A solid cylinder of mass `3 kg` is rolling on a horizontal surface with velocity `4 ms^(-1)`. It collides with a horizontal spring of force constant `200 Nm^(-1)`. The maximum compression producec in the spring will be `:`

Options

A. `0.5 m`
B. `0.6 m`
C. `0.7 m`
D. `0.2 m`

Explanation

To solve the problem of finding the maximum compression produced in the spring when a solid cylinder rolls into it, we can follow these steps:

### Step 1: Identify the kinetic energy of the cylinder
The total kinetic energy (KE) of the rolling cylinder consists of both translational and rotational kinetic energy. The formulas for these are:
- Translational KE: \( KE_{trans} = \frac{1}{2} m v^2 \)
- Rotational KE: \( KE_{rot} = \frac{1}{2} I \omega^2 \)

For a solid cylinder, the moment of inertia \( I \) about its axis is given by:
\[ I = \frac{1}{2} m r^2 \]
And the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ \omega = \frac{v}{r} \]

### Step 2: Substitute the values into the kinetic energy equations
Substituting \( \omega \) into the rotational kinetic energy formula:
\[ KE_{rot} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{4} m v^2 \]

Now, the total kinetic energy of the cylinder is:
\[ KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2 \]

### Step 3: Set the kinetic energy equal to the potential energy of the spring
The potential energy stored in the spring when compressed by a distance \( x \) is given by:
\[ PE_{spring} = \frac{1}{2} k x^2 \]

At maximum compression, all kinetic energy will be converted into potential energy:
\[ \frac{3}{4} m v^2 = \frac{1}{2} k x^2 \]

### Step 4: Solve for \( x \)
Rearranging the equation to solve for \( x^2 \):
\[ x^2 = \frac{3 m v^2}{2 k} \]

### Step 5: Substitute the known values
Given:
- Mass \( m = 3 \, \text{kg} \)
- Velocity \( v = 4 \, \text{m/s} \)
- Spring constant \( k = 200 \, \text{N/m} \)

Substituting these values into the equation:
\[ x^2 = \frac{3 \times 3 \times (4)^2}{2 \times 200} \]
\[ x^2 = \frac{3 \times 3 \times 16}{400} \]
\[ x^2 = \frac{144}{400} \]
\[ x^2 = 0.36 \]

Taking the square root to find \( x \):
\[ x = \sqrt{0.36} = 0.6 \, \text{m} \]

### Conclusion
The maximum compression produced in the spring is \( 0.6 \, \text{m} \).

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