Edexcel - London Examinations GCE, May 1997 - Paper 2 Question 1

 

1.

 a) i) a = (v - u ) / t = (26 m/s - 0 m/s) / 2.5 s

= 10.4 m / s2 Answer a) i)

 ii) F = m . a = 750 kg x 10.4 m/s2

 = 7 800 N Answer a) ii)

 iii) K.E. = (m v2) / 2 = (750 kg x 26 m/s x 26 m/s) / 2

 = 253 500 J Answer a) iii)

 iv) P.E. gained at C = K.E. gained at B = 253 500 J

 but P.E. = m.g.h

 therefore h = 253 500 J / m . g

 = 253 500 / (750 x 10)

 = 33.8 m Answer a) iv)

 The vertical height is actually less because some of the K.E. possessed by the car is used up in overcoming frictional forces and air resistance (drag forces) while moving. (11 marks)

 b)

 Experiment to investigate the relationship between unbalanced force and the acceleration of a trolley - given a newton-balance

 Extra apparatus needed:

 ticker tape and timer, ruler.

 

 Method:

 1. Place the trolley at the top of the runway. (see diagram above)

 2. Switch on the timer and pull the trolley with a constant force down the runway. Make sure force is kept constant while the trolley is moving (say 1 N by looking at the reading on the newton-balance.

 3. Find the acceleration from the tape.

 4. Repeat the experiment, increasing the force used e.g. 2 N.

 5. Keep repeating this procedure, each time increasing the force used and measuring the corresponding acceleration from the tape.

 Put down the results in a table as follows:

 

 

              
 
Force in newtons (from Newton-meter)
            acceleration in metre per second squared
            		             

             

            A graph of acceleration against force is then plotted. (9 marks)