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For verticsl motion of projectile h = (U sin \(\theta\)) t \(\frac{1}{2}\)gt\(^2\) = U sin \(\theta\) (\(\frac{Usin \theta}{g}\))\(\frac{1}{2}\)g (\(\frac{Usin \theta}{g}\)) = (\(\frac{U^2 \sin^2 \theta}{g}\))

(a) The molecule or atoms of a gas must be ionised before the gas can conduct electricity. Ionisation of the gas requires that light the gas pressure is greatly reduced within the enclosure and a high voltage is applied to the enclosed gas. (b) The electric charge carriers in gases are electrons and ions.

(a) Cathode rays: (i) Are negatively charged particles (ii) In field free space travel in straight line. (iii) Are deflected by electric and magnetic fields. (b) The intensity of the cathode rays may be increased by increasing the current through the filament. The energy of the cathode rays may be increased by increasing the p.d. between the anode and the cathode i.e. the anode potential.

\(\lambda = \frac{h}{p} = \frac{h}{mv}\) V = \(\frac{h}{m_2}\) Electron kitenic energy = \(\frac{1}{2} MV^2 = \frac{h^2}{2m \lambda^2}\) = \(\frac{(6.6 \times 10^{-34})^2}{2 \times 9.0 \times 10^{-37} \times 1.0 \times 10^{-20}}\) = 2.39 x 10\(^{-17}\)J

(i) Slope represents the Planck's constant, h. (ii) Intercept OC represents the work function of the metal. (iii) The intercept OK represents the threshold frequency of the radiation.

(a)(i) Stress is force per unit cross sectional area, while strain is extension over original length (ii) Stress has unit of Nm\(^{-2}\) while strain has no unit. (b) Work done on the cord = \(\frac{1}{2}\) Fe = \(\frac{1}{2}\) x 40 x 0.05 = 1.0J

(a) Range = \(\frac{U^2 sin 2\theta}{g}\) for first attempy, Range = \(\frac{8^2 \sin 60}{10} = \frac{64 x\times 0.8660}{10}\) = 5.5424m for second attemt; Range = \(\frac{8^2 \sin 90}{10} = \frac{64 \times 1}{10}\) = 6.4m gain 6.40 - 5.54 = 0.8576m (b)(i) Hooke's law states that provided the limit of proportionality is not exceeded, the force is proportional to the extension in an elastic material. (ii) Description: Suspend the spiral spring from a rigid support with a pointer attached at the free end. Suspend a stretching weight from the free end, and note the position of the pointer. Suspend a known mass, note the new position and determine the extension. Repeat the experiment using various known masses. Repeat the above procedure while unloading and determine mean extension. Plot a graph of load against mean extension. A straight line graph through the origin verifies Hooke's law. Precautions: (i) Loading beyond elastic limit should be avoided (ii) Pointer to be rigidly attached to the free end and made perpendicular to the spring. (iii) Avoid parallax error while taking readings. (c) mg = Ke, m = total mass If m = mass of scale pan then; (m + 40)g = 1.8k-----------(i) (m + 60) = 2.05k --------- (ii) Dividing (i) by (ii) \(\frac{m + 40}{m + 60} = \frac{1.8}{2.05}\) 2.05(m + 40) = 1.8 (m + 60) M = \(\frac{26}{0.25}\) = 104gm

(a) Specific heat capacity is the quantity of heat energy required to change the temperature of unit mass (1kg) of a substance by 1k. (b) Diagram Method Weigh empty calorimeter plus stirrer and record the mass M\(_1\), weigh empty calorimeter plus stirrer plus water and record the mass m. Weigh copper ball and record the mass Mc. Take the initial temperature of the water T\(_1\). Heat copper ball in water to temperature T and record T. Quickly transfer the copper ball into the calorimeter and stir. Take the final temperature of the mixture T\(_2\). Heat lost by copper ball = Heat gained by calorimeter; water and stirrer. McCc (T - T\(_2\)) = (M\(_2\) - M\(_1\))C\(_w\) (T\(_2\) - T\(_1\)) + M\(_1\) Cal(T\(_2\) - T\(_1\)) Were Cc, Cw and Cz = specific heat capacity of copper, water and calorimeter respectively. Cc = \(\frac{ (M - M_1C + M Cal (T - T)}{Mc(T - T_2)}\) Precautions: (i) Calorimeter should be lagged. (ii) Quickly transfer the hot ball to the calorimeter. (iii) Shake off water from hot calorimeter metal ball before transfer. (iv) Mix thoroughly the mixture with stirrer. (c) Heat lost = 0.024 x 400 (230 - T). Heat gained by water = 0.054 x 4200 (T - 31) Heat gained by calorimeter = 0.060 x 400 (T - 31) Heat lost = Heat gained = 0.024 x 400 (230 - T) = 0.054 x 4200 (T - 31) + 0.060 x 400 (T - 31) T = 38.34^\circC

(a)(i) An echo is a reflected sound. (ii) Echoes can be used to: - Determine the speed of sound in air. - Determine the depth of a sea bed. - Locate the positions of shoals of fish. - Echo sounding is used in mineral prospecting. (iii) Padding absorbs sound and minimizes reflections at the walls, floors or ceilings. This helps to minimize the effects of echoes. (b)(i) Timbre is the characteristic quality of a sound produced by a particular voice or instrument. Timbre of the note is pure and it comes out unaccompanied by undesired notes. Overtones are sound notes which accompany the fundamental notes. (ii) Resonance is said to occur whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other system which is vibrating with the same frequency. (c) Let S\(_1\) = distance i between the ship and the cliff in the first instance \(\frac{2S}{12}\) = 340; S\(_1\) = 2040m In the second instance \(\frac{2S_2}{8}\) = 340 S\(_2\) = 1360m Distance travelled by ship in 2.1 minute S\(_1\) - S\(_2\) = 2040 - 1360 = 680m Ship velocity V = \(\frac{680 }{2.1 \times 60}\) = 5.4ms\(^{-1}\)

(a) \(^{14}_7 + ^1_0n\) -> \(^4_2He + ^{11}_5X\) (b)(i) The radioactive atom disintegrates spontaneously while a stable atom does not disintegrate on its own. (ii) The time at which the number of radioactive atoms remaining is just one-half of the original number is called half-life of the element. The half-life T\(_{\frac{1}{2}}\) is related to the decay constant (\(\lambda\)) of the isotope by the equation. \(T \frac{1}{2} = \frac{0.693}{ \lambda}\) (iii) After 35days, 1/2 of it remains. After 70 days, 1/2 of 1/2 i.e. 1/4 remains. After 105 days, % of 1/4 i.e. 1/8 remains. (c)(i) The photon energy Ep = hf = \(\frac{hc}{f}\) = \(\frac{6.6 \times 10^{-34} x 3.0 \times 10^8}{ 5.0 x 10^{-7}}\) = 3.96x 10-19J (ii) The kinetic energy E\(_k\) = (Ep - W) E\(_k\) = hf - W = \(\frac{hc - W}{\lambda}\) W = 1.9eV = 1.6 x 10\(^{19}\) x 1.9 = 3.044 x 10\(^{-19}\)J. E\(_k\) = 3.96 x 10\(^{-19}\) - 3.044 x 10\(^{19}\) = 9.2 x 10\(^{-20}\)J. (iii) Stopping potential Vs = energy needed to stop most energetic electron. eVs =E\(_{kmax}\) Vs = \(\frac{E_{max}}{e}\) = \(\frac{9.20 \times 10^{-20}}{1.6 \times 10^{-19}}\) = 0.575 V

Table of values Vcm Ucm 10.0 47.0 15.0 48.3 20.0 49.4 25.0 51.20 30.0 52.50 35.0 54.00 Slope = \(\frac{Du}{DN} = \frac{13}{47}\) = 0.28 The intercept (c) on the vertical axis = 44cm (i) \(K_1\) = \(\frac{(1-2s)}{s}\)100 = \(\frac{1-0.56}{0.28}\)100 = 157 (ii) \(K_2\) = \(\frac{2c}{s}\) - 160 = \(\frac{2 \times 44}{0.28}\) - 160 = 154.3 Precautions: I Avoid draught. I Ensure mass does not rest on the table I avoid errors due to parallax when reading metre rule. (b)i. The conditions under which a rigid body remains in equilibrium when acted upon by three non-parallel coplanar forces are: The forces must be concurrent (i.e. they must meet at a point. The resultant of the forces must be equal to zero. The forces can be represented in magnitude and direction by the sides of a triangle taken in order. A body is in stable equilibrium if the vertical line through the centre of gravity falls outside the base of support when the body is slightly titled

The acceleration is not constant, i.e. its direction changes although the magnitude is constant (III is wrong), another thing that makes III wrong is the fact that the acceleration is not directed away from the centre but towards the centre. - No work is done on a body undergoing circular motion - Note that when a body is undergoing circular motion ], its velocity is not constant, i.e. the direction changes, although the magnitude is constant. - Since the acceleration is directed towards the centre, the centripetal force is directed towards the centre.

Efficiency = \(\frac{\text{M.A}}{\text{V.R}}\) 75% = \(\frac{\text{M.A}}{\text{V.R}}\) \(\frac{75}{100} = \frac{\text{M.A}}{5}\) M.A = \(\frac{75(5)}{100}\) = 3.75 But M.A = \(\frac{\text{Load}}{\text{Effort}}\) 3.75 = \(\frac{150}{\text{Effort}}\) Effort = \(\frac{150}{3.75}\) Effort = 40N

Given: mass of stone = 500g = 0.5kg, e = 20cm = 0.2m, velocity of stone on leaving the catapult = 40m/s The potential energy stored in the catapult when extended = the kinetic energy (KE) of the stone \(\frac{1}{2}\)Fe = \(\frac{1}{2}\)mv\(^2\) F x 0.2 = 0.5 x 40 x 40 F = \(\frac{800}{0.2}\) = 4000 N = 4 x 10\(^3\) N

Let the velocity of stream be Vs = 0.75m/s Angle at which the boat lands \theta = 30º the velocity of the boat = Vb \(\frac{0.75}{V_b}\) = Sin30º = 0.5 ∴ V\(_b\) = \(\frac{0.75}{0.5}\) = 1.5m/s