A machine gun fires a bullet of mass 40 g with a velocity `1200 ms^-1`. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can be fire per second at the most? a) 3 b) 5 c) 6 d) 9

Options

A. (a) 3
B. (b) 5
C. (c) 6
D. (d) 9

Explanation

To solve the problem, we need to determine how many bullets can be fired per second from the machine gun given the maximum force the man can exert on the gun.

### Step-by-Step Solution:

1. **Convert the mass of the bullet from grams to kilograms:**
\[ \text{Mass of bullet (m)} = 40 \text{ g} = \frac{40}{1000} \text{ kg} = 0.04 \text{ kg} \]

2. **Identify the velocity of the bullet:**
\[ \text{Velocity (v)} = 1200 \text{ m/s} \]

3. **Identify the maximum force exerted by the man:**
\[ \text{Maximum Force (F)} = 144 \text{ N} \]

4. **Calculate the change in momentum for one bullet:**
- The initial momentum of the bullet is zero (since it is at rest).
- The final momentum of the bullet when fired is given by:
\[ \text{Final momentum} = m \cdot v = 0.04 \text{ kg} \cdot 1200 \text{ m/s} = 48 \text{ kg m/s} \]
- Therefore, the change in momentum for one bullet is:
\[ \Delta p = \text{Final momentum} - \text{Initial momentum} = 48 \text{ kg m/s} - 0 = 48 \text{ kg m/s} \]

5. **Relate the force exerted to the change in momentum:**
- The force exerted by the man must balance the change in momentum of the bullets fired per second. If \( n \) is the number of bullets fired per second, then:
\[ F = n \cdot \Delta p \]
- Substituting the values we have:
\[ 144 \text{ N} = n \cdot 48 \text{ kg m/s} \]

6. **Solve for \( n \):**
\[ n = \frac{144 \text{ N}}{48 \text{ kg m/s}} = 3 \text{ bullets per second} \]

### Final Answer:
The maximum number of bullets that can be fired per second is **3**. Thus, the correct option is **(a) 3**.

---