A partical of mass `m` moving with velocity `1m//s` collides perfectly elastically with another particle of mass `2m`. If the incident particle is deflected by `90^(circ)`. The heavy mass will make and angle `theta` with the initial direction of `m` equal to:

Options

A. `60^(@)`
B. `45^(@)`
C. `15^(@)`
D. `30^(@)`

Explanation

To solve the problem, we will use the principles of conservation of momentum and conservation of kinetic energy, as the collision is perfectly elastic.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Mass of particle 1 (m) is moving with velocity \( v_1 = 1 \, \text{m/s} \).
- Mass of particle 2 (2m) is initially at rest, so its initial velocity \( v_2 = 0 \).
- After the collision, particle 1 is deflected by \( 90^\circ \).

2. **Set Up the Coordinate System:**
- Let the initial direction of particle 1 be along the x-axis.
- After the collision, particle 1 moves along the y-axis (since it is deflected by \( 90^\circ \)).
- Let the angle \( \theta \) be the angle that particle 2 makes with the x-axis after the collision.

3. **Apply Conservation of Momentum:**
- **In the x-direction:**
\[ m \cdot v_1 + 0 = 2m \cdot v_2 \cos(\theta) \]
Simplifying gives:
\[ 1 = 2v_2 \cos(\theta) \quad \text{(1)} \]

- **In the y-direction:**
\[ 0 + m \cdot v_1 = 2m \cdot v_2 \sin(\theta) \]
This simplifies to:
\[ 1 = 2v_2 \sin(\theta) \quad \text{(2)} \]

4. **Apply Conservation of Kinetic Energy:**
- The total kinetic energy before the collision equals the total kinetic energy after the collision:
\[ \frac{1}{2} m v_1^2 + 0 = \frac{1}{2} m v_1'^2 + \frac{1}{2} (2m) v_2^2 \]
Substituting \( v_1 = 1 \):
\[ \frac{1}{2} m (1)^2 = \frac{1}{2} m (0)^2 + \frac{1}{2} (2m) v_2^2 \]
This simplifies to:
\[ \frac{1}{2} = m v_2^2 \quad \Rightarrow \quad v_2^2 = \frac{1}{2} \quad \Rightarrow \quad v_2 = \frac{1}{\sqrt{2}} \quad \text{(3)} \]

5. **Substitute \( v_2 \) into Equations (1) and (2):**
- From equation (1):
\[ 1 = 2 \left(\frac{1}{\sqrt{2}}\right) \cos(\theta) \quad \Rightarrow \quad \cos(\theta) = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \theta = 45^\circ \]

- From equation (2):
\[ 1 = 2 \left(\frac{1}{\sqrt{2}}\right) \sin(\theta) \quad \Rightarrow \quad \sin(\theta) = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \theta = 45^\circ \]

6. **Conclusion:**
- The angle \( \theta \) that the heavy mass makes with the initial direction of mass \( m \) is \( 45^\circ \).

### Final Answer:
The angle \( \theta \) is \( 45^\circ \).

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