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Current Question (ID: 9565)

Question:
$\text{Two particles A and B, move with constant velocities } \vec{v_1} \text{ and } \vec{v_2}\text{. At the initial moment their position vector are } \vec{r_1} \text{ and } \vec{r_2} \text{ respectively. The conditions for particles A and B for their collision to happen will be:}$
Options:
  • 1. $\vec{r_1} \cdot \vec{v_1} = \vec{r_2} \cdot \vec{v_2}$
  • 2. $\vec{r_1} \times \vec{v_1} = \vec{r_2} \times \vec{v_2}$
  • 3. $\vec{r_1} - \vec{r_2} = \vec{v_1} - \vec{v_2}$
  • 4. $\frac{\vec{r_1} - \vec{r_2}}{|\vec{r_1} - \vec{r_2}|} = \frac{\vec{v_2} - \vec{v_1}}{|\vec{v_2} - \vec{v_1}|}$
Solution:
\text{Hint: For collision, their relative velocities should direct towards each other.} \text{Step 1: Find the relative position of one particle w.r.t. the other particle.} \text{For two particles A and B moving with constant velocities } v_1 \text{ and } v_2\text{,} \text{such that two particles collide, the direction of the relative velocity of one} \text{with respect to the other should be directed towards the relative position} \text{of the other particle.} \text{Direction of relative position of 1 w.r.t. 2 = } \frac{\vec{r_1} - \vec{r_2}}{|\vec{r_1} - \vec{r_2}|} \text{Step 2: Find the direction of the velocity of one particle w.r.t. the other particle.} \text{Similarly, direction of the velocity of 1 w.r.t. 2 = } \frac{\vec{v_1} - \vec{v_2}}{|\vec{v_1} - \vec{v_2}|} \text{Step 3: Find the condition of collision.} \text{So, for collision of A and B, we get:} \frac{\vec{r_1} - \vec{r_2}}{|\vec{r_1} - \vec{r_2}|} = \frac{\vec{v_1} - \vec{v_2}}{|\vec{v_1} - \vec{v_2}|} \text{Alternate Method:} \text{Step 1: Find the displacement of each particle.} \text{As the resultant displacement of the particles:} R = \vec{r_1} + \vec{v_1}t R' = \vec{r_2} + \vec{v_2}t \text{Step 2: Find the condition of collision.} \text{For collision, } R = R' \text{Therefore: } \vec{r_1} + \vec{v_1}t = \vec{r_2} + \vec{v_2}t \text{This gives us: } \vec{r_1} - \vec{r_2} = (\vec{v_2} - \vec{v_1})t \text{ ...(1)} \text{Step 3: As the options don't have the time variable, we need to find} \text{a clever way to get rid of t.} \text{As equal vectors have equal magnitude, from (1), we can take} \text{absolute value on both sides:} |\vec{r_1} - \vec{r_2}| = |(\vec{v_2} - \vec{v_1})|t \text{As the collision will happen in future, } t > 0 \text{ and therefore,} \text{we can take out t from right hand side of the equation:} |\vec{r_1} - \vec{r_2}| = |\vec{v_2} - \vec{v_1}|t \text{ ...(2)} \text{Now, dividing (1) by (2) we can get rid of t as below:} \frac{\vec{r_1} - \vec{r_2}}{|\vec{r_1} - \vec{r_2}|} = \frac{(\vec{v_2} - \vec{v_1})t}{|\vec{v_2} - \vec{v_1}|t} \text{Simplifying:} \frac{\vec{r_1} - \vec{r_2}}{|\vec{r_1} - \vec{r_2}|} = \frac{\vec{v_2} - \vec{v_1}}{|\vec{v_2} - \vec{v_1}|} \text{Hence, option (4) is the correct answer.}

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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
}