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Secondary 3 Physics Practice Paper 1

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TuitionGoWhere Practice Paper - Physics Secondary 3 (Version 1)

Answer Key and Marking Scheme

Section A: Multiple Choice [10 marks]

QuestionAnswerExplanation
1DVelocity is a vector quantity (has magnitude and direction). Mass, time, and speed are scalar quantities (magnitude only).
2BUsing a=v−ut=20−05=4 m s−2a = \frac{v-u}{t} = \frac{20-0}{5} = 4 \, \text{m s}^{-2}.
3AVertical forces: 50−50=050 - 50 = 0. Horizontal forces: 30−30=030 - 30 = 0. Resultant force = 0 N0 \, \text{N}. The object is in equilibrium.
4CAt maximum height, the ball momentarily stops (velocity = 0) but acceleration due to gravity (g=10 m s−2g = 10 \, \text{m s}^{-2} downward) still acts on it.
5DWeight is the gravitational force on a body, measured in newtons. Mass is measured in kg, density in kg m−3^{-3}, volume in m3^3.
6CUsing s=ut+12gt2=0+12×10×32=45 ms = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2} \times 10 \times 3^2 = 45 \, \text{m}.
7BMomentum = mass × velocity. This is a fundamental definition in mechanics.
8BThe principle of moments states that for equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments. A small effort at a large distance can balance a large load at a small distance.
9CUsing Pythagoras: displacement = 6002+8002=360000+640000=1000000=1000 m\sqrt{600^2 + 800^2} = \sqrt{360000 + 640000} = \sqrt{1000000} = 1000 \, \text{m}.
10CInitial momentum = 0.5×4=2 kg m s−10.5 \times 4 = 2 \, \text{kg m s}^{-1} (toward wall). Final momentum = 0.5×(−4)=−2 kg m s−10.5 \times (-4) = -2 \, \text{kg m s}^{-1} (away from wall). Change = (−2)−2=−4 kg m s−1(-2) - 2 = -4 \, \text{kg m s}^{-1}. Magnitude = 4 kg m s−14 \, \text{kg m s}^{-1}.

Section B: Structured Questions [32 marks]

Question 11 [5 marks]

(a) Uniform acceleration is constant rate of change of velocity / velocity changes by equal amounts in equal time intervals [1]

(b) Calculation:

  • a=v−uta = \frac{v-u}{t} [1]
  • a=25−050=0.5 m s−2a = \frac{25-0}{50} = 0.5 \, \text{m s}^{-2} [1]

(c) Calculation:

  • v=u+atv = u + at [1]
  • v=25+(0.5×30)=25+15=40 m s−1v = 25 + (0.5 \times 30) = 25 + 15 = 40 \, \text{m s}^{-1} [1]

[Total: 5 marks]

Question 12 [7 marks]

(a) Arrow drawn vertically downward from centre of car, labelled W [1]

(b) Weight = mg=0.2×10=2 Nmg = 0.2 \times 10 = 2 \, \text{N} [1]

(c) Explanation:

  • There is a component of the weight acting down the slope [1]
  • This component is unbalanced (or: net force down the slope), so by Newton's Second Law, the car accelerates down the slope [1]

(d) Average speed = distancetime=0.51.5=0.333 m s−1\frac{\text{distance}}{\text{time}} = \frac{0.5}{1.5} = 0.333 \, \text{m s}^{-1} [2] (Allow 0.33 m s−10.33 \, \text{m s}^{-1} or 13 m s−1\frac{1}{3} \, \text{m s}^{-1})

(e) Reason: friction between car and ramp / air resistance acting against motion [1]

[Total: 7 marks]

Question 13 [7 marks]

(a) 10 m s−110 \, \text{m s}^{-1} [1] (read directly from horizontal section of graph)

(b) Calculation:

  • a=ΔvΔt=10−05−0a = \frac{\Delta v}{\Delta t} = \frac{10-0}{5-0} [1]
  • =2 m s−2= 2 \, \text{m s}^{-2} [1]

(c) Calculation:

  • Distance = total area under graph [1]
  • Area 1 (0-5 s): 12×5×10=25 m\frac{1}{2} \times 5 \times 10 = 25 \, \text{m} [0.5]
  • Area 2 (5-12 s): 7×10=70 m7 \times 10 = 70 \, \text{m} [0.5]
  • Area 3 (12-20 s): 12×8×10=40 m\frac{1}{2} \times 8 \times 10 = 40 \, \text{m} [0.5]
  • Total distance = 25+70+40=135 m25 + 70 + 40 = 135 \, \text{m} [0.5]

(d) The bus is decelerating uniformly / slowing down at a constant rate (from 10 m s−110 \, \text{m s}^{-1} to rest) [1]

[Total: 7 marks]

Question 14 [7 marks]

(a) Newton's First Law: A body remains at rest, or continues to move with uniform velocity (constant speed in a straight line), unless acted upon by a resultant external force [2] (Mark breakdown: state change in motion [1]; state condition of resultant force/not acted upon by resultant force [1])

(b) Explanation:

  • By Newton's First Law, constant velocity means zero resultant force [1]
  • The applied force of 12 N is balanced by an equal and opposite frictional force of 12 N, so resultant force is zero [1]

(c) Frictional force = 12 N12 \, \text{N} [1] (by Newton's First Law, must equal applied force for constant velocity)

(d) Calculation:

  • Resultant force = 20−12=8 N20 - 12 = 8 \, \text{N} [1]
  • Direction of acceleration: in the direction of the pulling force / to the right / forward [1]

[Total: 7 marks]

Question 15 [6 marks]

(a) The moment of a force about a pivot is the product of the force and the perpendicular distance from the pivot to the line of action of the force [2] (Mark breakdown: product/force × distance [1]; perpendicular distance/from pivot to line of action [1])

(b) Calculation:

  • Moment = force × perpendicular distance [1]
  • Moment = 300×0.4=120 N m300 \times 0.4 = 120 \, \text{N m} (clockwise) [1]

(c) Calculation:

  • For equilibrium, sum of clockwise moments = sum of anticlockwise moments [0.5]
  • 120=F×2.0120 = F \times 2.0 [0.5]
  • F=60 NF = 60 \, \text{N} [1]

(d) Increase the effort arm / move the effort force further from the fulcrum / use a longer lever on the effort side [1]

[Total: 6 marks]

Section C: Longer Structured Question [18 marks]

Question 16 [18 marks]

(a)(i) Calculation:

  • GPE = mghmgh [1]
  • GPE = 60×10×40=24000 J60 \times 10 \times 40 = 24000 \, \text{J} [1]

(a)(ii) Principle of conservation of energy: Energy cannot be created or destroyed, but can only be converted from one form to another / the total energy in a closed system remains constant [2] (Mark breakdown: energy not created or destroyed [1]; converted from one form to another/total energy constant [1])

(b) Calculation:

  • By conservation of energy: loss in GPE = gain in KE [1]
  • mgh=12mv2mgh = \frac{1}{2}mv^2 [1]
  • 10×40=12×v210 \times 40 = \frac{1}{2} \times v^2 (mass cancels)
  • v2=800v^2 = 800
  • v=800=28.3 m s−1v = \sqrt{800} = 28.3 \, \text{m s}^{-1} [1] (accept 28 m s−128 \, \text{m s}^{-1} to 2 s.f.)

(c)(i) Calculation:

  • GPE at A = 24000 J24000 \, \text{J} [0.5]
  • GPE at D = 60×10×10=6000 J60 \times 10 \times 10 = 6000 \, \text{J} [1]
  • Energy lost = 24000−6000=18000 J24000 - 6000 = 18000 \, \text{J} [1.5] (Alternative: KE at B would be 24000 J; max possible height with no energy loss = 40 m; actual at 10 m, so energy lost = 60×10×30=1800060 \times 10 \times 30 = 18000 J)

(c)(ii) Explanation:

  • Energy is transferred from the skier to the surroundings [1]
  • Work is done against friction and air resistance, converting mechanical energy to thermal energy (and sound) [1]

(d)(i) Calculation:

  • F=maF = ma [1]
  • a=18060=3 m s−2a = \frac{180}{60} = 3 \, \text{m s}^{-2} [1] (magnitude; deceleration is 3 m s−23 \, \text{m s}^{-2} opposite to motion)

(d)(ii) Calculation:

  • Using v2=u2+2asv^2 = u^2 + 2as with v=0v = 0 [1]
  • 0=152+2(−3)s0 = 15^2 + 2(-3)s [1]
  • 0=225−6s0 = 225 - 6s
  • s=2256=37.5 ms = \frac{225}{6} = 37.5 \, \text{m} [1]

(e) Tight-fitting / streamlined clothing / aerodynamic helmet / smooth fabrics reduce turbulent airflow and drag [1]

[Total: 18 marks]

GRAND TOTAL: 60 marks

Secondary 3 Physics Quiz - Mechanics: Answer Key

Total Marks: 40

Section A: Multiple Choice [10 marks]

QuestionAnswerMarkExplanation
1D1Speed is a scalar (magnitude only). Displacement, force, and velocity are vectors (magnitude and direction).
2B1Time = 30 min = 1800 s; Distance = 40000 m; Average speed = 400001800=22.2 m s−1\frac{40000}{1800} = 22.2 \, \text{m s}^{-1}
3B1Gradient = 8−04−0=2.0 m s−2\frac{8-0}{4-0} = 2.0 \, \text{m s}^{-2}
4A1The book is in equilibrium. Weight (15 N down) is balanced by normal contact force from table (15 N up). Resultant force = 0 N.
5C1Newton's Second Law: resultant force = rate of change of momentum (F=dpdt=maF = \frac{dp}{dt} = ma). Option A is Newton's First Law; B is Newton's Third Law.
6B1F=ma⇒a=Fm=5010=5 m s−2F = ma \Rightarrow a = \frac{F}{m} = \frac{50}{10} = 5 \, \text{m s}^{-2}
7A1Moment = F×d=25×0.3=7.5 N mF \times d = 25 \times 0.3 = 7.5 \, \text{N m}
8B1v=u−gt⇒0=20−10t⇒t=2.0 sv = u - gt \Rightarrow 0 = 20 - 10t \Rightarrow t = 2.0 \, \text{s}
9B1In any collision (elastic or inelastic), total momentum is always conserved. Kinetic energy is only conserved in perfectly elastic collisions.
10B1For ideal single fixed + single movable pulley: MA = 2, so F=4002=200 NF = \frac{400}{2} = 200 \, \text{N}

Section B: Short Answer and Structured [18 marks]

Question 11 [3 marks]

(a) Distance is the total length of the path travelled by an object / the scalar quantity measuring how far an object has moved [1]

(b) Displacement is the straight-line distance from the initial to the final position, together with the direction / the vector quantity measuring change in position [1]

(c) Any situation where the object returns toward its starting point or follows a curved path, e.g.: A person walks 5 m east then 3 m west — distance = 8 m but displacement = 2 m east; or running around a track — distance is circumference but displacement is zero after one lap [1]

Question 12 [6 marks]

(a) Calculation:

  • v=u+atv = u + at [1]
  • v=0+(2.5×6)=15 m s−1v = 0 + (2.5 \times 6) = 15 \, \text{m s}^{-1} [1]

(b) Calculation:

  • s=ut+12at2s = ut + \frac{1}{2}at^2 or v2=u2+2asv^2 = u^2 + 2as [1]
  • s=0+12×2.5×62=12×2.5×36=45 ms = 0 + \frac{1}{2} \times 2.5 \times 6^2 = \frac{1}{2} \times 2.5 \times 36 = 45 \, \text{m} [1] (Or using s=u+v2t=0+152×6=45 ms = \frac{u+v}{2}t = \frac{0+15}{2} \times 6 = 45 \, \text{m})

(c) Explanation:

  • Air resistance provides a resultant force opposing the motion [1]
  • By Newton's Second Law, this resultant force produces a deceleration (negative acceleration), causing the bicycle to slow down [1]

Question 13 [5 marks]

(a) A uniform object has its mass evenly distributed throughout its volume, so the centre of gravity (the point where the whole weight appears to act) is at the geometric centre [1]

(b) Calculation by moments about X:

  • Anticlockwise moment from weight = 8000×3=24000 N m8000 \times 3 = 24000 \, \text{N m} [1]
  • Clockwise moment from force at Y = FY×4F_Y \times 4 [1]
  • For equilibrium: FY×4=24000F_Y \times 4 = 24000 [0.5]
  • FY=6000 NF_Y = 6000 \, \text{N} [0.5]

(c) Calculation:

  • Total upward force = total downward force [0.5]
  • FX+6000=8000⇒FX=2000 NF_X + 6000 = 8000 \Rightarrow F_X = 2000 \, \text{N} [0.5]

Question 14 [6 marks]

(a) Calculation:

  • Change in momentum = final momentum – initial momentum [0.5]
  • Taking direction toward wall as positive: Δp=(0.4×−2)−(0.4×3)\Delta p = (0.4 \times -2) - (0.4 \times 3) [0.5]
  • =−0.8−1.2=−2.0 kg m s−1= -0.8 - 1.2 = -2.0 \, \text{kg m s}^{-1} [0.5]
  • Magnitude of change = 2.0 kg m s−12.0 \, \text{kg m s}^{-1} (or state −2.0 kg m s−1-2.0 \, \text{kg m s}^{-1} with direction reversed) [0.5]

(b) Calculation:

  • F=ΔpΔtF = \frac{\Delta p}{\Delta t} or F=2.00.05F = \frac{2.0}{0.05} [1]
  • =40 N= 40 \, \text{N} [1]

(c) Statement and application:

  • Newton's Third Law: When body A exerts a force on body B, body B exerts a force of equal magnitude, opposite direction, on body A [1]
  • The wall exerts a force on the ball to reverse its motion; simultaneously the ball exerts an equal and opposite force on the wall [1]

Section C: Data Analysis and Extended Response [12 marks]

Question 17 [3 marks]

(a) Speed of A = gradient = 10010=10 m s−1\frac{100}{10} = 10 \, \text{m s}^{-1} [1]

(b) Speed of B = gradient = 608=7.5 m s−1\frac{60}{8} = 7.5 \, \text{m s}^{-1} [1]

(c) The gradient of line A is steeper than the gradient of line B / line A is steeper than line B / for the same time, cyclist A covers more distance [1]

Question 18 [7 marks]

(a) Independent variable: Force applied to trolley [1] Dependent variable: Acceleration of trolley [1]

(b) Method description:

  • Attach ticker tape to trolley; thread through ticker-tape timer [1]
  • Timer marks dots at 50 Hz (every 0.02 s); analyze spacing between dots to find velocity changes [1]
  • Or: use motion sensor connected to data logger to record position-time data, then software calculates velocity and acceleration [1]
  • Acceleration found from gradient of velocity-time graph or from Δv/Δt\Delta v/\Delta t [1] (Any coherent method with valid physics scores; maximum 3 marks)

(c) Sketch description: [2]

  • Straight line passing through origin [1]
  • Axes labelled: vertical acceleration/m s−2\text{m s}^{-2}, horizontal force/N [1]
  • (Note: a∝Fa \propto F when mass constant, so linear relationship through origin)

Question 19 [2 marks]

  • Pressure = force/area [1]
  • A sharp knife has a smaller contact area than a blunt knife, so for the same applied force, it produces greater pressure, cutting more easily [1]

Question 20 [5 marks]

(a) Calculation:

  • a=v−ut=0−105a = \frac{v-u}{t} = \frac{0-10}{5} [1]
  • =−2 m s−2= -2 \, \text{m s}^{-2}, so deceleration = 2 m s−22 \, \text{m s}^{-2} [1]

(b) Calculation:

  • s=u+v2t=10+02×5s = \frac{u+v}{2}t = \frac{10+0}{2} \times 5 [1]
  • =25 m= 25 \, \text{m} [1] (Or: s=ut+12at2=10×5+12×(−2)×25=50−25=25s = ut + \frac{1}{2}at^2 = 10 \times 5 + \frac{1}{2} \times (-2) \times 25 = 50 - 25 = 25 m)

(c) Explanation: reaction time of autonomous system / sensors may have detection delay / braking system response time / friction less than ideal / wet road conditions / not uniform deceleration [1]

QUIZ TOTAL: 40 marks