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Current Question (ID: 10107)

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 of the system 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} 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 of the system 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 centre of mass of the system is not moving.} \text{Therefore, the coordinates of the centre of mass of the system at } t = 3 \text{ seconds are the same as the initial i.e., } (2.25, 0) \text{Hence, option (4) is the correct answer.}

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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
}