Physics Paper 2, MAy/June. 2013  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main

General Comments

Question 11

  1. State the triangle law of vector addition (b) Name the four physical quantities that are associated with the equations of linear motion (c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.


Part (a):  This was attempted by many candidates but few failed this technically by  writing
               -   three forces can be represented by the three sides of a triangle taken in  order.
               -  Given three forces they can be represented in magnitude and direction by the three sides of a triangle taken in order.
                  While the first was incorrect outright, the second failed to include the necessary condition that the three forces must be in equilibrium.
Part (b):     This did not pose a problem to most responding candidates.  Performance was above average.
Part (c):     This was avoided by majority of the candidates.  Many got the shapes wrong and a few candidates used separate axes which was contrary to the instruction.  Performance was poor.
Part (d):     (i-iii)  was done correctly by most candidates however, a few candidates could not calculate (i-iii) because they applied the wrong formula.
Part(iv)      posed a challenge to some of the respondent candidates.  They approached it thus:
                  Ft  =  (Mp  +  MQ)v.  This was not a problem of impulse but that of conservation of momentum i.e
                      MpUp  + MQUQ  =  (Mp + MQ)v.
Performance was poor.
The expected answers are:
a)           Triangle law of vector  addition
               If three forces are in equilibrium, they can be represented in magnitude and direction by the  
               three sides of a triangle taken in order        
     (b)   Physical quantities associated with equations of motion


              e.g  -  distance/displacement
                     -   speed/velocity
                     -   acceleration
                     -   time.                                      





Both axes correctly distinguished                          
 Correct shape for potential energy                                                          
Correct shape for kinetic energy                                                               
(d)  (i)    W = ½ ke2                                                                                                
               10  =  ½ x k  x( )2                                                                             
                k   =  8000  Nm-1            


      (ii)     F = ke                                                                                                      
                  =  8000 x                                                                                   
                  =    400N                                                                                            

     (iii)     F = ma                                                                     a = ω2A =  A               
                400 =  2.0 x a                                  OR                    = x 5 x 10-2
                a =  200 ms-2                                                              = 200 ms-2        

     (iv)     Initial velocity of block P = Up = 200 x 0.25 = 50ms-1                       

                Mp Up + MQUQ  =   (Mp+MQ)v                                                           
                Where v is the common velocity after collision
                  (2 x 50) + (4.0 x 0)    =  (2 + 4) v                                                      
                    100                        =  6v
                             v   =   16.7 ms-1                                                                       





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