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Physics Semester 1  STPM  Chapter 1  Physical Quantities and Units

               Section  B   Structured Questions
              Question 4

                (a)  The force of attraction F between the Sun with mass M and a planet of mass
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                    m is related by the expression
                                             F =  GMm .
                                                  r 2
                      Where r is the distance of separation between the Sun and the planet. Use
                    the expression to find units for the gravitational constant G in terms of base
                    units.

         Semester
         1      (b)  The orbital period T of the planet about the Sun is related to the distance
                    r from the sun as shown in the equation below.
                                            x
                                                     y
                                           T  = (  4π 2  )M  r z
                                                G
                    Where M is the mass of the Sun and  G is the gravitational constant.
                    Determine the values of x, y and z.

              Answer:
                 (a)  F  =   GMm
                           r 2
                     G  =   Fr 2
                         Mm
                        Unit of F = kg m s kg 2 –2
                                    –2
          Penerbitan Pelangi           y  z
                                kg m s m
                     Unit of G  =

                              = m  kg s
                                 3


                                    –1 –2
                 (b)  [T]  = T
                              –1
                                 –2
                           3
                     [G]  = L  M  T
                     [r]  = L
                     [M] = M
                      x
                                  –2 –1
                     T   = (L  M  T ) M L
                               –1
                            3
                               1
                           –3
                                    y  z
                        = L  M  T M L
                                 2
                           z–3
                               y+1
                                   2
                        = L  M  T
                      Equating the coefficients:
                     x = 2
                     y + 1 = 0,  y = –1
                     z – 3 = 0,  z = 3
                                               4
         01 STPM Q&A PHY T1.indd   4                                         17/02/2022   10:03 AM
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