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Current Question (ID: 9909)

Question:

$\text{On a frictionless surface, a block of mass } M \text{ moving at speed } v \text{ collides elastically with another block of the same mass } M \text{ which is initially at rest. After the collision, the first block moves at an angle } \theta \text{ to its initial direction and has a speed } \frac{v}{3}. \text{ The second block's speed after the collision will be:}$

Options:

1. $\frac{2\sqrt{2}}{3}v$
2. $\frac{3}{4}v$
3. $\frac{v}{\sqrt{2}}$
4. $\frac{\sqrt{3}}{2}v$

Solution:

\text{According to the law of conservation of kinetic energy, we have:} \frac{1}{2}Mv^2 + 0 = \frac{1}{2}M\left(\frac{v}{3}\right)^2 + \frac{1}{2}Mv'^2 \Rightarrow v^2 = \frac{v^2}{9} + v'^2 \Rightarrow v^2 - \frac{v^2}{9} = v'^2 \Rightarrow v'^2 = \frac{8v^2}{9} \text{Velocity of second block after collision: } v' = \frac{2\sqrt{2}v}{3}

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
}