## What is Newton’s Law?

Newton’s laws of motion describe the relationship between the motion of an object and forces acting on it. Newton’s law of motion has two types as follows:

## Newton’s Second Law of Motion

Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

The second law of motion describes what happens to the massive body when acted upon by an external force. The 2nd law of motion states that the force acting on the body is equal to the product of its mass and acceleration.

Assume, P is momentum, $m$ is mass, $t$ is time, $u$ is initial velocity and $v$ is final velocity and $a$ is acceleration.

$\begin{array}{c}F\propto \frac{dP}{dt}\\ \propto \frac{mv-mu}{t}\\ \propto \frac{m(v-u)}{t}\\ \propto ma\end{array}$

$F=kma$

In the equation, k is the constant of proportionality, and it is equal to one when the values are taken in SI unit. Hence, the final expression will be,

$$F=ma$$From above description

Newton’s second law state that the rate of change of momentum is directly proportional to the applied force on the body.

$F\propto \frac{dP}{dt}$

$F\propto \frac{d(mv)}{dt}$

By Removing Constant of Proportionality Equation can be returned as follows

$$F=m\text{}\frac{dv}{dt}+v\text{}\frac{dm}{dt}$$

### For Constant Mass

Practical Illustration for constant mass

Whenever a bike is travelling from point $\text{1to2}$it has velocity and mass $\text{v1,v2}$and $\text{m1,m2}$

Since the mass of the vehicle is greater the mass of fuel mass is taken constantly.

$\text{}F=m\text{}\frac{dv}{dt}k$

Since the rate of change of velocity is called acceleration.

$k=1\text{for}m=1\text{and}a=1$

$$F=ma$$Where F is net force acting on the body, m is mass of the body, a is the acceleration of the body.

From Newton’s second law motion it can be said that acceleration of the body is directly proportional; net force acting inversely proportional to the mass of the body.

$$\begin{array}{l}a\propto F\\ a\propto \frac{1}{m}\end{array}$$

For varying mass

For Rocket or flight mass changes from one point to another point, constant mass cannot be taken.

$$P=mv$$Differentiating both sides with respect to t

$\frac{dp}{dt}=\frac{d(mv)}{dt}$

$F=m\frac{dv}{dt}+v\frac{dm}{dt}$

Here, the mass is not constant, mass is variable here so we use it. Generally, in the case of rocket propulsion, we see the concept of variable mass.

## Unit of Force

The definition of the standard metric unit of force is stated by the below equation. One Newton is defined as the amount of force required to give a one kg mass an acceleration of 1m/s^{2}.

$$\text{1Newton}=\text{1kg}\xb7{\text{m/s}}^{\text{2}}$$

## Direction of Net Force and Acceleration

Direction acceleration is the same as the direction of the net force

Ex: Whenever a car is at rest is throttled it tends to gain its speed resulting in an increase in acceleration in this case both net force and acceleration are in the forwarding direction.

### Results of Second law

- In the second law, if F=0 implies F=0 by this we can say that the second law follows the first law.

- The second law of motion is a vector law. We can resolve it for the three components.

$${F}_{x}=\frac{d{p}_{x}}{dt}=m{a}_{x},{F}_{y}=\frac{d{p}_{y}}{dt}=m{a}_{y},{F}_{z}=\frac{d{p}_{z}}{dt}=m{a}_{z}$$

This equation explains that if a force is not parallel to the velocity of the body, but makes some angle with it, as a result, it only changes the component of velocity along the direction of the force. The component of velocity normal to the force remains unchanged. To understand this we can take the projectile motion for which the horizontal component of velocity remains unchanged.

From the second law, impulse can be derived

## Practical Demonstration of Impulse

We sometimes encounter examples where a large force acts for a very short duration producing a finite change in momentum of the body. For example, when a cock hits a badminton bat and returns back, the force on the cock by the ball acts for a very short time when the two are in contact, yet the force is large enough to reverse the momentum of the cock. Often, in these situations, the force and the time duration are difficult to ascertain separately. However, the product of force and time, which is the change in momentum of the body, remains a measurable quantity. This product is called impulse:

A large force acting for a short time to produce a finite change in momentum is called an impulsive force

### The mathematical equation for impulse force

$${\int}_{1}^{2}dp=}{\displaystyle {\int}_{1}^{2}F.\text{\hspace{0.17em}}dt$$

Here F indicates vector force and dp indicates momentum

$$J={\displaystyle {\int}_{1}^{2}F.\text{\hspace{0.17em}}dt}$$

J indicates Impulse

Unit of Impulse

$J=1\text{N}\times 1\text{S}=1\text{NS}$

Newton second is unit of impulse)

## Newton’s Second Law Solved Examples

- If there is a block of mass 2kg , and a force of 20N is acting on it in the positive x-direction, and a force of 40N in the negative x-direction, then what would be its acceleration?

$${F}_{net}=20-40=-20\text{N}$$

(force applied is 20N in negative x-direction)

$$\text{m=2kg}$$(mass applied is$$\text{2kg}$$)

Newton second law states

$$\begin{array}{c}F=ma\\ -20=2\times a\\ a=\frac{-20}{2}\\ =-10{\text{m/s}}^{2}\end{array}$$

acceleration of the block is 10m/s2 in negative x-direction,negative sign indicates the direction of acceleration is the opposite direction of block which results in deceleration

2)A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of $15\text{}m\text{}/s$. If the mass of the ball is $0.10\text{}kg$, determine the impulse imparted to the ball. (Assume linear motion of the ball)

From Newton’s second law impulse is equal to change in momentum

$$J=\text{}{m}_{1}{v}_{1}-\text{}{m}_{2}{v}_{2}$$

In this example $\text{m1=m2=0}\text{.10}$${\text{v}}_{\text{1}}\text{=15m/s}$

Since $${\text{v}}_{\text{2}}$$is moving opposite to $${\text{v}}_{\text{1}}$$so

$$\begin{array}{l}{\text{v}}_{\text{2}}\text{=}-\text{15m/s}\hfill \\ \text{J=}\left(\text{0}\text{.10\xd715+0}\text{.10\xd715}\right)\hfill \\ \text{J=2}\text{.25NS}\hfill \end{array}$$

## Application of Newton’s Second Law

### Throwing a ball

When we throw a ball we exert force in a specific direction, which is the direction in which it will travel. In addition, the stronger the ball is thrown, the stronger the force we put on it and the further away it will travel.

### Pushing a fridge

It is easier to push an empty fridge in a home than it is to push a loaded one. More mass requires more force to accelerate.

### Two people running

Among the two people running, if one is heavier than the other then the one weighing heavier will run slower because the acceleration of the person weighing lighter is greater.

### Two balls thrown

If two balls, one tennis ball, and the other football, are dropped from the top of a building, a person on the ground will find it easier to catch the tennis than the football. The mass of a body is thus an important parameter that determines the effect of force on its motion

## Context and Applications

This Law is significant in everyday life which is incredibly valuable for scientists, Mechanical Engineers, inventors, etc.

- Bachelors in Science Physics
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