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Current Question (ID: 9905)

Question:

$\text{Two equal masses, } m_1 \text{ and } m_2\text{, moving in the same straight line at velocities } +3 \text{ m/s and } -5 \text{ m/s respectively, collide elastically. Their velocities after the collision will be:}$

Options:

1. $+4 \text{ m/s for both}$
2. $-3 \text{ m/s and } +5 \text{ m/s}$
3. $-4 \text{ m/s and } +4 \text{ m/s}$
4. $-5 \text{ m/s and } +3 \text{ m/s}$

Solution:

$\text{Given:}$ $m_1 = m_2 \text{ (equal masses)}$ $v_1 = +3 \text{ m/s}$ $v_2 = -5 \text{ m/s}$ $\text{For an elastic collision between two equal masses moving in the same straight line, the velocities get interchanged.}$ $\text{This is a fundamental property of elastic collisions between equal masses.}$ $\text{Therefore, after the collision:}$ $\text{velocity of mass } m_1 = -5 \text{ m/s}$ $\text{velocity of mass } m_2 = +3 \text{ m/s}$

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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
}