Two identical blocks `A` and `B` , each of mass `m` resting on smooth floor are connected by a light spring of natural length `L` and spring constant `k`, with the spring at its natural length. A third identical block `C` (mass `m`) moving with a speed `v` along the line joining `A` and `B` collides with `A`. The maximum compression in the spring is

Options

A. `v sqrt((m)/(2k))`
B. `msqrt((m)/(2k))`
C. `sqrt((mv)/(k))`
D. `(mv)/(2k)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the System
We have three blocks: A and B are at rest and connected by a spring, while block C is moving towards block A with speed \( v \). The goal is to find the maximum compression \( x \) of the spring when block C collides with block A.

### Step 2: Apply Conservation of Momentum
Since there are no external forces acting on the system, we can conserve momentum. The initial momentum of the system is given by the momentum of block C, as blocks A and B are at rest.

Initial momentum:
\[ p_{initial} = mv \]

After the collision, let the velocity of blocks A and B (which will be the same at maximum compression) be \( V \). The total mass of the system after the collision is \( 2m \) (mass of A and B).

Final momentum:
\[ p_{final} = mv + mv = 2mV \]

Setting initial and final momentum equal gives:
\[ mv = 2mV \implies V = \frac{v}{2} \]

### Step 3: Apply Conservation of Energy
At maximum compression of the spring, the kinetic energy of the system will be converted into potential energy stored in the spring.

Initial kinetic energy (only block C is moving):
\[ KE_{initial} = \frac{1}{2} mv^2 \]

Final kinetic energy (both blocks A and B are moving with velocity \( \frac{v}{2} \)):
\[ KE_{final} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 + \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \left(\frac{v^2}{4}\right) + \frac{1}{2} m \left(\frac{v^2}{4}\right) = \frac{mv^2}{4} \]

The potential energy stored in the spring at maximum compression \( x \) is given by:
\[ PE = \frac{1}{2} k x^2 \]

### Step 4: Set Up the Energy Conservation Equation
The loss in kinetic energy equals the potential energy stored in the spring:
\[ KE_{initial} - KE_{final} = PE \]
\[ \frac{1}{2} mv^2 - \frac{mv^2}{4} = \frac{1}{2} k x^2 \]

### Step 5: Simplify the Equation
Calculating the left side:
\[ \frac{1}{2} mv^2 - \frac{mv^2}{4} = \frac{2mv^2}{4} - \frac{mv^2}{4} = \frac{mv^2}{4} \]

Setting this equal to the potential energy:
\[ \frac{mv^2}{4} = \frac{1}{2} k x^2 \]

### Step 6: Solve for Maximum Compression \( x \)
Multiplying both sides by 2:
\[ \frac{mv^2}{2} = k x^2 \]

Rearranging gives:
\[ x^2 = \frac{mv^2}{2k} \]

Taking the square root:
\[ x = \sqrt{\frac{mv^2}{2k}} \]

### Final Answer
The maximum compression in the spring is:
\[ x_{max} = \sqrt{\frac{mv^2}{2k}} \]

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