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[
{
"slide": 1,
"fragments": [
{
"fragment_index": 1,
"text_description": "Law in Words",
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},
{
"fragment_index": 2,
"text_description": "Newton’s Second Law",
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},
{
"fragment_index": 3,
"text_description": "The rate of change of momentum of a body is directly proportional to the external force applied and takes the direction of that force.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Variables: Force (F), momentum \\(p = m v\\), and time interval \\(\\Delta t\\). Why does “rate of change” point to acceleration rather than speed?",
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}
]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Equation Reveal",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "\\[F = m\\,a\\]",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Variable Definitions\nF\nForce, \\(N\\)\nm\nMass, \\(kg\\)\na\nAcceleration, \\(m/s^{2}\\)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Applications\nLabel & Learn\nDrag each unit label onto F, m, and a.\nUnit Check\nConfirm \\(N = kg·m/s^{2}\\) so both sides match.\nQuick Example\nIf \\(m = 2\\,kg\\) and \\(a = 3\\,m/s^{2}\\), then \\(F = 6\\,N\\).\nSource: CBSE Grade 11 Textbook",
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}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Force vs Acceleration",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Force (N) vs Acceleration (m/s²)",
"image_description": "https://asset.sparkl.ac/pb/sparkl-vector-images/img_ncert/QjwWKAx0XxQqmPehMC4ii2POA4bddTDMRWt19IJW.png"
},
{
"fragment_index": 2,
"text_description": "Reading the F–a Graph\nLinear relationship: a straight line through the origin shows force is directly proportional to acceleration for a fixed mass.\nGraph interpretation: the slope \\( \\frac{F}{a} \\) equals the mass. Steeper slope ⇒ heavier body.\nKey Points:\nSlope gives mass in kilograms.\nDoubling force doubles acceleration when the mass is constant.\nQuiz: Read the slope—mass here is 2 kg.",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "From p to F = m a",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "1\n\\[\\vec{p}=m\\vec{v}\\]\nLinear momentum equals mass times velocity.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "2\n\\[\\vec{F}=\\frac{d\\vec{p}}{dt}\\]\nForce is the time rate of change of momentum.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "3\n\\[\\frac{d\\vec{p}}{dt}=m\\frac{d\\vec{v}}{dt}\\]\nAssuming \\(m\\) is constant, only velocity varies with time.",
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},
{
"fragment_index": 4,
"text_description": "4\n\\[\\vec{F}=m\\vec{a}\\]\nSince \\( \\frac{d\\vec{v}}{dt}=\\vec{a} \\), we reach the familiar form.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insight:\nConstant mass lets the momentum derivative collapse to acceleration, revealing \\( \\vec{F}=m\\vec{a} \\).",
"image_description": ""
}
]
},
{
"slide": 5,
"fragments": []
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Multiple Choice Question\nSubmit Answer\nCorrect!\nGreat job! \\(a = \\frac{12\\,\\text{N}}{3\\,\\text{kg}} = 4\\ \\text{m/s}^2\\).\nIncorrect\nRevisit \\(F = m a\\) and divide the force by the mass to find acceleration.\nconst correctOption = 2;\n const answerCards = document.querySelectorAll('.answer-card');\n const submitBtn = document.getElementById('submitBtn');\n const feedbackCorrect = document.getElementById('feedbackCorrect');\n const feedbackIncorrect = document.getElementById('feedbackIncorrect');\n\n let selectedOption = null;\n\n answerCards.forEach((card, index) => {\n card.addEventListener('click', () => {\n answerCards.forEach(c => c.classList.remove('border-blue-500', 'bg-blue-50'));\n card.classList.add('border-blue-500', 'bg-blue-50');\n selectedOption = index;\n });\n });\n\n submitBtn.addEventListener('click', () => {\n if (selectedOption === null) return;\n\n if (selectedOption === correctOption) {\n feedbackCorrect.classList.remove('hidden');\n feedbackIncorrect.classList.add('hidden');\n } else {\n feedbackIncorrect.classList.remove('hidden');\n feedbackCorrect.classList.add('hidden');\n }\n });",
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},
{
"fragment_index": 1,
"text_description": "Question\nA 3 kg drone experiences a net forward force of 12 N. What is its acceleration?",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "1\n2 m/s²",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "2\n3 m/s²",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "3\n4 m/s²",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "4\n6 m/s²",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Hint:\nApply \\(a = \\frac{F}{m}\\).",
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}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Takeaways",
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},
{
"fragment_index": 1,
"text_description": "Proportionality\nForce increases directly with acceleration: \\(F \\propto a\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Vector Direction\n\\( \\vec F \\) and \\( \\vec a \\) point the same way—both are vectors.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Mass from Slope\nSlope of an \\(F\\)–\\(a\\) graph equals the object’s mass.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Momentum Link\nSame law: \\(F = \\\\frac{dp}{dt}\\)—force is rate of momentum change.",
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}
]
}
]