A bullet is fired from a gun. The force on the bullet is given by `F=600-2xx10^(5)` t, where F is in newtons and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Options

A. 9Ns
B. zero
C. `0.9Ns`
D. `1.8Ns`

Explanation

To solve the problem of finding the average impulse imparted to the bullet, we can follow these steps:

### Step 1: Understand the Impulse Concept
Impulse is defined as the change in momentum of an object, and it can also be calculated as the integral of force over time. Mathematically, impulse \( J \) can be expressed as:
\[ J = \int F(t) \, dt \]

### Step 2: Identify the Force Function
The force acting on the bullet is given by:
\[ F(t) = 600 - 2 \times 10^5 t \]
where \( F \) is in newtons and \( t \) is in seconds.

### Step 3: Determine the Time Interval
The force on the bullet becomes zero when it leaves the barrel. To find the time when the force becomes zero, we set the force function to zero:
\[ 600 - 2 \times 10^5 t = 0 \]
Solving for \( t \):
\[ 2 \times 10^5 t = 600 \\ t = \frac{600}{2 \times 10^5} = 3 \times 10^{-3} \text{ seconds} \]
Thus, the bullet is acted upon by the force from \( t = 0 \) to \( t = 3 \times 10^{-3} \) seconds.

### Step 4: Set Up the Integral for Impulse
Now we can set up the integral for impulse from \( t = 0 \) to \( t = 3 \times 10^{-3} \):
\[ J = \int_0^{3 \times 10^{-3}} (600 - 2 \times 10^5 t) \, dt \]

### Step 5: Calculate the Integral
We can split the integral into two parts:
\[ J = \int_0^{3 \times 10^{-3}} 600 \, dt - \int_0^{3 \times 10^{-3}} 2 \times 10^5 t \, dt \]

Calculating the first integral:
\[ \int_0^{3 \times 10^{-3}} 600 \, dt = 600 \cdot t \bigg|_0^{3 \times 10^{-3}} = 600 \cdot (3 \times 10^{-3} - 0) = 600 \cdot 3 \times 10^{-3} = 1.8 \text{ Ns} \]

Calculating the second integral:
\[ \int_0^{3 \times 10^{-3}} 2 \times 10^5 t \, dt = 2 \times 10^5 \cdot \frac{t^2}{2} \bigg|_0^{3 \times 10^{-3}} = 10^5 \cdot (3 \times 10^{-3})^2 = 10^5 \cdot 9 \times 10^{-6} = 0.9 \text{ Ns} \]

### Step 6: Combine the Results
Now we can combine the results of the two integrals to find the total impulse:
\[ J = 1.8 \text{ Ns} - 0.9 \text{ Ns} = 0.9 \text{ Ns} \]

### Final Answer
The average impulse imparted to the bullet is:
\[ \boxed{0.9 \text{ Ns}} \]