Secondary 3 Physics Practice Paper 2

TuitionGoWhere Practice Paper - Physics Secondary 3 (Version 2)

Answer Key and Marking Scheme

Section A (40 marks)

Question 1 [2 marks]

Answer:

• A scalar quantity has magnitude (size) only. [½]
• A vector quantity has both magnitude and direction. [½]
• Example of scalar: speed, mass, distance, time, energy, temperature (any one) [½]
• Example of vector: velocity, displacement, force, acceleration, weight (any one) [½]

Teaching note: The key distinction is direction. Scalars are described by a number and unit alone; vectors need a direction as well. Common error: students confuse speed (scalar) with velocity (vector), or distance (scalar) with displacement (vector).

Question 2 [3 marks]

(a) [1 mark]

Answer: Precision is 0.1 cm (or 1 mm). [1]

Teaching note: A metre rule has the smallest division of 1 mm = 0.1 cm. All readings should be recorded to this precision.

(b) [2 marks]

Answer:

• The rod may not be perfectly uniform or straight [1]
• Taking measurements at different positions and finding a mean reduces the effect of random errors / gives a more reliable result [1]

Teaching note: This is an application of repeated measurements. Random errors arise from estimation between scale divisions or slight irregularities in the object. A mean value is more representative of the true length.

Question 3 [4 marks]

(a) [1 mark]

Answer: Total distance = $120 + 80 = 200 \text{ km}$ [1]

(b) [1 mark]

Answer: Displacement = $120 - 80 = 40 \text{ km}$ due east [1]

Common error: Forgetting direction for displacement. "40 km" alone loses the direction mark.

(c) [2 marks]

Answer: $\text{Average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{200 \text{ km}}{4.0 \text{ h}} = 50 \text{ km/h}$ [2]

Teaching note: Average speed uses total distance (scalar), not displacement. Average velocity would use displacement: $40/4.0 = 10$ km/h east.

Question 4 [4 marks]

(a) [1 mark]

Answer: Force $R$ is the normal contact force (or normal reaction force). [1]

(b) [1 mark]

Answer: The Newton's Third Law pair to $W$ is: the gravitational pull of the book on the Earth (or the book attracts Earth with equal and opposite force). [1]

Common error: Saying "reaction force from the table" — this is actually the pair to $R$, not $W$. The Third Law pair to weight must involve Earth as the other body.

(c) [2 marks]

Answer:

• The book is at rest, so the resultant force on it must be zero [1]
• This means $R$ and $W$ are equal in magnitude and opposite in direction, so they balance / the book is in equilibrium [1]

Teaching note: Equilibrium means zero resultant force. Students often confuse "balanced forces" with Newton's Third Law pairs. $R$ and $W$ act on the SAME body (the book), so they can balance. Third Law pairs always act on DIFFERENT bodies.

Question 5 [4 marks]

(a) [2 marks]

Answer: Using $v = u + at$: $v = 0 + (2.5 \times 6.0) = 15 \text{ m/s}$ [2]

(b) [2 marks]

Answer: Using $s = ut + \frac{1}{2}at^2$: $s = 0 + \frac{1}{2}(2.5)(6.0)^2 = \frac{1}{2}(2.5)(36) = 45 \text{ m}$ [2]

Or using $s = \frac{(u+v)}{2}t = \frac{(0+15)}{2} \times 6.0 = 45 \text{ m}$

Teaching note: Both kinematic equations give the same answer. Encourage students to write down the known values before selecting the appropriate equation.

Question 6 [7 marks]

(a) [2 marks]

Answer: $\text{acceleration} = \frac{\text{change in velocity}}{\text{time}} = \frac{15 - 0}{5 - 0} = \frac{15}{5} = 3.0 \text{ m/s}^2$ [2]

Teaching note: Gradient of velocity-time graph = acceleration. First segment: straight line from origin to (5, 15), so gradient = 15/5 = 3 m/s².

(b) [2 marks]

Answer: $\text{distance} = \text{velocity} \times \text{time} = 15 \times (15-5) = 15 \times 10 = 150 \text{ m}$ [2]

Teaching note: During constant velocity, distance = area of rectangle = base × height = 10 s × 15 m/s = 150 m.

(c) [3 marks]

Answer:

Total distance = $37.5 + 150 + 37.5 = 225 \text{ m}$ [3]

Teaching note: Total area under velocity-time graph = total distance. Since velocity is always positive, area gives distance (not displacement). Mark allocation: method (splitting into areas) [1], correct individual areas [1], final answer with unit [1].

Question 7 [5 marks]

(a) [2 marks]

Answer: Using $v^2 = u^2 + 2as$ with $v = 0$ at maximum height: $0 = (12)^2 + 2(-10)s$ $0 = 144 - 20s$ $s = \frac{144}{20} = 7.2 \text{ m}$ [2]

Teaching note: At max height, velocity is momentarily zero. Acceleration is $-10$ m/s² (or take $g = -10$ m/s²). Common error: forgetting acceleration is negative, or using wrong sign convention.

(b) [2 marks]

Answer: Using $v = u + at$: $0 = 12 + (-10)t$ $t = \frac{12}{10} = 1.2 \text{ s}$ [2]

Or using $s = \frac{(u+v)}{2}t$: $7.2 = \frac{(12+0)}{2}t$, so $t = 1.2$ s

(c) [1 mark]

Answer: Velocity = $12 \text{ m/s}$ downward (or $-12 \text{ m/s}$ if up is positive). [1]

Teaching note: With no air resistance, motion is symmetric. Returns with same speed as launch, opposite direction. Common error: saying "zero" (confusing with max height).

Question 8 [4 marks]

(a) [2 marks]

Answer: $F_{\text{horizontal}} = 25 \cos 30° = 25 \times 0.866 = 21.65 \approx 21.7 \text{ N}$ [2]

Teaching note: Resolving forces: adjacent side to angle uses cosine. The horizontal component pulls the box forward; vertical component $25 \sin 30° = 12.5$ N reduces the normal force.

(b) [2 marks]

Answer: Frictional force = 21.7 N [1]

Reason: The box moves at constant velocity, so resultant force is zero (equilibrium). Therefore friction equals the horizontal component of the applied force. [1]

Teaching note: "Constant velocity" is the key phrase — implies equilibrium, not acceleration. Friction opposes motion and exactly balances the forward horizontal component.

Question 9 [4 marks]

(a) [1 mark]

Answer: $r = 6.4 \times 10^6 + 400 \times 10^3 = 6.4 \times 10^6 + 0.4 \times 10^6 = 6.8 \times 10^6 \text{ m}$ [1]

Common error: Using 400 km directly without converting to metres, or forgetting to add Earth's radius.

(b) [3 marks]

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

$F = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(200)}{(6.8 \times 10^6)^2}$ [1]

$F = \frac{8.004 \times 10^{16}}{4.624 \times 10^{13}}$ [1]

$F = 1.73 \times 10^3 \text{ N} \approx 1730 \text{ N}$ [1]

Teaching note: $r$ is centre-to-centre distance, crucial for correct substitution. Always work in base units (kg, m, s). Final answer approximately 1700 N or 1730 N acceptable.

Question 10 [4 marks]

(a) [1 mark]

Answer: $W = mg = 0.40 \times 10 = 4.0 \text{ N}$ [1]

(b) [3 marks]

Answer: Given: $T - mg = \frac{mv^2}{r}$

$\frac{mv^2}{r} = \frac{0.40 \times (2.0)^2}{0.80} = \frac{0.40 \times 4.0}{0.80} = \frac{1.6}{0.80} = 2.0 \text{ N}$ [1]

$T - 4.0 = 2.0$ [1]

$T = 6.0 \text{ N}$ [1]

Teaching note: At the lowest point, tension must exceed weight to provide the centripetal force. The net force toward the centre (upward at this point) equals $mv^2/r$. So $T = mg + mv^2/r = 4.0 + 2.0 = 6.0$ N. This is larger than the weight because the string must both support the bob and pull it in a circle.

Section B (40 marks)

Question 11 [8 marks]

(a) [3 marks]

Answer: Using $v^2 = u^2 + 2as$ with $u = 0$, $a = g = 10 \text{ m/s}^2$, $s = 45 \text{ m}$:

$v^2 = 0 + 2(10)(45) = 900$ [1]

$v = \sqrt{900} = 30 \text{ m/s}$ [2]

[Final answer with unit, working shown: 3 marks. If no working but correct answer: 2 marks maximum.]

(b) [2 marks]

Answer: Using $v = u + at$:

$30 = 0 + 10t$ [1]

$t = 3.0 \text{ s}$ [1]

Or using $s = \frac{(u+v)}{2}t$: $45 = \frac{30}{2}t$, so $t = 3.0$ s

(c) [3 marks]

Answer:

Expected sketch:

• Straight line passing through origin with positive gradient [1]
• Correct labels: velocity (m/s) on y-axis, time (s) on x-axis [1]
• Line ends at $(3.0, 30)$ or indicates final values correctly [1]

Teaching note: Velocity-time graph for free fall from rest is a straight line through origin with gradient $g = 10$ m/s². The graph should NOT start below zero or have a curve.

Question 12 [7 marks]

(a) [2 marks]

Answer:

• Tilting compensates for friction between the trolley and the track [1]
• When tilted correctly, the component of the trolley's weight down the slope equals friction, so the trolley moves at constant velocity with no applied force, or acceleration is due only to the hanging mass [1]

Teaching note: This is called "friction compensation" or "tilting the track." Once compensated, any measured acceleration is caused by the applied force from the hanging mass only.

(b) [3 marks]

Answer: Total mass being accelerated, $M_{\text{total}} = 0.80 + 0.20 = 1.0 \text{ kg}$

Driving force = weight of hanging mass = $0.20 \times 10 = 2.0 \text{ N}$ [1]

Using $F = ma$: $a = \frac{F}{m} = \frac{2.0}{1.0} = 2.0 \text{ m/s}^2$ [2]

Teaching note: The hanging mass accelerates too, so total mass is trolley PLUS hanging mass. The driving force is the weight of the hanging mass. If students use 0.80 kg alone: 1 mark only for correct force calculation.

(c) [2 marks]

Answer:

• Friction would oppose the motion of the trolley [1]
• Therefore the resultant force on the system would be less than the weight of the hanging mass, producing a smaller acceleration for the same total mass [1]

Question 13 [9 marks]

(a) [4 marks]

Answer:

Method 1 — Using kinematics: $v^2 = u^2 + 2as$ $0 = (25)^2 + 2a(50)$ $0 = 625 + 100a$ $a = -6.25 \text{ m/s}^2$ [2]

Then $F = ma = 1200 \times 6.25 = 7500 \text{ N}$ [2]

Method 2 — Using work-energy: $\text{Work done by brakes} = \text{loss in KE}$ $F \times 50 = \frac{1}{2}(1200)(25)^2$ [2] $F \times 50 = 375000$ $F = 7500 \text{ N}$ [2]

Teaching note: Negative acceleration means deceleration. Magnitude of braking force is 7500 N. Both methods are valid. Award marks for correct physics principles applied.

(b) [2 marks]

Answer: $v = u + at$ $0 = 25 + (-6.25)t$ $t = \frac{25}{6.25} = 4.0 \text{ s}$ [2]

Or using average velocity: $s = \frac{(u+v)}{2}t$, so $50 = \frac{25}{2}t$, giving $t = 4.0$ s

(c) [3 marks]

Answer:

Expected sketch:

• Straight line with negative gradient from $(0, 25)$ to $(4.0, 0)$ [1]
• Correctly labeled axes: velocity (m/s) and time (s) [1]
• Correct values indicated: $v = 25$ m/s at $t = 0$, and $t = 4.0$ s when $v = 0$ [1]

Question 14 [8 marks]

(a) [1 mark]

Answer: $W = mg = 2.0 \times 10 = 20 \text{ N}$ [1]

(b) [2 marks]

Answer: $W_{\text{parallel}} = mg \sin \theta = 20 \times \sin 25° = 20 \times 0.423 = 8.46 \approx 8.5 \text{ N}$ [2]

Teaching note: The component down the slope uses sine. This is the component that tries to make the block slide. $\sin 25° \approx 0.423$.

(c) [2 marks]

Answer: Frictional force = 8.5 N [1]

Reason: The block is just about to slide / in limiting equilibrium, so friction (acting up the slope) exactly balances the component of weight down the slope. [1]

Teaching note: "Just about to slide" means friction is at its maximum (limiting friction) and equals the parallel component. If the angle were smaller, friction would be less than this maximum.

(d) [3 marks]

Answer: $W_{\text{perpendicular}} = mg \cos \theta = 20 \times \cos 25° = 20 \times 0.906 = 18.1 \approx 18 \text{ N}$ [2]

For equilibrium perpendicular to slope: $R = W_{\text{perpendicular}} = 18 \text{ N}$ [1]

Teaching note: There is no acceleration perpendicular to the slope, so forces balance in that direction. The normal force equals the perpendicular component of weight, not the full weight.

Question 15 [9 marks]

(a) [2 marks]

Answer: $\text{acceleration} = \text{gradient} = \frac{-15 - 15}{3.0 - 0} = \frac{-30}{3.0} = -10 \text{ m/s}^2$ [2]

Or from $t = 0$ to $t = 1.5$ s: $\frac{0 - 15}{1.5} = -10$ m/s²

The acceleration is $10 \text{ m/s}^2$ downward (or $-10 \text{ m/s}^2$ if up is positive). [2]

(b) [3 marks]

Answer:

Method — area under graph from $t = 0$ to $t = 1.5$ s:

$\text{Maximum height} = \text{area of triangle} = \frac{1}{2} \times 1.5 \times 15 = 11.25 \approx 11.3 \text{ m}$ [3]

Or using kinematics: $s = ut + \frac{1}{2}at^2 = 15(1.5) + \frac{1}{2}(-10)(1.5)^2 = 22.5 - 11.25 = 11.25 \text{ m}$

Or using $v^2 = u^2 + 2as$: $0 = 225 + 2(-10)s$, so $s = 11.25$ m

(c) [1 mark]

Answer: The gradient is constant because the acceleration is constant (due to gravity, $g$) throughout the flight. [1]

Teaching note: Air resistance is neglected, so only gravity acts. Constant acceleration means constant (negative) gradient on v-t graph.

(d) [3 marks]

Answer:

• From $t = 0$ to $t = 1.5$ s: velocity is positive (upward motion), so area is positive and gives upward displacement from starting point — this is the maximum height [1]
• From $t = 1.5$ s to $t = 3.0$ s: velocity is negative (downward motion), so area is negative [1]
• The positive and negative areas are equal in magnitude, so total displacement (net area) is zero, meaning the ball returns to its starting point [1]

Teaching note: Area under v-t graph gives displacement (vector), not distance. Above-axis area = positive displacement; below-axis = negative displacement. Total area = algebraic sum = displacement.

Question 16 [9 marks]

(a) [1 mark]

Answer: Reading = $mg = 55 \times 10 = 550$ N (or 55 kg if scale reads in kg, but normally scales read force in N). 550 N [1]

Teaching note: At rest, the scale reads the normal force equals weight. Some scales display mass; if so, 55 kg. Accept either with appropriate unit.

(b) [3 marks]

Answer: When accelerating upward: $R - mg = ma$ (Newton's Second Law, upward positive)

$R = m(g + a) = 55(10 + 2.0) = 55 \times 12 = 660 \text{ N}$ [3]

Mark breakdown:

• Correct equation/recognition of increased normal force [1]
• Correct substitution [1]
• Final answer with unit [1]

Teaching note: The scale reads more than weight when accelerating upward ("feeling heavier"). This is apparent weight. The net force is upward, so normal force exceeds weight.

(c) [2 marks]

Answer: Reading = 550 N [1]

Reason: Constant velocity means zero acceleration, so resultant force is zero. The normal force equals the weight. [1]

Teaching note: Just like the book on the table (Q4), constant velocity implies equilibrium. No acceleration means no extra force needed.

(d) [3 marks]

Answer:

Given $R = 440$ N:

$R - mg = ma$ $440 - 550 = 55a$ $-110 = 55a$ $a = -2.0 \text{ m/s}^2$ [2]

The negative sign indicates acceleration is downward. [1]

Or: magnitude is $2.0 \text{ m/s}^2$ directed downward.

Teaching note: Scale reads less than weight — the student feels lighter. This happens when accelerating downward (or decelerating while moving upward). The direction must be stated clearly.

END OF ANSWER KEY