A mass of `0.5 kg` moving with a speed of `1.5 m//s` on a horizontal smooth surface, collides with a nearly weightless spring of force constant `k =50 N//m` The maximum compression of the spring would be.

Options

A. `0.15 m`
B. `0.12m`
C. `0.5 m`
D. `0.25 m`

Explanation

To find the maximum compression of the spring when a mass collides with it, we can use the principle of conservation of energy. The kinetic energy of the mass will be converted into the potential energy stored in the spring at maximum compression.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass (m) = 0.5 kg
- Initial speed (v) = 1.5 m/s
- Spring constant (k) = 50 N/m

2. **Calculate the Initial Kinetic Energy (KE):**
The kinetic energy of the mass can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 0.5 \, \text{kg} \times (1.5 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 0.5 \times 2.25 = 0.5625 \, \text{J} \]

3. **Set Up the Equation for Potential Energy (PE) of the Spring:**
The potential energy stored in the spring at maximum compression (x) is given by:
\[ PE = \frac{1}{2} k x^2 \]

4. **Apply Conservation of Energy:**
At maximum compression, all the kinetic energy will be converted into potential energy:
\[ KE = PE \]
Therefore:
\[ 0.5625 = \frac{1}{2} \times 50 \times x^2 \]

5. **Solve for x:**
Rearranging the equation gives:
\[ 0.5625 = 25 x^2 \]
\[ x^2 = \frac{0.5625}{25} \]
\[ x^2 = 0.0225 \]
Taking the square root of both sides:
\[ x = \sqrt{0.0225} = 0.15 \, \text{m} \]

6. **Conclusion:**
The maximum compression of the spring is:
\[ x = 0.15 \, \text{m} \]