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

TuitionGoWhere Practice Paper - Physics Secondary 3 (SA2 Version 5) - Answer Key

Total Marks: 60

Section A: Multiple Choice Questions [10 marks]

1

Answer: B
Working:
Micrometer reading = main scale + thimble scale = 4.5 mm + (28 × 0.01 mm) = 4.5 + 0.28 = 4.78 mm
Actual diameter = measured reading − zero error = 4.78 mm − (+0.02 mm) = 4.76 mm

Wait, let me recalculate:
Zero error is +0.02 mm, meaning the reading is 0.02 mm too large.
Actual = measured − zero error = 4.78 − 0.02 = 4.76 mm → A

Correction: Answer: A
Marking note: Common error is adding the zero error instead of subtracting. Positive zero error means the instrument reads higher than actual.

2

Answer: B
Working:
For uniform acceleration from rest: $s = \frac{1}{2}at^2$ and $v = at$
$a = \frac{v}{t} = \frac{20}{8} = 2.5 \text{ m/s}^2$
$s = \frac{1}{2} \times 2.5 \times 8^2 = \frac{1}{2} \times 2.5 \times 64 = 80 \text{ m}$

Alternative: Average speed = $\frac{0 + 20}{2} = 10 \text{ m/s}$, distance = $10 \times 8 = 80 \text{ m}$

3

Answer: A
Working:
Resultant force = applied force − friction = 15 N − 5.0 N = 10 N
$a = \frac{F}{m} = \frac{10}{2.0} = 5.0 \text{ m/s}^2$

Wait: $F = ma$, so $a = F/m = 10/2 = 5.0 \text{ m/s}^2$ → B

Correction: Answer: B
Marking note: Some students forget to subtract friction or use the applied force directly.

4

Answer: C
Working:
Weight = mass × gravitational field strength = 500 kg × 4.0 N/kg = 2000 N

5

Answer: C
Working:
Impulse = change in momentum = force × time = 10 N × 3.0 s = 30 N·s = 30 kg·m/s
Since initial momentum = 0, final momentum = 30 kg·m/s

6

Answer: B
Working:
$v^2 = u^2 + 2as$, at max height $v = 0$
$0 = 15^2 + 2(-10)h$
$0 = 225 - 20h$
$h = \frac{225}{20} = 11.25 \text{ m}$

7

Answer: C
Working:
Constant speed → resultant force parallel to plane = 0
Component of weight down plane = $mg \sin 30° = 5.0 \times 10 \times 0.5 = 25 \text{ N}$
Pushing force = weight component + friction = 25 + 10 = 35 N

8

Answer: A
Working:
Take direction of 3.0 kg mass as positive.
Initial momentum = $(3.0 \times 4.0) + (5.0 \times -2.0) = 12 - 10 = 2.0 \text{ kg·m/s}$
Combined mass = 8.0 kg
Final velocity = $\frac{2.0}{8.0} = 0.25 \text{ m/s}$ in direction of 3.0 kg mass

9

Answer: B
Working:
Moment = force × perpendicular distance = 20 N × 0.40 m = 8.0 N·m

10

Answer: C
Working:
Take moments about pivot (40 cm mark).
Clockwise moment = $2.0 \text{ N} \times (40 - 10) \text{ cm} = 2.0 \times 30 = 60 \text{ N·cm}$
Anticlockwise moment = $3.0 \text{ N} \times (x - 40) \text{ cm}$
For balance: $3.0(x - 40) = 60$
$x - 40 = 20$
$x = 60 \text{ cm}$

Wait, that gives 60 cm which is option B. Let me recheck.
Weight at 10 cm is 30 cm from pivot (40-10). Moment = 2 × 30 = 60 N·cm clockwise.
3.0 N weight must provide 60 N·cm anticlockwise. Distance from pivot = 60/3 = 20 cm.
Position = 40 + 20 = 60 cm. → B

Correction: Answer: B

Section B: Short Answer and Structured Questions [30 marks]

11

(a) Average speed = $\frac{\text{total distance}}{\text{total time}} = \frac{1.2}{1.5} = 0.80 \text{ m/s}$ [1]

(b) For uniform acceleration from rest: $s = \frac{1}{2}at^2$
$1.2 = \frac{1}{2} \times a \times (1.5)^2$
$1.2 = 0.5 \times a \times 2.25 = 1.125a$
$a = \frac{1.2}{1.125} = 1.07 \text{ m/s}^2$ (or $1.1 \text{ m/s}^2$ to 2 s.f.) [2]
Marks: 1 for correct formula/substitution, 1 for correct answer

(c) $v = u + at = 0 + 1.07 \times 1.5 = 1.60 \text{ m/s}$ (or $v^2 = 2as = 2 \times 1.07 \times 1.2 = 2.57$, $v = 1.60 \text{ m/s}$) [1]

(d) $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.15 \times (1.60)^2 = 0.075 \times 2.56 = 0.192 \text{ J}$ [1]

(e) Loss in GPE = $mgh = 0.15 \times 10 \times 0.31 = 0.465 \text{ J}$ [1]

(f) Some gravitational potential energy is converted to heat and sound due to friction between the car and ramp, and air resistance. [1]
Accept: work done against friction / energy lost to surroundings

12

(a) Weight = $mg = 2000 \times 10 = 20,000 \text{ N}$ [1]

(b) Resultant force = thrust − weight = 30,000 − 20,000 = 10,000 N (upwards) [1]

(c) $a = \frac{F}{m} = \frac{10,000}{2000} = 5.0 \text{ m/s}^2$ [1]

(d) As the rocket rises, its mass decreases because fuel is burnt and ejected. Since thrust is constant and $a = F/m$, the acceleration increases as mass decreases. [2]
Marks: 1 for mass decreases, 1 for $a = F/m$ so acceleration increases
Also accept: gravitational field strength decreases slightly, but mass decrease is the main effect at this level

13

(a) Vertical forces balance (no vertical acceleration):
$N + F \sin 30° = mg$
$N + 50 \times 0.5 = 8.0 \times 10$
$N + 25 = 80$
$N = 55 \text{ N}$ [2]
Marks: 1 for resolving vertical component, 1 for correct answer

(b) Friction = $\mu N = 0.25 \times 55 = 13.75 \text{ N}$ (or 14 N to 2 s.f.) [1]

(c) Horizontal component = $F \cos 30° = 50 \times \frac{\sqrt{3}}{2} = 43.3 \text{ N}$ [1]

(d) Resultant horizontal force = $F \cos 30° - \text{friction} = 43.3 - 13.75 = 29.55 \text{ N}$
$a = \frac{F_{\text{net}}}{m} = \frac{29.55}{8.0} = 3.69 \text{ m/s}^2$ (or 3.7 m/s²) [2]
Marks: 1 for net force, 1 for acceleration

14

(a) The total momentum of a closed system remains constant if no external resultant force acts on it. [1]

(b) Initial total momentum = 0 (both at rest)
Final momentum: $m_A v_A + m_B v_B = 0$
$60 v_A + 40 \times 3.0 = 0$
$60 v_A = -120$
$v_A = -2.0 \text{ m/s}$ (i.e., 2.0 m/s in the opposite direction to skater B) [2]
Marks: 1 for conservation equation, 1 for correct magnitude and direction

(c) Momentum is a vector quantity; the skaters have equal and opposite momenta, so total momentum is zero. Kinetic energy is a scalar quantity; both skaters have positive kinetic energy, so total KE is the sum of their individual KEs and cannot be zero (unless both are at rest). [2]
Marks: 1 for vector vs scalar distinction, 1 for explaining KE adds as scalars

15

(a) Moment of beam's weight about A = $20 \text{ N} \times 1.0 \text{ m} = 20 \text{ N·m}$ (clockwise) [1]

(b) Moment of suspended weight about A = $50 \text{ N} \times 1.5 \text{ m} = 75 \text{ N·m}$ (clockwise) [1]

(c) Taking moments about A (anticlockwise positive):
Tension vertical component = $T \sin 30° = 0.5T$
Anticlockwise moment from tension = $0.5T \times 2.0 = T \text{ N·m}$
Clockwise moments = $20 + 75 = 95 \text{ N·m}$
For equilibrium: $T = 95 \text{ N}$ [3]
Marks: 1 for vertical component of tension, 1 for moment arm (2.0 m), 1 for correct equation and answer

16

(a) $v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 2.5 = 50$
$v = \sqrt{50} = 7.07 \text{ m/s}$ (downwards) [2]
Marks: 1 for correct formula/substitution, 1 for answer

(b) Rebound height 1.6 m: $v^2 = 2gh = 2 \times 10 \times 1.6 = 32$
$v = \sqrt{32} = 5.66 \text{ m/s}$ (upwards) [2]
Marks: 1 for correct formula/substitution, 1 for answer

(c) Change in momentum = $m(v - u)$ (taking upward as positive)
$u = -7.07 \text{ m/s}$, $v = +5.66 \text{ m/s}$
$\Delta p = 0.20 \times (5.66 - (-7.07)) = 0.20 \times 12.73 = 2.55 \text{ kg·m/s}$ [2]
Marks: 1 for correct signs/direction, 1 for correct calculation

(d) Average force = $\frac{\Delta p}{\Delta t} = \frac{2.55}{0.020} = 127.5 \text{ N}$ (upwards) [2]
Marks: 1 for formula, 1 for correct answer with unit

17

(a) Centripetal force = $\frac{mv^2}{r} = \frac{1200 \times 25^2}{80} = \frac{1200 \times 625}{80} = \frac{750,000}{80} = 9375 \text{ N}$ [2]
Marks: 1 for formula/substitution, 1 for answer

(b) Required centripetal force (9375 N) > maximum friction (9000 N), so the car will skid. [1]

(c) On a banked road, the normal reaction force has a horizontal component directed towards the centre of the circle. This component provides part (or all) of the required centripetal force, reducing the reliance on friction. [2]
Marks: 1 for normal reaction has horizontal component, 1 for this component provides centripetal force

18

(a) Extension for 2.0 N load: $x_1 = 12.0 - l_0$
Extension for 5.0 N load: $x_2 = 16.5 - l_0$
Hooke's law: $F = kx$
$5.0 - 2.0 = k(x_2 - x_1)$
$3.0 = k(16.5 - 12.0) = k \times 4.5$
$k = \frac{3.0}{4.5} = 0.667 \text{ N/cm} = 66.7 \text{ N/m}$ [2]
Marks: 1 for using difference in force/extension, 1 for correct answer with unit

(b) $F = kx$ → $2.0 = 0.667 \times (12.0 - l_0)$
$12.0 - l_0 = \frac{2.0}{0.667} = 3.0$
$l_0 = 12.0 - 3.0 = 9.0 \text{ cm}$ [1]

(c) At 5.0 N, extension $x = 16.5 - 9.0 = 7.5 \text{ cm} = 0.075 \text{ m}$
$EPE = \frac{1}{2} k x^2 = \frac{1}{2} \times 66.7 \times (0.075)^2 = 0.188 \text{ J}$
Or using $EPE = \frac{1}{2} F x = \frac{1}{2} \times 5.0 \times 0.075 = 0.1875 \text{ J}$ [2]
Marks: 1 for correct extension, 1 for correct formula and answer

19

(a) Vertical height dropped = $4.0 \times \sin 25° = 4.0 \times 0.4226 = 1.69 \text{ m}$
Loss in GPE = $mgh = 3.0 \times 10 \times 1.69 = 50.7 \text{ J}$ [2]
Marks: 1 for height calculation, 1 for GPE

(b) Gain in KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 3.0 \times 4.0^2 = 1.5 \times 16 = 24 \text{ J}$ [1]

(c) Work done against friction = loss in GPE − gain in KE = 50.7 − 24 = 26.7 J [1]

(d) Work done against friction = friction × distance
$26.7 = F_f \times 4.0$
$F_f = \frac{26.7}{4.0} = 6.68 \text{ N}$ [2]
Marks: 1 for formula, 1 for answer

20

(a) The particle accelerates uniformly from rest to 8 m/s in 4 seconds. [1]

(b) Acceleration = gradient = $\frac{8 - 0}{4 - 0} = 2.0 \text{ m/s}^2$ [1]

(c) Distance = area under v-t graph
Area 1 (0-4 s): $\frac{1}{2} \times 4 \times 8 = 16 \text{ m}$
Area 2 (4-10 s): $6 \times 8 = 48 \text{ m}$
Area 3 (10-14 s): $\frac{1}{2} \times 4 \times 8 = 16 \text{ m}$
Total distance = $16 + 48 + 16 = 80 \text{ m}$ [3]
Marks: 1 for each area, 1 for total

(d) Average speed = $\frac{\text{total distance}}{\text{total time}} = \frac{80}{14} = 5.71 \text{ m/s}$ [1]

End of Answer Key