A toy gun uses a spring of spring constant 1000 N/m . If spring is compressed by an amount 10 cm before fire then how high can be bullet be fired if mass of each bullet is 10 g .

Options

A. 30 m
B. 40 m
C. 50 m
D. 60 m

Explanation

To solve the problem, we will apply the law of conservation of energy. The potential energy stored in the compressed spring will be converted into gravitational potential energy when the bullet is fired. Let's break it down step by step.

### Step 1: Identify the given values
- Spring constant, \( k = 1000 \, \text{N/m} \)
- Compression of the spring, \( x = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Mass of the bullet, \( m = 10 \, \text{g} = 0.01 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the elastic potential energy stored in the spring
The elastic potential energy (EPE) stored in a compressed spring is given by the formula:
\[ \text{EPE} = \frac{1}{2} k x^2 \]

Substituting the values:
\[ \text{EPE} = \frac{1}{2} \times 1000 \, \text{N/m} \times (0.1 \, \text{m})^2 \]
\[ = \frac{1}{2} \times 1000 \times 0.01 \]
\[ = \frac{1}{2} \times 10 = 5 \, \text{J} \]

### Step 3: Set the elastic potential energy equal to the gravitational potential energy
When the bullet is fired, all the elastic potential energy will convert into gravitational potential energy (GPE) at the maximum height \( h \):
\[ \text{GPE} = mgh \]

Setting EPE equal to GPE:
\[ \frac{1}{2} k x^2 = mgh \]

### Step 4: Solve for height \( h \)
Rearranging the equation to solve for \( h \):
\[ h = \frac{\frac{1}{2} k x^2}{mg} \]

Substituting the values we calculated:
\[ h = \frac{5 \, \text{J}}{0.01 \, \text{kg} \times 10 \, \text{m/s}^2} \]
\[ = \frac{5}{0.1} \]
\[ = 50 \, \text{m} \]

### Final Answer
The bullet can be fired to a height of **50 meters**.

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