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Current Question (ID: 10112)

Question:

$\text{Two particles of masses 2 kg and 3 kg start to move towards each other due to mutual forces of attraction. The speed of the first particle is } v_1 \text{ and that of the second is } v_2 \text{ at a certain instant. The speed of the centre of mass is:}$

Options:

1. $\frac{v_1 + v_2}{2}$
2. $\frac{2v_1 + 3v_2}{5}$
3. $\frac{3v_1 + 2v_2}{5}$
4. $\text{zero}$

Solution:

\text{Hint: } \vec{v}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2} \text{Step 1: Find the external force on the system} \vec{F}_{ext} = 0 \vec{F}_{ext} = M_{sys} a_{CM} \Rightarrow a_{CM} = 0 \text{Step 2: Find the initial velocity of centre of mass} v_{CM,i} = 0 \text{Step 3: Find the velocity of centre of mass at time t} \text{As } a_{CM} = 0 \text{, the velocity of centre of mass at time t will be zero.}

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
}