A spherical ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest. After the collision, A and B move with velocities `v_1 m s^(-1) and v_2 m s^(-1)` respectively making angles of `30^@ and 60^@` with respect to the original direction of motion of A. The ratio `v_1/v_2` will be

Options

A. `sqrt3/4`
B. `4/sqrt3`
C. `1/sqrt3`
D. `sqrt3`

Explanation

To solve the problem, we will use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision for both the x and y components.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of ball A (m1) = 4 kg
- Mass of ball B (m2) = 1 kg
- Angle of ball A after collision = 30 degrees
- Angle of ball B after collision = 60 degrees
- Initial velocity of ball B = 0 m/s (at rest)

2. **Set Up the Momentum Conservation Equations:**
- Before the collision, the total momentum in the x-direction is only due to ball A:
\[ \text{Initial momentum} = m_1 \cdot v_{A_i} + m_2 \cdot 0 = m_1 \cdot v_{A_i} \]
- After the collision, the momentum in the x-direction is:
\[ \text{Final momentum in x} = m_1 \cdot v_1 \cdot \cos(30^\circ) + m_2 \cdot v_2 \cdot \cos(60^\circ) \]
- The momentum in the y-direction before the collision is 0. After the collision, it is:
\[ \text{Final momentum in y} = m_1 \cdot v_1 \cdot \sin(30^\circ) - m_2 \cdot v_2 \cdot \sin(60^\circ) \]

3. **Apply Conservation of Momentum:**
- For the x-direction:
\[ m_1 \cdot v_{A_i} = m_1 \cdot v_1 \cdot \cos(30^\circ) + m_2 \cdot v_2 \cdot \cos(60^\circ) \]
- For the y-direction:
\[ 0 = m_1 \cdot v_1 \cdot \sin(30^\circ) - m_2 \cdot v_2 \cdot \sin(60^\circ) \]

4. **Rearranging the y-direction Equation:**
- Rearranging gives:
\[ m_1 \cdot v_1 \cdot \sin(30^\circ) = m_2 \cdot v_2 \cdot \sin(60^\circ) \]

5. **Substituting Known Values:**
- Substitute \(m_1 = 4 \, \text{kg}\), \(m_2 = 1 \, \text{kg}\), \(\sin(30^\circ) = \frac{1}{2}\), and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ 4 \cdot v_1 \cdot \frac{1}{2} = 1 \cdot v_2 \cdot \frac{\sqrt{3}}{2} \]
- Simplifying this gives:
\[ 2 v_1 = \frac{\sqrt{3}}{2} v_2 \]

6. **Finding the Ratio \( \frac{v_1}{v_2} \):**
- Rearranging gives:
\[ \frac{v_1}{v_2} = \frac{\sqrt{3}}{4} \]

### Final Answer:
The ratio \( \frac{v_1}{v_2} = \frac{\sqrt{3}}{4} \).