The International System of Units (SI) is important because it provides a universal standard for measurement that is accepted and used worldwide. This system ensures consistency and accuracy in scientific, industrial, and everyday measurements.
For example, scientists in different countries can share data without confusion because they all use the same units like meters, kilograms, and seconds. SI units also help in reducing errors in calculations and making communication clear in research and engineering.
Moreover, it supports international trade, education, and technology by maintaining a uniform system of measurement that everyone understands and follows.
The following prefixes are used to indicate multiples and sub-multiples of base and derived units in the International System (SI):
For a body moving with uniform acceleration in
a straight line, we derive three basic equations of motion using
the definitions of speed, acceleration, and distance. Let:
- u = initial velocity,
- v = final velocity,
- a = acceleration,
- t = time taken,
- s = displacement.
First equation: v = u + at
This comes from the definition of acceleration: a = (v - u)/t
Second equation: s = ut + ½at²
This is derived by calculating the total displacement as initial
velocity plus the effect of acceleration over time.
Third equation: v² = u² + 2as
This is obtained by eliminating time (t) from the first two
equations and relating velocity and displacement directly.
These equations help us solve many problems related to objects moving with constant acceleration.
To solve word problems related to force and momentum, we use the
relation from Newton’s Second Law:
F = (mv - mu) / t
Where:
F = Force (N)
m = Mass (kg)
u = Initial velocity (m/s)
v = Final velocity (m/s)
t = Time (s)
Example:
A ball of mass 2 kg is moving with an initial velocity of 3 m/s.
It is hit and reaches a velocity of 9 m/s in 2 seconds. What is
the force applied?
Solution:
Initial momentum = 2 × 3 = 6 kg·m/s
Final momentum = 2 × 9 = 18 kg·m/s
Change in momentum = 18 - 6 = 12 kg·m/s
Time = 2 s
Force = 12 / 2 = 6 N
So, the applied force is 6 newtons.
Law of Conservation of Momentum:
The law states that:
"In a closed system, where no external forces are
acting, the total momentum before a collision is equal to
the total momentum after the collision."
Mathematically:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Where:
m₁ and m₂ = masses of the two bodies
u₁ and u₂ = initial velocities
v₁ and v₂ = final velocities
This principle is widely used in analyzing collisions and
explosions in physics.
Difference between Elastic and Inelastic
Collisions:
F = maF is force, m is mass, and
a is acceleration.Weight (W) = Mass (m) × Gravitational acceleration (g)Weight (W) = Mass (m) × Gravitational acceleration (g)W = 60 × 9.8 = 588 Nm = W / g = 490 / 9.8 = 50 kgDerivation of Expressions for Tension and Acceleration in a String During Motion of Bodies Connected by a String and Passing Over a Frictionless Pulley:
m₁g - T = m₁a ... (Equation 1)
T - m₂g = m₂a ... (Equation 2)
(m₁g - T) + (T - m₂g) = m₁a + m₂a(m₁ - m₂)g = (m₁ + m₂)aa = [(m₁ - m₂)g] / (m₁ + m₂)
T = m₁g - m₁aT = m₁g - m₁ × [(m₁ - m₂)g / (m₁ + m₂)]T = [2m₁m₂g] / (m₁ + m₂)
Thus, we have derived the expressions for both acceleration and tension in a string connecting two bodies over a frictionless pulley.
Fc = (mv²) / r
Fc = (2 × 4²) / 3 = (2 × 16) / 3 = 32 / 3 ≈ 10.67 N
Moment (Torque) = Force × Perpendicular Distance
Moment = 10 × 0.5 = 5 N·m
To solve such problems, apply the Principle of Moments:
Total Clockwise Moments = Total Anticlockwise Moments
Example:
A uniform meter rule is balanced at the 50 cm mark. A 2 kg mass
is hung at the 20 cm mark. Where should a 3 kg mass be hung to
balance the rule?
Solution:
Now apply the condition for balance:
2 × 30 = 3 × (x - 50)
60 = 3(x - 50)
60 = 3x - 150
3x = 210
x = 70
Answer: The 3 kg mass should be hung at the 70 cm mark.
The stability of an object depends greatly on the position of its centre of gravity and the size of its base. The following points explain this relationship:
Conclusion: To improve the stability of an object, lower its centre of gravity and widen its base.
Newton’s Law of Gravitation states that:
Mathematically:
F = G × (m₁ × m₂) / r²
According to Newton’s Third Law of Motion, “To every action, there is an equal and opposite reaction.”
This law also applies to gravitational forces:
Example:
The Earth pulls the Moon with a gravitational force, and the Moon pulls the Earth with an equal but opposite force.
This perfectly demonstrates the consistency of gravitational force with Newton’s Third Law.
A gravitational field is a region of space surrounding a mass where another mass experiences a force of attraction due to gravity.
Field Lines:
Gravitational Field Strength (g):
The strength of a gravitational field at a point is defined as the force experienced by a unit mass placed at that point.
g = F / m
Example: Earth has a gravitational field around it, so any object near the Earth experiences a downward force due to gravity.
Weight is the force exerted on a body due to gravity. It is the result of the gravitational attraction between the object and the Earth (or any other celestial body).
Mathematically, weight is given by the formula:
W = m × g
Key Points:
Example: An object with a mass of 10 kg on Earth has a weight of:
W = 10 × 9.8 = 98 N
Newton’s Law of Universal Gravitation states that every object in the universe attracts every other object with a force that is:
The formula is:
F = G × (m1 × m2) / r2
Example Problem:
Calculate the gravitational force between two objects with masses of 5 kg and 10 kg placed 2 meters apart.
Solution:
Using the formula:
F = G × (m1 × m2) / r2
F = (6.674 × 10-11) × (5 × 10) / (2)2
F = (6.674 × 10-11) × 50 / 4
F = 8.3425 × 10-10 N
Answer: The gravitational force between the two objects is approximately 8.34 × 10-10 N.
Note: Gravitational force is very small between ordinary objects but becomes significant with very large masses such as planets or stars.
We can calculate the mass of the Earth by rearranging Newton's Law of Gravitation and using the known values of:
Step-by-step Derivation:
Newton’s law of gravitation is:
F = G × (M × m) / R²
We also know that F = m × g, where g = 9.8 m/s²
Equating the two expressions for force:
m × g = G × (M × m) / R²
Cancel m from both sides:
g = G × M / R²
Rearrange to find M:
M = g × R² / G
Substitute values:
M = (9.8) × (6.371 × 106)² / (6.674 × 10-11)
M ≈ 5.97 × 1024 kg
Answer: The mass of Earth is approximately 5.97 × 1024 kilograms.
The acceleration due to gravity (g) on the surface of the Earth is approximately 9.8 m/s². However, as we move away from the Earth's surface (i.e., increase altitude), the value of g decreases. This is because gravitational force becomes weaker with distance from the center of the Earth.
Reason:
The formula for gravitational acceleration at a distance r from the center of the Earth is:
g = G × M / r²
As altitude increases, the distance r increases, and since r² is in the denominator, the value of g decreases.
Conclusion: The acceleration due to gravity decreases with altitude because gravitational force weakens as the distance from Earth's center increases.
Newton’s law of universal gravitation is fundamental in explaining and predicting the motion of satellites around the Earth and other celestial bodies.
Newton’s Law of Gravitation: Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
F = G × (m₁ × m₂) / r²
Importance in Satellite Motion:
Conclusion: Newton’s law of gravitation is essential to understand and calculate the forces involved in satellite motion, making it a cornerstone in the field of space science and satellite technology.
5.5.1 Explanation: Planets Have Moons and They Orbit Around Them
Moons (also called natural satellites) are celestial bodies that revolve around planets due to the gravitational attraction between the moon and the planet.
Example:
Conclusion: Moons orbit planets due to the gravitational force exerted by the planet. This natural motion follows the same physical laws that govern the movement of all celestial bodies, such as Newton’s law of gravitation and the concept of centripetal force.
Conclusion: Gravitational force is the key reason behind the orbital motion of moons, satellites, and planets. It acts as the necessary centripetal force that keeps these celestial objects in stable orbits.
The universe is unimaginably vast and contains everything that exists — including matter, energy, planets, stars, solar systems, galaxies, and space itself.
6.4.1 State the Law of Conservation of Energy
Electricity generation using fossil fuels involves multiple energy conversions. Below is a simple block diagram illustrating the process:
Energy Conversions Involved:
Q = (k × A × ΔT × t) / LWelcome to studynotes.pk – your trusted learning platform for high-quality AKUEB Class 9 Physics Notes. If you are preparing for your SSC-I exams under the Aga Khan University Examination Board (AKUEB), our physics notes are tailored specifically to match the curriculum, focusing on concept clarity, real-world application, and exam preparation.
These AKUEB Physics notes cover all chapters prescribed in the 9th class AKUEB syllabus:
These notes are ideal for:
We understand that AKUEB focuses more on conceptual understanding rather than rote learning. That’s why our notes are designed to explain every topic in a way that enhances analytical thinking. We ensure that every formula, diagram, and topic aligns with board requirements and exam trends.
Whether you're preparing for internal assessments, monthly tests, or board exams – our resources provide a solid foundation to help you achieve high scores.
Click below to download chapter-wise Physics notes in PDF format. These files are free to access and optimized for both mobile and desktop viewing.
We are constantly updating our website with latest AKUEB past papers, guess papers, quizzes, and additional support content. Bookmark this page: studynotes.pk/notes-ix-physics and never miss an update!