Newton's Laws and Momentum (Advanced)
Newton’s Second Law Of Motion
Building upon the concept of inertia introduced by the First Law, Newton's Second Law provides the quantitative relationship between force and the resulting change in motion. Unlike the First Law which describes the condition for no change in velocity (zero net force), the Second Law describes what happens when a non-zero net force acts on an object.
The Second Law introduces the concept of momentum, which is a measure of the "quantity of motion" an object possesses. For a single particle of mass $ m $ moving with velocity $ \vec{v} $, its linear momentum $ \vec{p} $ is defined as:
$ \vec{p} = m\vec{v} $
Momentum is a vector quantity, and its direction is the same as the velocity vector. The standard SI unit for momentum is kilogram-metre per second (kg m/s).
Statement of the Second Law
Newton's Second Law of Motion is formally stated in terms of momentum:
"The rate of change of the linear momentum of a body is directly proportional to the applied external unbalanced force acting on the body and takes place in the direction of the force."
This means that the faster an object's momentum changes, the larger the net force acting on it. Also, the change in momentum occurs in the direction of the net force.
Mathematical Formulation Of Second Law Of Motion
Mathematically, we can express this proportionality as:
$ \vec{F}_{net} \propto \frac{d\vec{p}}{dt} $
where $ \vec{F}_{net} $ is the net external force acting on the object, and $ \frac{d\vec{p}}{dt} $ is the instantaneous rate of change of momentum.
By choosing appropriate units for force, mass, and time (like the SI system), the constant of proportionality can be made equal to 1. Thus, the equation becomes:
$ \vec{F}_{net} = \frac{d\vec{p}}{dt} $
This is the fundamental form of Newton's Second Law. Now, let's substitute the definition of momentum, $ \vec{p} = m\vec{v} $:
$ \vec{F}_{net} = \frac{d(m\vec{v})}{dt} $
Using the product rule for differentiation, we get:
$ \vec{F}_{net} = m \frac{d\vec{v}}{dt} + \vec{v} \frac{dm}{dt} $
This is the general form of the Second Law, applicable even when the mass of the object changes over time (e.g., a rocket expelling fuel, a raindrop growing in size). The term $ \frac{d\vec{v}}{dt} $ is the acceleration $ \vec{a} $, and $ \frac{dm}{dt} $ is the rate of change of mass.
$ \vec{F}_{net} = m\vec{a} + \vec{v} \frac{dm}{dt} $
Special Case: Constant Mass
For the vast majority of problems in basic mechanics, the mass of the object remains constant during its motion ($ m = \text{constant} $). In this case, the rate of change of mass $ \frac{dm}{dt} = 0 $. The general equation simplifies to:
$ \vec{F}_{net} = m \frac{d\vec{v}}{dt} + \vec{v} (0) $
$ \vec{F}_{net} = m \frac{d\vec{v}}{dt} $
Since $ \vec{a} = \frac{d\vec{v}}{dt} $, we get the familiar form:
$ \vec{F}_{net} = m\vec{a} $
This simplified form is valid only when the mass of the object is constant. It states that the net force is directly proportional to the acceleration and the constant of proportionality is the mass. The direction of the acceleration is always the same as the direction of the net force.
Relation to First Law:
The First Law is a special case of the Second Law. If the net external force $ \vec{F}_{net} = 0 $, then from $ \vec{F}_{net} = m\vec{a} $ (for constant mass), we get $ 0 = m\vec{a} $. Since mass $ m $ cannot be zero for a physical object, the acceleration $ \vec{a} $ must be zero ($ \vec{a} = 0 $). Zero acceleration means the velocity $ \vec{v} $ is constant. This is precisely the statement of the First Law: an object maintains constant velocity (rest or uniform motion) if the net force is zero.
Impulse and Momentum:
Integrating the fundamental form of the Second Law $ \vec{F}_{net} = \frac{d\vec{p}}{dt} $ over a time interval from $ t_1 $ to $ t_2 $ gives:
$ \int_{t_1}^{t_2} \vec{F}_{net} \, dt = \int_{t_1}^{t_2} \frac{d\vec{p}}{dt} \, dt $
$ \int_{t_1}^{t_2} \vec{F}_{net} \, dt = \vec{p}(t_2) - \vec{p}(t_1) = \Delta \vec{p} $
The integral of the net force over time is called the Impulse ($ \vec{I} $) of the net force.
$ \vec{I} = \int_{t_1}^{t_2} \vec{F}_{net} \, dt $
For a constant net force, the impulse is simply $ \vec{I} = \vec{F}_{net} \Delta t $.
The equation $ \vec{I} = \Delta \vec{p} $ is known as the Impulse-Momentum Theorem. It states that the impulse of the net force acting on an object is equal to the change in its momentum. This theorem is particularly useful in analysing situations involving forces that act for a very short duration, such as impacts and collisions, where the force magnitude may vary rapidly with time.
Example (Impulse-Momentum Theorem):
Example 1. A cricket ball of mass 0.15 kg is bowled at a speed of 40 m/s. The batsman hits it straight back towards the bowler with a speed of 60 m/s. If the impact lasts for 0.01 seconds, what is the average force exerted by the bat on the ball?
Answer:
Let's choose the direction of the bowled ball as the positive direction. The initial velocity of the ball is $ \vec{v}_i = +40 $ m/s. The final velocity is in the opposite direction, so $ \vec{v}_f = -60 $ m/s.
Initial momentum: $ \vec{p}_i = m \vec{v}_i = (0.15 \, \text{kg}) \times (+40 \, \text{m/s}) = +6.0 \, \text{kg m/s} $.
Final momentum: $ \vec{p}_f = m \vec{v}_f = (0.15 \, \text{kg}) \times (-60 \, \text{m/s}) = -9.0 \, \text{kg m/s} $.
Change in momentum: $ \Delta \vec{p} = \vec{p}_f - \vec{p}_i = (-9.0 \, \text{kg m/s}) - (+6.0 \, \text{kg m/s}) = -15.0 \, \text{kg m/s} $.
The duration of the impact is $ \Delta t = 0.01 $ s.
Using the Impulse-Momentum Theorem ($ \vec{I} = \Delta \vec{p} $), and assuming the average force $ \vec{F}_{avg} $ is constant over the short time interval:
$ \vec{F}_{avg} \Delta t = \Delta \vec{p} $
$ \vec{F}_{avg} = \frac{\Delta \vec{p}}{\Delta t} = \frac{-15.0 \, \text{kg m/s}}{0.01 \, \text{s}} = -1500 \, \text{N} $
The average force exerted by the bat on the ball is 1500 N in the negative direction (opposite to the bowled ball's initial direction, i.e., in the direction the ball is hit). This shows that even for light objects, large forces can be involved in rapid changes of momentum.
Newton’s Third Law Of Motion
Newton's Third Law addresses the nature of force as an interaction between two objects. It states that forces always occur in pairs, and these pairs have specific properties.
Statement of the Third Law
The formal statement of Newton's Third Law is:
"To every action, there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts."
In modern terms, this is usually expressed as:
"For every action force, there is an equal and opposite reaction force."
If object A exerts a force $ \vec{F}_{AB} $ on object B, then object B simultaneously exerts a force $ \vec{F}_{BA} $ on object A, such that:
$ \vec{F}_{AB} = - \vec{F}_{BA} $
Here, the magnitude of the force exerted by A on B is equal to the magnitude of the force exerted by B on A ($ |\vec{F}_{AB}| = |\vec{F}_{BA}| $), and these forces act in opposite directions.
Detailed Analysis of Action-Reaction Pairs
It is critical to understand that action and reaction forces have the following defining characteristics:
- Act on Different Bodies: This is the most crucial point. If the action force is on body B (exerted by A), the reaction force is on body A (exerted by B). They never act on the same body. This is why action-reaction pairs do not cancel each other out when considering the net force on a single body. They only "cancel" in the sense that their vector sum for the *entire system* of interacting bodies might be zero if they are the only forces.
- Equal in Magnitude: The strength of the action force is precisely equal to the strength of the reaction force.
- Opposite in Direction: The direction of the reaction force is exactly opposite to the direction of the action force.
- Simultaneous: The pair of forces exists only during the interaction. They appear and disappear together. One does not cause the other; they are two aspects of a single interaction.
- Same Nature: Both forces in the pair are of the same fundamental type (e.g., both are gravitational, both are electromagnetic, both are contact forces).
Example (Identifying Action-Reaction Pairs):
Example 1. A man is standing on the ground. Identify the action-reaction pairs involving the man and the Earth.
Answer:
There are two main interactions involving the man and the Earth:
1. Gravitational Interaction:
- Action: The Earth pulls the man downwards with the force of gravity (which we call the man's weight, $ \vec{W} = m\vec{g} $). This force acts on the man.
- Reaction: The man pulls the Earth upwards with a gravitational force of equal magnitude and opposite direction. This force acts on the Earth.
2. Contact Interaction (Normal Force): Assuming the ground is preventing the man from falling through:
- Action: The man pushes downwards on the ground (or Earth's surface) due to his weight. This is a contact force exerted by the man on the ground.
- Reaction: The ground pushes upwards on the man with the normal force ($ \vec{N} $). This force acts on the man.
Note that the weight of the man ($ \vec{W} $) and the normal force ($ \vec{N} $) acting on the man are not an action-reaction pair, even though they are equal and opposite when the man is at rest on a level surface. This is because both forces act on the same object (the man). They are part of the forces contributing to the net force on the man (which is zero when he is at rest), but they are not linked by Newton's Third Law.
The Third Law is fundamental to understanding how objects interact and forms the basis for the conservation of momentum in isolated systems.
Conservation Of Momentum
The Law of Conservation of Linear Momentum is a fundamental principle that arises directly from Newton's Laws, particularly the Third Law. It is an extremely useful concept for analyzing systems of interacting objects, especially during events like collisions and explosions.
Momentum of a System
For a system of multiple particles, the total linear momentum $ \vec{P}_{total} $ is the vector sum of the individual momenta of all the particles within the system. If a system consists of $ N $ particles with momenta $ \vec{p}_1, \vec{p}_2, \ldots, \vec{p}_N $, the total momentum is:
$ \vec{P}_{total} = \sum_{i=1}^{N} \vec{p}_i = \vec{p}_1 + \vec{p}_2 + \ldots + \vec{p}_N $
$ \vec{P}_{total} = \sum_{i=1}^{N} m_i \vec{v}_i = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \ldots + m_N \vec{v}_N $
Isolated System
The Law of Conservation of Momentum applies to an isolated system. An isolated system is a system where the net external force acting on it is zero ($ \vec{F}_{ext, net} = 0 $).
Note that this does not mean there are no forces acting within the system (internal forces). Objects within the system can exert forces on each other. However, these internal forces always occur in action-reaction pairs, which, according to Newton's Third Law, are equal in magnitude and opposite in direction. As we will see, this property of internal forces is key to the conservation of momentum.
Statement of Conservation of Momentum
The Law of Conservation of Linear Momentum states:
"If the net external force acting on a system of particles is zero, then the total linear momentum of the system remains constant."
Mathematically, this can be expressed as: If $ \vec{F}_{ext, net} = 0 $, then $ \frac{d\vec{P}_{total}}{dt} = 0 $, which implies $ \vec{P}_{total} = \text{constant} $.
This means the total momentum of the system before an interaction (like a collision or explosion) is equal to the total momentum of the system after the interaction, provided no external forces acted during the interaction (or the net external force was zero).
$ \vec{P}_{total, \text{initial}} = \vec{P}_{total, \text{final}} $
Derivation from Newton's Laws
Consider a system of $ N $ particles. The rate of change of the total momentum of the system is given by:
$ \frac{d\vec{P}_{total}}{dt} = \frac{d}{dt} \left( \sum_{i=1}^{N} \vec{p}_i \right) = \sum_{i=1}^{N} \frac{d\vec{p}_i}{dt} $
According to Newton's Second Law, the rate of change of momentum of the $ i^{th} $ particle is equal to the net force acting on that particle:
$ \frac{d\vec{p}_i}{dt} = \vec{F}_{i, net} $
The net force on the $ i^{th} $ particle can be divided into two parts: external forces ($ \vec{F}_{i, ext} $) exerted by agents outside the system, and internal forces ($ \vec{F}_{i, int} $) exerted by other particles within the system.
$ \vec{F}_{i, net} = \vec{F}_{i, ext} + \vec{F}_{i, int} $
So, $ \frac{d\vec{P}_{total}}{dt} = \sum_{i=1}^{N} (\vec{F}_{i, ext} + \vec{F}_{i, int}) = \sum_{i=1}^{N} \vec{F}_{i, ext} + \sum_{i=1}^{N} \vec{F}_{i, int} $.
The sum of all external forces on all particles is the net external force on the system:
$ \vec{F}_{ext, net} = \sum_{i=1}^{N} \vec{F}_{i, ext} $
Now consider the sum of all internal forces. Internal forces always occur in action-reaction pairs between particles within the system. For any pair of particles $ i $ and $ j $, the force exerted by $ i $ on $ j $ ($ \vec{F}_{ji} $) and the force exerted by $ j $ on $ i $ ($ \vec{F}_{ij} $) satisfy Newton's Third Law:
$ \vec{F}_{ij} = - \vec{F}_{ji} $
When we sum all the internal forces in the system, every action force ($ \vec{F}_{ij} $) has a corresponding equal and opposite reaction force ($ \vec{F}_{ji} $), which is also an internal force within the system. Therefore, the sum of all internal forces is zero:
$ \sum_{i=1}^{N} \vec{F}_{i, int} = 0 $
Substituting these back into the equation for the rate of change of total momentum:
$ \frac{d\vec{P}_{total}}{dt} = \vec{F}_{ext, net} + 0 $
$ \frac{d\vec{P}_{total}}{dt} = \vec{F}_{ext, net} $
This equation is essentially Newton's Second Law applied to the entire system of particles: the rate of change of the system's total momentum is equal to the net external force acting on the system.
From this equation, if the net external force $ \vec{F}_{ext, net} = 0 $, then $ \frac{d\vec{P}_{total}}{dt} = 0 $, which means the total momentum $ \vec{P}_{total} $ is constant. This proves the Law of Conservation of Linear Momentum from Newton's Second and Third Laws.
Conservation of Momentum in Different Dimensions:
Momentum is a vector. In vector form, the conservation of momentum equation is $ \vec{P}_{initial} = \vec{P}_{final} $. This vector equation is equivalent to three separate scalar equations, one for each dimension (x, y, z). If the net external force is zero, then the total momentum is conserved along each coordinate axis independently:
$ P_{total, x, initial} = P_{total, x, final} $ $ P_{total, y, initial} = P_{total, y, final} $ $ P_{total, z, initial} = P_{total, z, final} $
This means if the net external force in, say, the x-direction is zero, the total momentum in the x-direction is conserved, even if there is a net external force (and thus momentum change) in the y or z directions.
Examples (More Advanced):
Example 1. A nucleus at rest spontaneously breaks into two fragments. If the nucleus has mass $ M $ and the two fragments have masses $ m_1 $ and $ m_2 $, with $ m_1 + m_2 = M $. If the fragment with mass $ m_1 $ moves with velocity $ \vec{v}_1 $, find the velocity of the fragment with mass $ m_2 $.
Answer:
Consider the nucleus as the system. Just before the decay (explosion), the nucleus is at rest, so the total initial momentum of the system is zero:
$ \vec{P}_{initial} = M \times 0 = 0 $
After the decay, the system consists of two fragments with masses $ m_1 $ and $ m_2 $ and velocities $ \vec{v}_1 $ and $ \vec{v}_2 $, respectively. The total final momentum is:
$ \vec{P}_{final} = m_1 \vec{v}_1 + m_2 \vec{v}_2 $
The decay process involves only internal nuclear forces. Assuming no significant external forces act during the instantaneous decay, the system is isolated, and momentum is conserved.
$ \vec{P}_{initial} = \vec{P}_{final} $
$ 0 = m_1 \vec{v}_1 + m_2 \vec{v}_2 $
Solving for $ \vec{v}_2 $:
$ m_2 \vec{v}_2 = - m_1 \vec{v}_1 $
$ \vec{v}_2 = - \frac{m_1}{m_2} \vec{v}_1 $
This shows that the second fragment moves in the opposite direction to the first fragment, and its speed is related by the inverse ratio of their masses. This is an example of recoil.
Example 2. A bomb initially at rest explodes into three fragments. Two fragments, of equal mass $ m $, fly off perpendicular to each other with a speed $ v $. What is the mass and speed of the third fragment if the original bomb had mass $ M $?
Answer:
Initial momentum of the bomb at rest is zero:
$ \vec{P}_{initial} = 0 $
After the explosion, there are three fragments. Let the masses be $ m_1 = m $, $ m_2 = m $, and $ m_3 $. The total mass $ M = m_1 + m_2 + m_3 = 2m + m_3 $. Thus, the mass of the third fragment is $ m_3 = M - 2m $.
Let the velocities of the fragments be $ \vec{v}_1 $, $ \vec{v}_2 $, and $ \vec{v}_3 $. The total final momentum is:
$ \vec{P}_{final} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3 $
By conservation of momentum:
$ \vec{P}_{initial} = \vec{P}_{final} $
$ 0 = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3 $
$ m_3 \vec{v}_3 = - (m_1 \vec{v}_1 + m_2 \vec{v}_2) $
Let the two equal mass fragments fly off along the positive x and positive y axes. So, $ \vec{v}_1 = v \hat{i} $ and $ \vec{v}_2 = v \hat{j} $. The masses are $ m_1 = m $ and $ m_2 = m $.
$ m_3 \vec{v}_3 = - (m (v \hat{i}) + m (v \hat{j})) = -mv (\hat{i} + \hat{j}) $
The vector $ \hat{i} + \hat{j} $ has a magnitude of $ \sqrt{1^2 + 1^2} = \sqrt{2} $. The direction is at 45 degrees from both the positive x and positive y axes.
The velocity of the third fragment is $ \vec{v}_3 = - \frac{mv}{m_3} (\hat{i} + \hat{j}) $.
The direction of $ \vec{v}_3 $ is opposite to the direction of $ \hat{i} + \hat{j} $, meaning it is directed at 45 degrees below the negative x-axis (or 225 degrees from the positive x-axis).
The magnitude of the velocity of the third fragment is:
$ |\vec{v}_3| = \frac{mv}{m_3} |\hat{i} + \hat{j}| = \frac{mv}{m_3} \sqrt{2} $
Substitute $ m_3 = M - 2m $:
$ |\vec{v}_3| = \frac{\sqrt{2} mv}{M - 2m} $
So, the third fragment has mass $ M-2m $ and moves with a speed $ \frac{\sqrt{2} mv}{M - 2m} $ in the direction opposite to the resultant velocity of the first two fragments.
The Law of Conservation of Momentum is a very general law and is valid in situations where energy may or may not be conserved (e.g., inelastic collisions). Its application relies on identifying the system correctly and determining if the net external force on that system is zero.