Newton's Second Law and Impulse

Momentum

While the First Law describes motion in the absence of a net force, the Second Law provides a quantitative relationship between force, mass, and the resulting change in motion. To understand this, we first need to define momentum.

Linear momentum ( $\vec{p}$ ) of a body is defined as the product of its mass (m) and velocity ( $\vec{v}$ ): $\vec{p} = m\vec{v}$

It is a vector quantity, and its direction is the same as the direction of the velocity.
Its SI unit is kg⋅m/s.
A heavy truck and a small car moving at the same velocity have very different momentums. The truck, having more mass, has more momentum and requires a much larger force to stop in the same amount of time.

Newton's Second Law of Motion

The Second Law, in its most general form, states: "The rate of change of momentum of a body is directly proportional to the applied external force and takes place in the direction in which the force acts." $\vec{F}_{ext} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}$ For a body of constant mass, this simplifies to the more common and well-known form: $\vec{F}_{ext} = m \frac{d\vec{v}}{dt} = m\vec{a}$ This is a vector equation, which means it can be broken down into three component equations: $F_x = ma_x$ , $F_y = ma_y$ , $F_z = ma_z$ .

Mathematically, $\vec{F} = \frac{d\vec{p}}{dt} = m\vec{a}$ Where:

$\vec{F}$ is the force
$\vec{p} = m\vec{v}$ is the momentum
$m$ is mass, $\vec{a}$ is acceleration

Interpretation:

This law gives the quantitative definition of force.
The greater the force, the greater the acceleration for a given mass.
The law explains the relationship between force, mass, and acceleration.

Force and acceleration diagram

Units of Force:

SI Unit: Newton (N) ⇒ $1\, \text{N} = 1\, \text{kg} \cdot \text{m/s}^2$
CGS Unit: dyne ⇒ $1\, \text{dyne} = 10^{-5}\, \text{N}$

Dimensional Formula: $[F] = [M^1 L^1 T^{-2}]$

Applications:

Calculating tension in strings
Predicting acceleration from known forces
Analyzing motion on inclined planes

Impulse

Impulse ( $\vec{J}$ ) is defined as the total effect of a force acting over a period of time. It is a vector quantity equal to the change in momentum of the object. From the Second Law, $d\vec{p} = \vec{F}dt$ . Integrating this from an initial time $t_i$ to a final time $t_f$ gives the Impulse-Momentum Theorem: $\vec{J} = \int_{t_i}^{t_f} \vec{F}(t) \,dt = \vec{p}_f - \vec{p}_i = \Delta\vec{p}$

For a constant force, this simplifies to $\vec{J} = \vec{F}_{avg} \Delta t$ .
Graphically, the impulse is the area under the Force-Time graph.
Impulse is particularly useful for analyzing situations involving large forces acting for very short durations, like a bat hitting a ball.

Example 1:

Question: A 2 kg object is acted upon by a force of 10 N. What is its acceleration?

Solution: Using $F = ma$ : $a = \frac{F}{m} = \frac{10}{2} = 5\, m/s^2$

Example 2: Constant Force

A body of mass 5 kg is acted upon by two perpendicular forces of 8 N and 6 N. Find the magnitude and direction of the acceleration.

Solution:

Find the Net Force: The two forces are perpendicular. We find the magnitude of the net force using the Pythagorean theorem. $F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\ N$
Apply Newton's Second Law: $a = \frac{F_{net}}{m} = \frac{10\ N}{5\ kg} = 2\ m/s^2$
Find the Direction: The direction of the acceleration is the same as the direction of the net force. Let $\theta$ be the angle the resultant force makes with the 8 N force. $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{6}{8} = 0.75 \implies \theta = \tan^{-1}(0.75) \approx 36.87^\circ$

Example 2: Impulse

A batsman hits back a ball of mass 0.15 kg straight in the direction of the bowler without changing its initial speed of $12\ m/s$ . If the ball is in contact with the bat for 0.001 s, what is the magnitude of the average force exerted by the bat?