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Current Question (ID: 9868)

Question:

$\text{A block of mass } M \text{ moving on the frictionless horizontal surface collides with the spring of spring constant } k \text{ and compresses it by length } L. \text{ The maximum momentum of the block after the collision will be:}$

Options:

1. $\text{zero}$
2. $\frac{ML^2}{k}$
3. $\sqrt{MkL}$
4. $\frac{kL^2}{2M}$

Solution:

$\text{When block of mass } M \text{ collides with the spring its kinetic energy gets converted into elastic potential energy of the spring.}$ $\text{From the law of conservation of energy:}$ $\frac{1}{2}Mv^2 = \frac{1}{2}KL^2$ $\text{Solving for velocity:}$ $v = \sqrt{\frac{K}{M}}L$ $\text{Where } v \text{ is the velocity of block by which it collides with spring. So, its maximum momentum is:}$ $P = Mv = M\sqrt{\frac{K}{M}}L = \sqrt{MKL}$ $\text{After collision the block will rebound with same linear momentum.}$

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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
}