Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WH5  CHAPTERS  12-­‐14]   1    +WHW5: Chapter 12: 7,43,57, Chapter 13: 55,81, Chapter 14: 23,25,37,50,51  Chapter  12  7.     Picture  the  Problem:  The  spaceship  is  attracted  gravitationally  to  both  the  Earth  and  the  Moon.  

  Strategy:  Use  the  Universal  Law  of  Gravity  (equation  12-­‐1)  to  relate  the  attractive  forces  from  the  Earth  and  the  Moon.    Set  the  force  due  to  the  Earth  equal  to  twice  the  force  due  to  the  Moon  when  the  spaceship  is  at  a  distance  r  from  the  center  of  the  Earth.    Let   R = 3.84 ×108 m,  the  distance  between  the  centers  of  the  Earth  and  Moon.    Then  solve  the  expression  for  the  distance  r.  

 Solution:  1.  (a)  Set   FE = 2FM  using  equation  12-­‐1  and  solve  for  r:  

GmsmE

r 2 = 2GmsmM

R − r( )2

mE R − r( )2= 2mMr 2

R − r = 2mM mE r

r = R1+ 2mM mE

= 3.84 ×108 m

1+ 2 × 7.35×1022 kg( ) 5.97 ×1024 kg( )= 3.32 ×108 m

 

  2.  (b)  The  answer  to  part  (a)  is  independent  of  the  mass  of  the  spaceship  because  the  spaceship’s  mass  is  included  in  the  force  between  it  and  both  the  Moon  and  the  Earth,  and  so  its  value  cancels  out  of  the  expression.  

  Insight:  The  distance  in  part  (a)  is  the  same  for  any  mass,  and  corresponds  to  about  52  Earth  radii  or  about  86%  of  the  distance  R  between  the  Earth  and  the  Moon.    The  two  forces  are  equal  at  3.46×108  m  or  about  90%  of  R.  

 43.     Picture  the  Problem:  An  object  is  located  at  the  surface  of  the  Earth  and  later  at  an  altitude  of  350  km.     Strategy: Use equation 12-8 to find the gravitational potential energy of the object as a function of its distance

r = RE + h from the center of the Earth. Then take the difference between the values at h = 0 and h = 350 km and compare it with the approximate change in potential, ΔU = mgh.

  Solution:  1.  (a)  Calculate  

U = −G

MEmRE + h

 at  h  =  0:   U0 = − 6.67 ×10−11 N ⋅m2 /kg2( ) 5.97 ×1024 kg( ) 8.8 kg( )

6.37 ×106 m= −5.5×108 J  

 2.  (b)  Calculate  U  at  h  =  350  km:  

Uh = − 6.67 ×10−11 N ⋅m2 /kg2( ) 5.97 ×1024 kg( ) 8.8 kg( )

6.37 ×106 + 350 ×103 m= −5.2 ×108 J  

  3.  (c)  Take  the  difference ΔU :   ΔU = Uh −U0 = −5.50 ×108 J( ) − −5.21×108 J( ) = 2.9 ×107 J  

  4.  Compare  with  mgh:   ΔU = mgh = 8.8 kg( ) 9.81 m/s2( ) 350 ×103 m( ) = 3.0 ×107 J  

  Insight:  The  two  calculations  of   ΔU (without  any  rounding)  differ  by  about  5%.    The  mgh  calculation  is  an  approximation  because  it  assumes  the  value  of  g  is  constant  over  the  350  km,  when  in  fact  it  gets  smaller  as  the  distance  from  the  Earth’s  center  increases.  

 57.     Picture  the  Problem:  The  planet  is  ten  times  more  massive  and  has  one-­‐tenth  the  radius  of  Earth.    A  

projectile  at  its  surface  is  given  sufficient  kinetic  energy  to  escape  the  planet.     Strategy:  Use  a  ratio  to  compare  the  escape  speed  on  the  new  planet  with  the  escape  speed  on  Earth.    Use  

equation    12-­‐13  to  form  the  ratio.  

  2  

 Solution:  Write  a  ratio  of  the  escape  speeds:  

ve,new

ve,Earth

=2GMnew Rnew

2GME RE

=Mnew RE

ME Rnew

=10ME( )RE

ME1

10 RE( ) = 100 = 10

ve,new = 10 ve,Earth

    Insight:  Although  the  speed  only  needs  to  be  increased  by  a  factor  of  10,  the  initial  kinetic  energy  of  a  

projectile  needs  to  be  100  times  larger  on  this  new  planet  in  order  for  the  projectile  to  escape  that  planet’s  gravity.  

 Chapter  13    55.     Picture  the  Problem:  A  block  and  spring  are  initially  at  rest  as  a  bullet  is  fired  at  high  speed  directly  

toward  them.    The  bullet  then  embeds  in  the  block  and  compresses  the  spring.  

  Strategy:  The  bullet  and  block  first  undergo  an  inelastic  collision.    Then  they  jointly  compress  the  spring,  converting  their  kinetic  energy  into  potential  energy  of  the  spring.    Use  conservation  of  energy  to  relate  the  speed  v  of  the  block  and  bullet  to  the  compression  distance  x.    Finally,  use  conservation  of  momentum  to  find  the  initial  speed  of  the  bullet  v0  from  the  combined  speed  of  bullet  and  block.    The  time  elapsed  from  impact  to  rest  is  one-­‐quarter  of  a  period.  

  Solution:  1.  (a)  Set  the  initial  kinetic    energy  of  the  block  and  bullet  to  the    final  potential  energy  of  the  spring:  

K i = U f

12

M + m( )v2 = 12

kA2

 

 2.    Solve  for  the  speed  of  the  bullet  and  block:  

v = kA2

M + m=

785 N/m( ) 0.0588 m( )2

1.500 kg + 0.00225 kg= 1.344 m/s

 

  3.    Using  conservation  of  momentum  write  the  initial  speed  of  the  bullet  in  terms  of  the  final  speed  of  bullet  and  block:  

mv0 = ( M + m)v

v0 =M + m

m⎛⎝⎜

⎞⎠⎟

v

 

 4.    Calculate  initial  speed  of  bullet:  

v0 =

1.500 kg + 0.00225 kg0.00225 kg

⎛⎝⎜

⎞⎠⎟

1.344 m/s( ) = 897 m/s  

 5.  (b)    Calculate  one-­‐quarter  period:  

T4= π

2M + m

k= π

21.50225 kg

785 N/m= 0.0687 s  

  Insight:  The  initial  kinetic  energy  of  the  bullet  does  not  equal  the  final  energy  of  the  compressed  spring.    Some  of  the  initial  kinetic  energy  is  lost  due  to  the  inelastic  collision  with  the  block.  

 81.     Picture  the  Problem:  A  mass  that  is  attached  to  a  spring,  displaced  from  equilibrium,  and  released,  will  

oscillate  with  simple  harmonic  motion.    The  maximum  speed  occurs  as  the  mass  passes  through  the  equilibrium  position.    The  maximum  force  occurs  when  the  spring  is  stretched  its  maximum  distance  from  equilibrium.  

  Strategy: Find the amplitude from the maximum speed and maximum acceleration. Then determine the maximum acceleration from the maximum force and Newton’s Second Law. Then obtain the angular speed ωfrom the maximum speed and acceleration, and finally calculate the force constant and frequency of oscillation fromω .

  Solution:  1.  (a)  Combine  the  maximum    velocity  and  maximum  acceleration  equations    

vmax = Aω , amax = Aω 2  

Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WH5  CHAPTERS  12-­‐14]   3    

to  solve  for  the  amplitude:      

vmax2

amax

=Aω( )2

Aω 2 = A  

  2.    Using  Newton’s  Second  Law  write  ampli-­‐  tude  as  a  function  of  the  maximum  force:  

A =

vmax2

amax

=vmax

2

Fmax / m=

mvmax2

Fmax

 

 3.  Solve  for  the  amplitude:  

A =

3.1 kg( ) 0.68 m/s( )2

11 N= 0.13 m  

  4.  (b)  Combine  maximum  velocity  and    acceleration  to  find  angular  speed:  

ω =

amax

vmax

=Fmax

mvmax

= 11 N3.1 kg( ) 0.68 m/s( ) = 5.2 rad/s

 

  5.    Solve  equation  13-­‐10  for  k:   k = mω 2 = 3.1 kg( ) 5.2 rad/s( )2

= 84 N/m  

 6.  (c)  Use  the  angular  speed  to  determine  f:  

f = ω

2π= 5.2 rad/s

2π= 0.83 Hz  

  Insight:  Another  way  to  solve  this  problem  is  to  first  calculate  the  angular  speed  from  the  maximum  force  and  velocity.    The  amplitude  can  then  be  found  from  the  maximum  speed  divided  by  the  angular  speed.    

 Chapter  14  23.     Picture  the  Problem:  We  are  given  the  equation  describing  a  wave  and  wish  to  determine  the  amplitude,  

wavelength,  period,  speed,  and  direction  of  travel.  

 Strategy:  The  general  form  of  a  wave  is  given  by

y = Acos 2π

λx − 2π

Tt

⎛⎝⎜

⎞⎠⎟

.    Compare  this  equation  to  

y = 15 cm( )cos π

5.0 cmx − π

12 st

⎛⎝⎜

⎞⎠⎟

, the  equation  given  in  the  problem,  to  identify  the  wave  parameters.    

Use  equation  14-­‐1  and  the  definition  of  frequency  to  calculate  the  wave  speed.      

  Solution:  1.  (a)  Identify  the  amplitude  as  A:   A  =  15  cm  

 2.  (b)  Identify  the  wavelength  as  λ:  

2πλ

= π5.0 cm

,  so   λ = 10 cm = 0.10 m  

 3.  (c)  Identify  the  period  as  T:  

2πT

= π12 s

,  so  T  =  24  s  

 4.  (d)  Use  equation14-­‐1  to  calculate  the  speed:  

v = λ f = λ

T= 10 cm

24 s= 0.42 cm/s  

  5.  (e)    The  wave  travels  to  the  right,  because  the  t-­‐term  and  x-­‐term  have  opposite  signs.  

  Insight:    The  wave  equation  is  a  compact  way  of  completely  describing  a  wave,  because  it  is  possible  to  extract  all  of  the  wave  properties  from  the  equation.  

 25.     Picture  the  Problem:  Equations  for  four  different  waves  are  given.    From  these  equations  we  need  to  

determine  the  directions  the  waves  travel,  which  waves  have  the  highest  frequency,  the  largest  wavelength,  and  the  greatest  speed.  

  Strategy: When x and t have opposite signs, the wave travels to the left. When they have the same sign, the wave travels to the right. Examine the equations to determine which waves travel in each direction. The coefficient of the t term is proportional to the frequency. Find the equation with the largest t-coefficient, and it

  4  

will have the largest frequency. The coefficient of the x term is inversely proportional to the wavelength. Find the equation with the smallest x-coefficient and it will have the largest wavelength. The speed is the t-coefficient divided by the x-coefficient. Divide the coefficients and find the largest speed.

  Solution:  1.  (a)  Find  the  waves  for  which  x  and  t  have  opposite  signs:   Waves  

yA and yC  travel  to  the  right.  

  2.  (b)  Find  the  waves  for  which  x  and  t  have  the  same  sign:  

Waves  

yB and yD  travel  to  the  left.  

  3.  (c)  Find  the  wave  with  the  largest  t-­‐coefficient:   Wave  

yC  has  the  largest  frequency.  

  4.  (d)  Find  the  wave  with  the  smallest  x-­‐coefficient:   Wave  

yA  has  the  largest  wavelength.  

  5.  (e)    Find  the  wave  for  which  the  magnitude  of  the    t-­‐coefficient  divided  by  the  x-­‐coefficient  is  greatest:   Wave  

yC  has  the  greatest  speed.  

  Insight:  The  wave  speeds  are:   vA = 1.33 cm/s, vB = 0.8 cm/s,   vC = 6.0 cm/s,  and   vD = 5.0 cm/s  

 37.     Picture  the  Problem:  The  intensity  level  at  a  distance  of  2.0  meters  is  given.    We  want  to  find  the  

intensity  levels  at    12  m  and  21  meters.    We  also  want  to  find  the  distance  for  which  the  intensity  level  is  0,  the  farthest  point  at  which  the  siren  can  be  heard.  

  Strategy:    Insert  equation  14-­‐6  into  equation  14-­‐8  to  create  an  equation  for  intensity  level  in  relation  to  distance.    Use  this  relationship  to  calculate  the  intensity  levels  at  12  m  and  21  m.      To  calculate  the  farthest  distance  at  which  the  siren  can  be  heard,  set  the  intensity  level  to  zero,  and  solve  the  relation  for  distance.  

 Solution:  1.    Insert  equation  14-­‐6  into  equation  14-­‐8:  

β = 10 logI2

I0

⎛

⎝⎜⎞

⎠⎟= 10 log

r1

r2

⎛

⎝⎜⎞

⎠⎟

2I1

I0

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 10 logI1

I0

⎛

⎝⎜⎞

⎠⎟+10 log

r1

r2

⎛

⎝⎜⎞

⎠⎟

2

 

 2.  (a)  Insert  β  at  r1  =  2.0  m  and  set  r2=12  m:  

β = 120 +10 log 2.0 m

12 m⎛⎝⎜

⎞⎠⎟

2

= 104 dB  

 3.  (b)    Repeat  for  r2=21  m:    

β = 120 +10 log 2.0 m

21 m⎛⎝⎜

⎞⎠⎟

2

= 99.6 dB  

 4.  (c)  Set  the  intensity  level  equal  to  zero:  

β = 10 logI1

I0

⎛

⎝⎜⎞

⎠⎟+10 log

r1

r2

⎛

⎝⎜⎞

⎠⎟

2

0 = 120 + 10 log 2.0 mr

⎛⎝⎜

⎞⎠⎟

2

 

  5.    Solve  for  r:  

−10 log 2.0 mr

⎛⎝⎜

⎞⎠⎟

2

= 120

2.0 mr

⎛⎝⎜

⎞⎠⎟

2

= 10−12 ⇒ r = 2.0 m10−6 = 2.0 ×106 m

 

  Insight:  This  is  a  theoretical  limit  that  could  be  realized  in  an  ideal  case.  In  a  more  realistic  scenario,  

Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WH5  CHAPTERS  12-­‐14]   5    

ambient  noise,  as  well  as  energy  losses  when  the  sound  waves  are  reflected  or  absorbed  by  surfaces,  would  prevent  us  from  hearing  the  sound  2000  km  away.  Sometimes  the  real-­‐world  factors  we  ignore  make  a  huge  difference!  

 50.     Picture  the  Problem:  The  image  depicts  

a  person  in  a  stationary  vehicle  listening  to  the  siren  of  a  fire  engine  as  it  passes.    From  the  frequencies  heard  as  the  fire  engine  approaches  and  leaves  we  want  to  calculate  the  time  it  takes  to  reach  a  fire  5.0  km  away.    

 

  Strategy:    The  frequency  f  emitted  by  the  fire  engine  remains  constant.    Use  equation  14-­‐10  to  write  the  observed  frequencies  in  terms  of  the  emitted  frequency  and  speed,  because  the  source  (fire  engine)  is  moving  and  the  observer  (you)  is  stationary.    The  negative  sign  is  used  when  the  engine  is  approaching  and  the  positive  sign  when  the  engine  is  moving  away.    Combine  the  two  equations  to  eliminate  the  emitted  frequency  and  solve  for  the  speed  of  the  fire  engine.    Divide  the  distance  by  the  speed  to  calculate  the  time  to  reach  the  fire.  

  Solution:  1.  Write  equation  14-­‐10  for  the    approaching  fire  engine  and  solve  for  f:      

′f1 =1

1− uv

⎛

⎝⎜⎞

⎠⎟f ⇒ f = ′f1 1− u

v⎛⎝⎜

⎞⎠⎟  

  2.  Write  equation  14-­‐10  for    the  receding  fire  engine  and  solve    for  the  emitted  frequency:      

′f2 =1

1+ uv

⎛

⎝⎜⎞

⎠⎟f ⇒ f = ′f2 1+ u

v⎛⎝⎜

⎞⎠⎟  

  3.  Set  the  equations  for    the  emitted  frequencies  equal    and  solve  for  the  speed:    

′f1 v − ′f2 v = ′f1 u + ′f2 u

u =′f1 − ′f2

′f1 + ′f2

⎛

⎝⎜⎞

⎠⎟v = 460 Hz − 410 Hz

460 Hz + 410 Hz⎛⎝⎜

⎞⎠⎟

343 m/s( ) = 19.7 m/s

   

4.  Solve  for  the  time  of  arrival:   t = d

u= 5.0 ×103 m

19.7 m/s= 254 s = 4.2 min  

  Insight:  The  frequency  of  the  stationary  fire  truck  is  434  Hz.    Note  that  the  frequency  of  the  approaching  engine  increases  more  than  the  frequency  of  the  receding  engine  decreases.  

 51.     Picture  the  Problem:    As  you  approach  a  source  of  sound,  the  increase  in  observed  frequency  is  related  

to  your  speed.         Strategy:    We  want  to  calculate  the  speed  necessary  for  the  observed  frequency  to  be  1.15  times  the  

emitted  frequency.    This  problem  has  a  stationary  source  and  moving  observer,  so  employ  equation  14-­‐9  to  calculate  the  observed  frequency,  using  the  plus  sign  because  the  observer  is  moving  toward  the  source.    Set  the  observed  frequency  equal  to  1.15  times  the  emitted  frequency  and  solve  for  the  speed.  

  Solution:  1.  Set  the  observed  frequency    in  equation  14-­‐9  equal  to  1.15  f  :  

′f = 1+ u v( ) f

1.15 f = 1+ u v( ) f  

 2.  Solve  for  the  speed:    

uv= 1.15−1= 0.15 ⇒ u = 0.15v = 0.15 343 m/s( ) = 51 m/s

    Insight:  Note  that  the  fractional  increase  in  frequency  is  equal  to  the  ratio  of  your  speed  to  the  speed  of  

sound.