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S5 Physics Practice Paper: Mechanics, Electricity and Waves
S5 Physics: Mechanics, Electricity and Waves
S5 Physics paper: mechanics, momentum, circular motion, parallel circuits, electrical power, Coulomb law, induction, sound, refraction and lenses. 12 questions, 100 marks, 75 minutes.
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Paper details Level: S5 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Mechanics (Q1-Q4)
Section A
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A trolley starts from rest and accelerates uniformly at 2.5 m s-2 for 6.0 s, then travels at constant speed for 4.0 s. (a) Find the final speed of the accelerating stage. (b) Find the displacement during that stage. (c) Find the total displacement in 10.0 s. (d) State what the gradient and the area under a velocity-time graph represent. | (a) v = u + at = 0 + 2.5 x 6.0 = 15.0 m s-1. (b) s = 0.5 a t2 = 0.5 x 2.5 x 6.0^2 = 45.0 m. (c) Constant-speed part = 15.0 x 4.0 = 60.0 m, so the total is 45.0 + 60.0 = 105 m. (d) The gradient is the acceleration and the area under the graph is the displacement. Marking: formula, substitution, units, graph meaning. |
| 2 | A block of mass 4.0 kg rests on a horizontal table and is pulled by a horizontal force of 24.0 N. The coefficient of kinetic friction is 0.25 and g = 9.81 m s-2. (a) Calculate the friction force. (b) Calculate the acceleration. (c) Find the speed after 3.0 s. (d) Describe what happens if the pull is reduced to 9.0 N. | (a) f = mu N = mu m g = 0.25 x 4.0 x 9.81 = 9.81 N. (b) Net force = 24.0 – 9.81 = 14.19 N, so a = 3.55 m s-2. (c) v = at = 3.55 x 3.0 = 10.64 m s-1. (d) 9.0 N is less than the maximum static friction of 9.81 N, so the block stays at rest and friction balances the pull. Marking: friction formula, net force, acceleration, static friction reasoning. |
| 3 | A trolley of mass 2.0 kg moving at 3.0 m s-1 collides with a stationary trolley of mass 1.0 kg and the two stick together. (a) Find the common speed. (b) Find the total kinetic energy before the collision. (c) Find the total kinetic energy after the collision and the energy lost. (d) State where the lost energy goes. | (a) Momentum is conserved: 2.0 x 3.0 = (2.0 + 1.0) x v, so v = 2.0 m s-1. (b) Initial kinetic energy = 0.5 x 2.0 x 3.0^2 = 9.0 J. (c) Final kinetic energy = 0.5 x 3.0 x 2.0^2 = 6.0 J, so 3.0 J is lost. (d) The lost energy becomes heat, sound and deformation of the trolleys. Marking: conservation equation, both energies, energy transfer. |
| 4 | A ball of mass 0.50 kg moves in a horizontal circle on a string of length 1.2 m at a speed of 3.0 m s-1, the string being horizontal. (a) Find the centripetal acceleration. (b) Find the tension in the string. (c) Find the time for one revolution. (d) If the speed doubles, by what factor does the tension change? | (a) a = v2 / r = 3.0^2 / 1.2 = 7.5 m s-2. (b) Tension supplies the centripetal force: T = ma = 0.50 x 7.5 = 3.75 N. (c) Period = 2 pi r / v = 2 pi x 1.2 / 3.0 = 2.51 s. (d) a is proportional to v2, so doubling v quadruples a and the tension becomes 15.0 N, four times larger. Marking: centripetal formula, force source, period, square relation. |
Section B: Electricity (Q5-Q8)
Section B
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A 6 ohm resistor is in parallel with a 3 ohm resistor, and this combination is in series with a 4 ohm resistor across a 12 V battery of negligible internal resistance. (a) Find the equivalent resistance of the parallel part. (b) Find the total resistance and the total current. (c) Find the voltage across the parallel part. (d) Find the current in the 6 ohm resistor and state how the two branch currents compare. | (a) In parallel 1/Rp = 1/6 + 1/3, so Rp = 2.0 ohm. (b) Total R = Rp + 4 = 6.0 ohm, and I = V / R = 12 / 6.0 = 2.0 A. (c) Voltage across the parallel part = I x Rp = 2.0 x 2.0 = 4.0 V. (d) Current in 6 ohm = 4.0 / 6 = 0.667 A and in 3 ohm = 1.33 A. The smaller resistance carries the larger current and the two currents add to 2.0 A. Marking: parallel formula, series sum, current splitting. |
| 6 | An electric kettle rated at 1500 W is connected to a 220 V supply and used for 8.0 minutes. (a) Find the current. (b) Find the resistance of the heating element. (c) Find the energy used in joules and in kWh. (d) At 1.2 dollars per kWh, find the cost of this heating. | (a) I = P / V = 1500 / 220 = 6.82 A. (b) R = V2 / P = 220^2 / 1500 = 32.3 ohm. (c) E = Pt = 1500 x 480 = 720000 J = 0.20 kWh. (d) Cost = 0.20 x 1.2 = 0.24 dollars. Marking: power formula, resistance form, joule to kWh, cost. |
| 7 | Two point charges of +2.0 microcoulomb and +3.0 microcoulomb are 0.30 m apart in air. Take the Coulomb constant as 9.0 x 10^9 N m2 C-2. (a) Calculate the magnitude of the electrostatic force. (b) State the direction of the force. (c) If the separation doubles, what fraction of the original force remains? (d) How do magnitude and direction change if one charge becomes negative? | (a) F = k q1 q2 / r2 = 9.0 x 10^9 x 2.0 x 10^-6 x 3.0 x 10^-6 / 0.30^2 = 0.60 N. (b) Both charges are positive, so the force is repulsive and acts outward along the line joining them. (c) Doubling the separation makes the denominator four times larger, so one quarter remains: 0.15 N. (d) The force becomes attractive with the same magnitude of 0.60 N, directed towards the other charge. Marking: Coulomb law, inverse square, like charges repel, opposite charges attract. |
| 8 | A coil of 200 turns has the flux through each turn increasing uniformly from zero to 2.0 x 10^-4 Wb in 0.050 s. The total resistance of the coil is 4.0 ohm. (a) Find the induced emf. (b) Find the induced current. (c) Use Lenz law to state the direction of the induced current. (d) What happens to emf and current if the flux changes twice as fast? | (a) emf = N dPhi/dt = 200 x 2.0 x 10^-4 / 0.050 = 0.80 V. (b) I = emf / R = 0.80 / 4.0 = 0.20 A. (c) The flux is increasing, so the induced current produces a field that opposes the increase and points opposite to the original field. (d) Twice the rate of change gives an emf of 1.60 V and a current of 0.40 A. Marking: Faraday law, Ohm law, Lenz law, proportionality. |
Section C: Waves (Q9-Q12)
Section C
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A source produces sound of frequency 250 Hz with a wavelength of 1.36 m in air. (a) Find the speed of sound. (b) Find the period in milliseconds. (c) If the frequency rises to 500 Hz at the same speed, find the new wavelength. (d) State what mainly determines the speed of sound and explain why. | (a) v = f lambda = 250 x 1.36 = 340 m s-1. (b) T = 1 / f = 1 / 250 = 4.0 ms. (c) lambda = v / f = 340 / 500 = 0.68 m. (d) The speed depends mainly on the medium (its elasticity and density) and on temperature; it is about 340 m s-1 in air and higher in solids. Changing the frequency changes only the wavelength. Marking: wave equation, period, medium dependence. |
| 10 | A ray of light enters water from air at an angle of incidence of 40 degrees. The refractive index of water is 1.33. (a) Find the angle of refraction. (b) Find the speed of light in water (take c = 3.0 x 10^8 m s-1). (c) State the two conditions for total internal reflection. (d) Find the critical angle for water to air. | (a) n1 sin theta1 = n2 sin theta2: 1.00 x sin 40 = 1.33 x sin theta2, so sin theta2 = 0.4833 and theta2 = 28.9 degrees. (b) v = c / n = 3.0 x 10^8 / 1.33 = 2.26 x 10^8 m s-1. (c) The light must travel from a denser to a less dense medium and the angle of incidence must exceed the critical angle. (d) sin C = 1 / n = 1 / 1.33 = 0.7519, so C = 48.8 degrees. Marking: Snell law, speed and refractive index, two conditions, critical angle. |
| 11 | An object is placed 15 cm in front of a convex lens of focal length 10 cm. (a) Find the image distance. (b) Find the magnification. (c) Describe the image (real or virtual, upright or inverted, magnified or diminished). (d) Describe the image if the object is moved to 5 cm from the lens. | (a) 1/f = 1/u + 1/v, so 1/10 = 1/15 + 1/v and v = 30 cm. (b) m = v / u = 30 / 15 = 2.0. (c) The positive image distance means a real, inverted image 2.0 times the object size, on the far side of the lens. (d) 5 cm is less than the focal length, so the rays diverge and the image is virtual, upright and magnified twice at -10 cm on the same side. Marking: lens formula, magnification, real versus virtual, object inside focus. |
| 12 | A string of length 1.2 m is fixed at both ends and vibrates at 180 Hz in its third harmonic (three antinodes). (a) Find the wavelength on the string. (b) Find the wave speed. (c) Find the fundamental frequency. (d) Find the difference between adjacent harmonic frequencies and relate it to the fundamental. | (a) The third harmonic has three antinodes, so lambda = 2L / 3 = 2 x 1.2 / 3 = 0.8 m. (b) v = f lambda = 180 x 0.8 = 144 m s-1. (c) f1 = v / 2L = 144 / 2.4 = 60 Hz. (d) Adjacent harmonics differ by the fundamental 60 Hz, and every harmonic is an integer multiple of it. Marking: wavelength from antinode count, wave equation, fundamental, harmonic series. |
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