A solid sphere of mass 2 kg is rolling on a frictionless horizontal surface with velocity `6 m//s`. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be
(Force constant of the spring = 36 N/m)

Options

A. `sqrt14m`
B. `sqrt2.8m`
C. `sqrt1.4m`
D. `sqrt0.7m`

Answer

To solve the problem, we need to find the maximum compression of the spring when a solid sphere collides with it. We will use the principles of conservation of energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the sphere, \( m = 2 \, \text{kg} \)
- Initial velocity of the sphere, \( v = 6 \, \text{m/s} \)
- Spring constant, \( k = 36 \, \text{N/m} \)

2. **Calculate the Initial Kinetic Energy:**
The sphere has both translational and rotational kinetic energy. However, since it is rolling without slipping, we can express the total kinetic energy as:
\[ KE_{\text{initial}} = KE_{\text{translational}} + KE_{\text{rotational}} \]
The translational kinetic energy is given by:
\[ KE_{\text{translational}} = \frac{1}{2} mv^2 \]
The moment of inertia \( I \) for a solid sphere is:
\[ I = \frac{2}{5} m r^2 \]
The angular velocity \( \omega \) can be related to the linear velocity \( v \) by:
\[ v = r \omega \implies \omega = \frac{v}{r} \]
Thus, the rotational kinetic energy is:
\[ KE_{\text{rotational}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{5} mv^2 \]
Therefore, the total kinetic energy becomes:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2 \]

3. **Substitute the Values:**
\[ KE_{\text{initial}} = \frac{7}{10} \times 2 \, \text{kg} \times (6 \, \text{m/s})^2 = \frac{7}{10} \times 2 \times 36 = \frac{7 \times 72}{10} = 50.4 \, \text{J} \]

4. **Set Up the Energy Conservation Equation:**
At maximum compression \( x \), all kinetic energy is converted into potential energy stored in the spring:
\[ KE_{\text{initial}} = PE_{\text{spring}} \implies \frac{1}{2} k x^2 \]
Thus, we have:
\[ 50.4 = \frac{1}{2} \times 36 \times x^2 \]

5. **Solve for \( x \):**
Rearranging gives:
\[ 50.4 = 18 x^2 \implies x^2 = \frac{50.4}{18} = 2.8 \implies x = \sqrt{2.8} \approx 1.673 \, \text{m} \]

6. **Final Result:**
The maximum compression produced in the spring is approximately:
\[ x \approx 1.673 \, \text{m} \]