Secondary 3 Physics Practice Paper 4

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answers)

Version: 4 of 5
Subject: Physics
Level: Secondary 3
Topic: Mechanics

Section A: Multiple Choice & Short Structured Questions

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

2. A
Explanation:
Total displacement = $\sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100$ km.
Total time = 1 + 1 = 2 hours.
Average velocity = Displacement / Time = $100 / 2 = 50$ km/h. [1]

3. B
Explanation: Acceleration = Gradient of v-t graph = $\Delta v / \Delta t = (10 - 0) / (5 - 0) = 2.0$ m/s². [1]

4. B
Explanation: At the highest point, the vertical velocity is momentarily zero. However, gravity still acts on the ball, so acceleration is $g$ (10 m/s²) downwards. [1]

5. An object remains at rest or continues to move at a constant velocity in a straight line unless acted upon by a resultant external force. [2]
(1 mark for "rest or constant velocity", 1 mark for "unless acted on by resultant force")

6.
(a) 50 N [1]
(b) Since the velocity is constant, the acceleration is zero. According to Newton's First Law, the resultant force is zero. Therefore, the forward applied force must be equal in magnitude and opposite in direction to the frictional force. [1]

7. Weight = $mass \times g = 80 \times 1.6 = 128$ N. [1]

8.
Pivot at 50 cm.
Load 1: 2 N at 20 cm. Distance from pivot $d_1 = 50 - 20 = 30$ cm.
Moment$_1$ (Anticlockwise) = $2 \times 30 = 60$ N cm.
Load 2: 3 N at distance $d_2$ from pivot.
For equilibrium: Clockwise Moment = Anticlockwise Moment.
$3 \times d_2 = 60$
$d_2 = 20$ cm.
Position on rule = $50 + 20 = 70$ cm mark. [2]

9. Pressure is defined as the force acting perpendicularly per unit area. [1]
(Formula $P=F/A$ is acceptable if defined in words)

10.
$P = h \rho g$
$P = 20 \times 1030 \times 10$
$P = 206,000$ Pa (or 206 kPa). [2]

11.
(a) Work Done = Force $\times$ Distance.
Force required to lift = Weight = $mg = 500 \times 10 = 5000$ N.
Work Done = $5000 \times 12 = 60,000$ J. [2]
(b) Power = Work / Time
Power = $60,000 / 30 = 2,000$ W. [1]

12.
$KE = \frac{1}{2} mv^2$
$KE = 0.5 \times 1000 \times 20^2$
$KE = 500 \times 400 = 200,000$ J. [2]

Section B: Structured Questions

13.
(a) The trolley undergoes uniform acceleration (or constant acceleration). [1]
(Accept: speed increases at a constant rate)

(b) Acceleration = Gradient = $\frac{\Delta v}{\Delta t}$
From graph: at $t=2$, $v=4.0$ m/s (assuming linear slope to peak).
$a = \frac{4.0 - 0}{2 - 0} = 2.0$ m/s². [2]

(c) Distance = Area under the velocity-time graph.
Area = Area of triangle (0-2s) + Area of trapezium/rectangle (2-5s).
Note: Graph shows constant velocity from t=2 to t=5 at 4.0 m/s.
Area 1 (Triangle) = $\frac{1}{2} \times 2 \times 4.0 = 4.0$ m.
Area 2 (Rectangle) = $(5 - 2) \times 4.0 = 3 \times 4.0 = 12.0$ m.
Total Distance = $4.0 + 12.0 = 16.0$ m. [3]

(d) $F = ma$
$m = 0.5$ kg, $a = 2.0$ m/s².
$F = 0.5 \times 2.0 = 1.0$ N. [2]

14.
(a) Free-body diagram:

• Weight ($W$ or $mg$) acting vertically downwards from center of mass. [1]
• Normal Contact Force ($N$ or $R$) acting perpendicular to the slope, outwards from the surface. [1]
• Tension ($T$) acting parallel to the slope, upwards. [1]
• Friction ($f$) acting parallel to the slope, downwards (opposing potential motion up, or if held stationary against gravity pulling down, friction acts up? Correction: The problem says "held in equilibrium by a string". Usually, without the string, the block would slide down. So friction acts up the slope to help the string, or down if the string pulls hard? Standard context: Block tends to slide down. String holds it. Friction acts up the slope if tension is low, or down if tension is high. However, usually in these "held by string" questions without specific tension values, we assume the string prevents sliding down. Let's assume standard case: Weight component down > Tension? No, "held by string" implies string provides main support. Let's look at part (b). It asks for weight component. Part (c) says "if string breaks... friction 5N up". This implies friction opposes motion down. In equilibrium (a), if we don't know T, we just draw forces.
  Correct Diagram Labels:

1. Weight (vertical down).
2. Normal Force (perpendicular to slope).
3. Tension (parallel to slope, up).
4. Friction (parallel to slope). Direction depends on tendency. If not specified, drawing it up the slope is standard for "preventing sliding down", or down if "preventing being pulled up". Given part (c) implies sliding down when broken, the natural tendency is down. So static friction acts UP the slope. [3]

(b) Component of weight down the slope = $mg \sin(\theta)$
$W = 4.0 \times 10 = 40$ N.
Component = $40 \times \sin(30^\circ)$
$\sin(30^\circ) = 0.5$
Component = $40 \times 0.5 = 20$ N. [2]

(c) When string breaks, forces along the slope are:

• Component of weight down slope = 20 N.
• Friction up slope = 5.0 N.
  Resultant Force $F_{net} = 20 - 5.0 = 15.0$ N (down the slope).
  $F = ma \Rightarrow 15.0 = 4.0 \times a$
  $a = 15.0 / 4.0 = 3.75$ m/s². [3]

15.
(a) Pressure = Force / Area
$P = 200 / 0.01 = 20,000$ Pa. [2]

(b) In a hydraulic system, pressure is transmitted equally.
Pressure at large piston = 20,000 Pa.
Force = Pressure $\times$ Area
$F = 20,000 \times 0.5 = 10,000$ N.
Max Weight = 10,000 N. [2]

(c) Liquids are virtually incompressible, whereas gases are compressible. This ensures that the force applied to the small piston is transmitted effectively to the large piston without significant loss of energy in compressing the fluid. [2]

16.
(a) $GPE = mgh$
$GPE = 600 \times 10 \times 30 = 180,000$ J. [2]

(b) Energy cannot be created or destroyed, only converted from one form to another. The total energy of an isolated system remains constant. [1]

(c) At Point A: Total Energy = GPE = 180,000 J (KE = 0).
At Point B: Height = 5 m.
$GPE_B = mgh = 600 \times 10 \times 5 = 30,000$ J.
By conservation of energy:
$KE_B + GPE_B = Total Energy_A$
$KE_B + 30,000 = 180,000$
$KE_B = 150,000$ J.
$KE = \frac{1}{2} mv^2$
$150,000 = 0.5 \times 600 \times v^2$
$150,000 = 300 v^2$
$v^2 = 150,000 / 300 = 500$
$v = \sqrt{500} \approx 22.36$ m/s.
Speed = 22.4 m/s (to 3 s.f.). [4]

(d) Energy is lost to the surroundings as heat and sound due to friction between the wheels and track, and air resistance. This means not all GPE is converted to KE. [2]

17.
(a) Weight = $mg = 70 \times 10 = 700$ N. [1]

(b)

1. Initially, the skydiver accelerates downwards because his weight (700 N) is greater than the air resistance (which is zero at start). [1]
2. As his speed increases, the air resistance (drag) increases. [1]
3. This reduces the resultant force ($Weight - Air Resistance$), so his acceleration decreases. [1]
4. Eventually, air resistance equals his weight. The resultant force becomes zero, and he falls at a constant speed called terminal velocity. [1]

(c) His terminal velocity decreases (becomes lower). [1]
Opening the parachute significantly increases the surface area, which causes a large increase in air resistance/drag. [1]
This upward drag force is now much larger than his weight, causing him to decelerate until a new, lower terminal velocity is reached where drag again equals weight. [1]