WAEC Physics Practical 2025 (Alternative B)

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.

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You can watch the full class in the video below:

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https://www.youtube.com/watch?v=PdrM6jnuJOU&t=599s

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.

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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=?

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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

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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.

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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.

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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).

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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

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Watch more educational videos and past questions: https://youtube.com/@allcbts

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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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