Two particles having position verctors `vecr_(1)=(3hati+5hatj)` metres and `vecr_(2)=(-5hati-3hatj)` metres are moving with velocities `vecv_(1)=(4hati+3hatj)m//s and vecv_(2)=(alphahati+7hatj)m//s`. If they collide after 2 seconds, the 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 start by determining the position vectors of both particles at that time.

### Step-by-Step Solution:

1. **Identify the Initial Position Vectors:**
- Particle 1: \(\vec{r_1} = 3\hat{i} + 5\hat{j}\) meters
- Particle 2: \(\vec{r_2} = -5\hat{i} - 3\hat{j}\) meters

2. **Identify the Velocity Vectors:**
- Particle 1: \(\vec{v_1} = 4\hat{i} + 3\hat{j}\) m/s
- Particle 2: \(\vec{v_2} = \alpha\hat{i} + 7\hat{j}\) m/s

3. **Determine the Position Vectors at \(t = 2\) seconds:**
- The position vector of Particle 1 at \(t = 2\) seconds:
\[ \vec{r_1}(t) = \vec{r_1} + \vec{v_1} \cdot t = (3\hat{i} + 5\hat{j}) + (4\hat{i} + 3\hat{j}) \cdot 2 \]
\[ = (3 + 8)\hat{i} + (5 + 6)\hat{j} = 11\hat{i} + 11\hat{j} \]

- The position vector of Particle 2 at \(t = 2\) seconds:
\[ \vec{r_2}(t) = \vec{r_2} + \vec{v_2} \cdot t = (-5\hat{i} - 3\hat{j}) + (\alpha\hat{i} + 7\hat{j}) \cdot 2 \]
\[ = (-5 + 2\alpha)\hat{i} + (-3 + 14)\hat{j} = (2\alpha - 5)\hat{i} + 11\hat{j} \]

4. **Set the Position Vectors Equal for Collision:**
Since the particles collide, their position vectors at \(t = 2\) seconds must be equal:
\[ 11\hat{i} + 11\hat{j} = (2\alpha - 5)\hat{i} + 11\hat{j} \]

5. **Equate the Components:**
- For the \(\hat{j}\) component:
\[ 11 = 11 \quad \text{(This is always true)} \]
- For the \(\hat{i}\) component:
\[ 11 = 2\alpha - 5 \]

6. **Solve for \(\alpha\):**
\[ 2\alpha - 5 = 11 \]
\[ 2\alpha = 11 + 5 \]
\[ 2\alpha = 16 \]
\[ \alpha = \frac{16}{2} = 8 \]

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