A machine gun is mounted on a 2000kg car on a horizontal frictionless surface. At some instant, the gun fires 10 bullets/second and each of mass 10g with a velocity of `500 ms^-1` the acceleration of the car is

Options

A. `0.025 ms^-2`
B. `0.25 ms^-2`
C. `0.50 ms^-2`
D. `500 ms^-2`

Explanation

To solve the problem of finding the acceleration of the car when the machine gun fires bullets, we can follow these steps:

### Step 1: Understand the System
The car has a mass of 2000 kg and is on a frictionless surface. The machine gun fires 10 bullets per second, each with a mass of 10 grams (0.01 kg), at a velocity of 500 m/s.

### Step 2: Calculate the Total Momentum Change of the Bullets
The momentum of one bullet is given by the formula:
\[ \text{Momentum (p)} = \text{mass (m)} \times \text{velocity (v)} \]

For one bullet:
\[ p = 0.01 \, \text{kg} \times 500 \, \text{m/s} = 5 \, \text{kg m/s} \]

Since 10 bullets are fired per second, the total momentum change per second (ΔP_bullets) is:
\[ \Delta P_{\text{bullets}} = 10 \times 5 \, \text{kg m/s} = 50 \, \text{kg m/s} \]

### Step 3: Apply Conservation of Momentum
According to the conservation of momentum, the total momentum of the system before firing must equal the total momentum after firing. Initially, the momentum is zero (the car and gun are at rest). Thus, the final momentum must also equal zero.

Let \( P_{\text{car}} \) be the momentum of the car after firing. Hence:
\[ P_{\text{car}} + \Delta P_{\text{bullets}} = 0 \]
\[ P_{\text{car}} = -\Delta P_{\text{bullets}} \]
\[ P_{\text{car}} = -50 \, \text{kg m/s} \]

### Step 4: Calculate the Force Acting on the Car
The change in momentum of the car (ΔP_car) is equal to the force (F) acting on it over time (Δt):
\[ F = \frac{\Delta P_{\text{car}}}{\Delta t} \]

Since the bullets are fired at a rate of 10 per second, Δt = 1 second:
\[ F = \frac{-50 \, \text{kg m/s}}{1 \, \text{s}} = -50 \, \text{N} \]

### Step 5: Calculate the Acceleration of the Car
Using Newton's second law, \( F = m \cdot a \), we can find the acceleration (a) of the car:
\[ -50 \, \text{N} = 2000 \, \text{kg} \cdot a \]
\[ a = \frac{-50}{2000} \]
\[ a = -0.025 \, \text{m/s}^2 \]

### Step 6: Interpret the Result
The negative sign indicates that the acceleration is in the opposite direction to the motion of the bullets. If we only need the magnitude, we take:
\[ |a| = 0.025 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the car is \( 0.025 \, \text{m/s}^2 \).

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