At time `t=0`, particle `A` is at `(1m,2m)` and `B` is at `(5m,5m)`. Velociyt of `B` is `(2hati+4hatj)m//s` Velocity of particle `A` is `sqrt(2)v)` at `45^(@)` with `x`-axis. A collides with `B` .
Value of `v` is ………..m/s

Options

A. 5
B. 15
C. 25
D. 10

Answer

To solve the problem, we need to find the value of \( v \) such that particle A collides with particle B. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Initial Positions
- Particle A is at the coordinates \( (1 \, \text{m}, 2 \, \text{m}) \).
- Particle B is at the coordinates \( (5 \, \text{m}, 5 \, \text{m}) \).

### Step 2: Determine the Velocities
- The velocity of particle B is given as \( \vec{V_B} = 2 \hat{i} + 4 \hat{j} \, \text{m/s} \).
- The velocity of particle A is given as \( \sqrt{2} v \) at an angle of \( 45^\circ \) with the x-axis.

### Step 3: Calculate the Components of Velocity for A
Since particle A moves at an angle of \( 45^\circ \):
- The horizontal component of the velocity \( V_{Ax} = \sqrt{2} v \cdot \cos(45^\circ) = \sqrt{2} v \cdot \frac{1}{\sqrt{2}} = v \).
- The vertical component of the velocity \( V_{Ay} = \sqrt{2} v \cdot \sin(45^\circ) = \sqrt{2} v \cdot \frac{1}{\sqrt{2}} = v \).

Thus, the velocity vector for particle A is:
\[ \vec{V_A} = v \hat{i} + v \hat{j} \]

### Step 4: Calculate the Relative Velocity
To find the relative velocity of A with respect to B:
\[ \vec{V_{AB}} = \vec{V_A} - \vec{V_B} = (v - 2) \hat{i} + (v - 4) \hat{j} \]

### Step 5: Determine the Direction of Approach
For A to collide with B, the direction of the relative velocity must point towards B. We can find the slope of the line connecting A and B:
- The difference in y-coordinates: \( 5 - 2 = 3 \)
- The difference in x-coordinates: \( 5 - 1 = 4 \)

Thus, the slope (or tangent of the angle) is:
\[ \tan(\theta) = \frac{3}{4} \]

### Step 6: Set Up the Equation
The slope of the relative velocity must equal the slope of the line connecting A to B:
\[ \frac{V_{Ay}}{V_{Ax}} = \frac{v - 4}{v - 2} = \frac{3}{4} \]

### Step 7: Cross Multiply and Solve for \( v \)
Cross-multiplying gives:
\[ 4(v - 4) = 3(v - 2) \]
Expanding both sides:
\[ 4v - 16 = 3v - 6 \]
Rearranging gives:
\[ 4v - 3v = 16 - 6 \]
Thus:
\[ v = 10 \, \text{m/s} \]

### Final Answer
The value of \( v \) is \( 10 \, \text{m/s} \).

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