To solve the problem, we will use the principle of conservation of momentum.
### Step-by-Step Solution:
1. **Identify the Given Data:**
- Mass of the vehicle, \( M = 200 \, \text{kg} \)
- Mass of each bullet, \( m_b = 10 \, \text{g} = 0.01 \, \text{kg} \) (converted to kg)
- Velocity of each bullet, \( v_b = 500 \, \text{m/s} \)
- Rate of firing bullets, \( n = 10 \, \text{bullets/s} \)
2. **Calculate the Total Mass of Bullets Fired per Second:**
\[
\text{Total mass of bullets fired per second} = n \times m_b = 10 \, \text{bullets/s} \times 0.01 \, \text{kg/bullet} = 0.1 \, \text{kg/s}
\]
3. **Calculate the Momentum of the Bullets Fired per Second:**
\[
\text{Momentum of bullets per second} = \text{mass} \times \text{velocity} = 0.1 \, \text{kg/s} \times 500 \, \text{m/s} = 50 \, \text{kg m/s}
\]
4. **Apply Conservation of Momentum:**
- Initially, the system (vehicle + gun + bullets) is at rest, so the initial momentum \( P_i = 0 \).
- When the bullets are fired, the momentum of the bullets will be in one direction, and the vehicle will move in the opposite direction to conserve momentum.
- Let \( V \) be the velocity of the vehicle after firing the bullets.
- The momentum of the vehicle after firing is \( M \times V \).
Using conservation of momentum:
\[
P_i = P_f \implies 0 = -50 + M \times V
\]
Rearranging gives:
\[
M \times V = 50
\]
5. **Solve for the Velocity of the Vehicle:**
\[
V = \frac{50}{M} = \frac{50}{200} = 0.25 \, \text{m/s}
\]
6. **Calculate the Acceleration of the Vehicle:**
- Since the bullets are fired continuously at a rate of 10 bullets per second, the change in momentum per second (which is the force) is equal to the momentum of the bullets fired per second.
- The acceleration \( a \) of the vehicle can be calculated using Newton's second law:
\[
F = M \times a \implies a = \frac{F}{M}
\]
Where \( F = 50 \, \text{kg m/s} \) (momentum change per second):
\[
a = \frac{50}{200} = 0.25 \, \text{m/s}^2
\]
### Final Answer:
The acceleration produced in the vehicle is \( 0.25 \, \text{m/s}^2 \).
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