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Current Question (ID: 19228)

Question:

$\text{In a collinear collision, a particle with an initial speed } v_0 \text{ strikes a stationary particle of the same mass.}$ $\text{If the final total kinetic energy is } 50\% \text{ greater than the original kinetic energy,}$ $\text{the magnitude of the relative velocity between the two particles, after collision, is:}$

Options:

1. $\frac{v_0}{4}$
2. $\sqrt{2}v_0$
3. $\frac{v_0}{2}$
4. $\frac{v_0}{\sqrt{2}}$

Solution:

$\text{Hint: Apply the conservation of energy.}$ $mv_0 = mv_1 + mv_2 \quad \text{(momentum conservation)} \quad \ldots (1)$ $\text{KE}_f = \text{KE}_i + 0.5\text{KE}_i$ $\text{KE}_f = 1.5\text{KE}_i$ $\frac{1}{2}m(v_1^2 + v_2^2) = \frac{1}{2}mv_0^2 \times 1.5$ $\text{From } (1), \quad v_1 + v_2 = v_0$ $\text{From } (2), \quad v_1^2 + v_2^2 = 1.5 \quad v_2^2$ $v_1 - v_2 = \sqrt{(v_1 + v_2)^2 - 4v_1v_2}$ $(v_1 + v_2)^2 = v_1^2 + v_2^2 + 2v_1v_2$ $v_0^2 = 1.5v_0^2 + 2v_1v_2$ $2v_1v_2 = -0.5v_0^2$ $v_1 - v_2 = \sqrt{v_0^2 - 2(-0.5v_0^2)}$ $= \sqrt{2}v_0$

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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
}