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HKDSE Physics Practice Paper: Mechanics, Momentum and Energy (12 Questions, 100 Marks, 75 Minutes, with Answers)
HKDSE Physics: Mechanics, Momentum and Energy
HKDSE Physics mechanics practice paper: 12 original questions on motion, momentum, friction, power, energy, circular motion, moments, density. Worked answers included.
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Paper details Level: S5 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Kinematics and Force (32 marks)
Answer all questions in Section A.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (Uniform acceleration) A car starts from rest and accelerates at 2.5 m/s2 along a straight line for 8 seconds. (a) Find the speed at the end of 8 seconds (3 marks). (b) Find the distance travelled in these 8 seconds (3 marks). (c) Explain why the average speed is not the same as the final speed (2 marks). | Answer: (a) v = u + at = 0 + 2.5 x 8 = 20.0 m/s. (b) s = 0.5 x 2.5 x 64 = 80.0 m. (c) The speed rises uniformly from zero to 20.0 m/s, so the average is 10.0 m/s, which is only the instantaneous speed halfway through the interval. Marking: 1 mark for units, 2 marks for full working. |
| 2 | (Conservation of momentum) A 0.5 kg body moving at 6 m/s hits a stationary 1.5 kg body and the two stick together. (a) Find the common velocity after the collision (3 marks). (b) Find the kinetic energy lost and its percentage of the initial kinetic energy (3 marks). (c) Explain why momentum is conserved while kinetic energy is not (2 marks). | Answer: (a) 0.5 x 6 = (0.5 + 1.5) x v, so v = 3 / 2.0 = 1.5 m/s. (b) Initial KE = 0.5 x 0.5 x 36 = 9.0 J; final KE = 0.5 x 2.0 x 2.25 = 2.25 J; loss = 6.75 J, that is 75.0 per cent. (c) In a perfectly inelastic collision momentum is conserved, but some kinetic energy becomes heat and sound. Marking: 1 mark for combining the masses. |
| 3 | (Force and motion) A 4 kg box on a horizontal floor is pulled by a 20 N horizontal force; the coefficient of kinetic friction is 0.25 and g = 10 m/s2. (a) Find the friction force (2 marks). (b) Find the acceleration of the box (3 marks). (c) Find its speed after 3 seconds (3 marks). | Answer: (a) f = 0.25 x 4 x 10 = 10.0 N. (b) Net force = 20 – 10.0 = 10.0 N, so a = 10.0 / 4 = 2.5 m/s2. (c) v = a x t = 2.5 x 3 = 7.5 m/s. Marking: the opposite directions of friction and pull must be stated; no marks in (b) and (c) if the net force is wrong. |
| 4 | (Power and efficiency) A crane lifts a 50 kg load 8 m in 10 seconds at constant speed; g = 10 m/s2 and the efficiency is 80 per cent. (a) Find the useful work done (3 marks). (b) Find the useful output power (3 marks). (c) Find the input power of the crane (2 marks). | Answer: (a) W = 50 x 10 x 8 = 4,000 J. (b) P = W / t = 4,000 / 10 = 400.0 W. (c) Input power = 400.0 / 0.8 = 500.0 W. Marking: useful work and total work must not be mixed up; an efficiency below 1 means the input exceeds the output. |
Section B: Energy, Momentum and Circular Motion (36 marks)
Answer all questions in Section B.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (Conservation of energy) A ball falls freely from 20 m and rebounds to 12.8 m; g = 10 m/s2 and air resistance is neglected. (a) Find the speed on landing (3 marks). (b) Find the time of fall (3 marks). (c) Find the rebound speed and the coefficient of restitution (3 marks). | Answer: (a) v = sqrt(2 x 10 x 20) = 20.0 m/s. (b) t = sqrt(2 x 20 / 10) = sqrt(4) = 2.0 s. (c) Rebound speed = sqrt(2 x 10 x 12.8) = 16.0 m/s, so e = 16.0 / 20.0 = 0.8. Marking: the coefficient is a ratio of speeds, not of heights (the height ratio is 0.64). |
| 6 | (Circular motion) A 0.2 kg ball moves at 6 m/s in a horizontal circle of radius 0.8 m. (a) Find the centripetal acceleration (3 marks). (b) Find the centripetal force needed (3 marks). (c) Find the period, taking pi as 3.14 (3 marks). | Answer: (a) a = 36 / 0.8 = 45.0 m/s2. (b) F = 0.2 x 45.0 = 9.0 N. (c) T = 2 x 3.14 x 0.8 / 6 = 0.84 s. Marking: the centripetal force points to the centre and is perpendicular to the velocity, so the speed stays constant. |
| 7 | (Moments) A light uniform rod rests on a pivot. A 20 N load hangs 1.2 m from the pivot and a force is applied 0.8 m away on the other side to balance it. (a) Find the force needed (3 marks). (b) Find the total downward force on the pivot (3 marks). (c) Find the mechanical advantage of this arrangement (3 marks). | Answer: (a) F x 0.8 = 20 x 1.2, so F = 24 / 0.8 = 30.0 N. (b) Total downward force = 20 + 30.0 = 50.0 N. (c) Mechanical advantage = 1.2 / 0.8 = 1.5. Marking: moments are taken about the pivot and the moment arm is the perpendicular distance; the light rod has no weight. |
| 8 | (Springs and stored energy) A spring of stiffness 200 N/m is stretched by 0.15 m within its elastic limit. (a) Find the spring force (3 marks). (b) Find the elastic potential energy stored (3 marks). (c) State the energy stored when the extension is doubled (3 marks). | Answer: (a) F = 200 x 0.15 = 30.0 N. (b) E = 0.5 x 200 x 0.0225 = 2.25 J. (c) The energy goes with the square of the extension, so doubling it gives 9.0 J, four times as much. Marking: use 0.5 k x squared, not k x. |
Section C: Moments, Buoyancy and Experimental Analysis (32 marks)
Answer all questions in Section C.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | (Density and upthrust) A block of volume 0.2 cubic metres and mass 160 kg is placed in water of density 1,000 kg/m3; g = 10 m/s2. (a) Find the density of the block (3 marks). (b) Find the percentage of its volume below the surface when it floats (3 marks). (c) Find the upthrust (2 marks). | Answer: (a) density = 160 / 0.2 = 800 kg/m3. (b) 800 / 1,000 = 0.8, so 80.0 per cent of the volume is submerged. (c) Upthrust = weight = 160 x 10 = 1,600 N. Marking: for a floating body the upthrust equals the weight, not the largest possible upthrust. |
| 10 | (Inclined plane) A 200 N load is pushed at constant speed up a 3.0 m long incline of height 1.5 m; the efficiency is 60 per cent. (a) Find the useful work done (3 marks). (b) Find the ideal push needed without friction (3 marks). (c) Find the actual push (2 marks). | Answer: (a) Useful work = 200 x 1.5 = 300.0 J. (b) Ideal push = 300.0 / 3.0 = 100.0 N. (c) Actual push = 100.0 / 0.6 = 166.7 N. Marking: with an efficiency below 1 the actual push must exceed the ideal one; an incline saves force but costs distance. |
| 11 | (Motion graphs) A car speeds up uniformly from rest to 20 m/s in 5 seconds, travels at 20 m/s for 10 seconds and then slows uniformly to rest in 4 seconds. (a) Find the acceleration in the first 5 seconds (3 marks). (b) Find the total distance travelled (3 marks). (c) Find the deceleration and the average speed of the whole journey (2 marks). | Answer: (a) a = 20 / 5 = 4.0 m/s2. (b) Distance = 0.5 x 20 x 5 + 20 x 10 + 0.5 x 20 x 4 = 290.0 m. (c) Deceleration = -20 / 4 = -5.0 m/s2; average speed = 290.0 / 19 = 15.3 m/s. Marking: the area under a velocity-time graph is the displacement, and slowing down means negative acceleration. |
| 12 | (Experimental analysis) A student times a ball falling freely from 1.8 m; the mean of five readings is 0.60 s and the accepted value is g = 9.81 m/s2. (a) Find the experimental value of g (3 marks). (b) Find the percentage error (3 marks). (c) Suggest two ways to reduce the error (2 marks). | Answer: (a) g = 2 x 1.8 / 0.36 = 10.0 m/s2. (b) Error = (10.0 – 9.81) / 9.81 x 100 = 1.9 per cent. (c) Use light gates instead of hand timing, and drop from a greater height with repeated trials. Marking: the error is written as a positive percentage; the square of t magnifies timing error. |
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