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Current Question (ID: 19334)

Question:

$\text{A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of } 0.70 \text{ ms}^{-1} \text{ with respect to the man. The speed of the man with respect to the surface is:}$

Options:

1. $0.28 \text{ ms}^{-1}$
2. $0.47 \text{ ms}^{-1}$
3. $0.20 \text{ ms}^{-1}$
4. $0.14 \text{ ms}^{-1}$

Solution:

$\text{Hint: Apply conservation of momentum.}$ $\vec{V}_{B/A} = 0.7 \text{ m/s}$ $\Rightarrow \vec{V}_{B/S} - \vec{V}_{A/S} = 0.7$ $\Rightarrow V_B - (-V_A) = 0.7$ $\Rightarrow V_B + V_A = 0.7$ $\text{Momentum conservation } \left( \vec{F}_{\text{ext}} = 0 \right)$ $P_C = P_f$ $\Rightarrow 0 = 20(+V_B) + 50(-V_A)$ $\Rightarrow 0 = 2V_B - 5V_A$ $\Rightarrow V_B = \frac{5V_A}{2}$ $\Rightarrow V_A + \frac{5V_A}{2} = 0.7$ $\Rightarrow \frac{7V_A}{2} = 0.7$ $\Rightarrow V_A = 0.7 \left( \frac{2}{7} \right)$ $\Rightarrow V_A = 0.2 \text{ m/s}$

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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
}