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Current Question (ID: 19231)

Question:
$\text{A body of mass } m_1 \text{ moving with an unknown velocity of } v_1 \hat{i}, \text{ undergoes a collinear collision with a body of mass } m_2 \text{ moving with a velocity } v_2 \hat{i}. \text{ After collision, } m_1 \text{ and } m_2 \text{ move with velocities of } v_3 \hat{i} \text{ and } v_4 \hat{i}, \text{ respectively. If } m_2 = 0.5m_1 \text{ and } v_3 = 0.5v_1, \text{ then } v_1 \text{ is:}$
Options:
  • 1. $v_4 - \frac{v_2}{4}$
  • 2. $v_4 + v_2$
  • 3. $v_4 - \frac{v_2}{2}$
  • 4. $v_4 - v_2$
Solution:
$\text{Hint: Apply the conservation of momentum.}$ $m_1v_1\hat{i} + m_2v_2\hat{i} = m_1\left(0.5v_1\right)\hat{i} + m_2v_4\hat{i}$ $m_1v_1\hat{i} + \frac{m_1v_2}{2}\hat{i} = \frac{m_1v_1}{2}\hat{i} + \frac{m_1v_4}{2}\hat{i}$ $v_1 = v_4 - v_2$

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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
}