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Current Question (ID: 19236)

Question:

$\text{A particle of mass } m \text{ with an initial velocity } u\hat{i} \text{ collides perfectly}$ $\text{elastically with a mass } 3m \text{ at rest. It moves with a velocity } v\hat{j} \text{ after}$ $\text{collision, then, } v \text{ is given by:}$

Options:

1. $v = \sqrt{\frac{2}{3}} u$
2. $v = \frac{1}{\sqrt{6}} u$
3. $v = \frac{u}{\sqrt{3}}$
4. $v = \frac{u}{\sqrt{2}}$

Solution:

$\text{Hint: Apply the conservation of momentum.}$ $\text{From momentum conservation}$ $\vec{P}_i = \vec{P}_f$ $M(u\hat{i}) = mv_j\hat{j} + 3m \frac{v_1}{v_1}$ $mui = -mv_j = 3m \frac{ui - v_j}{3} = 3m \frac{\sqrt{u^2 + v^2}}{3}$ $v_1^2 = \frac{u^2 + v^2}{9} \ldots (1)$ $\text{As collision is perfectly elastic hence } k_i = k_j$ $\frac{1}{2} mu^2 = \frac{1}{2} mv^2 + \frac{1}{2} 3mv_1^2$ $u^2 = v^2 + \frac{3(u_1^2 + v^2)}{9}$ $3u^2 = 3v^2 + u^2 + v^2$ $2u^2 = 4v^2$ $v = \frac{u}{\sqrt{2}}$

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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
}