A spring gun of spring constant 90 N/cm is compressed 12 cm by a ball of mass 16 g. If the trigger is pulled, the velocity of the ball is

Options

A. `50 ms^(-1)`
B. `9 ms^(-1)`
C. `40 ms^(-1)`
D. `90 ms^(-1)`

Explanation

To solve the problem step by step, we will follow the principles of energy conservation and the formulas related to springs and kinetic energy.

### Step 1: Convert all units to SI units
- **Spring constant (k)**: Given as 90 N/cm. Convert to N/m:
\[ k = 90 \, \text{N/cm} = 90 \times 100 \, \text{N/m} = 9000 \, \text{N/m} \]

- **Compression (x)**: Given as 12 cm. Convert to meters:
\[ x = 12 \, \text{cm} = 12 \times 10^{-2} \, \text{m} = 0.12 \, \text{m} \]

- **Mass (m)**: Given as 16 g. Convert to kg:
\[ m = 16 \, \text{g} = 16 \times 10^{-3} \, \text{kg} = 0.016 \, \text{kg} \]

### Step 2: Calculate the potential energy stored in the spring
The potential energy (PE) stored in a compressed spring can be calculated using the formula:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values:
\[ PE = \frac{1}{2} \times 9000 \, \text{N/m} \times (0.12 \, \text{m})^2 \]
Calculating:
\[ PE = \frac{1}{2} \times 9000 \times 0.0144 = 64.8 \, \text{J} \]

### Step 3: Apply the conservation of energy principle
When the spring is released, all the potential energy converts into kinetic energy (KE) of the ball:
\[ KE = \frac{1}{2} m v^2 \]
Setting the potential energy equal to the kinetic energy:
\[ 64.8 = \frac{1}{2} \times 0.016 \times v^2 \]

### Step 4: Solve for the velocity (v)
Rearranging the equation to solve for \(v^2\):
\[ v^2 = \frac{64.8 \times 2}{0.016} \]
Calculating:
\[ v^2 = \frac{129.6}{0.016} = 8100 \]
Taking the square root to find \(v\):
\[ v = \sqrt{8100} = 90 \, \text{m/s} \]

### Final Answer
The velocity of the ball when the trigger is pulled is:
\[ \boxed{90 \, \text{m/s}} \]