A bullet is fired from a gun. The force on the bullet is given by :
F = 600 – 2 × 10⁵ t.
Where F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

To find the average impulse imparted to the bullet, we first need to understand the relationship between force, time, and impulse. Impulse is defined as the change in momentum of an object, which can also be calculated as the integral of force over time. In this case, we have a force function given by F(t) = 600 - 2 × 10^5 t, where F is in newtons and t is in seconds.

Understanding the Force Function

The force function indicates that the force acting on the bullet decreases linearly over time until it reaches zero. To find out when the force becomes zero, we can set the equation to zero:

600 - 2 × 10^5 t = 0

Solving for t gives:

2 × 10^5 t = 600

t = 600 / (2 × 10^5) = 0.003 seconds

Calculating the Impulse

Impulse (J) can be calculated using the formula:

J = ∫ F(t) dt

From t = 0 to t = 0.003 seconds, we will integrate the force function:

Setting Up the Integral

We need to evaluate:

J = ∫ (600 - 2 × 10^5 t) dt from 0 to 0.003

Evaluating the Integral

First, we find the antiderivative of the force function:

∫ (600 - 2 × 10^5 t) dt = 600t - (2 × 10^5 / 2) t^2 + C

Which simplifies to:

600t - 10^5 t^2 + C

Calculating the Definite Integral

Now we evaluate this from 0 to 0.003:

J = [600(0.003) - 10^5(0.003)^2] - [600(0) - 10^5(0)^2]

J = [1.8 - 10^5(0.000009)]

J = 1.8 - 0.9 = 0.9 Ns (newton-seconds)

Final Result

The average impulse imparted to the bullet is 0.9 Ns. This value represents the change in momentum of the bullet as it accelerates down the barrel of the gun, reflecting the effect of the force applied over the time interval until the bullet exits the barrel.