IGCSE PHYSICS Topic 1.4 : Forces   – Best Revision Guide Syllabus 2023-2024

An In-depth Revision Guide Covering Everything You Need to Know About Forces

Table of Contents

Hello fellow readers! It’s Zahra, and here we’ll be reviewing a core topic in IGCSE Physics: Forces. I have jotted down the important key concepts so feel free to return to this blog for a quick review as I’m going to give a run-down of the following topic.

Before we begin, I would like to start with an open-ended question. What are the forces at work in a roller-coaster? Take a look at the picture below:

A roller-coaster involves many rapid changes in speed. These accelerations and decelerations give the ride it’s thrill. The ride’s designers have calculated the accelerations carefully to avoid the car from coming off its track and ensuring that the riders will stay in the car. Changing directions also require a force. Give it a food for thought! Now with that aside, let’s begin! 😀


I. Important Formulas for This Topic

You should be familiar with the following formulas:

  1. F = ma

        F: Force (N)

        m: mass (kg)

        a: acceleration (m/s^2) 

2. P = mv

        P: Momentum (kg m/s)

        m: mass (kg)

        v: velocity (m/s)

3. I = Ft

         I: Impulse (Ns)

         F: force (N)

         t: time (seconds)

4. Ft = mv – mu

        Ft: Impulse (kg m/s)

        mv: Final momentum (kg m/s)

        mu: Initial momentum (kg m/s)

5. M = Fd

         M: Moment (Nm)

         F: Force (N)

         d: Perpendicular distance (meters)

6. F = kx

         F: Force (N)

         k: Constant force (N/m)

         x: Extension (meters)

7. p = F/A

        p: Pressure (Pa) or (N/m^2)

        F: Force (N)

        A: Area (m^2)

8. p = pgh

        p: Pressure (Pa) or (N/m^2)

        p: Density (kg/m^3)

        g: Gravitational field (m/s^2)

        h: Height (meters)

II. IGCSE Physics EXAM QUESTIONS

Now that you’ve got that covered, let’s apply these formulas with some questions.

QUESTION 1.4.1

A 2kg object experiences a 10N force from the right and a 6N force from the left. What will be the acceleration of the object? What will be the acceleration of the object?

SOLUTION:

F = ma

4N = (2kg)(a)

a = 4N/2kg

a = 2m/s^2

EXPLANATION:

First, calculate the force needed to move the object. 10N + (-6N). 6N is negative because it is moved in the opposing direction. Next, rearrange the F= ma formula, making acceleration the subject by dividing the force by mass. Finally, ensure the final answer is in its correct unit for acceleration.


QUESTION 1.4.2

A car of mass 600kg is moving at 15m/s. Calculate it’s momentum.

SOLUTION:

P = mv

P = (600kg)(15m/s)

P = 900 kg m/s

EXPLANATION:

This question is pretty self-explanatory, but you must keep in mind that not every question will be as direct as this one. (as in, you’ll often find that some questions won’t be direct with what you have to find). As a result, you would have to rearrange the equations or convert a certain unit to give an appropriate answer. A trick I would use to avoid any confusion is to simply focus on the components given in the question, then matching them with the formulas that I have memorised. (For example, this one having mass and velocity being the only components in the question). Do more practice, and you will strengthen your fundamentals. 


QUESTION 1.4.3

A driver accelerates gently so that a force of 30N acts on the car for 10 seconds. Calculate the impulse of the force.

SOLUTION:

I = Ft

I = (30N)(10s)

I = 300Ns

EXPLANATION:

Once again, this question is simple and straight to the point, but i think it’s worth explaining what exactly impulse means. The idea of impulse is that the more time a larger amount of force is exerted on an object, the more the object’s motion will change. In short, impulse of force is the change in momentum.


QUESTION 1.4.4

The initial momentum of the car is 900 kg m/s. After accelerating, a 300 Ns impulse of the force acts on the car. Calculate the momentum of the car after the accelerating force has acted on it.

SOLUTION:

Ft = mv – mu

300 Ns = 900 kg m/s – mu

mu = 900 kg m/s + 300 Ns

mu = 1200 kg m/s

EXPLANATION:

Rearrange the Ft = mv – mu formula, making the final momentum the subject by adding the impulse with initial momentum. Note that the unit of momentum is kg m/s; this is the same as Ns, the unit of impulse.


QUESTION 1.4.5

What is the maximum and minimum moment which a 10N force can produce from a 0.5m wrench?

SOLUTION:

M = Fd

Maximum moment: (10N)(0.5m)

                                 =5 Nm

Minimum moment: (10N)(0m)

                                 =0 Nm


QUESTION 1.4.6

Calculate the force required to extend a spring by 0.04m. The spring’s constant is 200 N/m.

SOLUTION:

F = kx

F = (200 N/m)(0.04m)

F = 8N

EXPLANATION:

A spring’s constant refers to its stiffness or elastic material. This is expressed by the formula shown above. The larger the value of k, the stiffer the spring is and the larger the force needed to compress or stretch it. This concept is also explained in Hooke’s law, which states that the extension of a spring is proportional to the load applied to it, provided the limits of proportions are not extended.


QUESTION 1.4.7

The weight of a woman’s heels are 600N on an area of 0.0001m^2. Calculate the pressure exerted.

SOLUTION:

p = F/A

= 600 N / 0.0001m^2

= 6 000 000 Pa or 6.0 MPa

EXPLANATION:

The smaller the area, the larger the pressure. Visualise this: squeezing your skin with your pointer finger and thumb does not hurt as much as pinching your skin with your nails. Hence, area is inversely proportional to pressure.


QUESTION 1.4.8

A diver is 50m below the sea level. What is the water pressure acting on the diver, given that the water density is 1000 kg/m^3 and the gravitational field strength is 10 N/kg.

SOLUTION:

p = pgh

= (1000 kg/m^3)(10N/kg)(50m)

= 500 000 Pa

EXPLANATION:

Pascal refers to the standard unit of pressure or stress in the International System of Units (SI), equal to one newton per square meter. Another important point would be Pascal’s law, which expresses that “the pressure or power of pressure at a point in a static fluid is equivalent toward all paths”. This means that in a confined or a uniform fluid when external pressure is applied, it will be transmitted uniformly in all directions. The pressure remains constant and is distributed evenly across the enclosure, acting at a right angle to the enclosure’s wall. Since the pressure applied here is constant.

Quantity SymbolSI unit
forceFnewton, N
massmkilogram, kg
acceleration ametres per second squares, m/s^2
areaAmetres squared, m^2
PressurepPascal, Pa
Helpful quantities and their respective units related to this topic

Hey buddy 😀 You made it to the end of the blog! I hope the information above will be a useful source for your knowledge on forces. If this helps, then sure to check our other blogs related to the physics topic, here. Thanks for reading, and good luck on your studies reader!

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