1. ai) The two nuclei are positively charged.
ii) The nuclei will collide and fusion will result.
iii) By conservation of momentum the two nuclei cannot stop at the same instant as the initial momentum is non zero.
b) The total final momentum is 5m v" . The total initial momentum is mv. Hence the final speed is 0.2 v.
2. The uniform field is E = V/d. E = 48 000 Vm-1
ii) the Work = qV where V is the p.d . W= q(900V) = 7.2 x 1016 J
b) The potential is zero in the centre line as the change in V with distance is constant. Moreover both the magnitudes of the plates are the same. Hence the centre is zero.
3. a)The ratio V to I of the graph is the resistance. At the knee of the graph you can find the lowest resistance .
b) The pd across C and 5 ohm is the same. Hence using 5 ohm , V = IR = 4.25 V.
The total current is obtained by adding current through 5 ohm and current through C . To find current through C at 4.25 V check graph . The total I = 1.3 +0.85 = 2.15 A. The E.m.f = 4.25 + p.d across internal resistance . Hence emf = 4.25 + (2.15 x 0.8) = 6.0 V .
The energy to C = VIt , but pd across C = 4.25V and current across C is 1.3 A . Hence E = 4.24 x 2.15x 20 x 60 = 6630 J
4. Discuss the three crucial positions where 1. the GPE is a max and KE = 0 2. KE is max 3. EPE is max and KE = 0
b) The BE of product is more than the BE of reactant and hence energy is released. The loss in mass is converted to energy by E=mc2. The energy is released in the form of KE of the product and gamma photon emitted.
