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Current Question (ID: 19264)

Question:

$\text{A particle of mass } 1 \text{ kg is moving with a velocity towards a stationary particle of mass } 3 \text{ kg.}$ $\text{After the collision, the lighter particle returns along the same path with speed } 2 \text{ m/s.}$ $\text{If the collision was elastic, then speed of } 1 \text{ kg particle before collision is:}$

Options:

1. $2 \text{ m/s}$
2. $6 \text{ m/s}$
3. $3 \text{ m/s}$
4. $4 \text{ m/s}$

Solution:

$m_1 = 1 \text{ kg and } m_2 = 3 \text{ kg}$ $\text{Conserving linear momentum}$ $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ $u_1 + 0 = 2 + 3v_2 \quad \text{---------(1)}$ $\text{For elastic collision}$ $e = \frac{\text{velocity of separation}}{\text{velocity of approach}} = 1$ $\Rightarrow \frac{v_1 + v_2}{u_1} = 1$ $\Rightarrow 2 + v_2 = u_1 \quad \text{---------(2)}$ $\text{From (1) and (2)}$ $u_1 = 4 \text{ m/s}$

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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
}