A block of mass .M. moving on the firctionless , horizontal surface collides with the spring of spring constant .k.. The other end of the spring is rigidly fixed to a vertical wall. If the spring is compressed by L, the maximum momentum of the block after collision is :

Options

A. `sqrt(Mk)L`
B. `(kL^(2))/(2M)`
C. Zero
D. `(ML^(2))/k`

Answer

To solve the problem, we need to apply the principle of conservation of mechanical energy. The block collides with the spring and compresses it, converting its kinetic energy into potential energy stored in the spring. Let's break down the solution step by step:

### Step 1: Understand the conservation of energy
The total mechanical energy of the system remains constant if only conservative forces are acting. In this case, the kinetic energy of the block is converted into the potential energy of the spring.

**Hint:** Remember that the total mechanical energy is the sum of kinetic and potential energy.

### Step 2: Write the expressions for kinetic and potential energy
- The initial kinetic energy (KE_initial) of the block is given by:
\[ KE_{\text{initial}} = \frac{1}{2} M v^2 \]
where \( v \) is the initial velocity of the block.

- The potential energy (PE) stored in the spring when compressed by a distance \( L \) is given by:
\[ PE = \frac{1}{2} k L^2 \]
where \( k \) is the spring constant.

**Hint:** Identify the forms of energy involved in the collision.

### Step 3: Apply conservation of energy
According to the conservation of energy:
\[ KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}} \]
Assuming the spring is uncompressed initially, we have:
\[ \frac{1}{2} M v^2 + 0 = 0 + \frac{1}{2} k L^2 \]

**Hint:** Set up the equation based on the initial and final states of the system.

### Step 4: Rearrange the equation
From the above equation, we can rearrange it to find the initial velocity \( v \):
\[ \frac{1}{2} M v^2 = \frac{1}{2} k L^2 \]
Cancelling the \( \frac{1}{2} \) from both sides gives:
\[ M v^2 = k L^2 \]
Now, solving for \( v^2 \):
\[ v^2 = \frac{k L^2}{M} \]

**Hint:** Isolate \( v^2 \) to find the relationship with spring constant and compression.

### Step 5: Find the maximum momentum
The momentum \( p \) of the block is given by:
\[ p = M v \]
Substituting the expression for \( v \):
\[ p = M \sqrt{\frac{k L^2}{M}} = L \sqrt{M k} \]

**Hint:** Remember that momentum is mass times velocity.

### Final Answer
Thus, the maximum momentum of the block after the collision is:
\[ p = L \sqrt{M k} \]