Question

6.   A   10.0-kg   mass   is   traveling   to   the   right   with   a   speed   of   2.00   m/s   on   a   frictionless   horizontal   surface   when   it   collides   with   and   sticks   to   a   second   10.0-kg   mass   that   is   initially   at   rest   but   is   attached   to   a   light   spring   that   is   neither   stretched   nor   compressed   with   a   force   constant   80.0   N/m.   The   system   undergoes   SHM.   A)   Find   the   frequency,   amplitude,   the   period   of   the   subsequent   oscillations   and   the   phase   angle.   B)   Find   the   maximum   and   minimum   velocities   and   acceleration   attained   by   the   oscillating   system   C)   How   long   does   it   take   the   system   to   return   the   first   time   to   the   position   it   had   immediately   after   the   collision?   D)   Write   the   equations   for   the   displacement,   velocity,   and   acceleration   of   the   system   as   function   of   time.  

7.   In   March   2006,   two   small   satellites   were   discovered   orbiting   Pluto,   one   at   a   distance   of   48,000   km   and   the   other   at   64,000   km.   Pluto   already   was   known   to   have   a   large   satellite   Charon,   orbiting   at   19,600   km   with   an   orbital   period   of   6.39   days.   Assuming   that   the   satellites   do   not   affect   each   other,   find   the   orbital   periods   of   the   two   small   satellites   without   the   mass   of   Pluto.  
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Answer 1

7) To figure out the periods of Pluto's additional satellites without using Pluto's mass, we can set up a proportionality constant between Charon and the other two satellites using Kepler's third law.

Using the following equation,

Recognize that besides the periods and radii of orbits, is going to be exactly the same for both the satellites (ie, constant).

Therefore, we can see for Charon is equal to for both of the satellites and solve what the periods of those satellites would be.

For the first satellite :

  

= 2.12 × 10⁶ seconds = 24.5 days

Similarly for second satellite,

= 3.26 × 10⁶ seconds = 37.7 days

For satellite of distance 48000 km, T =24.5 days and for satellite at distance 64000km, T = 37.7 days.