The practicals having position vectors `vecr_1= (6hat i + 10hat j)` metre and `vecr_2=(-10hat i - 6hat j)` metre are moving with velocities `vecV_1 = (8hat i + 6hat j) m/s` and `vecV_2= (2alpha hat j + 14hat j)` m/s. If they collide after 2 second, then value of `alpha` is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the two particles collide after 2 seconds. We will calculate the final position of both particles after 2 seconds and set them equal to each other.

### Step 1: Determine the final position of Particle 1
The initial position vector of Particle 1 is given as:
\[ \vec{r_1} = 6\hat{i} + 10\hat{j} \text{ meters} \]
The velocity vector of Particle 1 is given as:
\[ \vec{V_1} = 8\hat{i} + 6\hat{j} \text{ m/s} \]
The time \( t \) is given as 2 seconds. The final position vector \( \vec{r_{f1}} \) after 2 seconds can be calculated using the formula:
\[ \vec{r_{f1}} = \vec{r_1} + \vec{V_1} \cdot t \]
Substituting the values:
\[ \vec{r_{f1}} = (6\hat{i} + 10\hat{j}) + (8\hat{i} + 6\hat{j}) \cdot 2 \]
Calculating the velocity term:
\[ \vec{V_1} \cdot 2 = (8\hat{i} + 6\hat{j}) \cdot 2 = 16\hat{i} + 12\hat{j} \]
Now substituting back:
\[ \vec{r_{f1}} = (6 + 16)\hat{i} + (10 + 12)\hat{j} = 22\hat{i} + 22\hat{j} \]

### Step 2: Determine the final position of Particle 2
The initial position vector of Particle 2 is given as:
\[ \vec{r_2} = -10\hat{i} - 6\hat{j} \text{ meters} \]
The velocity vector of Particle 2 is given as:
\[ \vec{V_2} = 2\alpha\hat{i} + 14\hat{j} \text{ m/s} \]
Using the same formula for the final position vector \( \vec{r_{f2}} \):
\[ \vec{r_{f2}} = \vec{r_2} + \vec{V_2} \cdot t \]
Substituting the values:
\[ \vec{r_{f2}} = (-10\hat{i} - 6\hat{j}) + (2\alpha\hat{i} + 14\hat{j}) \cdot 2 \]
Calculating the velocity term:
\[ \vec{V_2} \cdot 2 = (2\alpha\hat{i} + 14\hat{j}) \cdot 2 = 4\alpha\hat{i} + 28\hat{j} \]
Now substituting back:
\[ \vec{r_{f2}} = (-10 + 4\alpha)\hat{i} + (-6 + 28)\hat{j} = (-10 + 4\alpha)\hat{i} + 22\hat{j} \]

### Step 3: Set the final positions equal to each other
Since the particles collide, their final positions must be equal:
\[ \vec{r_{f1}} = \vec{r_{f2}} \]
This gives us two equations:
1. For the \( \hat{i} \) components:
\[ 22 = -10 + 4\alpha \]
2. For the \( \hat{j} \) components (which are already equal):
\[ 22 = 22 \]

### Step 4: Solve for \( \alpha \)
From the first equation:
\[ 22 = -10 + 4\alpha \]
Adding 10 to both sides:
\[ 32 = 4\alpha \]
Dividing both sides by 4:
\[ \alpha = 8 \]

### Final Answer
The value of \( \alpha \) is \( 8 \).