUS dfdf du dxdudx 1dn x dxnx n daxax dxeae d ax dxx 1ln sincosd dxaxaax cossind dxaxaax 11 ,11 nn xdxxn n 1ax edxeax a lndx xa xa 1 cossinaxdxax a 1sin cos axdxax a
VECTOR PRODUCTS cosABAB q
sinABAB q
- -
QUESTIONS
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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2019
AP® PHYSICS C: MECHANICS FREE-RESPONSE
1. In an experiment, students used video analysis to track the motion of an object falling vertically through a fluid in a glass cylinder. The object of m = 12 g is released from rest at the top of the column of fluid, as shown above. The data for the speed v of the falling object as a function of time t are graphed on the grid below. The dashed curve represents the best fit chosen by the students for these data.
(a) i. Does the speed of the object increase, decrease, or remain the same? ____ Increase ____ Decrease ____ Remain the same
In a brief statement, describe the direction of the object’s acceleration and how the magnitude of this acceleration changed as the object fell.
Using the graph, calculate an approximate value for the magnitude of the acceleration of the object at t = 0.20 s. v Acceleration is down, magnitude of acceleration decrees as drag force increases. scope a w 08-06 ( n 3100 0.5) v 0.226 0136 1226-136) S =2.22 mj h
ii.
iii.
The students use the equation v = A (1 e Bt ) to model the speed of the falling object and find the best fit coefficients to be A =1.18 m s and B = 5s-1. (b) Use the above equation to: i. Derive an expression for the magnitude of the vertical displacement y(t) of the falling object as a function of time t . ii. Derive an expression for the magnitude of the net force F (t ) exerted on the object as it falls through the fluid as a function of time t . -t I 0 = A (1-541) => 98 = A (1-Ept).t at t ↑ (-) dt = a jt -AP at188=) f =>A (t) +A3 = A StT *A Et B 1831e]O = A Ct # 1 / 282 17=A /t (83tn) 7 -> C118) ++ E (55[D) acceleration: de)A (l-2-3-) 7 = A (0 + 23t ) = A eBt F=ma = mA [Bt = (012) (118)(5) 25 ->0.0712
The students repeat the experiment with a taller glass cylinder that is filled with the same fluid. The cylinder is tall enough so that the object reaches a constant speed. (c) i. Determine the constant speed of the object. Justify your answer. ii. Determine the force exerted by the fluid on the object at this time. Justify your answer. 0 = A (1-2-Bt), constant speed will be attained at t = a 0 = A (1-130) -A (1 to Cno . 118 m 5 " since it is moving with constant velocity then acceleration =0 Net force = 0 Net force: weight of object drag force. 0 = 12x 183 x 10 N- drag force drag force = 12x18"x10 -> 0 N
A tax
2. A pendulum of length L consists of block 1 of mass 3M attached to the end of a string. Block 1 is released from rest with the string horizontal, as shown above. At the bottom of its swing, block 1 collides with block 2 of mass M, which is initially at rest at the edge of a table of height 2L. Block 1 never touches the table. As a result of the collision, block 2 is launched horizontally from the table, landing on the floor a distance 4L from the base of the table. After the collision, block 1 continues forward and swings up. At its highest point, the string makes an angle qMAX to the vertical. Air resistance and friction are negligible. Express all algebraic answers in terms of M, L, and physical constants, as appropriate.
(a) Determine the speed of block 1 at the bottom of its swing just before it makes contact with block 2.
(b) On the dot below, which represents block 1, draw and label the forces (not components) that act on block 1 just before it makes contact with block 2. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot. Forces with greater magnitude should be represented by longer vectors.
Oegh
to
mxgL=you
2x10XL = 202 mil
FT=amgtohso
Firs
(c) Derive an expression for the tension FT in the string when the string is vertical just before block 1 makes contact with block 2. If you need to draw anything other than what you have shown in part (b) to assist in your solution, use the space below. Do NOT add anything to the figure in part (b). For parts (d)–(g), the value for the length of the pendulum is L = 75 cm. (d) Calculate the time between the instant block 2 leaves the table and the instant it first contacts the floor.
=3mg + 6 mg = 9mg consider vertical motion * between starting point and final bout 26-0-18+2 2 L to a= 3see 100x10 Th -> 0.55 sec
(e) Calculate the speed of block 2 as it leaves the table. consider horizontal motion of Point 0 and P O -> U 46=4 (t), f 4 L 4L= U 4 g P g n = F9L 9 (42) = 10xx15330 = 5.47 M5 "
(f) Calculate the speed of block 1 just after it collides with block 2
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*pply conservation of linear momentum 3M 29L + +*(0) = 31XN+(5.47) 3 PSE 3V' + 597 3V = 3 15-5.47 =3 (3.07) 5'47 =2.04 mst
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(g) Calculate the angle qMAX that the string makes with the vertical, as shown in the original figure, when block 1 is at its highest point after the collision. x F or block. O ↓(**) (2.02) = gh donl z = 1=1108 = I (h-c0s0) (1-2000) = 2.04X2.04 1100 = 0.2774 2 x 10 x 7 5 20S8=-1-02774 = 0.722. 0 = 205" (0.722) = 43.8970
-Ap
p * t t op * Em Wp = No +C (At)
* D = F
=
= FD
Wo =
Wp ==b)
3. A horizontal circular platform with rotational inertia IP rotates freely without friction on a vertical axis. A small motor-driven wheel that is used to rotate the platform is mounted under the platform and touches it. The wheel has radius r and touches the platform a distance D from the vertical axis of the platform, as shown above. The platform starts at rest, and the wheel exerts a constant horizontal force of magnitude F tangent to the wheel until the platform reaches an angular speed wP after time Dt . During time Dt , the wheel stays in contact with the platform without slipping. (a) Derive an expression for the angular speed wP of the platform. Express your answer in terms of IP , r, D, F, Dt , and physical constants, as appropriate. =
(c)
a
IP
O
(Dt)
wp =FPA Ip
(b) Determine an expression for the kinetic energy of the platform at the moment it reaches angular speed wP . Express your answer in terms of IP , r, D, F, Dt , and physical constants, as appropriate.
(c) Derive an expression for the angular speed of the wheel wW when the platform has reached angular speed wP . 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p
p
↳
No: Uw Dup = U Ww Ww=Dlp
tup KE= IpWp=1Ip (Fopt)2 I F'D<(t)" Ip w
Ap
in ear speed of platform linear speed of wheel
p
fas Me ⑧p
When the platform is spinning at angular speed wP , the motor-driven wheel is removed. A student holds a disk directly above and concentric with the platform, as shown above. The disk has the same rotational inertia IP as the platform. The student releases the disk from rest, and the disk falls onto the platform. After a short time, the disk and platform are observed to be rotating together at angular speed w f . (d) Derive an expression for wf . Express your answer in terms of wP , IP , and physical constants, as appropriate.
Apply conservation of angular momentum
to
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A *p
IpWp
to = (Ip +10) Nf Not-TWO =
A student now uses the rotating platform (I = 3.1 kg m2 ) to determine the rotational inertia I of an PU unknown object about a vertical axis that passes through the object’s center of mass. The platform is rotating at an initial angular speed wi when the unknown object is dropped with its center of mass directly above the center of the platform. The platform and object are observed to be rotating together at angular speed w f . Trials are repeated for different values of wi . A graph of w f as a function of wi is shown on the axes below.
(e)
i. On the graph on the previous page, draw a best-fit line for the data.
ii. Using the straight line, calculate the rotational inertia of the unknown object IU about a vertical axis passing through its center of mass.
P ⑭ ⑤ ofn
fP ⑭ f not * v Apply Conservation of angular no mentum IpWi = (tp+Fv) Nf I --Othosa if [v =3.78gmt ↑ 3. D 3.1 (8) = (3.1) (3.0) + 3'6 Fu = 35 u = 3.1) 10) (3.1) (3.6)
↑
(f) The kinetic energy of the spinning platform before the object is dropped on it is Ki . The total kinetic energy of the platform-object system when it reaches angular speed w f is K f . Which of the following expressions is true? ____Kf <Ki ____Kf =Ki ____Kf >Ki Justify your answer. (g) One of the students observes that the center of mass of the object is not actually aligned with the axis of the platform. Is the experimental value of IU obtained in part (e) greater than, less than, or equal to the actual value of the rotational inertia of the unknown object about a vertical axis that passes through its center of mass? ____ Greater than Justify your answer.
____
*
Because centre of mass of the object is off from the centre axis of
And as per parallel axes theorem I'11cm +122 From the calculation we get I' which is Greater than1.
Less than ____ Equal to of of -**K KI Kf> KL Kf < K1 this is inelastic collision the KE will be lost.
p
platform)
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