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(a) (i) Definition of force Action/push/pull that changes/tends to change the state of rest or of uniform motion in straight line. Its unit is newton (N) (ii) Types of solid friction - dynamic (friction) - static (friction) (b) (i) Velocity-time graph for journey

The expected answer is: (a) (i) Identification of labelled parts of photocell A - Emitter / cathode B - Collector / anode (ii) Function of part of photocell A - emits electrons B - attracts electrons (iii) Meaning of terms in Einstein’s photoelectric equation E - kinetic energy of photoelectrons hf - energy of incident photon Wo - work function of metal

K.E = mgh but h = 80m, m = 2kg, g = 10m/s^2 K.E = 2 x 10 x 80 = 1600J.

No expanded working has been attached for this question yet. The verified answer is Option C.

Elastic Force: This is the force exerted by the spring when it is compressed or stretched. It is the elastic force that directly acts to restore the spring to its equilibrium position.

Wavelength (\(\alpha\)) = \(\frac{Velocity (V)}{Frequency (f)}\) Velocity = \(\frac{Distance}{Time}\) = \(\frac{4.5}{0.6}\) = \(\frac{7.5}{9.0}\) = 0.83m

For a transformer with 100% efficiency, the input power equals the output power: E\(_p\)​I\(_p\) ​= E\(_s\)​I\(_s\)​

P = \(\frac{V^2}{R}\) R = \(\frac{120^2}{100}\) = 144.0\(\Omega\)

Energy is directly proportional to mass when mass is converted to energy. According to Albert Einstein’s mass-energy equivalence formula: E = mc\(^2\)

From momentum conservation (800 x 80) + (1200 x 40) = (800 + 1200) V = 64000 + 48000 = 2000V 112000 = 2000V V = \(\frac{112000}{2000}\) = 56kmh\(^-1\) The common velocity = 56kmhr\(^{-1}\)

Force = mass x acceleration but force was halved = 1/2 and mass doubled = 2 ( assuming that force and mass are 1N and 1kg respectively) f = ma = \(\frac{1}{2}\) = 2 x a dividing both sides by 2 a = \(\frac{1}{2} \times\frac{1}{2} = \frac{1}{4}\)

Centripetal force (Fr) = \(\frac{MV^2}{r}\) i.e. Fr = \(\frac{Mv^2}{r}\) = \(\frac{9.1 \times 10^{-31} \times 2.0 \times 10^6 \times 2.0 \times 10^6}{6.1 \times 10^{-11}}\) N = (\(\frac{36.4 \times 10^{-19}}{6.1 \times 10^{-11}}\)) N = 5.967 x 10\(^{-8}\) N ~ 6.0 x 10\(^{-8}\) N

No expanded working has been attached for this question yet. The verified answer is Option D.

(a) The emission of light from a substance when its excited electrons move to a lower energy level. OR Distinct/separate bright lines of light of definite frequency/wavelength (emitted from vapourised atoms) (ii) Emission spectra is the pattern of light displayed when electrons jump from a higher energy level to a lower energy level, while absorption spectra is the displayed pattern of wavelengths absorbed by the gas. (b) \(\Delta\)E = E\(_i\) - E\(_f\) = -1.51 + 13.6 = 12.09 eV = 12.09 x 1.6 x 10\(^{-19}\) J = 1.934 x 10\(^{-18}\) J (ii) \(\Delta\)E- hf f = \(\frac{\Delta E}{h}\) = \(\frac{1.934 \times 10^{-18}}{6.6 \times 10^{-34}}\) f = 2.93 x 10\(^{15}\)Hz (ii) c = f\(\lambda\) \(\lambda\) = \(\frac{3 \times 10^8}{2.93 x 10^{13}}\) = 1.023 x 10\(^{-7}\)m (c) (I) Soft x-rays are those with longer wavelengths/lowNer penetrating/less energy while hard x-rays are those with shorter wavelengths/higher penetrating power more energy. (iii) To increase the electrical conductivity/reduce electrical resistivity of the semiconductor.

Using the formula F = \(\frac{Gm_1m_2}{r^2}\) = \(\frac{6.67 \times 10^{-11} \times 3.0 \times 2.0}{(50)^2}\) = \(\frac{40.02 \times 10^{-11}}{2500}\) = 1.6 x 10\(^{-9}\)N