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Unformatted text preview: R Ramesh, Physics 7A Midterm 2. Please write out. all answers in blue books. Even if numbers
are given wait until the end to plug in values. Whenever possible, diagrams are recommended. Problem 1 (20 points) A In = 10 g bullet is ﬁred in the xdirection into a stationary block of
wood that has a mass of M = 5 kg (see picture). The speed of the bullet before entry into the
block is V0 = 500 111/3. 21) What is the speed of the block just after the bullet has become embedded? (10 pts) b) What distance will the block slide on the surface with a coefﬁcient of friction, u = .50?
(10 pts) Problem 2 (20 points). A mad scientist wants to use a large spring to launch a
projectile to the moon. She is trying to calculate the minimum spring constant (k) she
should use for her system (see Figure below). She assumes that the moon and earth
do not move relative to each other during this process. She also assumes that there is
no energy loss due to air resistance or friction. **Please note, that she is a bad scientist and has an incorrect value for the distance
between the moon and the earth (use her values)” a) [2 points] Write down an algebraic expression (do not plug in numbers) for the
total energy of the projectile before the launch in terms of k and known values.
b) [2 points] Write down an algebraic expression for the total energy of the projectile just after the launch in terms of Edam and known values. 0) [3 points] At some location between the earth and the moon, the projectile is
momentarily at rest ( F i) Find an algebraic expression for this fie“ in terms of res known values.
d) [2 points] Write down an algebraic expression for the total energy of the projectile
at F in terms of known values and 1" rest res! ' e) [3 points] Find an algebraic expression for ﬁlaunch in terms of known values. 0 [3 points] Find an algebraic expression for the spring constant k in terms of known
values. g) What are the numerical values of 13mm and k? [1 point each]
h) What are three things wrong with her assumptions/approach? [1 point each] Rm
RE 6)
ME Viaunch Min
Spring Launch System EarthMoon System Known Values Me (earth mass) : 5.97  1024 kg
Mm (moon mass) = 7.36  1022 kg
Re (earth radius) = 6.37  106 m
Rm (moon radius) = 1.74  106 m
Ax (spring compression) = lOOOm
D (m—e distance) :3.35  107 m m (projectile mass) = 1 kg G = 6.67 10'” m3 kg'1 s'2 Problem 3 (20 points) .A car enters a turn whose radius is R. The road is banked at an angle 8 and the coefﬁcient of
static friction is it. Use 9 as the magnitude of gravitational acceleration. front mew top—d own View i p
l~£ __________ ’_ f ________._i __________ (a). (Spts) Find the smallest speed Vmin for which the car will , A
not skid. Draw freebody diagram and show your work ”step by step (b). (Spts) Find the largest speed Vmax for which the car will not skid. Draw ﬂeebody diagram
and show your work step by step. (c). (4pts) For a large enough it, the car will not skid no matter what'the speed is. Solve the
minimal value of the coefﬁcient of static friction n min for this to happen. Problem 4 (20 points). Suppose we have a block of mass m attached to a spring of spring
constant k on a frictionless ramp with angle of inclination 6. The block starts from rest, and the
spring begins at its equilibrium length. As the block slides along the ramp, it is subject to the
drag force F = bv (bold lettering indicates a vector quantity) because of air resistance. The
block will initially slide down a distance L before momentarily coming "to rest. a) (6 points) Name all the forces on the block during its motion. Which ones do work? b) (7 points) What is the work done by the air from the time when the block starts sliding to
when it reaches the distance L? c) (2 point) Draw a qualitative positiontime graph for the subsequent motion of the block.
d) (5 points) Where does the block ultimately come to rest (permanently, not momentarily)? Problem 5 (20 points) A disk with surface density pl,_ and radius “R”, has whole cutout of it
with radius “r”, and in its place a second material of density p2 is inserted as shown. (pl and p2
have units of mass/area). Assume the center of the large disc is at position (x, y) = (0,0), and
directions x and y are mutually perpendicular and deﬁned in the picture. Given that the center of the small disk is at (x, y) = (a, b), write the Center of Mass VECTOR in
terms of p1, p2, r, R, and the position of the center of the small disk.
Partial credit given for correctly explained work. a rn = NJ; .e\o kg (5 N— Lo nsmolﬁo n 0 Q ”Me/‘3‘" "\ mib‘o : 0"“ W) VIE
' 2912—»
ﬂ> \‘Q; (”\u"mQ: H' _. . 1
=9) "" g A: 3bu*me§ VSht
=9 _ HUIVNMQDSB’J ’
1
1» __ V“
’LHj H
3 ﬂ 3 . Kn—own Values Me (earth mass) = 5.97  1024 kg Dx (spring compression) = 1000rn
Mm (moon mass) = 7.36  1022 kg D (me distance) =3.85  107 In Re (earth radius) = 6.37 ' 106 m m (projectile mass) = 1 kg Rm (moon radius) = 1.74  106 rn G = 6.67 ‘ 10'11 m3 . kgl  52 a) [2 points] Write down an algebraic expression (do not plug in numbers) for the
total energy of the projectile before the launch in terms of k and known values. Before the launch the total energy is spring potential energy and gravitational
potential energy, because nothing is moving. We need to take into account both the
gravitational potential energy of the earth and the moon: ksz + —GMem+ —GMm m
r D—rm E total =PE Spring +PE Ear! h+PE = Moan M>— 8 b) [2 points] Write down an algebraic expression for the total energy of the projectile
just after the launch in terms of tam, and known values. Just after the launch the total energy is the kinetic energy of the projectile and the
gravitational potential energy of the earth and the moon. We can neglect the change
in distance Ax in our calculations of gravitational potential energy because it is less
than 1% of the distances between the projectile and the earth and moon. E total _ i _—e'_ _L
=KE + PEEW+ +PE — 2 mvmmh + + c) [3 points] At some location between the earth and the moon, the projectile is
momentarily at rest (F ). Find an algebraic expression for this r 1 in terms of rest 1' es known values. There is a point between the earth and the moon when the force due to the
gravitational pull of the earth is equal to the force due to the gravitational pull of the
moon. This is location of maximum gravitational potential energy of the system. At
this location (if we want to minimize the spring constant needed for the system) the
projectile should be momentarily at rest (or very close to rest) We can ﬁnd this locationr ( rm) by looking at the gravitational forces of the moon and the earth (r re"
the distance of the project from the earth):
GMe m GMM —m ear = F mm =
m ( )2 M (—9 New) rI‘BSf We can solve rm by finding when these two forces are equal: GM mu GMM __‘_m _) M = MM
(1" rrest )2 =—(DF rresl )2 (r rresr )2 D 2 2D rresf +(rres 12)—> WEST ME(D2_2Drresr+( (Fresl)2=) MM (rzr'é’Sf) > Mc—Dz ZM'Dr +(Me WM )( 1'2““) Use the quandratic formula to solve equation for r rest ‘
DJ— D Me
F—— F° “WM The ﬁrst of these solutions rm? D, which is not between the moon and the earth, so
the solution we are looking for 1s: = 3.46  107 m d) [2 points] Write down an algebraic expression for the total energy of the projectile
at r in terms of known values and r res! ra'rl At rm the projectile is not moving, so the energy of the projectile is the gravitational potential energy of the moon and the earth:
+ FEW = W + ﬂu:
r D— —r rest res! Eola.’ = PEF f .ath We can plug 1n our expression for:7 rm (from part 0) into the above expression to
get E 1n terms of known values: GM m GM," m GMm GM m
Emm =— . e ——’"—=— ,lMe +./Mm ——B+——’"———.
’ ——r’i‘i“e,,. D nlf'——if7*—,,. ( D'x/Me 0 JMB +JMm tD'J—M =—(F F)[ 53% 5M7]=FF)[Gm—MLG Mm
=——(‘/E+ Mj=—1.128107 e) [3 points] Find an algebraic expression for 17mm in terms of known values. t The scientist assumes that energy is conserved. Therefore we can set the energy just
after the launch (from part b) to the energy atr (from part (1) and solve for ﬁlaunch : rife“ GMm MGMm]__(GM,l_m GMmm m) 2
Emmi = %mv!aunch _[ re D _ rm rrest D_ rresr
l l l
%vlaunch2 = GME — H — + GMm _ I
re rrest D _ rm D _ ”rm We know the total energy atr is in terms of known values, so alternatively we can rresr use that expression to ﬁnd 12mm in terms of known values: Elam! = _mviaunch2 " [ Gﬁfem + (EA/5mm ] = ‘G?m(\/A—J: 'l' \j‘l’dr—m)Z = 9989 m/s f) [3 points] Find an algebraic expression for the spring constant k in terms of known
values. The scientist assumes that energy is conserved. Therefore we can set the energy just
after the launch (from part b) to the energy just before launch (from part a): GM
Emmi = é—ksz '— [ GM m + D m ] = %mvmunck2 — ( GMem + iMmm]
re —1’ re _rm
2
vaHC
%ksz = %mvlaunch2 —) k = Alx2 h Using the expression we found for ﬂow}, (from part c) we can ﬁnd an expression for k in terms of known values: m = 99.8 kg /52 g) What are the numerical values of ﬁaunch and k? [I point each] Plugging in the numbers provided, we get:
v = 9989 m/ s
k = 99.8 kg / 52 h) What are three things wrong with her assumptions/approach? [1 point each]
The following are some problems with her plan: You can’t neglect friction, the friction between the air and a package moving at 10
kin/s would be enough to incinerate it before it left the earth’s atmosphere. You can’t neglect the motion of the moon and the earth with respect to each other,
that motion would change the direction of the gravitational pull during the package’s
travel from earth to the moon. The average acceleration of the package during the launch is 49,878 mfs2 which is
over 5000g. This is such a force that anything you would send would smashed by this
acceleration (the human body cannot withstand more than 10g) The energy required to compress the spring is: ékAx2 = 4.99 107J and so it would be
very hard to compress the spring to begin with. The velocity of the projectile once it gets to the moon is: 4460 m/s which means that
the package would probably be destroyed upon landing or kill the astronaut who was
trying to catch it. Physics 7A, Fall 2006, Lecture 100, llfidterm H 1 Problem 3: (20 points) A car enters a turn Whose radius is R. The road is
banked at an angle 9 and the coefﬁcient of static friction is it. Use 9 as the magnitude of gravitational acceleration.
top—down View \ front View (a). (8pts) Find the smallest speed 11min for which the car will not skid. (b). (Spts) Find the largest speed 711mm for which the car will not skid. (c). (4pts) For a large enough ,u, the car Will not skid no matter What the
speed is. Solve the minimal value of the coefﬁcient of static friction umm for
this to happen. Solution: Assume that the car has mass m. (a). [1pt] For smallest speed, the static frictional force will be ” up the ramp”, thus the free—body diagram must be
F N m 3
Where FN is the normal force, fS is the static frictional force, mg is gravity.
[1pt] We want the smallest speed, therefore the static frictional force must
have the maximal possible magnitude fSZFL'FN Deﬁne the horizontal—vertical coordinates system (you can use the slanted Physics 7A, Fall 2006, Lecture 100, Midterm II 2 system, the algebra will be more complicated).
[2pts] Write down Newton’s 2nd law for both :1: and y components If x—component : FN sin(f9) — fs cos(6) m  ea 0 y—component : FN cos(9) + f3 sin(6)  mg
[1pt] where ac is centripetal acceleration.
ac : ’02/R [lpt] From the y—component equation and f3 2 Ju.  FN, we can solve the
normal force FN : mg/[cos(6) + ,Lisin(6)] [1pt] Plug this and f3 : n  FN into the :c—component equation, we can solve
the centripetal acceleration sin(6) — ,u, cos(6) ac : gcosw) + asin(6) [1pt] Finally use do = vQ/R we can solve 11min _ sin(9) — hcosw) cos(9) + asinw) (b). closely parallel the solution to part (a).
[1pt] For largest speed, the static frictional force will be ”' down the ramp”,
thus the free—body diagram must be m 8'
where FN is the normal forée, f5 is the static frictional force, mg is gravity. r’_\l Physics 7A, Fall 2006, Lecture 100, Midterm H 3 [lpt] We want the largest speed, therefore the static frictional force must
have the maximal possible magnitude fSZIU’IFN Deﬁne the horizontalvertical coordinates system (you can use the slanted
system, the algebra will be more complicated).
[2pts] Write down Newton’s 2nd law for both CC and 3; components micomponent : FN sin(6) + fS cos(6) m  ac 0 y—component : FN cos(9) — f5 sin(6) — mg
[lpt] where ac is centripetal acceleration.
ac : vg/R [lpt] From the y—component equation and f5 = u  FN, we can solve the
normal force FN : 7719/ [008(9)  MsinUElll [1pt] Plug this and f5 = a  FN into the x—component equation1 we can solve
the centripetal acceleration sin(6) + u 008(6) ac : 9005(6) — nsin(9) [lpt] Finally use ac = “HQ/R we can solve Umax _ Sin(9) + noos(6)
Umax # QR 003(6) — ,usin(6) (c). [1.5pts] From the solution of (a) we see that if sin(6) — ,u, 005(6) 3 0 there
will be no lower bound for Speed. 11mm —> 0. This condition is n 2 tan(tl).
[1.5pts] From the solution of (b) we see that if cos(6l) — Ju,sin(19) g 0 there
will be no upper bound for speed. vmax —> 00. This condition is a Z cot(6). [1pt] Both conditions must be satisﬁed, then we get ,umm : larger one in the two numbers tan(6) and cot(6) ' 3 .5 3 J: games/3 Mm/fejcm #5} EM/ 06 _ ,
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V" , ‘Q‘Om fie 4M20H_ogﬂm . 7 «IiafJ" 3 :35 3 5 23573‘3'3 ‘3 3 Physics 7A, Fall 2006, Lecture 100, Midterm II 1 Problem 5: (20 points) Solution [5 pts] We need to divide the object into several parts, each part should have
a high symmetry so its center of mass can be easily determined. There are
several different possibilities. Following is the simplest one. It is convinient to think of this object as one complete (no hole) large disk
with density p1 and radius R, pl'us a small complete disk with density p2 * p1 and radius 7". (following picture)
." )9 [2 pts] Center of mass of the large complete uniform disk is the center of
large disk 771 : 0:3“ + 01}. [2 pts] Center of mass of the small complete uniform disk is the center of
small disk 772 = (if; i by. [2 pts] Mass of the large disk is density times area m; : p1 as WRZ.
[2 pts] Mass of the small disk is density times area m2 : (p2 — p1) >1< 7W2. [5 pts] By deﬁnition of center of mass 2 m2 : FCM =
2 m1 [2 pts] Finally plug in m1, m2, F1, F2, _, W177i + m2F2 0 + (p2 — p1) >1: 7W2(a3“: — by)
7" : 2
CM m1+ m2 p1 * 7TB2 + (p2 — pl) >l< 7TT'2
a.>1<(p2—p1)>{<7rr2 "Jr —b>i<(p2—p1)>:<7rr
:13 : p1$ﬁR2+(p2—pl)*7rr2 p1>I<7rR2+(p2—p1)*7rr2y 2 A A You can also write down the vector in components form without a, y. D ...
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