Physics Paper 2, Nov/Dec. 2012  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments

Question 11

 Question 11

  1.     Define:
  2. work;
  3. energy.
  4.    A pump is used to raise water from a depth of 20 m to fill a reservoir of volume

   1800 m3 in 5 hours. Calculate the power of the pump.
         ( Density of water = 1000 kg m-3; g = 10 ms-2)

  1.   (i)   What is terminal velocity?

  (ii)    Explain the action of a parachute.

    •    Why is it important that a defender in football should have a lot of mass?
  1.  Most candidates gave the correct definition of energy but could not define work   correctly.  The phrase “ in the direction of the applied force “ was missing in their definition of work.  Some candidates mistook the definition of energy for work.
  2.  The calculation of the power of the pump was well tackled by most respondent candidates.  However, very few candidates had problems deriving the mass of water m from volume v and density ρ of water.  Some candidates did not convert the time from hours to seconds before substitution while few other candidates could not relate  Pt  =  mgh.
  1.  Many candidates were not able to explain the concept of terminal velocity nor the action of a parachute as many of them failed to establish the fact that the amount of air resistance experienced is increased by the umbrella of the parachute. Performance was low.


  1.  The reason why a defender in football should have a lot of mass was poorly tackled by many candidates as they could not relate mass to inertia.

The expected answers are:
a)(i)     Work is the product of  the force and  displacement/ distance in the
           direction of the force.                                                                                             
                          Accept W = Fs  provided  the letters  are correctly defined.

               (ii)     Energy is the capacity / ability to do work.                                                          

(b)       m                         =      D x V                                                                                     
                                        =      1000 x 1800                                                                         
                                        =      1,800,000                                                                             
             Pt                      =      mgh                                                                                       
           P x 5 x 60 x 60   =      1,800,000 x 10 x 20                                                                        
            P                         =      20,000W                                                                               

(c) (i)   The steady/constant speed with which an object/body moves freely
           downward through a fluid when the resultant force on it is zero                                                                                                      
The steady/constant speed attained by an object/body falling freely through a fluid when the viscous force and upthrust of the fluid on it are (jointly) equal to the weight of the object/body

(ii)     The umbrella of the parachute increases the amount of air resistance
         experienced  by the sky driver and this in turn slows down the (rate of )fall
         until terminal velocity is attained.

(d)       An attacker runs towards a defender’s goal with high velocity that
           generates high momentum in order to stop/block the attacker the defender 
           must have great inertia which is a measure of mass so the defender must
           have a lot of mass.


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