Good day, students! My name is Cyril, and today we’ll be exploring WAEC Physics Practical Alternative B past questions, focusing specifically on Mechanics — a core topic that often appears in the exams.
So please, grab your writing materials and calculator, and let’s jump right in.
You can watch the full class in the video below:
Question 1: Maximum Height Reached by a Thrown Ball
Question:
A ball is thrown vertically upward with a velocity of 20 m/s. Calculate the maximum height reached.
Step-by-Step Solution:
Step 1: Identify your parameters
- Initial velocity, u=20 m/su = 20 \, \text{m/s}u=20m/s
- Final velocity at maximum height, v=0 m/sv = 0 \, \text{m/s}v=0m/s
- Acceleration due to gravity, g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2 (opposing motion, so we use negative)
- Height, h=?h = ?h=?
Concept Reminder:
At the maximum height, the ball momentarily comes to rest. Therefore, final velocity v=0v = 0v=0. We’re going to use one of the equations of motion: v2=u2−2ghv^2 = u^2 – 2ghv2=u2−2gh
Substituting the values: 0=202−2×9.8×h⇒0=400−19.6h⇒19.6h=400⇒h=40019.6=20.4 m0 = 20^2 – 2 \times 9.8 \times h \Rightarrow 0 = 400 – 19.6h \Rightarrow 19.6h = 400 \Rightarrow h = \frac{400}{19.6} = 20.4 \, \text{m}0=202−2×9.8×h⇒0=400−19.6h⇒19.6h=400⇒h=19.6400=20.4m
Final Answer: 20.4 meters
Question 2: Define the Moment of a Force
Definition:
The moment of a force is the turning effect of a force about a pivot or point.
This is a direct theory question that is often repeated in WAEC and NECO Physics Practical papers.
Question 3: Find the Weight of a Uniform Meter Rule in Equilibrium
Question:
A uniform meter rule is balanced at 45 cm with a 50 g mass at 10 cm. Find the weight of the meter rule.
Step-by-Step Analysis:
Step 1: Understand the situation
- The meter rule is uniform → Total length = 100 cm
- Balanced at 45 cm → Pivot is at 45 cm
- A 50 g mass is placed at 10 cm
Let WWW be the weight of the meter rule, acting at the center (50 cm mark since it’s uniform).
Use the Principle of Moments:
Clockwise Moment=Anticlockwise Moment\text{Clockwise Moment} = \text{Anticlockwise Moment}Clockwise Moment=Anticlockwise Moment
Clockwise moment from 50 g mass: Force=50 g=0.05 kg×9.8=0.49 NDistance from pivot=45−10=35 cm⇒0.49×35=17.15 Ncm\text{Force} = 50 \, \text{g} = 0.05 \, \text{kg} \times 9.8 = 0.49 \, \text{N} \text{Distance from pivot} = 45 – 10 = 35 \, \text{cm} \Rightarrow 0.49 \times 35 = 17.15 \, \text{Ncm}Force=50g=0.05kg×9.8=0.49NDistance from pivot=45−10=35cm⇒0.49×35=17.15Ncm
Let the weight of the meter rule = WWW
Distance from 50 cm to 45 cm = 5 cm W×5=17.15⇒W=17.155=3.43 NW \times 5 = 17.15 \Rightarrow W = \frac{17.15}{5} = 3.43 \, \text{N}W×5=17.15⇒W=517.15=3.43N
Final Answer: 3.43 N
Watch more educational videos and past questions: https://youtube.com/@allcbts
Why You Should Practice WAEC Physics Past Questions
This practical problem taps into:
- Equations of motion (kinematics)
- Center of gravity and moments
- Conversion between grams and Newtons
- Real-life application of physics laws
These skills are tested every year in WAEC and NECO practical papers.
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