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 will use the principle of conservation of energy and the relationship between the spring potential energy and the kinetic energy of the blocks.

### Step 1: Calculate the potential energy stored in the spring for the first block
The potential energy (PE) stored in a compressed spring is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the compression distance.

For the first block of mass \( m \) that is compressed by a distance \( x \):
\[ PE_1 = \frac{1}{2} k x^2 \]

### Step 2: Relate the potential energy to the kinetic energy of the first block
When the block is released, all the potential energy is converted into kinetic energy (KE):
\[ KE_1 = \frac{1}{2} m v^2 \]

Setting the potential energy equal to the kinetic energy:
\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]

### Step 3: Simplify the equation for the first block
We can cancel out the \(\frac{1}{2}\) from both sides:
\[ k x^2 = m v^2 \]

### Step 4: Calculate the potential energy stored in the spring for the second block
For the second block of mass \( 4m \) that is projected with speed \( 3v \), the potential energy stored in the spring when compressed by a distance \( x_2 \) is:
\[ PE_2 = \frac{1}{2} k x_2^2 \]

### Step 5: Relate the potential energy to the kinetic energy of the second block
The kinetic energy of the second block is:
\[ KE_2 = \frac{1}{2} (4m) (3v)^2 = \frac{1}{2} (4m) (9v^2) = 18m v^2 \]

Setting the potential energy equal to the kinetic energy:
\[ \frac{1}{2} k x_2^2 = 18m v^2 \]

### Step 6: Simplify the equation for the second block
Again, we can cancel out the \(\frac{1}{2}\):
\[ k x_2^2 = 36m v^2 \]

### Step 7: Relate the two equations
From the first block, we have:
\[ k x^2 = m v^2 \quad \text{(1)} \]
From the second block, we have:
\[ k x_2^2 = 36m v^2 \quad \text{(2)} \]

### Step 8: Divide equation (2) by equation (1)
\[ \frac{k x_2^2}{k x^2} = \frac{36m v^2}{m v^2} \]
This simplifies to:
\[ \frac{x_2^2}{x^2} = 36 \]

### Step 9: Solve for \( x_2 \)
Taking the square root of both sides:
\[ \frac{x_2}{x} = 6 \]
Thus,
\[ x_2 = 6x \]

### Final Result
The distance the spring was compressed in the second case is:
\[ x_2 = 6x \]