Import Question JSON

Current Question (ID: 9926)

Question:

$\text{A body of mass } m \text{ moving with speed } v \text{ collides head-on elastically with another identical body at rest. The percentage loss in kinetic energy of the first body will be:}$

Options:

1. $0\%$
2. $25\%$
3. $50\%$
4. $100\%$

Solution:

\text{When a body of mass } m \text{ moving with speed } v \text{ collides head-on elastically with another identical body (mass } m\text{) at rest, the first body comes to rest, and the second body moves with the initial speed } v \text{ of the first body.} \text{The initial kinetic energy of the first body is } KE_{\text{initial}} = \frac{1}{2}mv^2. \text{After the collision, the first body is at rest, so its final kinetic energy is } KE_{\text{final}} = 0. \text{The loss in kinetic energy of the first body is:} KE_{\text{loss}} = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2 \text{The percentage loss in kinetic energy of the first body is:} \text{Percentage Loss} = \left(\frac{KE_{\text{loss}}}{KE_{\text{initial}}}\right) \times 100\% \text{Percentage Loss} = \left(\frac{\frac{1}{2}mv^2}{\frac{1}{2}mv^2}\right) \times 100\% \text{Percentage Loss} = 1 \times 100\% = 100\% \text{Therefore, the percentage loss in kinetic energy of the first body is } 100\%.

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}