Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 5

TuitionGoWhere Practice Paper - Physics Secondary 3: Answer Key (Version 5)

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SECTION A (20 marks)

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1. Answer: C (Velocity)

Explanation: A vector quantity has both magnitude and direction. Velocity includes both speed (magnitude) and direction of motion. Mass, time, and energy are scalar quantities—they have magnitude only.

Common mistake: Confusing speed and velocity. Speed is scalar; velocity is vector.

Mark: [1]

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2. Answer: 4

Working: $a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4 \text{ m/s}^2$

Explanation: Acceleration is the rate of change of velocity. Using $a = \frac{\Delta v}{t}$ with initial velocity $u = 0$ (from rest), final velocity $v = 20 \text{ m/s}$, and time $t = 5 \text{ s}$.

Mark: [1]

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3. Answer: An object will remain at rest or continue moving with constant velocity (constant speed in a straight line) unless acted upon by a resultant (unbalanced) force.

Explanation: Newton's First Law describes inertia—the tendency of objects to resist changes to their motion. "Constant velocity" means both constant speed AND constant direction (straight line). The key phrase is "resultant force" or "unbalanced force" (not just "force"—balanced forces allow constant velocity too).

Mark: [1]

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4. Answer: 80

Working: $W = mg = 8 \times 10 = 80 \text{ N}$

Explanation: Weight is the gravitational force on a mass, calculated by $W = mg$. Mass $m = 8 \text{ kg}$, $g = 10 \text{ N/kg}$.

Note: Weight is measured in newtons (N), mass in kilograms (kg).

Mark: [1]

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5. Answer: 3

Working: $F = ma \Rightarrow a = \frac{F}{m} = \frac{12}{4} = 3 \text{ m/s}^2$

Explanation: Newton's Second Law states $F = ma$. Rearranging for acceleration: $a = F/m$.

Mark: [1]

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6. Answer: Energy cannot be created or destroyed, only converted from one form to another. The total energy in a closed system remains constant.

Explanation: This principle means that in any energy transfer or transformation, the total amount of energy before equals the total after. Energy may change form (e.g., kinetic to potential, or to thermal energy due to friction), but the total is conserved.

Mark: [1]

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7. Answer: B

Explanation: At the highest point, the ball momentarily stops (velocity = 0) before falling back down. However, acceleration remains constant at $g = 10 \text{ m/s}^2$ downwards throughout the flight (ignoring air resistance). Gravity still acts on the ball even when it is instantaneously at rest.

Common mistake: Thinking acceleration is zero when velocity is zero. Acceleration is due to gravity and is constant.

Mark: [1]

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8. Answer: 500

Working: Displacement forms the hypotenuse of a right-angled triangle: $s = \sqrt{300^2 + 400^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \text{ m}$

Explanation: Displacement is the shortest distance from start to finish (straight line), not the total path travelled. Using Pythagoras' theorem for the perpendicular displacements north and east.

Mark: [1]

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9. Answer: Any two from:

• (i) Speed of the vehicle (greater speed → longer braking distance)
• (ii) Road surface condition (wet/icy roads increase braking distance)
• (iii) Condition of brakes/worn brake pads
• (iv) Tire condition/tread depth
• (v) Mass of vehicle

Explanation: Braking distance is the distance travelled while decelerating to a stop. It depends on factors affecting the deceleration achievable. Thinking distance (distance before brakes are applied) depends mainly on reaction time and speed.

Mark: [1] each, max [2]

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10. Answer: 2

Working: Using $s = ut + \frac{1}{2}at^2$ with $u = 0$, $a = g = 10 \text{ m/s}^2$, $s = 20 \text{ m}$: $20 = 0 + \frac{1}{2}(10)t^2$ $20 = 5t^2$ $t^2 = 4$ $t = 2 \text{ s}$

Explanation: For free-fall from rest, use the equation of motion $s = \frac{1}{2}gt^2$. Take care with signs—if downward is negative, both $s$ and $g$ are negative, giving the same result.

Mark: [1]

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SECTION B (24 marks)

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11.

(a) Deceleration = 2.5 m/s² [2]

Working: $a = \frac{v - u}{t} = \frac{0 - 25}{10} = -2.5 \text{ m/s}^2$

Deceleration is $2.5 \text{ m/s}^2$ (or $-2.5 \text{ m/s}^2$ if deceleration direction is specified).

Mark breakdown:

• [1] Correct substitution or method
• [1] Correct answer with unit

(b) Braking force = 3000 N [2]

Working: $F = ma = 1200 \times 2.5 = 3000 \text{ N}$ (Using magnitude of deceleration: $F = 1200 \times 2.5$)

Or using resultant force: The braking force causes deceleration, so $F = 3000 \text{ N}$ opposite to motion.

Mark breakdown:

• [1] Correct substitution or method
• [1] Correct answer with unit

(c) Velocity-time graph [2]

Expected features:

• Straight line starting at $(0, 25)$
• Sloping downward with negative gradient
• Reaching $(10, 0)$ on time axis
• Axes labelled: velocity (m/s) vertical, time (s) horizontal
• Straight line, not curved (uniform deceleration)

Mark breakdown:

• [1] Correct shape (straight line, proper start/end points)
• [1] Correctly labelled axes with units

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12.

(a) [2]

Answer: "Constant velocity" means the block moves with unchanging speed in a straight line (constant speed AND direction). The net (resultant) force on the block is zero.

Explanation: By Newton's First Law, if velocity is constant, acceleration is zero, so resultant force must be zero. The block is in equilibrium—forces balance in both horizontal and vertical directions.

Mark breakdown:

• [1] Correct statement about constant velocity (must mention constant speed in straight line, or no acceleration)
• [1] Net force = 0 (or forces are balanced)

(b) Horizontal component = 25.98 N ≈ 26 N [2]

Working: $F_{\text{horizontal}} = F \cos \theta = 30 \times \cos 30° = 30 \times 0.866 = 25.98 \text{ N}$

Or exactly: $15\sqrt{3} \text{ N}$ or approximately $26 \text{ N}$

Mark breakdown:

• [1] Correct formula or method (cosine of angle, not sine)
• [1] Correct answer with unit

(c) Frictional force = 25.98 N ≈ 26 N [2]

Working/Reasoning: Since velocity is constant, horizontal forces balance: $f = F_{\text{horizontal}} = 30 \cos 30° = 25.98 \text{ N}$

Explanation: For constant velocity, resultant horizontal force must be zero (Newton's First Law). Therefore friction exactly opposes and equals the horizontal component of the applied force.

Mark breakdown:

• [1] Correct reasoning (constant velocity means forces balanced, or resultant force = 0)
• [1] Correct value with unit

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13.

(a) Gravitational potential energy = 1.0 J [2]

Working: $GPE = mgh = 0.2 \times 10 \times 0.5 = 1.0 \text{ J}$

Mark breakdown:

• [1] Correct formula or substitution
• [1] Correct answer with unit

(b) Maximum speed = 3.16 m/s ≈ 3.2 m/s [3]

Working: By conservation of energy: GPE at top = KE at bottom $mgh = \frac{1}{2}mv^2$

$gh = \frac{1}{2}v^2$

$v^2 = 2gh = 2 \times 10 \times 0.5 = 10$

$v = \sqrt{10} = 3.16 \text{ m/s}$

Or using values: $1.0 = \frac{1}{2} \times 0.2 \times v^2$, so $v^2 = 10$, $v = 3.16$ m/s

Mark breakdown:

• [1] States conservation of energy or correct energy equation
• [1] Correct substitution and manipulation
• [1] Correct final answer with unit

Alternative using kinematics: $v^2 = u^2 + 2as = 0 + 2(10)(0.5) = 10$, $v = 3.16$ m/s — accept if correct.

(c) Energy is lost to air resistance/drag [1]

Explanation: Some mechanical energy is converted to thermal energy (and sound) due to air resistance acting on the bob. This means not all GPE becomes KE on the downward swing, and on the upward swing, some KE is lost to air resistance before becoming GPE.

Mark: [1]

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14.

(a) [2]

Answer: The gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Formula: $F = \frac{Gm_1m_2}{r^2}$

where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between centres.

Mark breakdown:

• [1] Correct statement of proportionality relationships (or correct formula with all variables defined)
• [1] Complete definition with "distance between centres" or correctly defined variables

(b) Gravitational force = 1250 N (accept 1251 N) [3]

Working: $F = \frac{Gm_1m_2}{r^2}$

Given: $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}$, $m_1 = 6.0 \times 10^{24} \text{ kg}$, $m_2 = 200 \text{ kg}$, $r = 8000 \text{ km} = 8.0 \times 10^6 \text{ m}$

$F = \frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 200}{(8.0 \times 10^6)^2}$

$F = \frac{8.004 \times 10^{16}}{6.4 \times 10^{13}}$

$F = 1250.6 \text{ N} \approx 1250 \text{ N} \text{ or } 1251 \text{ N}$

Mark breakdown:

• [1] Correct formula with correct substitution (including conversion of km to m)
• [1] Correct calculation of numerator or denominator
• [1] Correct final answer with unit

Note: $r$ must be in metres: $8000 \text{ km} = 8.0 \times 10^6 \text{ m}$. Common error: forgetting to convert km to m.

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SECTION C (16 marks)

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15.

(a) Resultant force = 6.0 × 10⁵ N [1]

Working: $F_{\text{resultant}} = F_{\text{drive}} - F_{\text{resist}} = 8.0 \times 10^5 - 2.0 \times 10^5 = 6.0 \times 10^5 \text{ N}$

Mark: [1] — correct answer with unit

(b) Acceleration = 1.5 m/s² [2]

Working: $F = ma \Rightarrow a = \frac{F}{m} = \frac{6.0 \times 10^5}{4.0 \times 10^5} = 1.5 \text{ m/s}^2$

Mark breakdown:

• [1] Correct substitution (using resultant force, not driving force)
• [1] Correct answer with unit

Common error: Using $8.0 \times 10^5$ N instead of $6.0 \times 10^5$ N gives $a = 2.0$ m/s² — incorrect.

(c) Distance = 675 m [3]

Working: Using $s = ut + \frac{1}{2}at^2$ with $u = 0$, $a = 1.5$ m/s², $t = 30$ s:

$s = 0 + \frac{1}{2} \times 1.5 \times (30)^2$

$s = 0.5 \times 1.5 \times 900$

$s = 0.75 \times 900 = 675 \text{ m}$

Mark breakdown:

• [1] Correct formula selected
• [1] Correct substitution
• [1] Correct answer with unit

Alternative using kinematic equations or average velocity: Average velocity = $(0 + 45)/2 = 22.5$ m/s, distance = $22.5 \times 30 = 675$ m — accept if correct.

(d) [2]

Answer: The train will decelerate (slow down) because the resistive force now exceeds the forward force (which is zero). The train will continue to slow down until it comes to rest.

Explanation: With no driving force, the resultant force is just the resistive force ($2.0 \times 10^5$ N) acting opposite to motion. By Newton's Second Law, this causes deceleration. The deceleration $a = -F_{\text{resist}}/m = -2.0 \times 10^5 / 4.0 \times 10^5 = -0.5$ m/s².

Mark breakdown:

• [1] Train decelerates/slows down/stops (stated or described)
• [1] Explanation: resistive force acts as resultant force opposing motion (or equivalent reasoning mentioning unbalanced force opposing motion)

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16.

(a) Terminal velocity = 50 m/s [1] (accept value read from graph: 50–52 m/s)

Mark: [1] — correct value with unit from graph

(b) [2]

Answer: As the skydiver falls, air resistance (drag) increases with speed. When air resistance equals the skydiver's weight, the resultant force becomes zero. By Newton's First Law, the skydiver then falls at constant (terminal) velocity.

Mark breakdown:

• [1] Air resistance increases with speed and eventually equals weight
• [1] Resultant force becomes zero, so no further acceleration / constant velocity reached

(c) [3]

Answer for A to ~25 s: The parachute opens, causing a large increase in air resistance (drag). This creates a large resultant force upwards, causing rapid deceleration. The gradient is steep and negative.

Answer for ~25 s to B: The skydiver slows until air resistance (now with parachute) again equals her weight. She reaches a new, lower terminal velocity (about 5 m/s). The graph becomes horizontal again. Finally, she lands at zero velocity.

Mark breakdown:

• [1] Parachute increases air resistance dramatically, causing rapid deceleration (steep negative gradient)
• [1] New lower terminal velocity reached where air resistance (with parachute) = weight
• [1] Landing at B with zero velocity / coming to rest

(d) Distance ≈ 500 m [2]

Working: The area under the velocity-time graph gives distance. For 0–20 s:

• Approximate as triangle or trapezium/triangle + rectangle
• Area ≈ $\frac{1}{2} \times 20 \times 50 = 500$ m

Or more accurately: area ≈ area of triangle from 0 to ~10 s + rectangle from ~10 s to 20 s at 50 m/s

• $\approx \frac{1}{2} \times 10 \times 50 + 10 \times 50 = 250 + 500 = 750$ m — but this overestimates

Better estimate using triangle to terminal velocity:

• $\frac{1}{2} \times 20 \times 50 = 500$ m is acceptable as approximation

Or using area of trapezium with visible curve: accept 400–600 m with clear working shown.

Mark breakdown:

• [1] Method: states area under graph = distance, or appropriate counting of squares
• [1] Reasonable estimate with working shown (accept 400–650 m depending on method; must show some calculation)

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17.

(a) [2]

Thinking distance: The distance travelled by the vehicle during the driver's reaction time (the time between seeing a hazard and applying the brakes). It depends on speed and reaction time.

Braking distance: The distance travelled by the vehicle while actually braking (decelerating) to a stop. It depends on speed, road conditions, brakes, and vehicle condition.

Mark breakdown:

• [1] Correct description of thinking distance (distance during reaction time/delay)
• [1] Correct description of braking distance (distance while brakes are applied/decelerating)

(b) Reaction time = 0.60 s [2]

Working: $\text{reaction time} = \frac{\text{thinking distance}}{\text{speed}} = \frac{6}{10} = 0.60 \text{ s}$

Check with other values: $\frac{12}{20} = 0.60$ s, $\frac{18}{30} = 0.60$ s, $\frac{24}{40} = 0.60$ s

Mark breakdown:

• [1] Correct method (thinking distance ÷ speed, or equivalent)
• [1] Correct answer with unit or consistent seconds

(c) [2]

Answer: Thinking distance is directly proportional to speed (linear relationship: double speed → double thinking distance). However, braking distance depends on $v^2$ (kinetic energy = $\frac{1}{2}mv^2$ which must be dissipated by braking work = force × braking distance). Since kinetic energy increases with speed squared, braking distance increases with speed squared (quadruples when speed doubles).

Mark breakdown:

• [1] Thinking distance ∝ speed (linear) — correct relationship stated
• [1] Braking distance ∝ $v^2$ (or related to kinetic energy $\frac{1}{2}mv^2$) — correct reasoning that KE or stopping work relates to squared relationship

Alternative acceptable reasoning: Work done = $F \times d$ = loss in KE = $\frac{1}{2}mv^2$. If $F$ is constant (max braking force), then $d \propto v^2$.

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TOTAL: 60 MARKS