CONTENTS
5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem
5.3 Work
5.4 Kinetic Energy
5.5 Work Done by a Variable Force
5.6 The Work-Energy Theorem for a Variable Force
5.7 The Concept of Potential Energy
5.8 The Conservation of Mechanical Energy
5.9 The Potential Energy of a Spring
5.10 Power
5.11 Collisions
5.1 Introduction
๐ Diving into Physics: Work, Energy, and Power
Welcome to the start of an essential journey in Physics! While we use the words "work" and "energy" every day to describe everything from studying for exams to feeling tired, Physics gives them a very specific, mathematical "personality."
Before we can master the concepts of Chapter 5, we need to sharpen one specific tool in our mathematical toolkit: The Scalar Product.
1. The "Intro" Vibe: Physics vs. Real Life
In everyday language, if you stand still holding a heavy box for an hour, you'd say you worked hard. In Physics? You did zero work. * Work: Isn't just about effort; it’s about a force causing a displacement.
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Energy: This is your "capacity" to do that work.
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Power: This is all about how fast you get that work done (think of a "powerful" boxer's punch).
5.1.1 The Scalar Product
2. The Scalar (Dot) Product: $A \cdot B$
When we multiply two vectors, we have two choices. One choice results in a Scalar (just a number, no direction). This is the Scalar Product, often called the Dot Product.
The Formula
The scalar product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is defined as:
$$\mathbf{A} \cdot \mathbf{B} = AB \cos \theta$$-
$A$ and $B$: The magnitudes (lengths) of the vectors.
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$\theta$: The angle between them.
Why a Scalar? Even though $\mathbf{A}$ and $\mathbf{B}$ have directions, their "dot product" is just a magnitude. It’s like mixing two ingredients to get a completely different final dish!
3. Key Properties to Remember
To solve problems quickly, keep these rules in your back pocket:
| Property | Rule | Why it matters |
| Commutative | $\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$ | Order doesn't matter! |
| Distributive | $\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{B} + \mathbf{A} \cdot \mathbf{C}$ | Works just like regular algebra. |
| Perpendicular | If $\theta = 90^\circ$, then $\mathbf{A} \cdot \mathbf{B} = 0$ | Since $\cos 90^\circ = 0$, perpendicular vectors always "dot" to zero. |
| Parallel | If $\theta = 0^\circ$, then $\mathbf{A} \cdot \mathbf{B} = AB$ | Max value occurs when they point the same way. |
4. Working with Components (The $i, j, k$ method)
If you are given vectors in unit vector notation, the dot product is actually quite simple. You just multiply the "like" terms and add them up:
If $\mathbf{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\mathbf{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$, then:
$$\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$$Quick Tip:
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$\hat{i} \cdot \hat{i} = 1$
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$\hat{i} \cdot \hat{j} = 0$ (because they are $90^\circ$ apart!)
๐ง Knowledge Check
Example: Find the angle between Force $\mathbf{F} = (3\hat{i} + 4\hat{j} - 5\hat{k})$ and displacement $\mathbf{d} = (5\hat{i} + 4\hat{j} + 3\hat{k})$.
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Find $\mathbf{F} \cdot \mathbf{d}$: $(3 \times 5) + (4 \times 4) + (-5 \times 3) = 15 + 16 - 15 = \mathbf{16}$.
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Find Magnitudes: Both vectors have a magnitude of $\sqrt{50}$ in this case.
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Solve for $\cos \theta$: $16 = \sqrt{50} \sqrt{50} \cos \theta \rightarrow 16 = 50 \cos \theta \rightarrow \cos \theta = 0.32$.
5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem
⚡ The "Speed-Work" Connection: The Work-Energy Theorem
Ever wonder why a car needs a much longer distance to stop if it’s going just a little bit faster? Or why a tiny bullet can punch through solid wood? It all comes down to a beautiful mathematical relationship between how hard you push (Work) and how fast things move (Kinetic Energy).
1. The Core Idea: What is Kinetic Energy?
If an object of mass $m$ is moving with a velocity $\mathbf{v}$, it possesses Kinetic Energy ($K$). It is the "energy of motion."
$$K = \dfrac{1}{2} mv^2$$-
It’s a Scalar: Even though velocity has direction, Kinetic Energy does not.
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The "v-squared" Rule: Because velocity is squared, if you double your speed, you have four times the energy.
2. The Big Reveal: The Work-Energy (WE) Theorem
The Work-Energy Theorem is the "Bridge" between forces and motion. It states:
The change in kinetic energy of a particle is equal to the work done on it by the net force.
In simple math:
$$K_f - K_i = W$$(Where $K_f$ is final energy and $K_i$ is initial energy)
๐ ️ The Derivation (How we get there)
From your previous lessons on motion, you know the equation:
$$v^2 - u^2 = 2as$$If we multiply both sides by $\frac{m}{2}$:
$$\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas$$Since $F = ma$ (Newton’s Second Law), then $mas = Fs$. And since $Fs$ is Work ($W$):
$\Delta K = W$
3. Real-World Case Study: The Raindrop ๐ง️
Imagine a 1.00g raindrop falling from 1.00 km.
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Gravity does positive work on it ($W_g = mgh$).
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Air Resistance does negative work on it ($W_r$), trying to slow it down.
According to the Work-Energy Theorem:
$$\Delta K = W_g + W_r$$This is why raindrops don't hit us like bullets! The negative work done by air resistance cancels out most of the kinetic energy gained from gravity.
4. Why this matters (The "Hook")
The Work-Energy Theorem allows us to solve complex problems without knowing the acceleration at every second.
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If you know the initial speed and the total work done, you can predict the final speed instantly.
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It explains why safety features like "crumple zones" in cars work—they increase the distance over which work is done to remove your kinetic energy, reducing the force on you!
๐ Quick Comparison: Work vs. Energy
| Feature | Work (W) | Kinetic Energy (K) |
| Definition | Force applied over a distance. | Energy due to motion. |
| Nature | A process (something you do). | A property (something you have). |
| Formula | $W = \mathbf{F} \cdot \mathbf{d}$ | $K = \frac{1}{2}mv^2$ |
๐ง Check Your Understanding
Scenario: A cyclist skids to a stop. The road exerts a friction force of 200 N over 10 m.
The Question: How much did the cyclist's kinetic energy change?
Step 1: Calculate Work ($W = Fd \cos \theta$). Since friction opposes motion, $\theta = 180^\circ$.
Step 2: $W = 200 \times 10 \times (-1) = -2000 \text{ J}$.
Result: The kinetic energy decreased by 2000 J.
5.3 Work
๐ ️ Decoding "Work": It's Not What You Think!
In everyday life, "work" is anything that makes you tired—studying for a physics final, holding a heavy grocery bag, or even thinking really hard.
In Physics, however, Work has a very strict, no-nonsense definition. You could be sweating and exhausted, but if you haven't moved an object, a physicist would say you've done zero work.
1. The Physics Definition of Work
Work is done when a Force ($F$) acting on an object causes a Displacement ($d$). But there’s a catch: only the part of the force that points in the direction of the motion counts!
The Formula
$$W = (F \cos \theta) d = \mathbf{F} \cdot \mathbf{d}$$-
$F \cos \theta$: The component of the force acting in the direction of the displacement.
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$d$: The magnitude of the displacement.
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$\theta$: The angle between the Force vector and the Displacement vector.
The Wall Test: If you push against a brick wall with all your might but the wall doesn't move ($d = 0$), the Work done is 0. Your muscles are using energy internally, but you aren't doing "Work" on the wall!
2. Three "Flavors" of Work
The value of Work depends entirely on the angle $\theta$:
| Angle (ฮธ) | Type of Work | Real-World Example |
| $0^\circ \leq \theta < 90^\circ$ | Positive | You pull a toy car forward; it moves forward. |
| $\theta = 90^\circ$ | Zero | You carry a bucket horizontally. Gravity pulls down, but you move sideways. Work = 0. |
| $90^\circ < \theta \leq 180^\circ$ | Negative | Friction! It pulls backward while the object slides forward. |
3. Units and Dimensions
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SI Unit: The Joule (J).
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$1 \text{ J} = 1 \text{ Newton} \times 1 \text{ meter} = 1 \text{ kg m}^2/\text{s}^2$.
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Dimensions: $[ML^2T^{-2}]$.
Other Common Units
While we love Joules, different fields use different "languages" for energy:
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Erg: $10^{-7} \text{ J}$ (Used in CGS units).
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Electron volt (eV): $1.6 \times 10^{-19} \text{ J}$ (Used in atomic physics).
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Calorie (cal): $4.186 \text{ J}$ (Used in nutrition and heat).
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Kilowatt-hour (kWh): $3.6 \times 10^6 \text{ J}$ (Check your electricity bill!).
4. Pro-Tips for Problem Solving
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Check for Displacement: If $d=0$, stop calculating. The work is zero.
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Identify the Angle: Always draw a quick arrow for Force and an arrow for Displacement. The angle between their "tails" is your $\theta$.
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Net Work: If multiple forces act on an object, the total work is the sum of work done by each force: $W_{net} = W_1 + W_2 + ...$
๐ง Quick Concept Check
Scenario: A waiter holds a tray weighing 20 N perfectly still while walking 10 meters across a level floor.
Question: How much work did the waiter do on the tray?
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Force: Upward (to counteract gravity).
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Displacement: Horizontal.
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Angle: $90^\circ$.
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Calculation: $W = 20 \times 10 \times \cos 90^\circ = \mathbf{0 \text{ J}}$.
5.4 Kinetic Energy
๐♂️ Kinetic Energy: The Power of Motion
Have you ever wondered why a cricket ball hurts more than a tennis ball thrown at the same speed? Or why a car crash at 60 km/h is much more than twice as destructive as one at 30 km/h?
The answer lies in Kinetic Energy ($K$)—the energy an object possesses simply because it is moving.
1. What is Kinetic Energy?
If an object has mass ($m$) and is moving with a velocity ($v$), it has the "capacity to do work" by virtue of its motion. In the language of Physics, we define it as:
$$K = \frac{1}{2} mv^2$$-
It’s a Scalar: It doesn't matter if you are running North, South, or in circles—your kinetic energy remains a simple number (Joules).
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The Velocity "Multiplier": Notice that $v$ is squared. This means if you double your speed, your kinetic energy doesn't just double—it quadruples! ๐
2. How Much Energy are we Talking About?
To give you a sense of scale, let’s look at some typical kinetic energies found in our world:
| Object | Mass (approx.) | Speed (m/s) | Kinetic Energy (J) |
| Meteors | $10^5 \text{ kg}$ | $3 \times 10^4$ | $4.5 \times 10^{13}$ |
| Car at Highway Speed | $2000 \text{ kg}$ | $25$ | $6.3 \times 10^5$ |
| Running Athlete | $70 \text{ kg}$ | $10$ | $3500$ |
| Fired Bullet | $50 \text{ g}$ | $200$ | $1000$ |
| Falling Raindrop | $3.5 \times 10^{-5} \text{ kg}$ | $9$ | $1.4 \times 10^{-3}$ |
3. The Work-Energy Connection
Kinetic energy isn't just a random formula; it is a measure of the work an object can do before it comes to a stop.
Think about it: To stop a moving car, the brakes must do an amount of work exactly equal to the car's kinetic energy. This is why high-speed chases are so dangerous—the "work" required to stop grows exponentially with speed!
4. Pro-Tip: The "10% Rule" (Ballistics Example)
In a famous demonstration (from the
Wait! Because $K \propto v^2$, the speed actually only drops by about 68%. Math can be surprising!
๐งช Quick Challenge
If a 2 kg object is moving at 4 m/s, what is its kinetic energy?
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$m = 2 \text{ kg}$
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$v = 4 \text{ m/s}$
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$K = \frac{1}{2} \times 2 \times 4^2 = \mathbf{16 \text{ J}}$
What if we double the speed to 8 m/s? * $K = \frac{1}{2} \times 2 \times 8^2 = \mathbf{64 \text{ J}}$
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(Notice: $16 \text{ J} \times 4 = 64 \text{ J}$. Doubling speed = 4x the energy!)
5.5 Work Done by a Variable Force
๐ Work Done by a Variable Force: The Calculus of Power
In our previous sections, we looked at work done by a constant force (like a steady push). But in the real world, forces are rarely constant. Think of a spring getting harder to pull the more you stretch it, or a rocket losing weight as it burns fuel—these are variable forces.
When the force changes at every point, we can't just multiply $F \times d$. We need a more sophisticated tool.
1. The Concept: Small Steps, Big Result
Imagine a force $F(x)$ that changes as an object moves along the x-axis. To calculate the total work, we break the displacement into tiny, microscopic intervals called $\Delta x$.
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In each tiny interval, the force is almost constant.
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The work for that tiny slice is: $\Delta W = F(x) \Delta x$.
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To get the total work, we add up all these tiny slices.
2. The Mathematical Tool: Integration
As we make those slices ($\Delta x$) infinitely small, our sum turns into a definite integral. This is the most accurate way to calculate work for a changing force:
$$W = \int_{x_i}^{x_f} F(x) \, dx$$-
$x_i$: Starting position.
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$x_f$: Ending position.
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Area Under the Curve: Geometrically, the work done by a variable force is exactly equal to the area under the Force-Displacement graph.
3. Real-World Example: The Tired Pusher
Let's look at a scenario from the
A woman pushes a trunk on a platform.
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First 10m: She applies a constant force of 100 N.
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Next 10m: She gets tired, and her force drops linearly from 100 N down to 50 N.
Calculating the Work:
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Part 1 (Rectangle): $100\text{ N} \times 10\text{ m} = 1000\text{ J}$.
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Part 2 (Trapezium): $\frac{1}{2} \times (100 + 50) \times 10\text{ m} = 750\text{ J}$.
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Total Work: $1000 + 750 = \mathbf{1750\text{ J}}$.
Note on Friction: If there was a constant friction of 50 N acting against her, the work done by friction would be $W_f = -50\text{ N} \times 20\text{ m} = -1000\text{ J}$.
4. Summary Table
| Force Type | Formula | Graphical Representation |
| Constant | $W = F \cdot d$ | A simple rectangle. |
| Variable | $W = \int F(x) \, dx$ | The area under the curve. |
๐ง Quick Brain Teaser
If you have a Force-Displacement graph and the "curve" is a triangle starting at $0$ and reaching $10\text{ N}$ over a distance of $2\text{ m}$, what is the work done?
(Hint: Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height}$)
5.6 The Work-Energy Theorem for a Variable Force
๐ The Work-Energy Theorem: Variable Force Edition
In the previous sections, we saw how the Work-Energy Theorem works for a constant force. But what happens when the force is constantly changing—like a spring being compressed or a magnet pulling an object closer?
Does the relationship between Work and Kinetic Energy still hold up? Absolutely. Here is how we prove it using the power of Calculus.
1. The Mathematical Proof (The "Time-Rate" Approach)
To prove the theorem for a variable force, we look at how Kinetic Energy ($K$) changes over a tiny sliver of time ($dt$):
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Start with the definition of K: $K = \frac{1}{2}mv^2$
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Take the derivative with respect to time:
$$\frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2}mv^2 \right) = m \cdot v \cdot \frac{dv}{dt}$$ -
Apply Newton’s Second Law ($F = ma$): Since $\frac{dv}{dt}$ is acceleration ($a$), we get:
$$\frac{dK}{dt} = m \cdot v \cdot a = F \cdot v$$ -
Rewrite velocity: Since $v = \frac{dx}{dt}$, we can say:
$$\frac{dK}{dt} = F \frac{dx}{dt}$$ -
Cancel the $dt$: This leaves us with the fundamental relationship:
$dK = F \, dx$
2. The Final Result: Integration
By integrating both sides from the initial position ($x_i$) to the final position ($x_f$), we sum up all those tiny changes in energy:
$$\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F(x) \, dx$$Which leads us back to the golden rule:
$$K_f - K_i = W$$The Big Takeaway: Even if the force is swinging wildly or changing every millimeter, the Total Work Done (the area under the $F-x$ graph) will always equal the Change in Kinetic Energy.
3. Why is this so useful?
Newton’s Second Law ($F=ma$) tells us what is happening at one specific instant. The Work-Energy Theorem is like a "summary report"—it tells us what happened over an entire interval of time or space.
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Vector vs. Scalar: Newton’s Law is a vector equation (direction matters). The Work-Energy Theorem is a scalar equation (much easier to calculate!).
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Missing Info: You don't need to know the acceleration at every single microsecond to find the final speed; you only need the total work.
๐งช Example from the
NCERT Textbook
The Case of the Retarding Force:
A block ($1 \text{ kg}$) moving at $2 \text{ m/s}$ hits a "rough patch" where the force is $F = -k/x$.
By integrating this variable force over the distance of the patch, you can find exactly how much kinetic energy was "stolen" by friction and calculate the final exit speed without ever knowing the exact acceleration at any point.
๐ง Check Your Logic
If a variable force does positive work on an object, its speed must:
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Increase
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Decrease
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Stay the same
Answer: 1. Increase! Positive work always adds to the Kinetic Energy ($K_f > K_i$).
Based on the
5.7 The Concept of Potential Energy
Potential Energy (PE) is defined as the "stored energy" an object possesses by virtue of its position or configuration. The word "potential" suggests a capacity for action; when a system is released, this stored energy can transform into kinetic energy.
1. Key Characteristics
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Storage: Energy is stored when work is done against a conservative force (like gravity or a spring).
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Examples:
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A stretched bow-string.
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Fault lines in the earth’s crust (acting like compressed springs).
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A ball raised to a certain height.
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Dimensions: $[ML^2T^{-2}]$ (The same as work and kinetic energy).
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Unit: Joule (J).
2. Gravitational Potential Energy
For an object of mass $m$ at a height $h$ (where $h$ is much smaller than the Earth's radius), the gravitational force $mg$ is considered constant.
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Formula: The potential energy $V(h)$ is given by:
$$V(h) = mgh$$ -
Relationship with Force: The force $F$ is the negative derivative of the potential energy with respect to height:
$$F = -\frac{d}{dh}V(h) = -mg$$The negative sign indicates the gravitational force acts downward.
3. Mathematical Definition
Mathematically, for a one-dimensional conservative force $F(x)$, the potential energy $V(x)$ is defined such that:
$$F(x) = -\frac{dV}{dx}$$This leads to the relation where the work done by a conservative force is the negative of the change in potential energy:
$$\Delta V = -F(x) \Delta x$$4. Conservative vs. Non-Conservative Forces
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Conservative Forces: The work done depends only on the initial and final positions, not the path taken. Examples include gravity and spring forces.
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Example: A mass sliding down a smooth inclined plane of height $h$ will have the same speed $\sqrt{2gh}$ at the bottom regardless of the angle of inclination.
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Non-Conservative Forces: The work done depends on the path taken or velocity (e.g., friction).
Summary Table: Kinetic vs. Potential Energy
| Feature | Kinetic Energy (K) | Potential Energy (V) |
| Origin | Motion of the object | Position or configuration |
| Formula (Gravity) | $K = \frac{1}{2}mv^2$ | $V = mgh$ |
| Work-Energy Link | $W = \Delta K$ | $W = -\Delta V$ (for conservative forces) |
Note: Potential energy is only defined for conservative forces. If the work done depended on the path taken, the concept of potential energy would not apply.
Based on the
5.8 The Conservation of Mechanical Energy
The Principle of Conservation of Mechanical Energy states that the total mechanical energy of a system remains constant if the forces doing work on it are conservative.
1. Fundamental Concepts
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Total Mechanical Energy ($E$): The sum of the kinetic energy ($K$) and the potential energy ($V$) of an object.
$$E = K + V(x)$$ -
The Derivation: From the Work-Energy Theorem, $\Delta K = F(x) \Delta x$. For a conservative force, $F(x) \Delta x = -\Delta V$. Therefore:
$$\Delta K + \Delta V = 0 \implies \Delta(K + V) = 0$$ -
Conservation Equation: For any two points in motion (initial $i$ and final $f$):
$$K_i + V(x_i) = K_f + V(x_f)$$
2. Defining Conservative Forces
A force $F(x)$ is considered conservative if it meets these criteria:
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It can be derived from a scalar potential: $F(x) = -dV/dx$.
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The work done depends only on the endpoints (initial and final positions), not the path taken.
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The work done by the force in a closed path is zero.
3. Practical Example: A Freely Falling Body
Consider a ball of mass $m$ dropped from a height $H$. We can track its energy at three stages:
| Position | Height | Kinetic Energy (K) | Potential Energy (V) | Total Energy (E) |
| Top (At rest) | $H$ | $0$ | $mgH$ | $mgH$ |
| Middle | $h$ | $\frac{1}{2}mv^2$ | $mgh$ | $mgh + \frac{1}{2}mv^2$ |
| Ground | $0$ | $\frac{1}{2}mv_f^2$ | $0$ | $\frac{1}{2}mv_f^2$ |
Conclusion: At the maximum height, the energy is purely potential. As the ball falls, $V$ is transformed into $K$. Just before hitting the ground, the energy is purely kinetic. In all instances, $E = mgH$.
4. Application: Vertical Circular Motion
For a bob of mass $m$ on a string of length $L$ completing a vertical circle:
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At the lowest point (A): The velocity $v_0$ must be sufficient to maintain tension.
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At the highest point (C): The string may slacken ($T=0$), but the bob must have a minimum speed of $v_c = \sqrt{gL}$ to complete the loop.
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Energy Conservation: The energy at the bottom must equal the energy at the top:
$$\frac{1}{2}mv_0^2 = \frac{1}{2}mv_c^2 + mg(2L)$$This results in the minimum required velocity at the bottom: $v_0 = \sqrt{5gL}$.
Summary Points
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Mechanical energy is not conserved if non-conservative forces (like friction or air resistance) do significant work.
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In the presence of friction, some mechanical energy is transformed into heat energy.
This study guide breaks down
Example 5.7
from the
๐ง Neuro-Friendly Strategy: The "Big Picture"
Before diving into math, visualize the "Energy Handshake." As the bob goes up, Kinetic Energy (speed) is handed over to Potential Energy (height). At the top, the bob must have just enough "swing" left so the string doesn't go floppy (slack).
๐ The Three Key Points
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Point A (Bottom): Max speed ($v_0$), Zero height ($h=0$).
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Point B (Side): Medium speed, Height is $L$.
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Point C (Top): Minimum speed ($v_c$), Max height ($2L$).
๐ Step-by-Step Solution
Part (i): Finding the Launch Velocity ($v_0$)
The Goal: Find the slowest speed at the bottom that keeps the string tight at the top.
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Condition at the Top (Point C): For the string to just "become slack," the tension $T_C$ becomes $0$. Gravity alone provides the centripetal force.
$$\frac{mv_c^2}{L} = mg \implies v_c = \sqrt{gL}$$ -
Conservation of Energy (A to C): Energy at Bottom = Energy at Top.
$$\frac{1}{2}mv_0^2 = \frac{1}{2}mv_c^2 + mg(2L)$$ -
Substitute and Solve: Plug in $v_c^2 = gL$.
$$\frac{1}{2}mv_0^2 = \frac{1}{2}mgL + 2mgL = \frac{5}{2}mgL$$Result: $v_0 = \sqrt{5gL}$
Part (ii): Speeds at Points B and C
We already found the speed at C in the previous step:
Speed at C: $v_c = \sqrt{gL}$
To find the speed at B (Horizontal position):
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Energy Handshake: Energy at A = Energy at B.
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Equation: $\frac{1}{2}mv_0^2 = \frac{1}{2}mv_B^2 + mgL$
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Substitute $v_0^2 = 5gL$:
$$\frac{5}{2}mgL = \frac{1}{2}mv_B^2 + mgL$$ $$\frac{5}{2}gL - gL = \frac{1}{2}v_B^2 \implies \frac{3}{2}gL = \frac{1}{2}v_B^2$$Result: $v_B = \sqrt{3gL}$
Part (iii): Ratio of Kinetic Energies ($K_B / K_C$)
The Goal: Compare how much "moving energy" is left at the side vs. the top.
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$K_B = \frac{1}{2}mv_B^2 = \frac{1}{2}m(3gL)$
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$K_C = \frac{1}{2}mv_C^2 = \frac{1}{2}m(gL)$
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Ratio: $\frac{K_B}{K_C} = \frac{3gL}{gL} = \mathbf{3:1}$
The "What Happens Next?" (Point C Trajectory)
If the string is cut exactly at Point C:
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The bob has horizontal velocity to the left ($v_c = \sqrt{gL}$).
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Gravity pulls it down.
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The Path: It will follow a Parabolic Trajectory (Projected Motion), just like a ball thrown horizontally off a cliff.
๐ก Quick Summary for Revision
| Point | Height | Velocity (v) | Total Energy (E) |
| A (Bottom) | $0$ | $\sqrt{5gL}$ | $\frac{5}{2}mgL$ |
| B (Side) | $L$ | $\sqrt{3gL}$ | $\frac{5}{2}mgL$ |
| C (Top) | $2L$ | $\sqrt{gL}$ | $\frac{5}{2}mgL$ |
Based on the
5.9 The Potential Energy of a Spring
The spring force is a classic example of a variable conservative force. When a spring is compressed or stretched, it exerts a restoring force that depends on the displacement from its equilibrium position.
1. Hooke’s Law
For an ideal spring, the spring force $F_s$ is proportional to the displacement $x$ of the block from its equilibrium position ($x = 0$):
$$F_s = -kx$$-
$k$ (Spring Constant): A measure of the spring's stiffness. Unit: N/m.
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Stiff spring: Large $k$.
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Soft spring: Small $k$.
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Negative Sign: Indicates that the force is a restoring force, always acting in the direction opposite to the displacement.
2. Work Done by the Spring Force
When a spring is pulled from $x = 0$ to a maximum displacement $x_m$, the work done by the spring force ($W_s$) is calculated by integrating the force over the displacement:
$$W_s = \int_{0}^{x_m} F_s dx = \int_{0}^{x_m} (-kx) dx = -\frac{1}{2}kx_m^2$$Conversely, the work done by the external pulling force ($W_{ext}$) is positive because it overcomes the spring force:
$$W_{ext} = \frac{1}{2}kx_m^2$$3. Elastic Potential Energy ($V$)
The work done by the external agency is stored as Elastic Potential Energy $V(x)$ in the spring.
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Formula: $V(x) = \frac{1}{2}kx^2$
-
Relationship to Force: Just like gravity, the spring force is the negative derivative of its potential energy:
$$F_s = -\frac{dV}{dx} = -kx$$
4. Conservation of Energy in a Spring System
In a frictionless system (a block attached to a spring on a smooth surface), the total mechanical energy $E$ is conserved. At any displacement $x$ and velocity $v$:
$$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{Constant}$$-
At Maximum Displacement ($x_m$): The block momentarily stops ($v = 0$). All energy is potential: $E = \frac{1}{2}kx_m^2$.
-
At Equilibrium ($x = 0$): Potential energy is zero. All energy is kinetic: $E = \frac{1}{2}mv_{max}^2$.
-
At any intermediate point: $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$.
Key Summary Table
| State | Displacement (x) | Potential Energy (V) | Kinetic Energy (K) |
| Equilibrium | $0$ | $0$ (Minimum) | $\frac{1}{2}mv_{max}^2$ (Maximum) |
| Max Extension | $+x_m$ | $\frac{1}{2}kx_m^2$ (Maximum) | $0$ (Minimum) |
| Max Compression | $-x_m$ | $\frac{1}{2}kx_m^2$ (Maximum) | $0$ (Minimum) |
Crucial Note: Since the work done by the spring force depends only on the initial and final positions (and is zero for a round trip), the spring force is a conservative force.
This study material is based on
Section 5.9
of the
5.9 The Potential Energy of a Spring
A spring force is a classic example of a variable conservative force. When a spring is distorted, it exerts a restoring force to return to its equilibrium position ($x = 0$).
1. Hooke’s Law (The Force Law)
For an ideal spring, the spring force $F_s$ is proportional to the displacement $x$:
$$F_s = -kx$$-
$k$ (Spring Constant): Represents the stiffness of the spring (Unit: $N m^{-1}$).
-
Large $k$: Stiff spring.
-
Small $k$: Soft spring.
-
-
The Negative Sign: Indicates it is a restoring force—it always points toward the equilibrium position ($x=0$).
๐ง Neuro-Friendly Approach: Understanding the Work Done
Instead of just memorizing the formula, visualize the "Area under the Curve."
Step 1: The Integration
If we pull a block to a maximum displacement $x_m$, the work done by the spring force ($W_s$) is the integral of force over displacement:
$$W_s = \int_{0}^{x_m} F_s dx = \int_{0}^{x_m} (-kx) dx = -\frac{1}{2}kx_m^2$$Step 2: The Visual Anchor
If you plot Force ($F_s$) vs. Displacement ($x$), the work done is the area of the triangle formed under the line.
-
Base = $x_m$
-
Height = $kx_m$
-
Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}kx_m^2$.
2. Elastic Potential Energy ($V$)
The work done by an external force ($W_{ext} = \frac{1}{2}kx^2$) is stored in the spring as potential energy.
-
Formula: $V(x) = \frac{1}{2}kx^2$
-
Property: Like gravity, the spring force is the negative derivative of this potential energy: $F_s = -\frac{dV}{dx}$.
3. Conservation of Mechanical Energy
In a frictionless system, the total energy ($E$)—the sum of Kinetic ($K$) and Potential ($V$)—remains constant throughout the oscillation.
$$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{Constant}$$| Position | Displacement (x) | Potential Energy (V) | Kinetic Energy (K) |
| Equilibrium | $0$ | Zero | Maximum ($\frac{1}{2}mv_m^2$) |
| Max Extension | $+x_m$ | Maximum ($\frac{1}{2}kx_m^2$) | Zero |
| Max Compression | $-x_m$ | Maximum ($\frac{1}{2}kx_m^2$) | Zero |
๐ Step-by-Step Problem Solving: Maximum Speed
Problem: A block of mass $m$ is pulled to $x_m$ and released. Find its maximum speed $v_m$.
-
Identify the Energy: At $x_m$, all energy is Potential ($V = \frac{1}{2}kx_m^2$).
-
State the Conversion: At the equilibrium point ($x=0$), all that energy turns into Kinetic ($K = \frac{1}{2}mv_m^2$).
-
Equate them:
$$\frac{1}{2}mv_m^2 = \frac{1}{2}kx_m^2$$ -
Solve for $v_m$:
$$v_m = x_m \sqrt{\frac{k}{m}}$$
๐ก Key Takeaways for Exams
-
The spring force is conservative because the work done in a closed path is zero.
-
The zero point for potential energy ($V=0$) is arbitrarily chosen at the equilibrium position ($x=0$).
-
If friction is present, mechanical energy is not conserved, and you must use the Work-Energy Theorem: $\Delta E = W_{friction}$.
Based on Example 5.8 from the
๐ The Question
Example 5.8: To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant $5.25 \times 10^3\text{ N m}^{-1}$. What is the maximum compression of the spring?
๐ง Neuro-Friendly Strategy: The "Energy Transfer"
Visualize a car zooming toward a giant spring. At the moment it hits the spring and squishes it completely, the car momentarily stops.
-
Before the hit: The car has 100% Kinetic Energy.
-
At max compression: The car has 0% Kinetic Energy; all that energy has been "pushed" into the spring as Potential Energy.
๐ Step-by-Step Solution
Step 1: List the Given Values (and Convert Units)
Standard physics formulas require SI units (meters and seconds).
-
Mass ($m$): $1000\text{ kg}$
-
Initial Velocity ($v$): $18.0\text{ km/h}$
-
Conversion: $18 \times \dfrac{5}{18} = \mathbf{5\text{ m s}^{-1}}$
-
-
Spring Constant ($k$): $5.25 \times 10^3\text{ N m}^{-1}$
Step 2: Calculate Initial Kinetic Energy ($K$)
This is the total "energy budget" we have to compress the spring.
$$K = \dfrac{1}{2}mv^2$$ $$K = \dfrac{1}{2} \times 1000 \times (5)^2$$ $$K = 500 \times 25 = \mathbf{1.25 \times 10^4\text{ J}}$$Step 3: Apply Conservation of Mechanical Energy
At maximum compression ($x_m$), the Potential Energy ($V$) of the spring equals the initial Kinetic Energy of the car.
$$V = K$$ $$\dfrac{1}{2}kx_m^2 = 1.25 \times 10^4\text{ J}$$Step 4: Solve for Maximum Compression ($x_m$)
Now, we just isolate $x_m$:
$$x_m^2 = \dfrac{2 \times (1.25 \times 10^4)}{k}$$ $$x_m^2 = \dfrac{2.5 \times 10^4}{5.25 \times 10^3}$$ $$x_m^2 \approx 4.76$$ $$x_m = \sqrt{4.76}$$Result: $x_m \approx 2.18\text{ m}$ (Note: The textbook approximates this to $2.00\text{ m}$ depending on the rounding of the spring constant calculation).
๐ก Quick Summary
-
The "Why": Energy cannot be destroyed; the car's motion energy just changes into "spring-stretch" energy.
-
The "Condition": This only works because the road is smooth (no friction). If there were friction, the compression would be smaller because some energy would be lost to heat.
Following the neuro-friendly approach, here is the solution for
Example 5.9
from the
๐ The Question
Example 5.9: Consider Example 5.8 (a 1000 kg car hitting a spring at 18 km/h), but now take the coefficient of friction ($\mu$) to be 0.5. Calculate the maximum compression of the spring.
๐ง Neuro-Friendly Strategy: The "Energy Leak"
In the previous example, 100% of the car's energy went into the spring. Now, we have a "leak." As the car slides, friction steals some of that energy and turns it into heat.
-
Initial Energy: Kinetic Energy ($K$) of the car.
-
The Work Done: The car has to do work against two things:
-
The Spring (storing potential energy).
-
Friction (wasting energy as heat).
-
๐ Step-by-Step Solution
Step 1: List Known Values
-
Mass ($m$): $1000\text{ kg}$
-
Velocity ($v$): $5\text{ m s}^{-1}$ (from $18\text{ km/h}$)
-
Spring Constant ($k$): $5.25 \times 10^3\text{ N m}^{-1}$
-
Coefficient of Friction ($\mu$): $0.5$
-
Gravity ($g$): $10\text{ m s}^{-2}$ (approx.)
Step 2: Identify the Energy Equation
Since friction is a non-conservative force, we use the Work-Energy Theorem:
$$\Delta K = W_{net}$$Because the car stops at maximum compression, the change in kinetic energy is:
$$0 - \dfrac{1}{2}mv^2 = W_{spring} + W_{friction}$$Or, more simply: Initial Kinetic Energy = Energy in Spring + Energy lost to Friction.
$$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx_m^2 + \mu mgx_m$$Step 3: Plug in the Numbers
-
Initial Kinetic Energy: $1.25 \times 10^4\text{ J}$ (calculated in Ex 5.8).
-
Frictional Force ($f$): $\mu mg = 0.5 \times 1000 \times 10 = 5000\text{ N}$.
-
The Equation:
$$12500 = \dfrac{1}{2}(5250)x_m^2 + (5000)x_m$$Divide by 10 to simplify:
$$262.5x_m^2 + 500x_m - 1250 = 0$$
Step 4: Solve the Quadratic Equation
We use the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
-
$a = 262.5$
-
$b = 500$
-
$c = -1250$
After calculation (taking the positive root since compression distance can't be negative):
Result: $x_m = 1.35\text{ m}$
๐ง Comparison & Conclusion
-
Without Friction (Ex 5.8): Compression was 2.00 m.
-
With Friction (Ex 5.9): Compression is only 1.35 m.
Why? Because friction "helped" stop the car, so the spring didn't have to do all the work alone. This is why brakes (friction) are quite important in a real car!
๐ก Key Concept for Exams
When you see friction or air resistance, you can no longer use the simple "Conservation of Mechanical Energy" ($K_i + V_i = K_f + V_f$). You must use the Work-Energy Theorem to account for the energy lost to the environment.
Would you like to move on to Section 5.10: Power, or would you like some practice problems on these spring examples?
This study material is based on
Section 5.10
of the
5.10 Power
While Work tells us how much energy is transferred, Power tells us how fast that transfer happens. It is the time rate at which work is done.
1. Key Definitions
-
Average Power ($P_{av}$): The ratio of total work ($W$) to the total time ($t$) taken.
$$P_{av} = \dfrac{W}{t}$$ -
Instantaneous Power ($P$): The limiting value of average power as the time interval approaches zero. It is the derivative of work with respect to time:
$$P = \dfrac{dW}{dt}$$
2. Power in Terms of Velocity
If a constant force $\vec{F}$ causes a small displacement $d\vec{r}$, the work done is $dW = \vec{F} \cdot d\vec{r}$. Substituting this into the power formula:
$$P = \vec{F} \cdot \dfrac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}$$Concept: Instantaneous power is the dot product of the Force and the Instantaneous Velocity.
3. Units and Dimensions
-
Dimensions: $[ML^2T^{-3}]$
-
SI Unit: Watt (W), named after James Watt. $1\text{ W} = 1\text{ J s}^{-1}$.
-
Horsepower (hp): A common unit used in the automotive and motor industry.
$$1\text{ hp} = 746\text{ W}$$ -
Kilowatt-hour (kWh): This is a unit of Energy, not power. It is commonly used in electricity bills.
$$1\text{ kWh} = 3.6 \times 10^6\text{ J}$$
๐ง Neuro-Friendly Approach: Step-by-Step Problem Solving
Let’s look at how to solve power problems using the logic from Example 5.10.
The Problem: An elevator (total mass $1800\text{ kg}$) moves up at a constant speed of $2\text{ m s}^{-1}$. A frictional force of $4000\text{ N}$ opposes its motion. Find the power delivered by the motor.
Step 1: Identify Total Downward Force ($F$)
The motor has to fight two things:
-
Gravity: $mg = 1800 \times 10 = 18000\text{ N}$
-
Friction: $4000\text{ N}$
-
Total Force $= 18000 + 4000 = \mathbf{22000\text{ N}}$
Step 2: Use the Power Formula
Since the speed is constant, we use $P = F \times v$.
-
$P = 22000\text{ N} \times 2\text{ m s}^{-1} = \mathbf{44000\text{ W}}$
Step 3: Convert to Horsepower (Optional but common)
-
$P\text{ (in hp)} = \dfrac{44000}{746} \approx \mathbf{59\text{ hp}}$
๐ก Summary Table: Power at a Glance
| Feature | Description |
| Type of Quantity | Scalar |
| Formula | $P = \dfrac{W}{t}$ or $P = \vec{F} \cdot \vec{v}$ |
| SI Unit | Watt ($1\text{ J/s}$) |
| Commercial Unit | Horsepower ($746\text{ W}$) |
| Key Distinction | Power is a rate, Work is a quantity. |
Based on the
5.11 Collisions
In physics, a collision occurs when two or more bodies exert relatively strong forces on each other for a relatively short time. While the total linear momentum is always conserved in all collisions, the total kinetic energy may or may not be conserved.
1. The Physics of Momentum Conservation
In any collision, the impulsive forces $F_{12}$ and $F_{21}$ are internal to the system. According to Newton's Third Law ($F_{12} = -F_{21}$), the total change in momentum is zero:
$$\Delta p_1 + \Delta p_2 = 0 \implies p_{initial} = p_{final}$$5.11.1 Elastic and Inelastic Collisions
Collisions are categorized based on whether kinetic energy ($K$) is conserved:
| Type of Collision | Momentum | Kinetic Energy | Characteristics |
| Elastic | Conserved | Conserved | No energy lost to heat/sound (e.g., subatomic particles). |
| Inelastic | Conserved | Not Conserved | Some $K$ is lost to deformation, heat, or sound. |
| Completely Inelastic | Conserved | Max Loss | The two bodies stick together after impact. |
3. Completely Inelastic Collision (One Dimension)
Imagine mass $m_1$ moving with velocity $v_{1i}$ hits a stationary mass $m_2$. After the collision, they move together with a common velocity $v_f$.
-
Momentum Balance: $m_1 v_{1i} = (m_1 + m_2) v_f$
-
Final Velocity:
$$v_f = \dfrac{m_1}{m_1 + m_2} v_{1i}$$ -
Energy Loss: The loss in kinetic energy ($\Delta K$) is always positive, confirming that energy is transformed into other forms (like heat).
5.11.2 Collisions in One Dimension
For an elastic collision between $m_1$ (moving) and $m_2$ (at rest), both momentum and kinetic energy equations must be satisfied.
The Final Velocities:
-
$v_{1f} = \dfrac{(m_1 - m_2)}{(m_1 + m_2)} v_{1i}$
-
$v_{2f} = \dfrac{2m_1}{(m_1 + m_2)} v_{1i}$
Special Cases to Remember:
-
Equal Masses ($m_1 = m_2$): The masses simply exchange velocities. The first mass stops ($v_{1f} = 0$) and the second mass moves off with the initial speed of the first.
-
Heavy Target ($m_2 \gg m_1$): The heavy mass stays at rest, and the light mass bounces back with the same speed ($v_{1f} \approx -v_{1i}$).
-
Heavy Projectile ($m_1 \gg m_2$): The heavy mass keeps moving at its original speed, and the light mass is "kicked" forward at twice that speed ($v_{2f} \approx 2v_{1i}$).
๐ง Neuro-Friendly Approach: Collision Logic
-
Always start with Momentum: $m_1v_1 + m_2v_2 = \text{constant}$. This works every single time.
-
Check the "Stickiness": If they stick, it's Completely Inelastic. If they bounce and the problem says "elastic," then use the $K$ conservation.
-
Visual Anchor: Think of an elastic collision like a perfect "spring" that compresses and then perfectly pushes back. Think of an inelastic collision like "chewing gum" that hits a wall and stays there.
๐ก Summary Table for Quick Revision
| Feature | Elastic Collision | Inelastic Collision |
| Total Momentum | Conserved | Conserved |
| Total Energy | Conserved | Conserved |
| Kinetic Energy | Conserved | Not Conserved |
| Forces Involved | Conservative | Non-conservative |
This proof follows the mathematical derivation found in
Section 5.11.2
of the
๐ Proof: Elastic Collision in One Dimension
The Setup:
Imagine a mass $m_1$ moving with initial velocity $v_{1i}$ hitting a stationary mass $m_2$ ($v_{2i} = 0$). After the collision, $m_1$ moves with $v_{1f}$ and $m_2$ moves with $v_{2f}$.
Step 1: Conservation of Linear Momentum
In any collision, the total momentum before equals the total momentum after.
$$m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$$Rearrange to group $m_1$ terms:
$$m_1 (v_{1i} - v_{1f}) = m_2 v_{2f} \quad \text{--- (Eq. A)}$$Step 2: Conservation of Kinetic Energy
Since the collision is elastic, kinetic energy is also conserved.
$$\dfrac{1}{2}m_1 v_{1i}^2 = \dfrac{1}{2}m_1 v_{1f}^2 + \dfrac{1}{2}m_2 v_{2f}^2$$Multiply by 2 and rearrange to group $m_1$ terms:
$$m_1 (v_{1i}^2 - v_{1f}^2) = m_2 v_{2f}^2$$Using the algebraic identity $(a^2 - b^2) = (a-b)(a+b)$:
$$m_1 (v_{1i} - v_{1f})(v_{1i} + v_{1f}) = m_2 v_{2f}^2 \quad \text{--- (Eq. B)}$$Step 3: Finding the Velocity Relationship
Divide Eq. B by Eq. A:
$$\dfrac{m_1 (v_{1i} - v_{1f})(v_{1i} + v_{1f})}{m_1 (v_{1i} - v_{1f})} = \dfrac{m_2 v_{2f}^2}{m_2 v_{2f}}$$This simplifies to a very important relationship:
$$v_{1i} + v_{1f} = v_{2f} \quad \text{--- (Eq. C)}$$Neuro-friendly tip: This shows that the relative velocity of approach equals the relative velocity of separation.
Step 4: Solving for Final Velocity $v_{2f}$
Substitute the value of $v_{1f}$ from Eq. C ($v_{1f} = v_{2f} - v_{1i}$) into the original momentum equation (Eq. A):
$$m_1 v_{1i} = m_1 (v_{2f} - v_{1i}) + m_2 v_{2f}$$ $$m_1 v_{1i} = m_1 v_{2f} - m_1 v_{1i} + m_2 v_{2f}$$ $$2m_1 v_{1i} = (m_1 + m_2) v_{2f}$$Result 1:
$$v_{2f} = \left( \dfrac{2m_1}{m_1 + m_2} \right) v_{1i}$$Step 5: Solving for Final Velocity $v_{1f}$
Now substitute $v_{2f}$ from Eq. C into Eq. A:
$$m_1 (v_{1i} - v_{1f}) = m_2 (v_{1i} + v_{1f})$$ $$m_1 v_{1i} - m_1 v_{1f} = m_2 v_{1i} + m_2 v_{1f}$$ $$(m_1 - m_2) v_{1i} = (m_1 + m_2) v_{1f}$$Result 2:
$$v_{1f} = \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right) v_{1i}$$๐ง Summary of Special Case Proofs
Based on these formulas, we can prove the common observations:
-
Equal Masses ($m_1 = m_2$): If you plug this into Result 2, $v_{1f} = 0$. If you plug it into Result 1, $v_{2f} = v_{1i}$. The balls exchange velocities.
-
Heavy Target ($m_2 \gg m_1$): The denominator becomes huge. $v_{1f}$ becomes approximately $-v_{1i}$ (it bounces back), and $v_{2f}$ becomes approximately $0$.
Based on the
Example 5.11: Slowing Down of Neutrons
Question: In a nuclear reactor, a neutron of high speed (typically $10^7$ m/s) must be slowed to $10^3$ m/s to interact with isotope $^{235}_{92}U$ and cause fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with light nuclei like deuterium or carbon, which have masses only a few times that of the neutron.
Step-by-Step Solution
To solve this, we use the principles of elastic collision where both momentum and kinetic energy are conserved.
1. Define the Variables
-
Let $m_1$ be the mass of the neutron and $v_{1i}$ be its initial velocity.
-
Let $m_2$ be the mass of the moderating nucleus (at rest, so $v_{2i} = 0$).
-
Let $v_{1f}$ be the final velocity of the neutron after the collision.
2. Kinetic Energy Ratio
The initial kinetic energy of the neutron is:
$$K_{1i} = \dfrac{1}{2} m_1 v_{1i}^2$$From the equations of a one-dimensional elastic collision (Eq. 5.26 in the text), the final velocity of the neutron $v_{1f}$ is:
$$v_{1f} = \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right) v_{1i}$$Therefore, the final kinetic energy $K_{1f}$ is:
$$K_{1f} = \dfrac{1}{2} m_1 v_{1f}^2 = \dfrac{1}{2} m_1 \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right)^2 v_{1i}^2$$3. Fractional Energy Remaining ($f_1$)
The fraction of the original kinetic energy retained by the neutron is:
$$f_1 = \dfrac{K_{1f}}{K_{1i}} = \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right)^2$$4. Fractional Energy Lost ($f_2$)
The fraction of energy transferred to the moderating nucleus is:
$$f_2 = 1 - f_1 = \dfrac{4 m_1 m_2}{(m_1 + m_2)^2}$$Application to Specific Moderators
Case A: Deuterium ($m_2 \approx 2m_1$)
Substituting $m_2 = 2m_1$ into the formulas:
-
Retained ($f_1$): $(\dfrac{1 - 2}{1 + 2})^2 = (\dfrac{-1}{3})^2 = 1/9 \approx 11.1\%$
-
Lost ($f_2$): $1 - 1/9 = 8/9 \approx \mathbf{88.9\%}$
Result: Almost 90% of the neutron's energy is transferred to deuterium in a single head-on collision.
Case B: Carbon ($m_2 \approx 12m_1$)
Substituting $m_2 = 12m_1$ into the formulas:
-
Retained ($f_1$): $(\dfrac{1 - 12}{1 + 12})^2 = (\dfrac{-11}{13})^2 \approx 71.6\%$
-
Lost ($f_2$): $1 - 0.716 = \mathbf{28.4\%}$
Conclusion
The calculation shows that light nuclei are highly effective at "absorbing" kinetic energy from neutrons. While heavier nuclei exist, using a moderator with a mass close to the neutron (like Hydrogen or Deuterium) ensures the maximum energy transfer per collision, effectively slowing the neutron down for fission.
The core of this problem is to show how much energy a neutron transfers to a target nucleus during an elastic collision. Here is the formal step-by-step derivation and solution for Example 5.11.
Alternative Explanations
Example 5.11: Slowing Down of Neutrons
The Question:
In a nuclear reactor, a neutron of high speed (typically $10^7$ m s⁻¹) must be
slowed to $10^3$ m s⁻¹ so that it can have a high probability of interacting
with isotope $^{235}_{92}\text{U}$ and causing it to fission. Show that a
neutron can lose most of its kinetic energy in an elastic collision with a
light nuclei like deuterium or carbon which has a mass of only a few times the
neutron ma
1. Identify the Physics Principles
We assume a one-dimensional elastic collision where:
-
Mass $m_1$: The neutron (initially moving at $v_{1i}$).
-
Mass $m_2$: The moderator nucleus (initially at rest, $v_{2i} = 0$).
-
Conservation of Momentum and Kinetic Energy are both applicable.
2. Formulate the Energy Equations
The initial kinetic energy of the neutron ($K_{1i}$) is:
$$K_{1i} = \dfrac{1}{2} m_1 v_{1i}^2$$
From the standard equations for elastic collisions (as derived in
Therefore, the final kinetic energy of the neutron ($K_{1f}$) is:
$$K_{1f} = \dfrac{1}{2} m_1 v_{1f}^2 = \dfrac{1}{2} m_1 \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right)^2 v_{1i}^2$$3. Calculate the Fractional Energy
-
Fraction of energy retained by the neutron ($f_1$):
$$f_1 = \dfrac{K_{1f}}{K_{1i}} = \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right)^2$$ -
Fraction of energy lost (transferred to the moderator) ($f_2$):
Since energy is conserved, $f_2 = 1 - f_1$:
$$f_2 = \dfrac{4 m_1 m_2}{(m_1 + m_2)^2}$$
4. Solve for Specific Moderators
Case A: Deuterium ($m_2 \approx 2m_1$)
Substitute $m_2 = 2m_1$ into the $f_1$ formula:
$$f_1 = \left( \dfrac{m_1 - 2m_1}{m_1 + 2m_1} \right)^2 = \left( \dfrac{-m_1}{3m_1} \right)^2 = \dfrac{1}{9} \approx 11.1\%$$Energy Lost ($f_2$): $1 - \dfrac{1}{9} = \dfrac{8}{9} \approx \mathbf{88.9\%}$
Conclusion: Nearly 90% of the neutron's energy is transferred to the deuterium in a single head-on collision.
Case B: Carbon ($m_2 \approx 12m_1$)
Substitute $m_2 = 12m_1$ into the $f_1$ formula:
$$f_1 = \left( \dfrac{m_1 - 12m_1}{m_1 + 12m_1} \right)^2 = \left( \dfrac{-11}{13} \right)^2 \approx 71.6\%$$Energy Lost ($f_2$): $1 - 0.716 = \mathbf{28.4\%}$
Summary Table
| Moderator | Mass Ratio (m2/m1) | % Energy Retained | % Energy Lost |
| Deuterium | 2 | 11.1% | 88.9% |
| Carbon | 12 | 71.6% | 28.4% |
This proves that light nuclei are much more effective "moderators" than heavy ones, as they allow for a significant energy transfer in just a few collisions.
This example demonstrates a fascinating property of elastic collisions between
two objects of equal mass. Here is the step-by-step breakdown of
Example 5.12
from the
5.11.3 Collisions in Two Dimensions
Example 5.12: The Billiard Ball Collision
Question: Consider a collision between two billiard balls with equal masses ($m_1 = m_2$). The first ball (the cue) hits the second ball (the target), which is initially at rest. The player wants to ‘sink’ the target ball in a corner pocket at an angle $\theta_2 = 37^\circ$. Assuming the collision is elastic and neglecting friction and rotational motion, obtain the angle $\theta_1$ at which the cue ball moves after the impact.
Step 1: Conservation of Linear Momentum
Since there are no external forces, the total momentum before the collision must equal the total momentum after the collision. Because the masses are equal ($m_1 = m_2 = m$), the mass terms cancel out in the vector equation:
$$\vec{v}_{1i} = \vec{v}_{1f} + \vec{v}_{2f}$$To relate the magnitudes and the angle between the final velocity vectors, we square both sides (take the dot product of the vector with itself):
$$(\vec{v}_{1i})^2 = (\vec{v}_{1f} + \vec{v}_{2f}) \cdot (\vec{v}_{1f} + \vec{v}_{2f})$$ $$v_{1i}^2 = v_{1f}^2 + v_{2f}^2 + 2\vec{v}_{1f} \cdot \vec{v}_{2f}$$Using the definition of a dot product ($\vec{A} \cdot \vec{B} = AB \cos\theta$):
$$v_{1i}^2 = v_{1f}^2 + v_{2f}^2 + 2v_{1f}v_{2f} \cos(\theta_1 + \theta_2) \quad \text{--- (Eq. 1)}$$Step 2: Conservation of Kinetic Energy
The problem states the collision is elastic, meaning kinetic energy is conserved. Again, since $m_1 = m_2$, the $\dfrac{1}{2}m$ terms cancel out:
$$\dfrac{1}{2}mv_{1i}^2 = \dfrac{1}{2}mv_{1f}^2 + \dfrac{1}{2}mv_{2f}^2$$ $$v_{1i}^2 = v_{1f}^2 + v_{2f}^2 \quad \text{--- (Eq. 2)}$$Step 3: Comparing the Equations
Now, compare Eq. 1 and Eq. 2. For both to be true simultaneously, the extra term in Eq. 1 must be zero:
$$2v_{1f}v_{2f} \cos(\theta_1 + \theta_2) = 0$$Since the balls are moving ($v_{1f} \neq 0$ and $v_{2f} \neq 0$), the cosine term must be zero:
$$\cos(\theta_1 + \theta_2) = 0$$Step 4: Calculate the Angle
We know that $\cos(90^\circ) = 0$. Therefore:
$$\theta_1 + \theta_2 = 90^\circ$$Given that the target ball angle $\theta_2 = 37^\circ$:
$$\theta_1 + 37^\circ = 90^\circ$$ $$\theta_1 = 90^\circ - 37^\circ$$ $$\mathbf{\theta_1 = 53^\circ}$$The "Billiard Rule"
This result proves a fundamental rule in physics: When two equal masses undergo a glancing elastic collision (with one initially at rest), they will always move at right angles ($90^\circ$) to each other after the impact.
Would you like me to summarize the "Points to Ponder" section from the end of this chapter to help with your exam revision?
