Question

A bullet is fired from a gun, the force on the bullet is given by$F=600-2\times {10}^{5}t$ 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 a bullet?

Answer 1

Step 1: Given data

1. The force exerted on the bullet is $F=600-2\times {10}^{5}t$.
2. The force exerted on the bullet outside of the barrel is zero.

Step 2: Concept and formulae

1. Impulse is the change in momentum of a body when a force is exerted on it.
2. impulse is a vector quantity. The unit of impulse is N.s.
3. Impulse is defined by the form, $I=F.t$, where, F is the force applied to a body, I is the impulse and t is the time.

Step 3: Finding the average impulse of the bullet

The force exerted on the bullet is,

$F=600-2\times {10}^{5}t$.

Now let's say at time T the bullet leaves the barrel. And we know outside the barrel exerted force on the bullet is zero.

So,

$$\begin{gathered} 0=600-2\times {10}^{5}T \\ orT=\frac{600}{2\times {10}^5}=3\times {10}^{-3} \\ orT=3\times {10}^{-3}. \end{gathered}$$

Now, the impulse imparted by the bullet is,

$$\begin{gathered} I={\int }_0^{3\times {10}^{-3}}\left(600-2\times {10}^{5}t\right)dt \\ orI={\left[600t-2\times {10}^{-3}\frac{t^2}{2}\right]}_0^{3\times {10}^{-3}} \\ orI=0.9N.s \end{gathered}$$

Therefore, the average impulse imparted by the bullet is $0.9N.s$