A machine gun is mounted on a 200kg vehicle on a horizontal smooth road ( friction negligible). The gun fires 10 bullets `//`s with a velocity of `500ms^(-1)`. If the mass of each bullet be 10g, what is the acceleration produced in the vehicle ?

Options

A. `25 cms^(-2)`
B. `30cms^(-2)`
C. `50 cm^(-2)`
D. `50 cm^(-2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Mass of the vehicle (M) = 200 kg
- Mass of each bullet (m) = 10 g = 0.01 kg (since 1 g = 0.001 kg)
- Velocity of each bullet (v) = 500 m/s
- Number of bullets fired per second (n) = 10 bullets/s

### Step 2: Calculate the rate of change of mass (dm/dt)
The rate of change of mass per unit time when bullets are fired can be calculated as:
\[ \frac{dm}{dt} = n \times m \]
Substituting the values:
\[ \frac{dm}{dt} = 10 \, \text{bullets/s} \times 0.01 \, \text{kg/bullet} = 0.1 \, \text{kg/s} \]

### Step 3: Calculate the thrust force (F)
The thrust force produced by the bullets can be calculated using the formula:
\[ F = \frac{dm}{dt} \times v \]
Substituting the values:
\[ F = 0.1 \, \text{kg/s} \times 500 \, \text{m/s} = 50 \, \text{N} \]

### Step 4: Calculate the acceleration (a) of the vehicle
Using Newton's second law, the acceleration can be calculated as:
\[ a = \frac{F}{M} \]
Substituting the values:
\[ a = \frac{50 \, \text{N}}{200 \, \text{kg}} = 0.25 \, \text{m/s}^2 \]

### Step 5: Convert the acceleration to cm/s²
Since the options are in cm/s², we convert:
\[ 0.25 \, \text{m/s}^2 = 0.25 \times 100 \, \text{cm/s}^2 = 25 \, \text{cm/s}^2 \]

### Final Answer:
The acceleration produced in the vehicle is **25 cm/s²**.

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