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Physics - Laws of Motion (Newton's Laws, Inertia, Gravitation)

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Newton's First Law of Motion (Law of Inertia): A body continues to be in its state of rest or of uniform motion in a straight line unless compelled by an external force to change that state. Inertia is the inherent property of an object to resist any change in its state of motion.

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Types of Inertia: 1. Inertia of Rest (e.g., a person falling backwards when a bus starts suddenly). 2. Inertia of Motion (e.g., a person falling forward when a bus stops suddenly). 3. Inertia of Direction (e.g., passengers being thrown outwards when a car takes a sharp turn).

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Newton's Second Law of Motion: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. It leads to the mathematical expression F=maF = ma.

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Momentum (pp): It is the product of mass and velocity of a body. It is a vector quantity with SI unit kgβ‹…m/skg \cdot m/s.

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Newton's Third Law of Motion: To every action, there is always an equal and opposite reaction. Action and reaction forces act on two different bodies.

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Newton's Universal 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 (m1,m2m_1, m_2) and inversely proportional to the square of the distance (rr) between them.

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Acceleration due to Gravity (gg): The uniform acceleration produced in a freely falling body due to the gravitational pull of the Earth. Its average value on Earth's surface is approximately 9.8 m/s29.8 \, m/s^2.

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Mass vs. Weight: Mass is the quantity of matter in a body (scalar, constant everywhere), while Weight is the force with which the Earth attracts the body (vector, varies with gg).

πŸ“Formulae

p=mΓ—vp = m \times v

F=mΓ—aF = m \times a

F=Ξ”pΞ”t=m(vβˆ’u)tF = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{t}

F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

g=GMR2g = \frac{GM}{R^2}

W=mΓ—gW = m \times g

πŸ’‘Examples

Problem 1:

A constant force acts on an object of mass 5 kg5 \, kg for a duration of 2 s2 \, s. It increases the object's velocity from 3 m/s3 \, m/s to 7 m/s7 \, m/s. Find the magnitude of the applied force.

Solution:

Given: m=5 kgm = 5 \, kg, u=3 m/su = 3 \, m/s, v=7 m/sv = 7 \, m/s, t=2 st = 2 \, s. Using the formula F=m(vβˆ’u)tF = \frac{m(v - u)}{t}, we get F=5(7βˆ’3)2=5Γ—42=10 NF = \frac{5(7 - 3)}{2} = \frac{5 \times 4}{2} = 10 \, N.

Explanation:

The force is calculated by determining the rate of change of momentum over the given time interval.

Problem 2:

Calculate the gravitational force between a ball of mass 2 kg2 \, kg and the Earth. (Mass of Earth M=6Γ—1024 kgM = 6 \times 10^{24} \, kg, Radius of Earth R=6.4Γ—106 mR = 6.4 \times 10^6 \, m, and G=6.67Γ—10βˆ’11 Nβ‹…m2/kg2G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2)

Solution:

Using F=GMmR2F = G \frac{Mm}{R^2}, F=(6.67Γ—10βˆ’11)Γ—(6Γ—1024)Γ—2(6.4Γ—106)2F = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times 2}{(6.4 \times 10^6)^2}. Fβ‰ˆ19.6 NF \approx 19.6 \, N.

Explanation:

This force is also the weight of the ball on the surface of the Earth, calculated using the Universal Law of Gravitation.

Problem 3:

An object has a mass of 10 kg10 \, kg on Earth. What will be its mass and weight on the Moon, where acceleration due to gravity is 1/6th1/6^{th} of that on Earth?

Solution:

Mass remains constant, so mass on Moon = 10 kg10 \, kg. Weight on Earth We=mΓ—g=10Γ—9.8=98 NW_e = m \times g = 10 \times 9.8 = 98 \, N. Weight on Moon Wm=16Γ—We=986β‰ˆ16.33 NW_m = \frac{1}{6} \times W_e = \frac{98}{6} \approx 16.33 \, N.

Explanation:

Mass is an intrinsic property and does not change with location, whereas weight depends on the local acceleration due to gravity (gg).

Laws of Motion (Newton's Laws, Inertia, Gravitation) Revision - Class 9 Science ICSE