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Unformatted text preview: ”fem?“ $3“? ave put A, n2 is Model: For the marble spinning around the inside of a smooth pipe, the sum of the kinetic and gravita~ potential energy does not change. : o ' y1,v1=3.0 mls
use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius (R) of the pipe is @ cm, and therefore ylop = y2 = 2R = 0.20 m. At an arbitrary angle 9, measured from the bottom of the circle, y = :R cos 6. 1018 Chapter 10 Solve: (a) The energy conservation equation K2 + U32 = KI + U 8, 18 => lmvzz + mgy2= 2mv,21 + mgyl => v2 = v12+2g(y,  )=.1’(3 0 m/s)2 +2(9. 8 m/s2 )(0 m— 0.20 m)= 2. 25 m/s (b) Expressing the energy conservation equation as a function of 9: K(9)+Ug (9): K +Ug Izzémv (9)+mgy(9)=— —mv,2 +0] =>v(9)=1/vl2——)2gy(9 =1’V1 ——2gR(1—cos9) Using v,= —.3 0m/s, g= 9.8 m/sz, and R: 0.10 m, we get v(9)= 1/9—1.96(1—cos9)(m/s) (c) The accompanying ﬁgure shows a graph of v for a complete revolution (0°_ < 9 < 360°).
v (In/s) 3 / 3.0 m/s at bottom 2.25 m/s at top 0 90° 180° 270° 360° Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’ s potential energy increases and kinetw energy decreases as it gets toward the top of the circle. At the top, its speed is 2. 25 m/s. This IS reasonable sine
some of the kinetic energy has been transformed into the marble’ s potential energy. 1110: ”a . . 3’
(V0)m = 3’0 Before
After
y
(b)
~5— 5 After
0 : :
: For a package with mass m the conservation of energy equation is 1 1
K1 + U3, = K, + Ugo => 5mm )ﬁ, + mgyl = 5mm, )1 + mgy0 sing 010),” =0 m/s andy, =0 m, a) For the perfectly inelastic collision the oonservation of momentum equation is
Pa = Pk => (m + 2m)(V2)3,., = m0. ),,, + (ZMXVJM
Using (v,)2m = 0 m/s, we get (v2)3m = (vI )m [3 = 2.56 m/s ’(b) For the elastic collision, the mass m package rebounds with velocity (V3)m= m — 2m ——————)(vl m = ———(7. 668 m ls): ——2.56 m/s
m+2m $11202l )3" = mgyo => (vl )m = ,IZgyo = 1/2(9.8 m/sz)(3.0 m) = 7.668 m/s 10—22 Chapter 10 The negative sign with (v3),,, shows that the package with mass m rebounds and goes to the position y4. We can
determine y4 by applying the conservation of energy equation as follows. For a package of mass m: l l
Kf + Ugr = Kl + Us, => Emma?" + mgy4 = §m(v3),2n + mgy3
Using (v3),,, = —2.55 m/s, y3 = 0 m, and (114)," = 0 m/s, we get mgy4 = %m(—2.56 m/s)2 => y4 = 33.3 cm \ ; ..... 10.54. Model: This is a twopart problem. In the ﬁrst part, we will find the critical velocity for the ball to go over the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining
the angle 9. ‘ Visualize: y
i _..__..,.L_ _ r
I
1
r
I
l
I
\ ,_.L_
_3 We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitational
potential energy. For 9 to be minimum, the ball will just go over the top of the peg. 10—30 Chapter 10 Solve: The two forces in the free~body force diagram provide the centripetal acceleration at the top of the circle.
Newton’s second law at this point is 2 mv
w+T=
r
where T is the tension in the string. The critical speed vc at which the string goes slack is found when T ——) O. In
this case,
_ mvz mg =>vé=gr=gL/3 The ball should have kinetic energy at least equal to l 2 1 (L)
—mvc = —mg —
2 2 3 for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get
the minimum angle 9. The equation for the conservation of energy is Kf + Ugf = K, +Ugi => émvf2 +mgyf == émvf +mgyi Using vf = vc, yf = §L, vi = O, and the above value for v3, we get
i E + m A  m =>  ~14
2 g 3 g 3 gyi y, 2 That is, the ball is a vertical distance {L above the peg's location or a distance of
( 2L L) __ L 3 2 6
below the point of suspension of the pendulum, as shown in the ﬁgure on the right. Thus,
‘ Ll6 l cosG=———=—::6=80.4°
L 6 3 a
he 0.70. ‘Motiel: Assume an ideal, massless spring that obeys Hooke’s law. Let us also assume that the cannon )fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion.
isnaliz : g? Before (U _ The before—and—after pictorial representation is shown, with the origin of the coordinate system located at the spring’s free end when the spring is neither compressed nor stretched. This free end of the spring is just behind the
; cannon. , Solve: The momentum conservation equation pit = pLt is "1130):: )B + "100’ka = ma (vlr )3 + mc (vi! )c 10—40 Chapter 10 Since the initial momentum is zero, (Viv )B = _nﬂC_(v£t )C = —(%.%:'{g§)(v va)c= ”20(Vtx)c 1B The mechanical energy conservation equation for the cannon + spring Kf + Usf = Kl + Usi is é—mm): + ikamz = —l—m(v,)g + 0 J = 0 J + %k(Ax)2 = £11201“): 3 (V5,)c = aim: + ’£_’___ZO::§]:/m) ———————(0.5 In) = i5.0 m/s To make this velocity physically correct, we retain the minus sign with (v&)c. Substituting into the momentum con
servation equation yields: (er )B = —20(—5.0 m/s) = 100 m/s 11.47. Model: We will use the spring, the package,
a particle. Visualize: and the ramp as the system. We will model the package as Known xo=y0=0m m=2.0kg
xe—x0=30cm yl=1.0m
v0=0mls k=500Nlm Find
”1 sticky spot
We place the origin of our coordinate s
the package to be shot. Model: (a) The energy conservation equation is KI +Ugl +U51+AEIh =Ko +Ugo +Uso +Ww émvf + mgy, +—21—k(xe —xe )2 +12\E,,l = émv: + mgy0 + %k(Ax)2 + Wm ystem on the end of the spring when it is compressed and is in contact with Using y1 = 1 m, AE,h = 0 J (note the frictionless ramp), v0 = 0 m/s, y0 = 0 m, Ax = 30 cm, and Wm = 0 J, we get émvlz+mgﬂ m)+0J+OJ=0J+OJ+%k(O.3Om)2 +01 %(2.0 kg»? + (2.0 kg)(9.8 m/s2 )(1 m) = $4500 N/m)(0.30 m)2 = v1 = 1.70 m/s (b) How high can the package go after crossing the sticky spot? If the package can reach y,
(vI = 0), then it makes it. But if yI < 1.0 m when vl
energy 2 1.0 in before stopping
= 0, it does not. The friction of the sticky spot generates thermal AE‘h = —Wdiss = {ﬂ Ai‘° = ('.Ukmg)Ax = (0.30)(2.0 kg)(9.8 m/s2 )(0.50 m) = 2.94 J
The energy conservation equation is now
i mv,2 + mgy, + AEm = ~2Lk(Ax)2
If we set v1 = 0 m/s to ﬁnd the highest point the package can reach, we get
y1 = (ikmxf — AEm) / mg =(1L(500 N/m) (0.30 m)2 ~ 2.94 J) / (2.0 kg)(9.8 m/sz) = 0.998 m
The package does not make it. It just barely misses. Work 1 13 1 .74. Model: Assume an ideal spring that obeys Hooke’s law, the particle model for the ball, and the model
f kinetic friction. The ball, the spring, and the barrel comprise our system. the
v1 = 0 m/s
ply e placed the origin of the coordinate system on the free end of the spring which is in contact with the ball.
_ in olve: Force F does external work Wm. This work compresses the spring, gives the ball kinetic energy, and is
;gy artially dissipated to thermal energy by friction. _ a) The energy equation for our system is
' K1+Usl+AEm =K0+U50+Wm émv.2 +%k(xl ~26“)2 +(fk)(xl ——x0) = émvﬁ +—:~k(xe —.xe)2 + F(x‘ —x0) %(I .0 kg)(2.0 m/s)2 + —;—(3000 N/m)(0.3 m)2 + ,uk (mg)(0.30 m) = 0 J + 0 J + F (0.30 m) The friction force is fk = ,ukmg. With ilk = 0.30 and m = 1.0 kg, we can solve this equation to obtain F =460 N.
'(b) Using the energy equation for our system once again, we have rgy K2 + U52 + AEth = K, + US, + Wext émvi + %k(xe — )ce)2 +(fk)(xz — x,)=—12—mv12+—;—k(x, —Jr0)2 +0 I
ion 1 1
Emvz2 + ,ukmg(x2 — x,) = 0 J +Ek(xl — x0)2 ‘ %(1.o kg)v§ + (0.30)(1.o kg)(9.8 m/s2)(1.5 m) = ~£43000 N/m)(0.30 my )dy ,
(0.5 kg)v§ + 4.41 J = 135 J = v2 = 16.2 m/s Oscillations > 1423 6. Visualize: Please refertoFigure P1466.
ve: The potential energy curve of a simple harmonic oscillator is described by U=1}k(Ax)2, where .~= x — x0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is x0 = ‘ nm. We can find the bond’s spring constant by reading the value of the potential energy U at a displacement
’ and using the potential energy formula to calculate k. 1.9 x10"" J
3.4 X 10‘[9 J e three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring
nstant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be _ _27 = 7.9 x1013 Hz
27: m 27: 1.67X10 kg N is:
ad ['6 .— 14175. Model: The two springs obey Hooke’s law.
Visualize: x o
Equilibrium‘f Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces
have negative values: — one pushing, the other pulling — are directed to the left and ...
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 Winter '08
 Susskind,L
 Physics, mechanics

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