Secondary 4 Pure Physics Mechanics Quiz

Secondary 4 Pure Physics Quiz - Mechanics (Answer Key)

1. D
Reasoning: Acceleration has both magnitude and direction. Speed, distance, and mass are scalars. [1]

2. A
Reasoning: Average velocity = Total Displacement / Total Time. Displacement is 0 km (returns to start). $0/2 = 0$ km/h. [1]

3. C
Reasoning: Constant velocity implies zero acceleration, meaning forces are balanced (Weight = Air Resistance). This is terminal velocity. [1]

4. Newton metre (Nm)
[1]

5. $W = mg = 5 \times 10 = 50 \text{ N}$
[1]

6. Inertia is the resistance of an object to change its state of rest or uniform motion.
[1]

7. $\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \text{ N}$
[1]

8. Pascal’s Principle
[1]

9. $W = F \times d = 20 \times 5 = 100 \text{ J}$
[1]

10. A sharp knife has a smaller surface area. For the same force, Pressure = Force/Area is higher, allowing it to penetrate easier.
[1]

11.
(a) Initially, weight is greater than air resistance, so there is a resultant downward force. The skydiver accelerates downwards. As speed increases, air resistance increases. The resultant force decreases, so acceleration decreases. Eventually, air resistance equals weight, resultant force is zero, and he moves at constant terminal velocity. [3]
(b) Opening the parachute greatly increases the surface area, causing a large increase in air resistance. The upward air resistance becomes greater than the downward weight, creating a resultant upward force that decelerates the skydiver. [2]

12.
(a) Distance from pivot = $50 - 20 = 30 \text{ cm} = 0.3 \text{ m}$.
Moment = $4.0 \text{ N} \times 0.3 \text{ m} = 1.2 \text{ Nm}$. [2]
(b) Clockwise Moment = Anticlockwise Moment.
Distance of W from pivot = $80 - 50 = 30 \text{ cm} = 0.3 \text{ m}$.
$W \times 0.3 = 1.2$
$W = 1.2 / 0.3 = 4.0 \text{ N}$. [2]
(c) New pivot at 30 cm.
4.0 N weight is at 20 cm. Distance = $30 - 20 = 10 \text{ cm} = 0.1 \text{ m}$.
Anticlockwise Moment = $4.0 \times 0.1 = 0.4 \text{ Nm}$.
W must provide Clockwise Moment of 0.4 Nm.
$W = 4.0 \text{ N}$ (from previous part, assuming W is the same object).
$4.0 \times d = 0.4$
$d = 0.1 \text{ m} = 10 \text{ cm}$ from pivot.
Position = $30 \text{ cm} + 10 \text{ cm} = 40 \text{ cm}$ mark. [3]

13.
(a) Acceleration = Gradient = $\frac{\Delta v}{\Delta t} = \frac{8 - 0}{4 - 0} = 2 \text{ m/s}^2$. [2]
(b) Distance = Area under graph.
Area 1 (Triangle) = $\frac{1}{2} \times 4 \times 8 = 16 \text{ m}$.
Area 2 (Rectangle) = $(10 - 4) \times 8 = 6 \times 8 = 48 \text{ m}$.
Total Distance = $16 + 48 = 64 \text{ m}$. [3]
(c) Curve starting from origin with increasing gradient (parabolic shape) up to t=4s. [2]

14.
(a) $P = h \rho g = 20 \times 1030 \times 10 = 206,000 \text{ Pa}$ (or $2.06 \times 10^5 \text{ Pa}$). [2]
(b) Total Pressure = Atmospheric + Liquid Pressure
$P_{total} = 1.0 \times 10^5 + 2.06 \times 10^5 = 3.06 \times 10^5 \text{ Pa}$. [2]

15.
(a) 800 N. Since the car is moving at constant speed, acceleration is zero. By Newton's First Law, forces are balanced, so Resistive Force = Driving Force. [2]
(b) Resultant Force = Driving Force - Resistive Force = $1400 - 800 = 600 \text{ N}$.
$F = ma \Rightarrow 600 = 1200 \times a$
$a = 600 / 1200 = 0.5 \text{ m/s}^2$. [3]
(c) As speed increases, air resistance (a resistive force) increases. This reduces the resultant force ($F_{driving} - F_{resistive}$), and since $a = F/m$, the acceleration decreases. [2]

16.
(a) The point through which the entire weight of the object appears to act. [1]
(b) As the bottle tilts, the line of action of its weight (acting from the centre of gravity) moves towards the edge of the base. When the line of action falls outside the base area, there is no normal contact force to counteract the moment of the weight, causing the bottle to topple. [2]

17.
(a) Since speed is constant, acceleration is 0. Tension = Weight.
$T = mg = 500 \times 10 = 5000 \text{ N}$. [2]
(b) Work Done = Force $\times$ Distance.
$W = 5000 \times 12 = 60,000 \text{ J}$ (or $60 \text{ kJ}$). [2]

18.
(a) Circumference $C = 2 \pi r = 2 \times \pi \times 50 = 100\pi \text{ m}$.
Speed = Distance / Time = $100\pi / 40 = 2.5\pi \approx 7.85 \text{ m/s}$. [2]
(b) Velocity is a vector quantity consisting of speed and direction. Although the speed is constant, the direction of motion is continuously changing as the cyclist moves around the circle. Therefore, the velocity is changing. [2]

19.
(a) Hooke's Law: $F = kx$.
$x = 4 \text{ cm} = 0.04 \text{ m}$.
$k = F / x = 20 / 0.04 = 500 \text{ N/m}$. [2]
(b) Elastic Potential Energy $E = \frac{1}{2} k x^2$.
$E = 0.5 \times 500 \times (0.04)^2 = 250 \times 0.0016 = 0.4 \text{ J}$. [2]

20.
(a) Normal Reaction Force $R = mg = 10 \times 10 = 100 \text{ N}$.
Frictional Force $f = \mu R = 0.5 \times 100 = 50 \text{ N}$. [2]
(b) Resultant Force $F_{net} = F_{applied} - f = 60 - 50 = 10 \text{ N}$.
$F_{net} = ma \Rightarrow 10 = 10 \times a$.
$a = 1 \text{ m/s}^2$. [2]