# Laws of Motion ന്യ

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## **Newton's First Law of Motion (Law of Inertia)**

This law states that every body continues in its state of **rest** or of **uniform motion** in a straight line unless it is compelled by some external agency (force) to change that state.

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### **Inertia**
Inertia is the inability of a body to change its state of rest or of uniform motion in a straight line by itself. It's the inherent property of a body that resists a change in its state of motion.

#### **Types of Inertia**
1.  **Inertia of Rest**: The inability of a body to change its state of rest by itself.
    * **Example 1**: When a bus at rest suddenly starts moving, passengers fall backward.
    * **Example 2**: When we shake a tree, its fruits fall down.

2.  **Inertia of Motion**: The inability of a body to change its state of uniform motion by itself.
    * **Example 1**: When a moving bus suddenly stops, passengers fall forward.
    * **Example 2**: A person jumping from a moving bus falls forward.

3.  **Inertia of Direction**: The inability of a body to change its direction of motion by itself.
    * **Example 1**: When a moving bus takes a sharp turn, passengers are thrown outward.
    * **Example 2**: Mud flies tangentially off a spinning wheel.

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## **Momentum (p)**

Momentum is the product of the mass and velocity of a body. It is a measure of the quantity of motion contained in a body.
* **Formula**: $p = m \times v$
* **SI Unit**: $kg \cdot m/s$

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## **Newton's Second Law of Motion**

This law states that the rate of change of momentum of a body is directly proportional to the applied external force.

#### **Derivation of F = ma**
* From the statement of the law, Force (F) is proportional to the rate of change of momentum ($\frac{dp}{dt}$).
    $F \propto \frac{dp}{dt}$
* Introducing a proportionality constant, k:
    $F = k \frac{dp}{dt}$
* The value of k is chosen to be 1 in the SI system.
    $F = \frac{dp}{dt}$
* Since momentum $p = mv$:
    $F = \frac{d(mv)}{dt}$
* Using the product rule for differentiation:
    $F = m\frac{dv}{dt} + v\frac{dm}{dt}$

**Case 1: Mass is constant**
If the mass of the body is constant, then $\frac{dm}{dt} = 0$.
$F = m\frac{dv}{dt} + v(0)$
Since acceleration $a = \frac{dv}{dt}$,
$$ F = ma $$

**Case 2: Velocity is constant (Variable Mass System)**
If the velocity of the body is constant, then $\frac{dv}{dt} = 0$. This applies to systems like rockets ejecting fuel.
$F = m(0) + v\frac{dm}{dt}$
$$ F = v\frac{dm}{dt} $$

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## **Apparent Weight in a Lift ⚖️**

The apparent weight of a body is the reaction force (R) exerted by the weighing machine on the body.


* **Case 1: Body is at Rest or Moving with Constant Velocity**
    Here, acceleration $a = 0$.
    $R - mg = 0 \implies R = mg$
    **Apparent Weight = Real Weight**

* **Case 2: Body is Accelerating Upward**
    The net upward force is $R - mg = ma$.
    $R = mg + ma = m(g+a)$
    **Apparent Weight > Real Weight**

* **Case 3: Body is Accelerating Downward**
    The net downward force is $mg - R = ma$.
    $R = mg - ma = m(g-a)$
    **Apparent Weight < Real Weight**

* **Case 4: Body is in Freefall**
    In freefall, the downward acceleration is equal to gravity ($a=g$).
    $R = m(g - g) = m(0)$
    $R = 0$
    This is the condition of **weightlessness**.

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## **Newton's Third Law of Motion**

This law states that for every action, there is an equal and opposite reaction.

* **Example 1**: When we sit on a bench, our weight acts as the action, and the upward force from the bench is the reaction.
* **Example 2**: When we walk, we push the ground backward (action), and the ground pushes us forward (reaction).

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## **Why the Second Law is the "Real" Law of Motion**

Newton's first and third laws can be derived from the second law, which is why the second law is often considered the most fundamental or "real" law of motion.

#### **Derivation of the First Law from the Second Law**
* From the second law, we have $F = ma$.
* If the external force on a body is zero ($F=0$), then:
    $ma = 0 \implies a = 0$
* Since acceleration is the rate of change of velocity ($a = \frac{dv}{dt}$), zero acceleration means:
    $\frac{dv}{dt} = 0 \implies v = \text{constant}$
* This means the body will continue in its state of rest (if $v=0$) or uniform motion (if $v \neq 0$), which is the statement of the first law.

#### **Derivation of the Third Law from the Second Law**
* Consider an isolated system of two interacting bodies, A and B. The total momentum of the system is $\Delta P = \Delta P_A + \Delta P_B$.
* In the absence of an external force, the total change in momentum is zero ($\Delta P = 0$).
    $F_A \Delta t + F_B \Delta t = 0$
    $F_A \Delta t = -F_B \Delta t$
    $$ F_A = -F_B $$
* This shows that the force exerted by A on B is equal and opposite to the force exerted by B on A, which is the statement of the third law.

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## **Impulse (J)**

The product of the average force and the time interval over which the force acts is called impulse. Impulse is also equal to the change in momentum.

* **Formula**: $J = F_{avg} \times \Delta t$
* **Impulse-Momentum Theorem**: $J = \Delta p = p_2 - p_1$

#### **Derivation**
* From Newton's Second Law, $F = \frac{dp}{dt}$, which can be written as $F dt = dp$.
* To find the total impulse over a time interval from $t_1$ to $t_2$, we integrate:
    $J = \int_{t_1}^{t_2} F dt = \int_{p_1}^{p_2} dp$
    $J = [p]_{p_1}^{p_2}$
    $$ J = p_2 - p_1 $$

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## **Applications of Newton's Laws**

### **Mass-Pulley System (Atwood Machine)**

Consider two masses, $m_1$ and $m_2$ (with $m_1 > m_2$), connected by a string passing over a frictionless pulley. Mass $m_1$ moves downward and $m_2$ moves upward with a constant acceleration 'a'.


* **For mass $m_1$ (downward motion):**
    The net force is $m_1g - T = m_1a$ --- (i)
* **For mass $m_2$ (upward motion):**
    The net force is $T - m_2g = m_2a$ --- (ii)

* **To find Acceleration (a):**
    Add equations (i) and (ii):
    $(m_1g - T) + (T - m_2g) = m_1a + m_2a$
    $m_1g - m_2g = (m_1 + m_2)a$
    $$ a = \frac{(m_1 - m_2)g}{m_1 + m_2} $$

* **To find Tension (T):**
    Substitute the value of 'a' into equation (ii):
    $T = m_2g + m_2a = m_2g + m_2\left(\frac{(m_1 - m_2)g}{m_1 + m_2}\right)$
    $T = \frac{m_2g(m_1 + m_2) + m_2g(m_1 - m_2)}{m_1 + m_2}$
    $T = \frac{m_1m_2g + m_2^2g + m_1m_2g - m_2^2g}{m_1 + m_2}$
    $$ T = \frac{2m_1m_2g}{m_1 + m_2} $$

### **Horse-Cart Problem 🐎**

A horse pulls a cart by applying a force on the ground at an angle. The reaction force from the ground is resolved into two components:
* A vertical component ($R\sin\theta$) that balances the weight of the horse.
* A horizontal component ($R\cos\theta$) that is responsible for the forward motion.


* Let H be the horizontal component of the reaction force ($H = R\cos\theta$).
* Let T be the tension in the rope connecting the horse and cart.
* Let $f_r$ be the frictional force on the cart.
* The equation of motion for the horse is:
    $H - T = m_{horse}a$
* The equation of motion for the cart is:
    $T - f_r = m_{cart}a$

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## **Conservation of Linear Momentum**

This principle states that if no external force acts on a system, the total linear momentum of the system remains constant. In other words, the total momentum before a collision is equal to the total momentum after the collision.
* **Formula**: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

#### **Derivation from Newton's Laws**
Consider two bodies, A and B, with masses $m_1$ and $m_2$ and initial velocities $u_1$ and $u_2$. They collide for a time $\Delta t$ and their final velocities are $v_1$ and $v_2$.

* The force exerted by A on B during the collision ($F_{AB}$) is the rate of change of momentum of B:
    $F_{AB} = \frac{m_2v_2 - m_2u_2}{\Delta t}$
* The force exerted by B on A ($F_{BA}$) is the rate of change of momentum of A:
    $F_{BA} = \frac{m_1v_1 - m_1u_1}{\Delta t}$
* According to Newton's Third Law, the action-reaction forces are equal and opposite:
    $F_{AB} = -F_{BA}$
* Substituting the expressions for the forces:
    $\frac{m_2v_2 - m_2u_2}{\Delta t} = - \left( \frac{m_1v_1 - m_1u_1}{\Delta t} \right)$
* Canceling $\Delta t$ from both sides and distributing the negative sign:
    $m_2v_2 - m_2u_2 = -m_1v_1 + m_1u_1$
* Rearranging the terms to group initial and final momenta:
    $$ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 $$