A particle `P` lying on a smooth horizontal `x-y` plane starts from `(3hati+4hatj)m` with velocity `(2hati)m//s`. Another particle `Q` is projected (horizontally from origin with velocity `(xhati+yhatj)` so that is strikes `P` after `2s`. Then

Options

A. `x=2.0`
B. `x=3.5`
C. `y=2.0`
D. `y=3.5`

Answer

To solve the problem step by step, we will analyze the motion of both particles \( P \) and \( Q \).

### Step 1: Determine the final position of particle \( P \)
- Particle \( P \) starts at position \( (3\hat{i} + 4\hat{j}) \) meters and moves with a velocity of \( (2\hat{i}) \) m/s.
- The time of motion is \( 2 \) seconds.
- The displacement of \( P \) can be calculated using the formula:
\[ \text{Displacement} = \text{Velocity} \times \text{Time} = (2\hat{i}) \times 2 = 4\hat{i} \text{ meters} \]
- Therefore, the final position of particle \( P \) after \( 2 \) seconds is:
\[ \text{Final Position of } P = (3\hat{i} + 4\hat{j}) + 4\hat{i} = (3 + 4)\hat{i} + 4\hat{j} = 7\hat{i} + 4\hat{j} \text{ meters} \]

### Step 2: Determine the final position of particle \( Q \)
- Particle \( Q \) is projected horizontally from the origin \( (0\hat{i} + 0\hat{j}) \) with a velocity of \( (x\hat{i} + y\hat{j}) \).
- The position of \( Q \) after \( 2 \) seconds can be calculated as:
\[ \text{Final Position of } Q = (x\hat{i} + y\hat{j}) \times 2 = (2x\hat{i} + 2y\hat{j}) \text{ meters} \]

### Step 3: Set the final positions equal
- Since both particles strike each other after \( 2 \) seconds, their final positions must be equal:
\[ 7\hat{i} + 4\hat{j} = 2x\hat{i} + 2y\hat{j} \]

### Step 4: Equate the coefficients
- By equating the coefficients of \( \hat{i} \) and \( \hat{j} \):
1. For \( \hat{i} \):
\[ 2x = 7 \implies x = \frac{7}{2} = 3.5 \]
2. For \( \hat{j} \):
\[ 2y = 4 \implies y = \frac{4}{2} = 2 \]

### Conclusion
- The values of \( x \) and \( y \) are:
\[ x = 3.5, \quad y = 2 \]