Import Question JSON

Current Question (ID: 19233)

Question:

$\text{A particle of mass } m \text{ is moving at a speed of } 2v \text{ and collides with another particle of mass } 2m, \text{ which is moving at a speed of } v \text{ in the same direction.}$ $\text{After the collision, the first mass comes to a complete stop, while the second mass splits into two particles, each of mass } m, \text{ which moves at an angle of } 45^\circ \text{ with respect to the original direction.}$ $\text{The speed of each of the moving particles will be:}$

Options:

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

Solution:

$\text{Hint: Apply conservation of linear momentum.}$ $\text{Step: Find the speed of each moving particle after collision.}$ $\text{The motion of the particles before collision and after collision is shown in the figure below;} \newline \text{By applying the conservation of linear momentum along the } x\text{-direction is given by;} \newline m(2v) + 2mv = \frac{2mv'}{\sqrt{2}} \newline \Rightarrow 2mv' = 4\sqrt{2}mv \newline \Rightarrow v' = 2\sqrt{2}v \newline \text{Therefore, the speed of each moving particle after collision is } 2\sqrt{2}v. \newline \text{Hence, option (4) is the correct answer.}$

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
}