Two particles having position vectors `vec(r )_(1) = ( 3 hat(i) + 5 hat(j)) meters` and `vec( r)_(2) = (- 5 hat(i) - 3 hat(j)) metres` are moving with velocities `vec(v)_(1) = ( 4 hat(i) + 3 hat(j)) m//s` and `vec(v)_(2) = (alpha hat(i) + 7 hat(j)) m//s`. If they collide after `2 s`, the value of `alpha` is

Options

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

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Write down the initial position vectors and velocities
We have the position vectors of two particles:
- \(\vec{r}_1 = 3\hat{i} + 5\hat{j}\) meters
- \(\vec{r}_2 = -5\hat{i} - 3\hat{j}\) meters

And their velocities:
- \(\vec{v}_1 = 4\hat{i} + 3\hat{j}\) m/s
- \(\vec{v}_2 = \alpha\hat{i} + 7\hat{j}\) m/s

### Step 2: Find the position vectors after 2 seconds
The position of each particle after \(t = 2\) seconds can be calculated using the formula:
\[ \vec{r} = \vec{r}_0 + \vec{v} \cdot t \]

For particle 1:
\[ \vec{r}_1(t) = \vec{r}_1 + \vec{v}_1 \cdot t = (3\hat{i} + 5\hat{j}) + (4\hat{i} + 3\hat{j}) \cdot 2 \]
Calculating this gives:
\[ \vec{r}_1(t) = (3 + 8)\hat{i} + (5 + 6)\hat{j} = 11\hat{i} + 11\hat{j} \]

For particle 2:
\[ \vec{r}_2(t) = \vec{r}_2 + \vec{v}_2 \cdot t = (-5\hat{i} - 3\hat{j}) + (\alpha\hat{i} + 7\hat{j}) \cdot 2 \]
Calculating this gives:
\[ \vec{r}_2(t) = (-5 + 2\alpha)\hat{i} + (-3 + 14)\hat{j} = (-5 + 2\alpha)\hat{i} + 11\hat{j} \]

### Step 3: Set the position vectors equal for collision
Since the particles collide after 2 seconds, their position vectors will be equal:
\[ \vec{r}_1(t) = \vec{r}_2(t) \]
This gives us the equation:
\[ 11\hat{i} + 11\hat{j} = (-5 + 2\alpha)\hat{i} + 11\hat{j} \]

### Step 4: Solve for \(\alpha\)
From the equation, we can equate the coefficients of \(\hat{i}\):
\[ 11 = -5 + 2\alpha \]
Rearranging this gives:
\[ 2\alpha = 11 + 5 \]
\[ 2\alpha = 16 \]
\[ \alpha = \frac{16}{2} = 8 \]

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

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