Physics Paper 3 May/June 2015

Question 1A


You are provided with a retort stand, boss head, clamp, stop watch slotted weights, hanger, grooved pulley, thread, measuring tape and other necessary materials


(i) Measure and record the radius R of the pulley.

  (ii) Set-up the apparatus as illustrated in the diagram above, such

               that the clamp is 1.5 m above the floor.

  (iii) Tie one end of the thread to the pulley.

(iv) Tie the other end of the thread to the hanger.

(v) Slot a mass m  =  50 g on the hanger.

(vi) Wind the thread around the groove of the pulley until the base of

                the hanger is at a height h = 1.4 m above the floor.

                Maintain this height h for every other value of m throughout the experiment.

(vii)Release the mass to unwind the thread

(viii) Determine and record the time t taken by the mass m to reach the floor.

(ix)  Evaluate t2

(x)  Also evaluate a  =  , T =  (10 – ) and  =

 (xi)  Repeat the procedure for four other values of m = 70 g,90 g,110 g and 130 g.     

(xii)     Tabulate your reading

   (xiii)     Plot a graph with Cl on the vertical axis and T on the horizontal
s .

   (xiv)     Determine the slope s, of the graph
  (xv)     Evaluate I =   

  (xvi)        State two precautions taken to obtain accurate results
                                                                                             [ 21 marks]
(b) (i)   Define centripetal force                                      [2 marks]                               

(ii)    An object drops to the ground from a height of 2.0 m, calculate the speed with
which it strikes the ground

                 [ g = 10ms-2 ]                                                         [ 2 marks ]


    Part (a) this question was not popular among the candidates.  Candidates are challenged in 
    measuring the time of  fall of the weight.  Most responding candidates failed to record the
    value of R as demanded in the question.  Few recorded the radius as 6m!  Some had problem
    in evaluating  T =  ( 10 – a)   However, the graph and slope were well attempted .
The precautions were well stated.
    Part (b) many candidates failed to do justice to this part.  Performance was poor.
    Candidates were expected to

  •      Measure and record Radius of pulley to at least 1 d. p. in cm          
  •      Record five values of m in grams                                                                     
  •      Measure and record five values of t  to at least 1 d.p in seconds and in t
  •            trend: as m increases t decreases                                                   

  •      Evaluated five values of t2 to at least 2 d. p.                                                               
  •      Evaluated five values of a =                            
  •      Evaluated five values of T =  ( 10 – a)                                 
  • Evaluated  five values of   =                                                             
  • Composite table  m, t, t2, a,T and   in distinguished both axes   
  •  Plot graph using reasonable scales
  • Draw line of best fit
  • Evaluate the slope as I =
  • State any two of the following precautions
  • Retort stand / pulley firmly clamped
  • Avoided parallax error in reading stop-watch/clock/metre rule/ tape rule
  • Noted / corrected zero error on stop watch/ clock/metre rule / tape rule
  • Repeated readings shown on table.
  •       The expected answers for part (b) are:

            b(i)     Centripetal force is the inward force / force acting towards the centre of the circle
    required to keep an object moving with a constant speed in a circular path                                       OR                             
    F  =                        Where the symbols have their usual meanings



     (ii)             v2 = u22gh                OR                   k.E. = P.E.
    v2  =02  +  2 x 10 x 2                                  ½ mv2  = mgh                                    
    =  40                                                              v2 =  2 x 10 x 2     
    v  =                                                         v =  6.32  ms-1      
    =   6.32 ms-1