IEMDaily - Video Lecture Notes

Definition

Derivation of newton's second law of motion

From newton's second law of motion,

\vec{F} \propto \frac{d \vec{p}}{d t}

\vec{F} = k \frac{d \vec{p}}{d t} = k m \vec{a}

For simplicity, the constant of proportionality is chosen to be

1

∴ \vec{F} = m \vec{a}

Definition

Qualitative meaning of force

Force is a quantity which changes the velocity of a body. So, a body that is sliding stops because frictional force opposes relative motion.

Definition

Applications of second law of motion

According to Newton's second law:

\vec{F} = \frac{d \vec{P}}{d t}

If time interval for the application of force is increased then the value of applied force will decrease. Cricketers use this while catching the ball. They pull their hands backwards so that time of contact with ball increase and less jerk they would experience due to motion of ball.

Definition

Second Law of Motion

A balloon with mass 'm' is descending down with an acceleration 'a' (where a< g). The mass that should be removed from it so that is starts moving up with an acceleration 'a' will be given by:

m g - F = m a

.... (1)

F - (m - m^{'}) g = (m - m^{'}) a

from (1)

F - m g + m^{'} g = m a - m^{'} a; m g - m a - m g + m^{'} g = m a - m^{'} a

m^{'} (g + a) = 2 m a

m^{'} = \frac{2 m a}{g + a}

Law

Third Law of Motion

To every action, there is always equal and opposite reaction, i.e.

\vec{F_{12}} = - \vec{F_{21}}

Definition

Applications of third law of motion

Newton's third law states that for every action there is an equal and opposite reaction. As you stand on the ground, your body push on the earth with a force, and the earth reacts on your body with the same force in opposite direction. This is an example of Newton's third law.

Result

Newton's Third Law from Law of Conservation of Momentum

Consider an isolated system consisting of two particles.

\vec{P}

= Momentum of the whole system, and

{\vec{p}}_{1}

and

{\vec{p}}_{2}

are the respective momentum of the particles.

∴ \vec{P} = {\vec{p}}_{1} + {\vec{p}}_{2}

On differentiating both sides with respect to time, we wiil get

\frac{\vec{d P}}{d t} = \frac{\vec{d p_{1}}}{d t} + \frac{\vec{d p_{2}}}{d t}

But from Newton Second's Law, we know that

\vec{F} = \frac{\vec{d} p}{d t}

∴ {\vec{F}}_{e x t} = {\vec{F}}_{1} + {\vec{F}}_{2}

But we know that the momentum remains conserved in an isolated system,

∴ \vec{P} = c o n s t a n t

⟹ {\vec{F}}_{e x t} = 0

{\vec{F}}_{1} + {\vec{F}}_{2} = 0

{\vec{F}}_{1} = - {\vec{F}}_{2}

which is Newton's third law.

Example

Newton's Law of motion to system of masses

Example: Two blocks A and B of mass

m

and

2 m

are connected together by a light spring of stiffness

k

. The system is lying on a smooth horizontal surface with the block

A

in contact with a fixed vertical wall as shown in the figure. The block

B

is pressed towards the wall by a distance

x_{0}

and then released. There is no friction anywhere. If spring takes time

Δ t

to acquire its natural length then what is the average force on the block

A

by the wall.

Solution:Let

v

is the velocity of mass

2 m

in natural length of spring, then from conservation of energy we have,

\frac{1}{2} k x_{0}^{2} = \frac{1}{2} (2 m) v^{2}

∴ v = \sqrt{\frac{k}{2 m}} x_{0}

Velocity of centre of mass at this instant,

v_{c m} = \frac{Total momentum}{Total mass}

v_{c m} = \frac{2 m v}{3 m}

v = \frac{1}{3} \sqrt{\frac{2 k}{m}} x_{0}

Now, impulse on system

=

change in momentum of system

∴ F_{a v} Δ t = (3 m) v_{C M} = (\sqrt{2 m k}) x_{0}

∴ F_{a v} = \frac{(\sqrt{2 m k}) x_{0}}{Δ t}

Example

Motion of object under constant force

A constant force (

F

) is applied on a stationary particle of mass

m

. The velocity attained by the particle in a certain displacement will be proportional to:

We know,

F = m a

a = \frac{F}{m}

Moreover,

v^{2} = u^{2} + 2 a S

\Rightarrow v^{2} = 2 \times \frac{F}{m} \times S

v = \sqrt{2 \times \frac{F}{m} \times S}

\Rightarrow v \propto \frac{1}{\sqrt{m}}

Definition

Vector Condition for equilibrium

If forces

{\vec{F}}_{1}, {\vec{F}}_{2}, {\vec{F}}_{3}

are acting on a particle then at equilibrium:

{\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = 0

Laws of Motion Concept Page - 3

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Laws of Motion Concept Page - 3

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BookMarks

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