A block of mass `m` is pushed up against a spring, compressing it a distance `x`, and the block is then released. The spring projects the block along a frictionaless horizontal surface, grving the block a speed `v`. The same spring projects a second block of mass `4m`, giving it a speed `3v`. What distance was the spring compressed in the second case ?

Options

A. `6x`
B. `x//6`
C. `36x`
D. `12x`

Explanation

To solve the problem, we need to apply the principles of energy conservation and the relationship between the spring's compression and the kinetic energy imparted to the blocks.

### Step-by-Step Solution:

1. **Initial Setup**:
- Let the spring constant be \( k \).
- When the first block of mass \( m \) is compressed by a distance \( x \), the potential energy stored in the spring is given by:
\[ PE = \frac{1}{2} k x^2 \]
- This potential energy is converted into kinetic energy when the block is released, giving it a speed \( v \):
\[ KE = \frac{1}{2} m v^2 \]

2. **Energy Conservation for the First Block**:
- Setting the potential energy equal to the kinetic energy, we have:
\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]
- Simplifying this gives:
\[ k x^2 = m v^2 \quad \text{(Equation 1)} \]

3. **Setup for the Second Block**:
- Now, consider the second block of mass \( 4m \) which is projected with a speed \( 3v \).
- The potential energy stored in the spring when compressed by a distance \( y \) is:
\[ PE_{2} = \frac{1}{2} k y^2 \]
- The kinetic energy of the second block when it is projected is:
\[ KE_{2} = \frac{1}{2} (4m) (3v)^2 = \frac{1}{2} (4m) (9v^2) = 18m v^2 \]

4. **Energy Conservation for the Second Block**:
- Setting the potential energy equal to the kinetic energy for the second block gives:
\[ \frac{1}{2} k y^2 = 18m v^2 \]
- Simplifying this gives:
\[ k y^2 = 36m v^2 \quad \text{(Equation 2)} \]

5. **Relating the Two Equations**:
- From Equation 1, we have \( k x^2 = m v^2 \).
- We can express \( v^2 \) from this equation:
\[ v^2 = \frac{k x^2}{m} \]
- Substitute \( v^2 \) into Equation 2:
\[ k y^2 = 36m \left(\frac{k x^2}{m}\right) \]
- This simplifies to:
\[ k y^2 = 36 k x^2 \]
- Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ y^2 = 36 x^2 \]
- Taking the square root of both sides gives:
\[ y = 6x \]

### Final Answer:
The distance the spring was compressed in the second case is \( y = 6x \).