5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳

2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.

14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.