Physics Paper 2, Nov/Dec. 2011  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments

Question 11

 (a)     (i)  Explain relative motion.

         (ii)  Two cars moving in opposite directions on the same straight road with velocities 80 km
                h-1 and 60 km h-1 respectively pass each other at a point.  Determine the velocity of
                the first car relative to the second.

(b)     (i)   Define force

         (ii)   Classify the following forces as either contact forces or field forces; push; tension;
                 gravitational; electrostatic; reaction and magnetic.

(c)      A body starts from rest and travels in a straight line for 2 seconds with uniform
           acceleration of 1 ms-2.  It then travels at a constant speed for some time before coming
           to rest with a uniform retardation of 2 ms-2.  The total distance covered by it is 1.5m

         (i) Draw and label a velocity – time graph for the motion.
        (ii)  Calculate the:
                                    (A)  velocity attained after the first 2 seconds;
                                    (B)   total time taken for the journey.

Part(b) many candidates were unable to explain relative motion.  Those that made effort lost marks because they did not mention the frame of reference in their explanation.  Performance was average.

Part(b) majority of the responding candidates were able to define correctly force and classify the given forces as  either contact forces or field forces.

Part(c) the drawing of the velocity –time graph did not pose any problem to the candidates; however, majority of those that attempted this question failed to calculate correctly the total time taken for the journey by the body.


The expected answers are:

a)(i)   This is a change in position measured from a point or frame of reference which may
           be stationary or moving e.g.
                      -      movement of a person walking inside a moving train
                      -      a person inside a moving train observing a stationery observer/tree outside the
                             train .
         (ii)        Relative velocity  =  V = Vfirst – V second                                                       
                                                     =  80 – (-60)                                                                        
                                                     = 140 kmh -1                                                                      

        (b)(i)    Force is an agent that changes or tends to change the state of rest or of uniform
                    motion in a straight line of a body.                                                               

Contact forces

Field forces


Gravitational force
Electrostatic force
Magnetic force.

       (c)   V/ms-1                                                              


                                                                                            Axes correctly distinguished                                    
                                                                                              origin indicated (020)  

   (ii)   (A)   Acceleration  a   =   V/t1                                                                                                      
                      V  =  at1
                      =  1  x  2                                                                                                                    
                      =   2ms-1                                                                                                                 

         Retardation a1  =  V/t3
          t3  =  V/a1       =   2/2                                                                                                              
                              =  Is
          x  =   area of trapezium
          15   =   ½  (t2 + t2 + 3) 2                                                                                                            
                 =  2t2  +  3
             t2  =  6s                                                                                                                         

       Total time required  =  6  + 3 = 9s

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