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Current Question (ID: 10227)

Question:
$\text{At } t = 0, \text{ the positions of the two blocks are shown. There is no external force acting on the system. Find the coordinates of the centre of mass of the system (in SI units) at } t = 3 \text{ seconds.}$
Options:
  • 1. $(1, 0)$
  • 2. $(3, 0)$
  • 3. $(4.5, 0)$
  • 4. $(2.25, 0)$
Solution:
\text{Hint: } x_{\text{com}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \text{Step: Find the coordinates of the centre of mass at } t = 3 \text{ seconds.} \text{The centre of mass of the system is given by:} x_{\text{com}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \Rightarrow x_{\text{com}} = \frac{4 \times 0 + 4 \times 4.5}{4 + 4} = \frac{18}{8} = 2.25 \text{ m} y_{\text{com}} = 0 \text{The coordinates of the centre of mass at } t = 0 \text{ s are } (2.25, 0). \text{The velocity of the centre of mass is given by:} v_{\text{com}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{4 \times 2 - 4 \times 2}{4 + 4} = 0 \text{ m/s} \text{This means the centre of mass is not moving.} \text{Therefore, the coordinates at } t = 3 \text{ seconds are } (2.25, 0) \text{Hence, option (4) is the correct answer.}

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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
}