Import Question JSON

Current Question (ID: 19246)

Question:

$\text{Particle } A \text{ of mass } m_1 \text{ moving with velocity } (\hat{i} + \hat{j}) \text{ ms}^{-1} \text{ collides}$ $\text{with another particle } B \text{ of mass } m_2 \text{ which is at rest initially. Let } \vec{v}_1$ $\text{and } \vec{v}_2 \text{ be the velocities of particles } A \text{ and } B \text{ after collision}$ $\text{respectively. If } m_1 = 2m_2 \text{ and after collision } \vec{v}_1 = (\hat{i} - \hat{j}) \text{ ms}^{-1}$ $\text{then the final velocity of the particle } B \text{ is:}$

Options:

1. $(2\hat{i} + \hat{j}) \text{ ms}^{-1}$
2. $(2\hat{i} - \hat{j}) \text{ ms}^{-1}$
3. $4\hat{j} \text{ ms}^{-1}$
4. $-4\hat{i} \text{ ms}^{-1}$

Solution:

$\text{Hint: Use conservation of momentum}$ $\text{Step: Find the final velocity of the particle } B.$ $\text{Since no external force is acting on the system, apply conservation}$ $\text{of momentum.}$ $\text{Initial momentum } \vec{P}_i = \text{ final momentum } \vec{P}_f$ $\Rightarrow \vec{P}_i = \vec{P}_f \Rightarrow m_1 v_i + m_2 v_i' = m_1 v_1 + m_2 v_2$ $\Rightarrow m_1 (\hat{i} + \hat{j}) + 0 = m_1 (\hat{i} - \hat{j}) + m_2 \vec{v}_2$ $\Rightarrow \vec{v}_2 = 2 (\hat{i} + \hat{j}) - 2 (\hat{i} - \hat{j})$ $\Rightarrow \vec{v}_2 = 4\hat{j} \text{ ms}^{-1}$ $\text{Hence, option (3) is the correct answer.}$

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
}