A spring is held compressed so that its stored energy is `2.4 J`. Its ends are in constant with masses `1 g and 48 g` placed on a friction less table. When the spring in released, the heavier mass will acquire a speed of:

Options

A. `(240)/(149)m//s`
B. `(24xx49)/(48)m//s`
C. `(10)/(7)m//s`
D. `(10^(4))/(7)m//s`

Answer

To solve the problem, we will use the principles of conservation of energy and conservation of momentum.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Stored energy in the spring (E) = 2.4 J
- Mass of the first object (m1) = 1 g = 0.001 kg
- Mass of the second object (m2) = 48 g = 0.048 kg

2. **Use Conservation of Energy:**
When the spring is released, the potential energy stored in the spring is converted into the kinetic energy of the two masses.
\[ E = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the value of E:
\[ 2.4 = \frac{1}{2} (0.001) v_1^2 + \frac{1}{2} (0.048) v_2^2 \]
Simplifying, we get:
\[ 2.4 = 0.0005 v_1^2 + 0.024 v_2^2 \quad \text{(Equation 1)} \]

3. **Use Conservation of Momentum:**
Since the initial velocities are zero, the momentum before release is zero. Therefore, the momentum after release must also be zero:
\[ m_1 v_1 + m_2 v_2 = 0 \]
This implies:
\[ m_1 v_1 = -m_2 v_2 \quad \Rightarrow \quad v_1 = -\frac{m_2}{m_1} v_2 \]
Substituting the values of m1 and m2:
\[ v_1 = -\frac{0.048}{0.001} v_2 = -48 v_2 \quad \text{(Equation 2)} \]

4. **Substituting Equation 2 into Equation 1:**
Substitute \( v_1 = -48 v_2 \) into Equation 1:
\[ 2.4 = 0.0005 (-48 v_2)^2 + 0.024 v_2^2 \]
Simplifying:
\[ 2.4 = 0.0005 \times 2304 v_2^2 + 0.024 v_2^2 \]
\[ 2.4 = 1.152 v_2^2 + 0.024 v_2^2 \]
\[ 2.4 = 1.176 v_2^2 \]

5. **Solving for \( v_2^2 \):**
\[ v_2^2 = \frac{2.4}{1.176} \approx 2.04 \]
\[ v_2 = \sqrt{2.04} \approx 1.43 \, \text{m/s} \]

### Final Answer:
The speed of the heavier mass (48 g) when the spring is released is approximately **1.43 m/s**.