A block of mass 200 gm is suspended through a vertical spring. The spring is stretched by 1 cm when the block is in equilibrium. A particle of mass 120 gm is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the System
We have a block of mass \( m_1 = 200 \, \text{g} = 0.2 \, \text{kg} \) suspended from a spring. The spring stretches by \( x_1 = 1 \, \text{cm} = 0.01 \, \text{m} \) when the block is in equilibrium. A second mass \( m_2 = 120 \, \text{g} = 0.12 \, \text{kg} \) is dropped from a height of \( h = 45 \, \text{cm} = 0.45 \, \text{m} \) and sticks to the block upon impact.

### Step 2: Calculate the Force on the Spring
The force exerted by the block on the spring at equilibrium can be calculated using the formula:
\[ F = m_1 \cdot g \]
where \( g \approx 9.81 \, \text{m/s}^2 \).

Calculating this gives:
\[ F = 0.2 \cdot 9.81 = 1.962 \, \text{N} \]

### Step 3: Determine the Spring Constant
Using Hooke's Law, which states that \( F = k \cdot x \), we can find the spring constant \( k \):
\[ k = \frac{F}{x_1} = \frac{1.962}{0.01} = 196.2 \, \text{N/m} \]

### Step 4: Calculate the Velocity of the Dropped Mass Just Before Impact
The velocity of the mass \( m_2 \) just before it hits the block can be found using the equation of motion:
\[ v = \sqrt{2gh} \]
Substituting the values:
\[ v = \sqrt{2 \cdot 9.81 \cdot 0.45} \approx \sqrt{8.829} \approx 2.97 \, \text{m/s} \]

### Step 5: Apply Conservation of Momentum
When the mass \( m_2 \) sticks to \( m_1 \), we can use the conservation of momentum:
\[ m_2 \cdot v = (m_1 + m_2) \cdot V \]
where \( V \) is the velocity of the combined mass after the collision. Rearranging gives:
\[ V = \frac{m_2 \cdot v}{m_1 + m_2} \]
Substituting the values:
\[ V = \frac{0.12 \cdot 2.97}{0.2 + 0.12} = \frac{0.3564}{0.32} \approx 1.11 \, \text{m/s} \]

### Step 6: Calculate the Maximum Extension of the Spring
At maximum extension, all the kinetic energy will be converted into potential energy stored in the spring. The kinetic energy just after the collision is:
\[ KE = \frac{1}{2} (m_1 + m_2) V^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \cdot (0.2 + 0.12) \cdot (1.11)^2 \approx \frac{1}{2} \cdot 0.32 \cdot 1.2321 \approx 0.196 \, \text{J} \]

The potential energy stored in the spring at maximum extension \( x \) is given by:
\[ PE = \frac{1}{2} k x^2 \]
Setting \( KE = PE \):
\[ 0.196 = \frac{1}{2} \cdot 196.2 \cdot x^2 \]
Solving for \( x^2 \):
\[ x^2 = \frac{0.196 \cdot 2}{196.2} \approx 0.001 \, \text{m}^2 \]
Thus,
\[ x \approx 0.0316 \, \text{m} = 3.16 \, \text{cm} \]

### Final Answer
The maximum extension of the spring is approximately \( 3.16 \, \text{cm} \).

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