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Current Question (ID: 9908)

Question:
$\text{A body of mass 2 kg moving with a velocity of 3 m/s collides with a body of mass of 1 kg moving with a velocity of 4 m/s in the opposite direction. If the collision is head-on and completely inelastic, then the wrong statement is:}$
Options:
  • 1. $\text{Both bodies move together with a velocity } \frac{2}{3} \text{ m/s.}$
  • 2. $\text{The momentum of the system is 2 kg-m/s throughout.}$
  • 3. $\text{The momentum of the system is 10 kg-m/s.}$
  • 4. $\text{The loss of KE for the system is } \frac{49}{3} \text{ J.}$
Solution:
\text{Hint: Apply conservation of linear momentum.} \text{Step 1: Draw the diagram} \text{Before: 2kg at 3m/s, 1kg at 4m/s opposite} \text{After: Both stick together, velocity = v} \text{Step 2: Conserve momentum} 2 \cdot 3 - 1 \cdot 4 = 3v v = 2/3 \text{ m/s} \text{Step 3: Find KE loss} K_i = (1/2)(2)(9) + (1/2)(1)(16) = 17 \text{ J} K_f = (1/2)(3)(4/9) = 2/3 \text{ J} \text{Loss} = 17 - 2/3 = 49/3 \text{ J} \text{Results:} \text{1. Final velocity: 2/3 m/s - CORRECT} \text{2. Momentum: 2 kg-m/s - CORRECT} \text{3. Momentum: 10 kg-m/s - WRONG} \text{4. KE loss: 49/3 J - CORRECT}

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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
}