Page 3 - 20. Sci Phy Answer Key 2023 (AK016)
P. 3

Section C Free response questions                 (c)  Average velocity = 58.2/0.75 = 77.6 km/h

              1.  (a) A  scalar  quantity  is  a  quantity  that  has  a   Revision 2
                     magnitude  only  while  a  vector  quantity  is  a
                     quantity  that  has  both  the  magnitude  and   1.  (a)   It is a rate of change of velocity.
                     direction.                                 (b)   An object experiences a change in the velocity./
                 (b) Total distance = 240 + 160 = 400 m             An object experiences a change in direction.
                     Take the direction towards east as positive.     (c)  Deceleration
                     Displacement = 240 + (–160)
                                                                   u
                                         = 80 m due east    2.  v -= 80 -   =45  35 km/h
                                                                         .
                                                                35 km/h = 9 722 m/s
              2.  (a)  Set  up  a  simple  pendulum  as  shown.  Make   v - u  9 722   2
                                                                          .
                     the pendulum swing (or oscillate). Record the   a =  t  =  4  =  243 m/s.
                     time t for 20 complete oscillations by using a
                     stopwatch. Repeat the experiment a few times   3.  v =+  at =  + 0  (.  )(  = 10  m/s
                                                                   u
                                                                             2 05)
                     to obtain the average value of t. Find the time
                     for one complete oscillation, T:           Distance travelled in 5 seconds  =  1 ´ ´  ´10 5  = 25 m
                     T= t/20 = Time for one complete oscillation                         2
                                                                Remaining distance of 10 m/s  =  - 125  25  = 100  m
                                                                                  100
                                                                Time taaken for 100 m =  =  10  s
                                                                                   10
                                                                Total time taken =+10  = 15  s
                                                                               5
                                                            Revision 3
                                                            1.  (a)   B           (b)   C
                                bob                             (c)   A           (d)   D

                                                            2.  (a)  C            (b) D

                 (b)  (i)  No           (ii)  Yes             (iii)  No  (c)  E   (d) B
                                                                (e)  A
              Unit 2: Kinematics                                                        8  - 0
              Revision 1                                    3.  (a)  v =  2 - 0  = 4 m/s

              1.  (a)   T [The object may be travelling in a circular   (b) He is at rest at a distance 8 m from the
                         path and returns to the original position.]  starting point.
                 (b)   T  [When  the  object  moves  in  the  opposite
                     direction to its original direction.]      (c)  v =  12 -8  =  4  =1 m/s
                 (c)   F  [It  is  defined  as  a  rate  of  change  of   10  -6  4
                     displacement.]                                 John’s speed is 1 m/s.
                 (d)   F [It measures the instantaneous speed.]
                 (e)   T                                    Revision 4
                 (f)   F [It is called motion.]
                 (g)   T                                    1.  (a)  D            (b) E
                                                                (c)  G            (d) C
              2.  (a)   Speed = 85 000/(35 × 60) = 40.5 m/s
                 (b)   Time = 30 000/11.9 = 2521 s = 42.0 min   (e)  A            (f) B
                 (c)   Total distance = 1200 × 8 = 9600 km      (g)  F

              3.  (a)   (i)   Distance  =  24  m/s  ×  (15  ×  60)  s  =   2.  (a)   I:   The  object  moves  with  a  non-uniform
                         21600 m = 21.6 km                             increasing acceleration.
                     (ii)   Distance  =  30  m/s  ×  (30  ×  60)  s  =      II:  The  object  moves  with  a  non-uniform
                         54000 m = 54.0 km                             decreasing acceleration.
                 (b)   Total distance = 21.6 + 54.0 = 75.6 km       III:   The  object  moves  with  a  uniform  speed
                     Total time taken = 45 minutes = 0.75 h            with zero acceleration.
                     Average speed = 75.6/0.75 = 101 km/h           IV:   The  object  moves  with  a  non-uniform
                                                                       decreasing deceleration.

            338                                                                              © GLM Pte Ltd
   1   2   3   4   5   6   7   8