Edexcel - London Examinations GCE, January 1999 - Paper 2 Question 1

1.

a) i) F = m . a = 7 000 kg x 16.7 m/s = 116 900 N Answer a) (i)

ii) The two forces are: Air resistance (or push of air on car)

and Frictional forces

iii) Push of car on air = [air resistance ] + [ frictional forces]

iv) a = ? u = 350 m/s v = 0 m/s t = 34 s

a = (v - u) / t

or a = (0 - 350 m/s) / 34 s

= 10.3 m/s² Answer (a)

v) distance = area under graph = area of trapezium

= ½ (sum of parallel sides) x height = ½ (60 + 5) x 350 = 11 375 m Answer (a) (v)

vi) The car exceeds the speed of sound (which is about 330 m/s). Hence it is said to be supersonic or faster than sound. (12 marks)

b) i) Work needed to be done = K.E. possessed (as work is converted to K.E) = 4.3 x 10⁸ J. Answer (b) (i)

ii) Work done = Force x distance to stop it

4.3 x 108 J = f x 5950 m (where f is force needed)

So, f = 4.3 x 108 J / 5950 m = 72 269 N Answer (b) (ii)

(iii) It changes to heat energy (as well as some sound energy).

(iv) The air resistance acting against the parachute helps to increase the opposing forces and hence the decelerating force (an unbalance force in the opposite direction to motion) increases in size, stopping the car more quickly. (8 marks)