Secondary 3 Physics Practice Paper 1

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

Version 1

Section A: Multiple Choice & Short Structured Questions

1. D
Explanation: Displacement has both magnitude and direction. Mass, speed, and distance are scalars. [1]

2. Total Distance = $120 + 60 = 180 \text{ km}$
Total Time = $2 + 1 = 3 \text{ hours}$
Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{180}{3} = 60 \text{ km/h}$
[2] (1 for distance/time, 1 for answer)

3. Distance = Area under graph
Area = Area of triangle (0-5s) + Area of rectangle (5-15s) + Area of triangle (15-20s)
$= (\frac{1}{2} \times 5 \times 10) + (10 \times 10) + (\frac{1}{2} \times 5 \times 10)$
$= 25 + 100 + 25 = 150 \text{ m}$
[2] (1 for method, 1 for answer)

4. Acceleration is the rate of change of velocity.
[1]

5. Resultant Force $F_{net} = ma = 50 \times 0.5 = 25 \text{ N}$
$F_{net} = F_{applied} - F_{friction}$
$25 = F_{applied} - 20$
$F_{applied} = 45 \text{ N}$
[2] (1 for net force, 1 for applied force)

6. An object remains at rest or moves with constant velocity in a straight line unless acted upon by a resultant external force.
[2] (1 for rest/constant velocity, 1 for resultant force condition)

7. The passenger has inertia [1]. When the bus brakes, the bus slows down, but the passenger’s body tends to continue moving forward at the original speed [1].

8. Moment clockwise = Moment anticlockwise
Pivot at 50 cm.
2 N weight is at 20 cm, so distance from pivot $d_1 = 50 - 20 = 30 \text{ cm}$.
Moment$_1 = 2 \times 30 = 60 \text{ N cm}$.
Let distance of 3 N weight from pivot be $d_2$.
$3 \times d_2 = 60$
$d_2 = 20 \text{ cm}$.
Position = $50 + 20 = 70 \text{ cm}$ mark (or $50 - 20 = 30$ cm mark, but usually opposite side). Assuming opposite side to balance: 70 cm mark.
[2] (1 for moment equation, 1 for position)

9. $P = h \rho g$
$P = 10 \times 1000 \times 10 = 100,000 \text{ Pa}$ (or $100 \text{ kPa}$)
[2] (1 for formula/substitution, 1 for answer with unit)

10. Pressure is transmitted equally.
$P_1 = P_2 \Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2}$
$\frac{50}{0.01} = \frac{F_2}{0.5}$
$5000 = \frac{F_2}{0.5}$
$F_2 = 2500 \text{ N}$
[2] (1 for principle/equation, 1 for answer)

Section B: Structured Questions

11.
(a) $v = u + at$
$v = 0 + (10 \times 4) = 40 \text{ m/s}$
[2]
(b) $s = ut + \frac{1}{2}at^2$
$s = 0 + \frac{1}{2}(10)(4^2) = 5 \times 16 = 80 \text{ m}$
[2]
(c) Graph: Curve starting from origin, getting steeper (parabola opening upwards).
Y-axis: Displacement (m), X-axis: Time (s).
[2] (1 for shape, 1 for labels)

12.
(a) Diagram showing:

• Weight ($mg$) acting vertically downwards.
• Normal contact force acting perpendicular to the plane.
• Applied force (80 N) acting up the slope.
• Friction acting down the slope.
  [2] (1 for correct directions, 1 for all 4 forces)
  (b) Component of weight down slope = $mg \sin \theta$
  $= 10 \times 10 \times \sin(30^\circ)$
  $= 100 \times 0.5 = 50 \text{ N}$
  [2]
  (c) Since speed is constant, acceleration is 0, so resultant force is 0.
  Forces up slope = Forces down slope
  $F_{applied} = F_{friction} + \text{Weight component}$
  $80 = F_{friction} + 50$
  $F_{friction} = 30 \text{ N}$
  [2] (1 for equilibrium condition, 1 for answer)

13.
(a) Work Done = Force $\times$ Distance
Force = Weight = $mg = 500 \times 10 = 5000 \text{ N}$
$W = 5000 \times 20 = 100,000 \text{ J}$
[2]
(b) Power = $\frac{\text{Work Done}}{\text{Time}}$
$P = \frac{100,000}{10} = 10,000 \text{ W}$ (or $10 \text{ kW}$)
[2]
(c) Efficiency = $\frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%$
$= \frac{100,000}{120,000} \times 100\% = 83.3\%$
[2] (1 for substitution, 1 for answer)

Section C: Free Response & Application

14.
(a)

1. Initially, weight is greater than air resistance [1].
2. There is a resultant downward force, so the skydiver accelerates downwards [1].
3. As speed increases, air resistance increases [1].
4. Eventually, air resistance equals weight. Resultant force is zero, and she falls at constant terminal velocity [1].
   (b)
5. Opening the parachute greatly increases the surface area, causing a large increase in air resistance [1].
6. Air resistance becomes much larger than weight [1].
7. There is a resultant upward force, causing deceleration (upward acceleration) [1].
   (c) Speed / Surface area / Shape / Density of air. (Any one) [1]

15.
(a) For an object in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot. [2]
(b)
(i) Pivot at 50 cm.
Mass 1 (100g = 0.1kg, Weight = 1N) at 10 cm. Distance $d_1 = 40 \text{ cm}$.
Moment$_1 = 1 \times 40 = 40 \text{ N cm}$ (Anticlockwise).
Mass 2 (200g = 0.2kg, Weight = 2N) at 80 cm. Distance $d_2 = 30 \text{ cm}$.
Moment$_2 = 2 \times 30 = 60 \text{ N cm}$ (Clockwise).
$40 \neq 60$, so not in equilibrium. [2]
(ii) Clockwise moment is larger, so it will rotate clockwise. [1]